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For this Question: Given an integer K, find a positive integer x such that <b>K = x<sup>2</sup> + 3*x</b>. If no such positive integer x exists, print -1.First and the only line of the input contains an integer K.
Constraints:
1 <= K <= 10<sup>18</sup>Print a positive integer x such that the above equation satisfies. If no such integer x exists, print -1.Sample Input:
28
Sample Output:
4
Explaination:
4<sup>2</sup> + 3*4 = 28
There is no other positive integer that will give such result., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define int long long
#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using T = pair<int, int>;
typedef long double ld;
const int mod = 1e9 + 7;
const int INF = 1e9;
void solve(){
int n;
cin >> n;
int l = 1, r = 1e9, ans = -1;
while(l <= r){
int m = (l + r)/2;
int val = m*m + 3*m;
if(val == n){
ans = m;
break;
}
if(val < n){
l = m + 1;
}
else r = m - 1;
}
cout << ans;
}
signed main(){
fast
int t = 1;
// cin >> t;
for(int i = 1; i <= t; i++){
solve();
if(i != t) cout << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer K, find a positive integer x such that <b>K = x<sup>2</sup> + 3*x</b>. If no such positive integer x exists, print -1.First and the only line of the input contains an integer K.
Constraints:
1 <= K <= 10<sup>18</sup>Print a positive integer x such that the above equation satisfies. If no such integer x exists, print -1.Sample Input:
28
Sample Output:
4
Explaination:
4<sup>2</sup> + 3*4 = 28
There is no other positive integer that will give such result., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
long K = Long.parseLong(br.readLine());
long ans = -1;
for(long x =0;((x*x)+(3*x))<=K;x++){
if(K==((x*x)+(3*x))){
ans = x;
break;
}
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of integers of length N, where N is even.
You need to split the array into two parts of equal size in such a way that the difference between the sums of the parts is maximised. The sum of a part is defined as the sum of the elements it contains.
Note that each element of the original array <b>must be in exactly one of the parts</b>.
Print the maximum possible difference.The first line of the input contains a single integer N (2 <= N <= 10<sup>4</sup>, N is even).
The second line of the input contains N space separated integers, the elements of the original array. Each element is within the range 0 to 10<sup>4</sup> inclusive.Print a single integer, the required maximum difference.Sample Input:
4
3 5 1 4
Sample Output:
5
Explanation:
It is optimal to make 2 parts as -- {4, 5} and {3, 1}., I have written this Solution Code: n = int(input())
elements = input()
array = str.split(elements)
for i in range(len(array)):
array[i] = int(array[i])
array.sort()
sum1 = 0
sum2 = 0
for x in range(int(n/2)):
sum1 += array[x]
for x in range(int(n/2)):
sum2 += array[x+int((n/2))]
diff = sum2 - sum1
print(diff), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of integers of length N, where N is even.
You need to split the array into two parts of equal size in such a way that the difference between the sums of the parts is maximised. The sum of a part is defined as the sum of the elements it contains.
Note that each element of the original array <b>must be in exactly one of the parts</b>.
Print the maximum possible difference.The first line of the input contains a single integer N (2 <= N <= 10<sup>4</sup>, N is even).
The second line of the input contains N space separated integers, the elements of the original array. Each element is within the range 0 to 10<sup>4</sup> inclusive.Print a single integer, the required maximum difference.Sample Input:
4
3 5 1 4
Sample Output:
5
Explanation:
It is optimal to make 2 parts as -- {4, 5} and {3, 1}., I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #include <sys/resource.h>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define f first
#define s second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now();
#define measure() auto end_time = std::chrono::high_resolution_clock::now(); cerr << (end_time - start_time)/std::chrono::milliseconds(1) << "ms" << endl;
typedef long long ll;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifdef LOCALY
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
// DEBUG FUNCTIONS END
// CUSTOM HASH TO SPEED UP UNORDERED MAP AND TO AVOID FORCED CLASHES
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count()); // FOR RANDOM NUMBER GENERATION
ll mod_exp(ll a, ll b, ll c)
{
ll res=1; a=a%c;
while(b>0)
{
if(b%2==1)
res=(res*a)%c;
b/=2;
a=(a*a)%c;
}
return res;
}
ll mymod(ll a,ll b)
{
return (((a = a%b) < 0) ? a + b : a);
}
ll gcdExtended(ll,ll,ll *,ll *);
ll modInverse(ll a, ll m)
{
ll x, y;
ll g = gcdExtended(a, m, &x, &y);
g++; //this line was added just to remove compiler warning
ll res = (x%m + m) % m;
return res;
}
ll gcdExtended(ll a, ll b, ll *x, ll *y)
{
if (a == 0)
{
*x = 0, *y = 1;
return b;
}
ll x1, y1;
ll gcd = gcdExtended(b%a, a, &x1, &y1);
*x = y1 - (b/a) * x1;
*y = x1;
return gcd;
}
struct Graph
{
vector<vector<int>> adj;
Graph(int n)
{
adj.resize(n+1);
}
void add_edge(int a, int b, bool directed = false)
{
adj[a].pb(b);
if(!directed) adj[b].pb(a);
}
};
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n;
cin >> n;
assert(n%2 == 0);
vll A(n);
REP(i, 0, n)
{
cin >> A[i];
}
sort(all(A));
ll big = 0, small = 0;
REP(i, 0, n)
{
if(i<n/2) small += A[i];
else big += A[i];
}
cout << big - small << "\n";
return 0;
}
/*
1. Check borderline constraints. Can a variable you are dividing by be 0?
2. Use ll while using bitshifts
3. Do not erase from set while iterating it
4. Initialise everything
5. Read the task carefully, is something unique, sorted, adjacent, guaranteed??
6. DO NOT use if(!mp[x]) if you want to iterate the map later
7. Are you using i in all loops? Are the i's conflicting?
*/
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of integers of length N, where N is even.
You need to split the array into two parts of equal size in such a way that the difference between the sums of the parts is maximised. The sum of a part is defined as the sum of the elements it contains.
Note that each element of the original array <b>must be in exactly one of the parts</b>.
Print the maximum possible difference.The first line of the input contains a single integer N (2 <= N <= 10<sup>4</sup>, N is even).
The second line of the input contains N space separated integers, the elements of the original array. Each element is within the range 0 to 10<sup>4</sup> inclusive.Print a single integer, the required maximum difference.Sample Input:
4
3 5 1 4
Sample Output:
5
Explanation:
It is optimal to make 2 parts as -- {4, 5} and {3, 1}., I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt();
long arr[] = sc.readArrayLong(n);
ruffleSort(arr);
long a = 0;
for(int i = 0; i<n/2; i++) {
a+=arr[i];
}
long b = 0;
for(int i = n/2; i<n; i++) {
b+=arr[i];
}
System.out.println(abs(b-a));
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main (String[] args)throws IOException {
Reader sc = new Reader();
int N = sc.nextInt();
int[] arr = new int[N];
for(int i=0;i<N;i++){
arr[i] = sc.nextInt();
}
int max=0;
if(arr[0]<arr[N-1])
System.out.print(N-1);
else{
for(int i=0;i<N-1;i++){
int j = N-1;
while(j>i){
if(arr[i]<arr[j]){
if(max<j-i){
max = j-i;
} break;
}
j--;
}
if(i==j)
break;
if(j==N-1)
break;
}
if(max==0)
System.out.print("-1");
else
System.out.print(max);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
/* For a given array arr[],
returns the maximum j – i such that
arr[j] > arr[i] */
int maxIndexDiff(int arr[], int n)
{
int maxDiff;
int i, j;
int *LMin = new int[(sizeof(int) * n)];
int *RMax = new int[(sizeof(int) * n)];
/* Construct LMin[] such that
LMin[i] stores the minimum value
from (arr[0], arr[1], ... arr[i]) */
LMin[0] = arr[0];
for (i = 1; i < n; ++i)
LMin[i] = min(arr[i], LMin[i - 1]);
/* Construct RMax[] such that
RMax[j] stores the maximum value from
(arr[j], arr[j+1], ..arr[n-1]) */
RMax[n - 1] = arr[n - 1];
for (j = n - 2; j >= 0; --j)
RMax[j] = max(arr[j], RMax[j + 1]);
/* Traverse both arrays from left to right
to find optimum j - i. This process is similar to
merge() of MergeSort */
i = 0, j = 0, maxDiff = -1;
while (j < n && i < n)
{
if (LMin[i] < RMax[j])
{
maxDiff = max(maxDiff, j - i);
j = j + 1;
}
else
i = i + 1;
}
return maxDiff;
}
// Driver Code
signed main()
{
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
int maxDiff = maxIndexDiff(a, n);
cout << maxDiff;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: n=int(input())
arr=list(map(int,input().split()))
rightMax = [0] * n
rightMax[n - 1] = arr[n - 1]
for i in range(n - 2, -1, -1):
rightMax[i] = max(rightMax[i + 1], arr[i])
maxDist = -2**31
i = 0
j = 0
while (i < n and j < n):
if (rightMax[j] >= arr[i]):
maxDist = max(maxDist, j - i)
j += 1
else:
i += 1
if maxDist==0:
maxDist=-1
print(maxDist), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: public static void For_Loop(int n){
for(int i=1;i<=n;i++){
if(i%2==1){System.out.print("odd ");}
else{
System.out.print("even ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: n = int(input())
for i in range(1, n+1):
if(i%2)==0:
print("even ",end="")
else:
print("odd ",end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A pair is called lucky if its sum is even and positive. Given three numbers find if there exists a lucky pair or not.The only line contains three integers a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>
<b>Constraints:</b>
-10<sup>9</sup> <= a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub> <= 10<sup>9</sup>Print "YES" without quotes if there exists a lucky pair otherwise print "NO" without quotes.Sample Input 1:
23 32 12
Sample Output 1:
YES
Sample Input 2:
1 -1 2
Sample Output 2:
NO, I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC target("popcnt")
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define fr first
#define sc second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
typedef long long ll;
typedef long long unsigned int llu;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
void solve(){
vector<ll>a(3);
for(ll i = 0;i<3;i++)cin >> a[i];
for(ll i = 0;i<3;i++){
for(ll j = i+1;j<3;j++){
if((a[i]+a[j])>0 && (a[i]+a[j])%2 == 0){
cout << "YES\n";
return;
}
}
}
cout << "NO\n";
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
// cout.tie(NULL);
#ifdef LOCALFLAG
freopen("Input.txt", "r", stdin);
freopen("Output2.txt", "w", stdout);
#endif
ll t = 1;
//cin >> t;
while(t--){
solve();
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given 2 non-negative integers m and n, find gcd(m, n)
GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n. Both m and n fit in a 32 bit signed integer.
NOTE: DO NOT USE LIBRARY FUNCTIONSInput contains two space separated integers, m and n
Constraints:-
1 < = m, n < = 10^18Output the single integer denoting the gcd of the above integers.Sample Input:
6 9
Sample Output:
3
Sample Input:-
5 6
Sample Output:-
1, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int n, m;
cin >> n >> m;
cout << __gcd(n, m);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given 2 non-negative integers m and n, find gcd(m, n)
GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n. Both m and n fit in a 32 bit signed integer.
NOTE: DO NOT USE LIBRARY FUNCTIONSInput contains two space separated integers, m and n
Constraints:-
1 < = m, n < = 10^18Output the single integer denoting the gcd of the above integers.Sample Input:
6 9
Sample Output:
3
Sample Input:-
5 6
Sample Output:-
1, I have written this Solution Code: def hcf(a, b):
if(b == 0):
return a
else:
return hcf(b, a % b)
li= list(map(int,input().strip().split()))
print(hcf(li[0], li[1])), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given 2 non-negative integers m and n, find gcd(m, n)
GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n. Both m and n fit in a 32 bit signed integer.
NOTE: DO NOT USE LIBRARY FUNCTIONSInput contains two space separated integers, m and n
Constraints:-
1 < = m, n < = 10^18Output the single integer denoting the gcd of the above integers.Sample Input:
6 9
Sample Output:
3
Sample Input:-
5 6
Sample Output:-
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] sp = br.readLine().trim().split(" ");
long m = Long.parseLong(sp[0]);
long n = Long.parseLong(sp[1]);
System.out.println(GCDAns(m,n));
}
private static long GCDAns(long m,long n){
if(m==0)return n;
if(n==0)return m;
return GCDAns(n%m,m);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: a="Hello World"
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
System.out.println("Hello World");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and a matrix of size (N x N). Check whether given matrix is identity or not.
<b> Note </b>
An identity matrix is a square matrix in which all the elements of principal diagonal are one, and all other elements are zeros.First line contain a single integer N
Next N line contain N space- separated integer i. e. elements of matrix.If the given matrix is the identity matrix. print "Yes" otherwise print "NO"
Constraints:
1<=N<=100Sample Input 1:
3
1 0 0
0 1 0
0 0 1
Sample Output 1:
Yes
Explanation:
Given matrix is an identity matrix because all main diagonal elements are 1 and rest of elements are 0., I have written this Solution Code: import java.util.*;
import java.io.*;
public class Main {
public static void main(String args[] ) throws Exception {
Scanner s = new Scanner(System.in);
int N = s.nextInt();
int [][]arr=new int[N][N];
for(int i=0;i<N;i++){
for(int j=0;j<N;j++)
arr[i][j] = s.nextInt();
}
boolean flag=false;
for(int i=0;i<N;i++){
for(int j=0;j<N;j++)
{
if(i==j && arr[i][j]!=1)flag=true;
else if(i!=j && arr[i][j]!=0)flag=true;
}
}
if(flag==true){
System.out.println("No");
}
else {
System.out.println("Yes");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
0, I have written this Solution Code: def lenOfLongSubarr(arr, N, K):
mydict = dict()
sum = 0
maxLen = 0
for i in range(N):
sum += arr[i]
if (sum == K):
maxLen = i + 1
elif (sum - K) in mydict:
maxLen = max(maxLen, i - mydict[sum - K])
if sum not in mydict:
mydict[sum] = i
return maxLen
if __name__ == '__main__':
T = int(input())
#N,K=list(map(int,input().split()))
for i in range(T):
N,k= [int(N)for N in input("").split()]
arr=list(map(int,input().split()))
N = len(arr)
print(lenOfLongSubarr(arr, N, k)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
unordered_map<long long,int> um;
int n,k;
cin>>n>>k;
long arr[n];
int maxLen=0;
for(int i=0;i<n;i++){cin>>arr[i];}
long long sum=0;
for(int i=0;i<n;i++){
sum += arr[i];
// when subarray starts from index '0'
if (sum == k)
maxLen = i + 1;
// make an entry for 'sum' if it is
// not present in 'um'
if (um.find(sum) == um.end())
um[sum] = i;
// check if 'sum-k' is present in 'um'
// or not
if (um.find(sum - k) != um.end()) {
// update maxLength
if (maxLen < (i - um[sum - k]))
maxLen = i - um[sum - k];
}
}
cout<<maxLen<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
0, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){
if(t%10==0){
System.gc();
}
int arrsize=sc.nextInt();
int k=sc.nextInt();
int[] arr=new int[arrsize];
for(int i=0;i<arrsize;i++){
arr[i]=sc.nextInt();
}
int subsize=0;
int sum=0;
HashMap<Integer, Integer> hash=new HashMap<>();
for(int i=0;i<arrsize;i++){
sum+=arr[i];
if(sum==k){
subsize=i+1;
}
if(!hash.containsKey(sum)){
hash.put(sum,i);
}
if(hash.containsKey(sum-k)){
if(subsize<(i-hash.get(sum-k))){
subsize=i-hash.get(sum-k);
}
}
}
System.out.println(subsize);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Complete the function <strong>handleCallback</strong> which has a single argument <code>callback</code>, which may or may not be a function.
Create a <strong>try</strong>, and <strong>catch</strong> block. Call the <code>callback</code> function inside the <strong>try</strong> block. In case <code>callback</code> is not a function, an error will be thrown which should be caught in the <strong>catch</strong> block. The <code>catch</code> block should print <code>"error thrown"</code> to the console.
<strong>Note:</strong> To check the expected output, we have already created a <code>anyCallback</code> function which prints <code>"No error thrown"</code> to the console. Just add <code>anyCallback</code> function without the parenthesis, to the input field to check the result.The <strong>handleCallback</strong> function takes in one argument, which may or may not be a function.The <strong>handleCallback</strong> function prints a string <code>"error thrown"</code> in the console, if the argument is not a function, otherwise nothing should be returned or printed.<strong>Example 1:</strong> Giving no function as an argument to the function
handleCallback() // prints "error thrown"
<strong>Example 2:</strong> Giving <code>anyCallback</code> function as an argument to the function
handleCallback(anyCallback) // prints "No error thrown", I have written this Solution Code: function handleCallback(callback) {
try {
callback();
} catch (error) {
console.log("error thrown");
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a Binary Tree, your task is to convert it to a Doubly Linked List. The left and right pointers in nodes are to be used as previous and next pointers respectively in converted Double linked list. The order of nodes in Double linked list must be same as Inorder of the given Binary Tree. The first node of Inorder traversal (left most node in Binary tree) must be head node of the Doubly linked list.<b>User Task:</b>
Since this will be a functional problem. You don't have to take input. You just have to complete the function <b>BToDLL()</b> that takes "root" node of binary tree as parameter.
Constraint:-
1 <= Number of Nodes <= 1000
1 <= Node.data <= 1000
The printing is done by the driver code you just need to complete the function.Sample Input:-
3
1 2 3
Sample Output:-
2 1 3
Sample Input:-
5
6 5 4 3 2
Sample Output:-
3 5 2 6 4 , I have written this Solution Code: static void BToDLL(Node root)
{
// Base cases
if (root == null)
return ;
// Recursively convert right subtree
BToDLL(root.right);
// insert root into DLL
root.right = head;
// Change left pointer of previous head
if (head != null)
(head).left = root;
// Change head of Doubly linked list
head = root;
// Recursively convert left subtree
BToDLL(root.left);
} , In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: def compound_interest(principle, rate, time):
Amount = principle * (pow((1 + rate / 100), time))
CI = Amount - principle
print( '%.2f'%CI)
principle,rate,time=map(int, input().split())
compound_interest(principle,rate,time), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: function calculateCI(P, R, T)
{
let interest = P * (Math.pow(1.0 + R/100.0, T) - 1);
return interest.toFixed(2);
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int p,r,t;
cin>>p>>r>>t;
double rate= (float)r/100;
double amt = (float)p*(pow(1+rate,t));
cout << fixed << setprecision(2) << (amt - p);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math;
class Main {
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] s= br.readLine().split(" ");
double[] darr = new double[s.length];
for(int i=0;i<s.length;i++){
darr[i] = Double.parseDouble(s[i]);
}
double ans = darr[0]*Math.pow(1+darr[1]/100,darr[2])-darr[0];
System.out.printf("%.2f",ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find the highest target achieved by the employee or employees who works under the manager id 13. Output the first name of the employee and target achieved. The solution should show the highest target achieved under manager_id=13 and which employee(s) achieved it.DataFrame/SQL Table with the following schema -
<schema>[{'name': 'salesforce_employees', 'columns': [{'name': 'id', 'type': 'int64'}, {'name': 'first_name', 'type': 'object'}, {'name': 'last_name', 'type': 'object'}, {'name': 'age', 'type': 'int64'}, {'name': 'sex', 'type': 'object'}, {'name': 'employee_title', 'type': 'object'}, {'name': 'department', 'type': 'object'}, {'name': 'salary', 'type': 'int64'}, {'name': 'target', 'type': 'int64'}, {'name': 'bonus', 'type': 'int64'}, {'name': 'email', 'type': 'object'}, {'name': 'city', 'type': 'object'}, {'name': 'address', 'type': 'object'}, {'name': 'manager_id', 'type': 'int64'}]}]</schema>Each row in a new line and each value of a row separated by a |, i.e.,
0|1|2
1|2|3
2|3|4-, I have written this Solution Code: df = salesforce_employees[salesforce_employees['manager_id']==13]
df = df[df['target']== df['target'].max()]
for i, r in df.iterrows():
print(f"{r['first_name']}|{r['target']}"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find the highest target achieved by the employee or employees who works under the manager id 13. Output the first name of the employee and target achieved. The solution should show the highest target achieved under manager_id=13 and which employee(s) achieved it.DataFrame/SQL Table with the following schema -
<schema>[{'name': 'salesforce_employees', 'columns': [{'name': 'id', 'type': 'int64'}, {'name': 'first_name', 'type': 'object'}, {'name': 'last_name', 'type': 'object'}, {'name': 'age', 'type': 'int64'}, {'name': 'sex', 'type': 'object'}, {'name': 'employee_title', 'type': 'object'}, {'name': 'department', 'type': 'object'}, {'name': 'salary', 'type': 'int64'}, {'name': 'target', 'type': 'int64'}, {'name': 'bonus', 'type': 'int64'}, {'name': 'email', 'type': 'object'}, {'name': 'city', 'type': 'object'}, {'name': 'address', 'type': 'object'}, {'name': 'manager_id', 'type': 'int64'}]}]</schema>Each row in a new line and each value of a row separated by a |, i.e.,
0|1|2
1|2|3
2|3|4-, I have written this Solution Code: SELECT first_name,
target
FROM salesforce_employees
WHERE target IN
(SELECT MAX(target)
FROM salesforce_employees
WHERE manager_id = 13)
AND manager_id = 13, In this Programming Language: SQL, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n candies put from left to right on a table. The candies are numbered from left to right. The i- th candy has weight w[i]. Alice and Bob eat candies.
Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row).
Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row).
Of course, if Alice ate a candy, Bob can't eat it (and vice versa).
They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total?First line contains a single integer n.
Next line contains n space separated integers.
<b> Constraints </b>
1 <= n <= 10<sup>5</sup>
1 <= w[i] <= 10<sup>9</sup>A single integer denoting maximum number of candies that can be eaten under given condition.Input:
6
2 1 4 3 4 3
Output:
5
Explanation :
Alice eats first 3 candies. Bob eats last 2 candies., I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
int n=Integer.parseInt(in.next());
Long a[] = new Long[n];
for(int i=0;i<n;i++){
a[i]=Long.parseLong(in.next());
}
long suf[] = new long[n];
suf[n-1]=a[n-1];
for(int i=n-2;i>=0;i--){
suf[i] = suf[i+1] + a[i];
}
int ans=0;
int pref=0;
for(int i=0;i<n-1;i++){
pref += a[i];
int l=i+1,r=n-1;
while(l<r){
int m=l+(r-l)/2;
if(suf[m] <= pref){
r=m;
}
else l=m+1;
}
if(suf[l] == pref){
ans=Math.max(ans,i+1+n-l);
}
}
out.print(ans);
out.close();
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joey and Chandler are super bored. So, Chandler makes up a game they can play. The game is called Chandy Game. Initially, Chandler has A candies and Joey has B candies.
In the first move Joey has to give Chandler 1 candy.
In the second move Chandler has to give Joey 2 candies.
In the third move Joey has to give Chandler 3 candies.
In the fourth move Chandler has to give Joey 4 candies.
In the fifth move Joey has to give Chandler 5 candy.
... and so on.
The game continues till one of the player can not make a move. The player who cannot make a move loses. Help them find who wins the game.Input contains two integers A and B.
Constraints:
0 <= A, B <= 10<sup>15</sup>Print "Chandler" (without quotes) if Chandler wins the game and "Joey" (without quotes) if Joey wins the game.Sample Input
2 1
Sample Output
Chandler
Explanation:
In first move Joey gives Chandler 1 candy so, Chandler has 3 candies and Joey has 0.
In second move Chandler gives Joey 2 candies so, Chandler has 1 candy and Joey has 2.
In third move Joey has to give Chandler 3 candies but he has only 2 candies so he loses., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main(String args[])throws Exception
{
BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb=new StringBuilder();
String s[]=bu.readLine().split(" ");
long a=Long.parseLong(s[0]),b=Long.parseLong(s[1]);
if(a>=b) sb.append("Chandler");
else sb.append("Joey");
System.out.print(sb);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joey and Chandler are super bored. So, Chandler makes up a game they can play. The game is called Chandy Game. Initially, Chandler has A candies and Joey has B candies.
In the first move Joey has to give Chandler 1 candy.
In the second move Chandler has to give Joey 2 candies.
In the third move Joey has to give Chandler 3 candies.
In the fourth move Chandler has to give Joey 4 candies.
In the fifth move Joey has to give Chandler 5 candy.
... and so on.
The game continues till one of the player can not make a move. The player who cannot make a move loses. Help them find who wins the game.Input contains two integers A and B.
Constraints:
0 <= A, B <= 10<sup>15</sup>Print "Chandler" (without quotes) if Chandler wins the game and "Joey" (without quotes) if Joey wins the game.Sample Input
2 1
Sample Output
Chandler
Explanation:
In first move Joey gives Chandler 1 candy so, Chandler has 3 candies and Joey has 0.
In second move Chandler gives Joey 2 candies so, Chandler has 1 candy and Joey has 2.
In third move Joey has to give Chandler 3 candies but he has only 2 candies so he loses., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int a,b;
cin>>a>>b;
if(a>=b)
cout<<"Chandler";
else
cout<<"Joey";
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joey and Chandler are super bored. So, Chandler makes up a game they can play. The game is called Chandy Game. Initially, Chandler has A candies and Joey has B candies.
In the first move Joey has to give Chandler 1 candy.
In the second move Chandler has to give Joey 2 candies.
In the third move Joey has to give Chandler 3 candies.
In the fourth move Chandler has to give Joey 4 candies.
In the fifth move Joey has to give Chandler 5 candy.
... and so on.
The game continues till one of the player can not make a move. The player who cannot make a move loses. Help them find who wins the game.Input contains two integers A and B.
Constraints:
0 <= A, B <= 10<sup>15</sup>Print "Chandler" (without quotes) if Chandler wins the game and "Joey" (without quotes) if Joey wins the game.Sample Input
2 1
Sample Output
Chandler
Explanation:
In first move Joey gives Chandler 1 candy so, Chandler has 3 candies and Joey has 0.
In second move Chandler gives Joey 2 candies so, Chandler has 1 candy and Joey has 2.
In third move Joey has to give Chandler 3 candies but he has only 2 candies so he loses., I have written this Solution Code: ch,jo=map(int,input().split())
if ch>=jo:
print('Chandler')
else:
print('Joey'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
String[] st = bf.readLine().split(" ");
if(Integer.parseInt(st[1])==0)
System.out.print(-1);
else {
int f = (Integer.parseInt(st[0])/Integer.parseInt(st[1]));
System.out.print(f);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: D,Q = input().split()
D = int(D)
Q = int(Q)
if(0<=D and Q<=100 and Q >0):
print(int(D/Q))
else:
print('-1'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: #include <iostream>
using namespace std;
int main(){
int n,m;
cin>>n>>m;
if(m==0){cout<<-1;return 0;}
cout<<n/m;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: static void printDataTypes(int a, long b, float c, double d, char e)
{
System.out.println(a);
System.out.println(b);
System.out.printf("%.2f",c);
System.out.println();
System.out.printf("%.4f",d);
System.out.println();
System.out.println(e);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: void printDataTypes(int a, long long b, float c, double d, char e){
cout<<a<<endl;
cout<<b<<endl;
cout <<fixed<< std::setprecision(2) << c << '\n';
cout <<fixed<< std::setprecision(4) << d << '\n';
cout<<e<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: a=int(input())
b=int(input())
x=float(input())
g = "{:.2f}".format(x)
d=float(input())
e = "{:.4f}".format(d)
u=input()
print(a)
print(b)
print(g)
print(e)
print(u), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Internet download speed is often expressed in bit per second whereas file size is expressed in Bytes. It is known that 1 Byte = 8 bits. Given file size in megabytes(MB) and internet speed in megabits per seconds(Mbps), find the time taken in seconds to download the file.The only line of input contains two integers denoting the file size in MB and download speed in Mbps.
1 <= file size <= 1000
1 <= download speed <= 1000Print a single integer denoting the time taken to download the file in seconds.
It is guaranteed that the result will be an integer.Sample Input:
10 16
Sample Output:
5, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
void solve(){
int f, s;
cin >> f >> s;
cout << (8*f)/s << endl;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Internet download speed is often expressed in bit per second whereas file size is expressed in Bytes. It is known that 1 Byte = 8 bits. Given file size in megabytes(MB) and internet speed in megabits per seconds(Mbps), find the time taken in seconds to download the file.The only line of input contains two integers denoting the file size in MB and download speed in Mbps.
1 <= file size <= 1000
1 <= download speed <= 1000Print a single integer denoting the time taken to download the file in seconds.
It is guaranteed that the result will be an integer.Sample Input:
10 16
Sample Output:
5, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int f = sc.nextInt();
int s = sc.nextInt();
int ans = (8*f)/s;
System.out.print(ans);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Internet download speed is often expressed in bit per second whereas file size is expressed in Bytes. It is known that 1 Byte = 8 bits. Given file size in megabytes(MB) and internet speed in megabits per seconds(Mbps), find the time taken in seconds to download the file.The only line of input contains two integers denoting the file size in MB and download speed in Mbps.
1 <= file size <= 1000
1 <= download speed <= 1000Print a single integer denoting the time taken to download the file in seconds.
It is guaranteed that the result will be an integer.Sample Input:
10 16
Sample Output:
5, I have written this Solution Code: a = list(map(int,input().strip().split()))[:2]
print(int((a[0]*8)/a[1])), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of heights of N buildings in a row. You can start from any building and jump to the <b>adjacent</b> right building till the height of the building to the right is less than or equal to the height of your current building. Find the maximum number of jumps you can make.The first line of input contains a single integer N.
The second line of input contains N integers, denoting the array height.
<b>Constraints:</b>
1 <= N <= 10<sup>5</sup>
1 <= height[i] <= 10<sup>9</sup>Print the maximum number of jumps you can make.Sample Input:-
5
5 4 1 2 1
Sample Output:-
2
<b>Explanation:</b>
We start from building with height 5 then jump right to building with height 4 then again to building with height 1 making a total of 2 jumps., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
#define rep(i,n) for (int i=0; i<(n); i++)
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int a[n];
for(int i=0;i<n;++i){
cin>>a[i];
}
int ans=0;
int v=0;
for(int i=1;i<n;++i){
if(a[i]>a[i-1])
v=0;
else
++v;
ans=max(ans,v);
}
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of heights of N buildings in a row. You can start from any building and jump to the <b>adjacent</b> right building till the height of the building to the right is less than or equal to the height of your current building. Find the maximum number of jumps you can make.The first line of input contains a single integer N.
The second line of input contains N integers, denoting the array height.
<b>Constraints:</b>
1 <= N <= 10<sup>5</sup>
1 <= height[i] <= 10<sup>9</sup>Print the maximum number of jumps you can make.Sample Input:-
5
5 4 1 2 1
Sample Output:-
2
<b>Explanation:</b>
We start from building with height 5 then jump right to building with height 4 then again to building with height 1 making a total of 2 jumps., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int num = Integer.parseInt(br.readLine());
String s= br.readLine();
int[] arr= new int[num];
String[] s1 = s.split(" ");
int ccount = 0, pcount = 0;
for(int i=0;i<(num);i++)
{
arr[i]=Integer.parseInt(s1[i]);
if(i+1 < num){
arr[i+1] = Integer.parseInt(s1[i+1]);}
if(((i+1)< num) && (arr[i]>=arr[i+1]) ){
ccount++;
}else{
if(ccount > pcount){
pcount = ccount;
}
ccount = 0;
}
}
System.out.print(pcount);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of heights of N buildings in a row. You can start from any building and jump to the <b>adjacent</b> right building till the height of the building to the right is less than or equal to the height of your current building. Find the maximum number of jumps you can make.The first line of input contains a single integer N.
The second line of input contains N integers, denoting the array height.
<b>Constraints:</b>
1 <= N <= 10<sup>5</sup>
1 <= height[i] <= 10<sup>9</sup>Print the maximum number of jumps you can make.Sample Input:-
5
5 4 1 2 1
Sample Output:-
2
<b>Explanation:</b>
We start from building with height 5 then jump right to building with height 4 then again to building with height 1 making a total of 2 jumps., I have written this Solution Code: N = int(input())
arr = iter(map(int,input().strip().split()))
jumps = []
jump = 0
temp = next(arr)
for i in arr:
if i<=temp:
jump += 1
temp = i
continue
temp = i
jumps.append(jump)
jump = 0
jumps.append(jump)
print(max(jumps)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix <b>A</b> of dimensions <b>n x m</b>. The task is to perform <b>boundary traversal</b> on the matrix in clockwise manner.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase has two lines of input. The first line contains dimensions of the matrix A, n and m. The second line contains n*m elements separated by spaces.
Constraints:
1 <= T <= 100
1 <= n, m <= 30
0 <= A[i][j] <= 100For each testcase, in a new line, print the boundary traversal of the matrix A.Input:
4
4 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
3 4
12 11 10 9 8 7 6 5 4 3 2 1
1 4
1 2 3 4
4 1
1 2 3 4
Output:
1 2 3 4 8 12 16 15 14 13 9 5
12 11 10 9 5 1 2 3 4 8
1 2 3 4
1 2 3 4
Explanation:
Testcase1: The matrix is:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
The boundary traversal is 1 2 3 4 8 12 16 15 14 13 9 5
Testcase 2: Boundary Traversal will be 12 11 10 9 5 1 2 3 4 8.
Testcase 3: Boundary Traversal will be 1 2 3 4.
Testcase 4: Boundary Traversal will be 1 2 3 4., I have written this Solution Code:
def boundaryTraversal(matrix, N, M): #N = 3, M = 4
start_row = 0
start_col = 0
count = 0
if N != 1 and M != 1:
total = 2 * (N - 1) + 2 * (M - 1)
else:
total = max(M, N)
#First row
for i in range(start_col, M, 1): #0,1, 2, 3
count += 1
print(matrix[start_row][i], end = " ")
if count == total:
return
start_row += 1
#Last column
for i in range(start_row, N, 1): # 1, 2
count += 1
print(matrix[i][M - 1], end = " ")
if count == total:
return
M -= 1
#Last Row
for i in range(M - 1, start_col - 1, -1): # 2, 1, 0
count += 1
print(matrix[N - 1][i], end = " ")
if count == total:
return
#First Column
N -= 1 # 2
for i in range(N - 1, start_row - 1, -1):
count += 1
print(matrix[i][start_col], end = " ")
start_col += 1
testcase = int(input().strip())
for test in range(testcase):
dim = input().strip().split(" ")
N = int(dim[0])
M = int(dim[1])
matrix = []
elements = input().strip().split(" ") #N * M
start = 0
for i in range(N):
matrix.append(elements[start : start + M]) # (0, M - 1) | (M, 2*m-1)
start = start + M
boundaryTraversal(matrix, N, M)
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix <b>A</b> of dimensions <b>n x m</b>. The task is to perform <b>boundary traversal</b> on the matrix in clockwise manner.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase has two lines of input. The first line contains dimensions of the matrix A, n and m. The second line contains n*m elements separated by spaces.
Constraints:
1 <= T <= 100
1 <= n, m <= 30
0 <= A[i][j] <= 100For each testcase, in a new line, print the boundary traversal of the matrix A.Input:
4
4 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
3 4
12 11 10 9 8 7 6 5 4 3 2 1
1 4
1 2 3 4
4 1
1 2 3 4
Output:
1 2 3 4 8 12 16 15 14 13 9 5
12 11 10 9 5 1 2 3 4 8
1 2 3 4
1 2 3 4
Explanation:
Testcase1: The matrix is:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
The boundary traversal is 1 2 3 4 8 12 16 15 14 13 9 5
Testcase 2: Boundary Traversal will be 12 11 10 9 5 1 2 3 4 8.
Testcase 3: Boundary Traversal will be 1 2 3 4.
Testcase 4: Boundary Traversal will be 1 2 3 4., I have written this Solution Code: import java.io.*;
import java.lang.*;
import java.util.*;
class Main
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0)
{
int n1 = sc.nextInt();
int m1 = sc.nextInt();
int arr1[][] = new int[n1][m1];
for(int i = 0; i < n1; i++)
{
for(int j = 0; j < m1; j++)
arr1[i][j] = sc.nextInt();
}
boundaryTraversal(n1, m1,arr1);
System.out.println();
}
}
static void boundaryTraversal( int n1, int m1, int arr1[][])
{
// base cases
if(n1 == 1)
{
int i = 0;
while(i < m1)
System.out.print(arr1[0][i++] + " ");
}
else if(m1 == 1)
{
int i = 0;
while(i < n1)
System.out.print(arr1[i++][0]+" ");
}
else
{
// traversing the first row
for(int j=0;j<m1;j++)
{
System.out.print(arr1[0][j]+" ");
}
// traversing the last column
for(int j=1;j<n1;j++)
{
System.out.print(arr1[j][m1-1]+ " ");
}
// traversing the last row
for(int j=m1-2;j>=0;j--)
{
System.out.print(arr1[n1-1][j]+" ");
}
// traversing the first column
for(int j=n1-2;j>=1;j--)
{
System.out.print(arr1[j][0]+" ");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code:
// X and Y are numbers
// ignore number of testcases variable
function pow(X, Y) {
// write code here
// console.log the output in a single line,like example
console.log(Math.pow(X, Y).toFixed(2))
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code: def power(N,K):
return ("{0:.2f}".format(N**K))
T=int(input())
for i in range(T):
X,N = map(float,input().strip().split())
print(power(X,N)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main (String[] args)throws Exception {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine());
while(t-- > 0)
{
String str[] = read.readLine().trim().split(" ");
double X = Double.parseDouble(str[0]);
int N = Integer.parseInt(str[1]);
System.out.println(String.format("%.2f", myPow(X, N)));
}
}
public static double myPow(double x, int n) {
if (n == Integer.MIN_VALUE)
n = - (Integer.MAX_VALUE - 1);
if (n == 0)
return 1.0;
else if (n < 0)
return 1 / myPow(x, -n);
else if (n % 2 == 1)
return x * myPow(x, n - 1);
else {
double sqrt = myPow(x, n / 2);
return sqrt * sqrt;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define ld long double
#define int long long int
#define speed ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define endl '\n'
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
ld power(ld x, ld n){
if(n == 0)
return 1;
else
return x*power(x, n-1);
}
signed main() {
speed;
int t; cin >> t;
while(t--){
double x; int n;
cin >> x >> n;
if(n < 0)
x = 1.0/x, n *= -1;
cout << setprecision(2) << fixed << power(x, n) << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a container with A cyan balls. Alexa will do the following operation as many times as he likes (possibly zero times):
Add B cyan balls and C red balls into the container.
Alexa's objective is to reach a situation where the number of cyan balls in the container is at most D times the number of red balls in it.
Determine whether the objective is achievable. If it is achievable, find the minimum number of operations needed to achieve it.The input contains 4 space-separated numbers:
A B C D
<b>Constraints</b>
1 ≤ A, B, C, D ≤ 10<sup>5</sup>
All values in the input are integers.If Alexa's objective is achievable, print the minimum number of operations needed to achieve it. Otherwise, print -1.<b>Sample Input 1</b>
5 2 3 2
<b>Sample Output 1</b>
2
<b>Sample Input 2</b>
6 9 2 3
<b>Sample Output 2</b>
-1, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
long long A,B,C,D; cin >> A >> B >> C >> D;
long long blue = A, red = 0;
for(int i = 1; i <= A; i++){
blue += B;
red += C;
if(blue <= D*red){
cout << i << endl;
return 0;
}
}
cout << -1 << endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted array your task is to merge these two arrays into a single array such that the merged array is also sortedFirst line contain two integers N and M the size of arrays
Second line contains N separated integers the elements of first array
Third line contains M separated integers elements of second array
<b>Constraints:-</b>
1<=N,M<=10<sup>4</sup>
1<=arr1[], arr2[] <=10<sup>5</sup>Output the merged arraySample Input:-
3 4
1 4 7
1 3 3 9
Sample Output:-
1 1 3 3 4 7 9
, I have written this Solution Code: n1,m1=input().split()
n=int(n1)
m=int(m1)
l1=list(map(int,input().strip().split()))
l2=list(map(int,input().strip().split()))
i,j=0,0
while i<n and j<m:
if l1[i]<=l2[j]:
print(l1[i],end=" ")
i+=1
else:
print(l2[j],end=" ")
j+=1
for a in range(i,n):
print(l1[a],end=" ")
for b in range(j,m):
print(l2[b],end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted array your task is to merge these two arrays into a single array such that the merged array is also sortedFirst line contain two integers N and M the size of arrays
Second line contains N separated integers the elements of first array
Third line contains M separated integers elements of second array
<b>Constraints:-</b>
1<=N,M<=10<sup>4</sup>
1<=arr1[], arr2[] <=10<sup>5</sup>Output the merged arraySample Input:-
3 4
1 4 7
1 3 3 9
Sample Output:-
1 1 3 3 4 7 9
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 10000001
int main(){
int n,m;
cin>>n>>m;
int a[n];
int b[m];
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<m;i++){
cin>>b[i];
}
int c[n+m];
int i=0,j=0,k=0;
while(i!=n && j!=m){
if(a[i]<b[j]){c[k]=a[i];k++;i++;}
else{c[k]=b[j];j++;k++;}
}
while(i!=n){
c[k]=a[i];
k++;i++;
}
while(j!=m){
c[k]=b[j];
k++;j++;
}
for(i=0;i<n+m;i++){
cout<<c[i]<<" ";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted array your task is to merge these two arrays into a single array such that the merged array is also sortedFirst line contain two integers N and M the size of arrays
Second line contains N separated integers the elements of first array
Third line contains M separated integers elements of second array
<b>Constraints:-</b>
1<=N,M<=10<sup>4</sup>
1<=arr1[], arr2[] <=10<sup>5</sup>Output the merged arraySample Input:-
3 4
1 4 7
1 3 3 9
Sample Output:-
1 1 3 3 4 7 9
, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int a[] = new int[n];
int b[] = new int[m];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
for(int i=0;i<m;i++){
b[i]=sc.nextInt();
}
int c[]=new int[n+m];
int i=0,j=0,k=0;
while(i!=n && j!=m){
if(a[i]<b[j]){c[k]=a[i];k++;i++;}
else{c[k]=b[j];j++;k++;}
}
while(i!=n){
c[k]=a[i];
k++;i++;
}
while(j!=m){
c[k]=b[j];
k++;j++;
}
for(i=0;i<n+m;i++){
System.out.print(c[i]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers <b>a</b> and <b>b</b>, your task is to calculate and print the following four values:-
a+b
a-b
a*b
a/bThe input contains two integers a and b separated by spaces.
<b>Constraints:</b>
1 ≤ b ≤ a ≤ 1000
<b> It is guaranteed that a will be divisible by b</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5
<b>Explanation:-</b>
First operation is a+b so 15+3 = 18
The second Operation is a-b so 15-3 = 12
Third Operation is a*b so 15*3 = 45
Fourth Operation is a/b so 15/3 = 5, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args){
FastReader read = new FastReader();
int a = read.nextInt();
int b = read.nextInt();
System.out.println(a+b);
System.out.println(a-b);
System.out.println(a*b);
System.out.println(a/b);
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader(){
InputStreamReader inr = new InputStreamReader(System.in);
br = new BufferedReader(inr);
}
String next(){
while(st==null || !st.hasMoreElements())
try{
st = new StringTokenizer(br.readLine());
}
catch(IOException e){
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
long nextLong(){
return Long.parseLong(next());
}
String nextLine(){
String str = "";
try{
str = br.readLine();
}
catch(IOException e){
e.printStackTrace();
}
return str;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers <b>a</b> and <b>b</b>, your task is to calculate and print the following four values:-
a+b
a-b
a*b
a/bThe input contains two integers a and b separated by spaces.
<b>Constraints:</b>
1 ≤ b ≤ a ≤ 1000
<b> It is guaranteed that a will be divisible by b</b>Print the mentioned operations each in a new line.Sample Input:
15 3
Sample Output:
18
12
45
5
<b>Explanation:-</b>
First operation is a+b so 15+3 = 18
The second Operation is a-b so 15-3 = 12
Third Operation is a*b so 15*3 = 45
Fourth Operation is a/b so 15/3 = 5, I have written this Solution Code: a,b=map(int,input().split())
print(a+b)
print(a-b)
print(a*b)
print(a//b), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the simple interest for given principal amount P, time Tm(in years) and rate R.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SimpleInterest()</b> that takes the principal amount P, rate R, and time Tm as a parameter.
Constraints:
1 <= P <= 10^3
1 <= Tm <= 20
1 <= R <= 20Return the floor value of the simple interest i.e. interest in integer format.Input:
42 15 8
Output:
50
Explanation:
Testcase 1: Simple interest of given principal amount 42, in 8 years at a 15% rate of interest is 50., I have written this Solution Code: import math
p,t,r = [int(x) for x in input().split()]
res=p*t*r
print(math.floor(res/100)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the simple interest for given principal amount P, time Tm(in years) and rate R.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SimpleInterest()</b> that takes the principal amount P, rate R, and time Tm as a parameter.
Constraints:
1 <= P <= 10^3
1 <= Tm <= 20
1 <= R <= 20Return the floor value of the simple interest i.e. interest in integer format.Input:
42 15 8
Output:
50
Explanation:
Testcase 1: Simple interest of given principal amount 42, in 8 years at a 15% rate of interest is 50., I have written this Solution Code: static int SimpleInterest(int P, int R, int Tm){
return (P*Tm*R)/100;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string S of length N consisting of lowercase characters 'a' and 'b'. In one operation, you can select a character and make it equal to one of its adjacent characters. For example, if S = "aab", in one operation you can convert it to any of the following:
1. "aab": By changing the 1<sup>st</sup> character to the 2<sup>nd</sup> character.
2. "aab": By changing the 2<sup>nd</sup> character to the 1<sup>st</sup> character.
3. "abb": By changing the 2<sup>nd</sup> character to the 3<sup>rd</sup> character.
4. "aaa": By changing the 3<sup>rd</sup> character to the 2<sup>nd</sup> character.
Find the minimum number of operations to make all the characters in the string equal.The first line of the input contains a single integer N.
The second line contains a string S of length N consisting of only 'a' and 'b'.
<b> Constraints: </b>
1 ≤ N ≤ 5000Print the minimum number of operations to make all the characters in the string equal.Sample Input 1:
4
abaa
Sample Output 1:
1
Sample Explanation 1:
You can replace the 'b' at the second position by the 'a' at the first position. Now the string becomes "aaaa".
Sample Input 2:
5
bbbaa
Sample Output 2:
2
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
int c = 0;
String str = br.readLine();
for(int i=0;i<n;i++){
if(str.charAt(i) == 'a'){
c++;
}
}
if((n-c)<c){
System.out.println(n-c);
}else{
System.out.println(c);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string S of length N consisting of lowercase characters 'a' and 'b'. In one operation, you can select a character and make it equal to one of its adjacent characters. For example, if S = "aab", in one operation you can convert it to any of the following:
1. "aab": By changing the 1<sup>st</sup> character to the 2<sup>nd</sup> character.
2. "aab": By changing the 2<sup>nd</sup> character to the 1<sup>st</sup> character.
3. "abb": By changing the 2<sup>nd</sup> character to the 3<sup>rd</sup> character.
4. "aaa": By changing the 3<sup>rd</sup> character to the 2<sup>nd</sup> character.
Find the minimum number of operations to make all the characters in the string equal.The first line of the input contains a single integer N.
The second line contains a string S of length N consisting of only 'a' and 'b'.
<b> Constraints: </b>
1 ≤ N ≤ 5000Print the minimum number of operations to make all the characters in the string equal.Sample Input 1:
4
abaa
Sample Output 1:
1
Sample Explanation 1:
You can replace the 'b' at the second position by the 'a' at the first position. Now the string becomes "aaaa".
Sample Input 2:
5
bbbaa
Sample Output 2:
2
, I have written this Solution Code: input()
s = input()
a = s.count("a")
b = s.count("b")
print(a if a<b else b), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string S of length N consisting of lowercase characters 'a' and 'b'. In one operation, you can select a character and make it equal to one of its adjacent characters. For example, if S = "aab", in one operation you can convert it to any of the following:
1. "aab": By changing the 1<sup>st</sup> character to the 2<sup>nd</sup> character.
2. "aab": By changing the 2<sup>nd</sup> character to the 1<sup>st</sup> character.
3. "abb": By changing the 2<sup>nd</sup> character to the 3<sup>rd</sup> character.
4. "aaa": By changing the 3<sup>rd</sup> character to the 2<sup>nd</sup> character.
Find the minimum number of operations to make all the characters in the string equal.The first line of the input contains a single integer N.
The second line contains a string S of length N consisting of only 'a' and 'b'.
<b> Constraints: </b>
1 ≤ N ≤ 5000Print the minimum number of operations to make all the characters in the string equal.Sample Input 1:
4
abaa
Sample Output 1:
1
Sample Explanation 1:
You can replace the 'b' at the second position by the 'a' at the first position. Now the string becomes "aaaa".
Sample Input 2:
5
bbbaa
Sample Output 2:
2
, I have written this Solution Code: //Author: Xzirium
//Time and Date: 03:04:29 27 December 2021
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t clk;
clk = clock();
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll a=0,b=0;
FORI(i,0,N)
{
if(S[i]=='a')
{
a++;
}
else
{
b++;
}
}
cout<<min(a,b)<<endl;
//-----------------------------------------------------------------------------------------------------------//
clk = clock() - clk;
cerr << fixed << setprecision(6) << "Time: " << ((double)clk)/CLOCKS_PER_SEC << endl;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
long z=0,x=0,y=0;
int choice;
Scanner in = new Scanner(System.in);
choice = in.nextInt();
String s="";
int f = 1;
while(f<=choice){
x = in.nextLong();
y = in.nextLong();
z = in.nextLong();
System.out.println((long)(Math.max((z-x),(z-y))));
f++;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: n = int(input())
for i in range(n):
l = list(map(int,input().split()))
print(l[2]-min(l)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
const ll mod2 = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
signed main()
{
read(t);
assert(1 <= t && t <= ll(1e4));
while (t--)
{
readc(x, y, z);
assert(1 <= x && x <= ll(1e15));
assert(1 <= y && y <= ll(1e15));
assert(max(x, y) < z && z <= ll(1e15));
int r = 2*z - x - y - 1;
int l = z - max(x, y);
print(r - l + 1);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
int t;
cin>>t;
while(t--)
{
int x, y, z;
cin>>x>>y>>z;
cout<<max(z - y, z- x)<<endl;
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen("INPUT.txt", "r", stdin);
freopen("OUTPUT.txt", "w", stdout);
}
#endif
solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
String[] st = bf.readLine().split(" ");
if(Integer.parseInt(st[1])==0)
System.out.print(-1);
else {
int f = (Integer.parseInt(st[0])/Integer.parseInt(st[1]));
System.out.print(f);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: D,Q = input().split()
D = int(D)
Q = int(Q)
if(0<=D and Q<=100 and Q >0):
print(int(D/Q))
else:
print('-1'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: #include <iostream>
using namespace std;
int main(){
int n,m;
cin>>n>>m;
if(m==0){cout<<-1;return 0;}
cout<<n/m;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the roots of a quadratic equation.
<b>Note</b>: Try to do it using Switch Case.
<b>Quadratic equations</b> are the polynomial equations of degree 2 in one variable of type f(x) = ax<sup>2</sup> + bx + c = 0 where a, b, c, ∈ R and a ≠ 0. It is the general form of a quadratic equation where 'a' is called the leading coefficient and 'c' is called the absolute term of f (x).The first line of the input contains the three integer values a, b, and c of equation ax^2 + bx + c.
<b>Constraints</b>
1 ≤ a, b, c ≤ 50Print the two roots in two different lines and for imaginary roots print real and imaginary part of one root with (+/- and i )sign in between in one line and other in next line. For clarity see sample Output 2.
<b>Note</b> Imaginary roots can also be there and roots are considered upto 2 decimal places.Sample Input 1 :
4 -2 -10
Sample Output 2 :
1.85
-1.35
Sample Input 2 :
2 1 10
Sample Output 2:
-0.25+i2.22
-0.25-i2.22, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s=br.readLine().toString();
String[] str = s.split(" ");
float a=Float.parseFloat(str[0]);
float b=Float.parseFloat(str[1]);
float c=Float.parseFloat(str[2]);
float div = (float)(b*b-4*a*c);
if(div>0.0){
float alpha= (-b+(float)Math.sqrt(div))/(2*a);
float beta= (-b-(float)Math.sqrt(div))/(2*a);
System.out.printf("%.2f\n",alpha);
System.out.printf("%.2f",beta);
}
else{
float rp=-b/(2*a);
float ip=(float)Math.sqrt(-div)/(2*a);
System.out.printf("%.2f+i%.2f\n",rp,ip);
System.out.printf("%.2f-i%.2f\n",rp,ip);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the roots of a quadratic equation.
<b>Note</b>: Try to do it using Switch Case.
<b>Quadratic equations</b> are the polynomial equations of degree 2 in one variable of type f(x) = ax<sup>2</sup> + bx + c = 0 where a, b, c, ∈ R and a ≠ 0. It is the general form of a quadratic equation where 'a' is called the leading coefficient and 'c' is called the absolute term of f (x).The first line of the input contains the three integer values a, b, and c of equation ax^2 + bx + c.
<b>Constraints</b>
1 ≤ a, b, c ≤ 50Print the two roots in two different lines and for imaginary roots print real and imaginary part of one root with (+/- and i )sign in between in one line and other in next line. For clarity see sample Output 2.
<b>Note</b> Imaginary roots can also be there and roots are considered upto 2 decimal places.Sample Input 1 :
4 -2 -10
Sample Output 2 :
1.85
-1.35
Sample Input 2 :
2 1 10
Sample Output 2:
-0.25+i2.22
-0.25-i2.22, I have written this Solution Code: import math
a,b,c = map(int, input().split(' '))
disc = (b ** 2) - (4*a*c)
sq = disc ** 0.5
if disc > 0:
print("{:.2f}".format((-b + sq)/(2*a)))
print("{:.2f}".format((-b - sq)/(2*a)))
elif disc == 0:
print("{:.2f}".format(-b/(2*a)))
elif disc < 0:
r1 = complex((-b + sq)/(2*a))
r2 = complex((-b - sq)/(2*a))
print("{:.2f}+i{:.2f}".format(r1.real, abs(r1.imag)))
print("{:.2f}-i{:.2f}".format(r2.real, abs(r2.imag))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the roots of a quadratic equation.
<b>Note</b>: Try to do it using Switch Case.
<b>Quadratic equations</b> are the polynomial equations of degree 2 in one variable of type f(x) = ax<sup>2</sup> + bx + c = 0 where a, b, c, ∈ R and a ≠ 0. It is the general form of a quadratic equation where 'a' is called the leading coefficient and 'c' is called the absolute term of f (x).The first line of the input contains the three integer values a, b, and c of equation ax^2 + bx + c.
<b>Constraints</b>
1 ≤ a, b, c ≤ 50Print the two roots in two different lines and for imaginary roots print real and imaginary part of one root with (+/- and i )sign in between in one line and other in next line. For clarity see sample Output 2.
<b>Note</b> Imaginary roots can also be there and roots are considered upto 2 decimal places.Sample Input 1 :
4 -2 -10
Sample Output 2 :
1.85
-1.35
Sample Input 2 :
2 1 10
Sample Output 2:
-0.25+i2.22
-0.25-i2.22, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-01-19 02:44:22
**/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
int main() {
float a, b, c;
float root1, root2, imaginary;
float discriminant;
scanf("%f%f%f", &a, &b, &c);
discriminant = (b * b) - (4 * a * c);
switch (discriminant > 0) {
case 1:
root1 = (-b + sqrt(discriminant)) / (2 * a);
root2 = (-b - sqrt(discriminant)) / (2 * a);
printf("%.2f\n%.2f", root1, root2);
break;
case 0:
switch (discriminant < 0) {
case 1:
root1 = root2 = -b / (2 * a);
imaginary = sqrt(-discriminant) / (2 * a);
printf("%.2f + i%.2f\n%.2f - i%.2f", root1, imaginary, root2, imaginary);
break;
case 0:
root1 = root2 = -b / (2 * a);
printf("%.2f\n%.2f", root1, root2);
break;
}
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Complete the function <code>countBy</code>
Such that it takes a initial number which is the defualt value of out counter. And returns
a function which also takes a number and returns the initialCount + number supplied to second function.
Ex:-
<code>
const count = countBy(4) // initial value of counter 4, returns a function <br>
console. log(count(2)) // prints 6 because 4 + 2 <br>
console. log(count(-4)) // prints 2 because 6 - 4 <br>
console. log(count(8)) // prints 10 because 2 + 8 <br>
</code>
You have to return implement countBy function such that it can be run like that.<code>countBy</code> will take one number as input which will be the initial count.<code>countBy</code> will return a function which can be run many times and takes a number as input and returns the sum of it with previously maintained counter values<code>
const count = countBy(4) // initial value of counter 4, returns a function <br>
console. log(count(2)) // prints 6 because 4 + 2 <br>
console. log(count(-4)) // prints 2 because 6 - 4 <br>
console. log(count(8)) // prints 10 because 2 + 8 <br>
</code>, I have written this Solution Code: function countBy(initial){
let copy = initial
return (y)=>{
copy += y
return copy
}
// return the output using return keyword
// do not console.log it
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: n = int(input())
if (n%4==0 and n%100!=0 or n%400==0):
print("YES")
elif n==0:
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code:
import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int n = scanner.nextInt();
LeapYear(n);
}
static void LeapYear(int year){
if(year%400==0 || (year%100 != 0 && year%4==0)){System.out.println("YES");}
else {
System.out.println("NO");}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: def findMin(arr, low, high):
if high < low:
return arr[0]
if high == low:
return arr[low]
mid = int((low + high)/2)
if mid < high and arr[mid+1] < arr[mid]:
return arr[mid+1]
if mid > low and arr[mid] < arr[mid - 1]:
return arr[mid]
if arr[high] > arr[mid]:
return findMin(arr, low, mid-1)
return findMin(arr, mid+1, high)
T =int(input())
for i in range(T):
N = int(input())
arr = list(input().split())
for k in range(len(arr)):
arr[k] = int(arr[k])
print(findMin(arr,0,N-1)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
int l = 0, h = n+1;
if(a[1] < a[n]){
cout << a[1] << endl;
continue;
}
while(l+1 < h){
int m = (l + h) >> 1;
if(a[m] >= a[1])
l = m;
else
h = m;
}
cout << a[h] << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main{
public static void main (String[] args) {
//code
Scanner s = new Scanner(System.in);
int t = s.nextInt();
for(int j=0;j<t;j++){
int al = s.nextInt();
int a[] = new int[al];
for(int i=0;i<al;i++){
a[i] = s.nextInt();
}
binSearchSmallest(a);
}
}
public static void binSearchSmallest(int a[]) {
int s=0;
int e = a.length - 1;
int mid = 0;
while(s<=e){
mid = (s+e)/2;
if(a[s]<a[e]){
System.out.println(a[s]);
return;
}
if(a[mid]>=a[s]){
s=mid+1;
}
else{
e=mid;
}
if(s == e){
System.out.println(a[s]);
return;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: // arr is the input array
function findMin(arr,n) {
// write code here
// do not console.log
// return the number
let min = arr[0]
for(let i=1;i<arr.length;i++){
if(min > arr[i]){
min = arr[i]
}
}
return min
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It's Solo's 1st birthday and everyone is gifting her chocolates. There are N guests invited, and the i<sup>th</sup> guest gives Solo C<sub>i</sub> chocolates.
Find the total number of chocolates that Solo receives.The first line of the input contains an integer N, the number of guests.
The second line of the input contains N integers C<sub>1</sub>, C<sub>2</sub>, ....C<sub>N</sub>
Constraints
1 <= N <= 100
1 <= C<sub>i</sub> <= 100Output a single integer, the total number of chocolates that Solo receives.Sample Input
5
1 2 4 3 2
Sample Output
12
Explanation: Solo receives a total of 1+2+4+3+2 = 12 chocolates.
Sample Input
1
2
Sample Output
2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
int ans = 0;
For(i, 0, n){
int a; cin>>a;
ans += a;
}
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It's Solo's 1st birthday and everyone is gifting her chocolates. There are N guests invited, and the i<sup>th</sup> guest gives Solo C<sub>i</sub> chocolates.
Find the total number of chocolates that Solo receives.The first line of the input contains an integer N, the number of guests.
The second line of the input contains N integers C<sub>1</sub>, C<sub>2</sub>, ....C<sub>N</sub>
Constraints
1 <= N <= 100
1 <= C<sub>i</sub> <= 100Output a single integer, the total number of chocolates that Solo receives.Sample Input
5
1 2 4 3 2
Sample Output
12
Explanation: Solo receives a total of 1+2+4+3+2 = 12 chocolates.
Sample Input
1
2
Sample Output
2, I have written this Solution Code: n = int(input())
chocolates = list(map(int, input().strip().split(" ")))
count = 0
for val in chocolates:
count += val
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It's Solo's 1st birthday and everyone is gifting her chocolates. There are N guests invited, and the i<sup>th</sup> guest gives Solo C<sub>i</sub> chocolates.
Find the total number of chocolates that Solo receives.The first line of the input contains an integer N, the number of guests.
The second line of the input contains N integers C<sub>1</sub>, C<sub>2</sub>, ....C<sub>N</sub>
Constraints
1 <= N <= 100
1 <= C<sub>i</sub> <= 100Output a single integer, the total number of chocolates that Solo receives.Sample Input
5
1 2 4 3 2
Sample Output
12
Explanation: Solo receives a total of 1+2+4+3+2 = 12 chocolates.
Sample Input
1
2
Sample Output
2, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int A[] = new int[n];
for(int i=0;i<n;i++){
A[i]=sc.nextInt();
}
int Total=0;
for(int i=0;i<n;i++){
Total+=A[i];
}
System.out.print(Total);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, print all balanced bracket strings of length 2*N.
A bracket string is a string that contains only '(' and ')' as its characters.
<ul>
<li>Empty string is a balanced bracket string</li>
<li>If <i>S</i> is a balanced bracket string, so is <i>(S)</i></li>
<li>If <i>S</i> and <i>T</i> are balanced bracket strings, so is <i>ST</i></li>
</ul>
Print in lexicographical order. '(' appears before ')' in lexicographical orderThe single line of input containing an integer N.
1 <= N <= 15Print all possible balanced bracket strings of length 2*N in a separate line.Sample Input 1:
1
Sample Output 1:
()
Sample Input 2:
3
Sample Output 2:
((()))
(()())
()(())
()()()
Explanation:
It is printed in lexicographical order ., I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
vector<string> rec(int n){
if(n == 1){
string t = "()";
vector<string> v;
v.push_back(t);
return v;
}
vector<string> v = rec(n-1);
vector<string> cur;
for(auto i: v){
string t = "(" + i + ")";
cur.push_back(t);
}
for(auto i: v){
string t = "()" + i;
cur.push_back(t);
}
return cur;
}
void solve(){
int n; cin >> n;
vector<string> v = rec(n);
for(auto i: v)
cout << i << endl;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
testcases();
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, print all balanced bracket strings of length 2*N.
A bracket string is a string that contains only '(' and ')' as its characters.
<ul>
<li>Empty string is a balanced bracket string</li>
<li>If <i>S</i> is a balanced bracket string, so is <i>(S)</i></li>
<li>If <i>S</i> and <i>T</i> are balanced bracket strings, so is <i>ST</i></li>
</ul>
Print in lexicographical order. '(' appears before ')' in lexicographical orderThe single line of input containing an integer N.
1 <= N <= 15Print all possible balanced bracket strings of length 2*N in a separate line.Sample Input 1:
1
Sample Output 1:
()
Sample Input 2:
3
Sample Output 2:
((()))
(()())
()(())
()()()
Explanation:
It is printed in lexicographical order ., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static ArrayList premutaionsB(ArrayList<String> arr, int n){
if(n == 1)
return arr;
else{
ArrayList<String> arr2 = new ArrayList<>();
for(int i=0;i<arr.size();i++)
arr2.add("(" + arr.get(i) + ")");
for(int i=0;i<arr.size();i++)
arr2.add("()" + arr.get(i));
return premutaionsB(arr2, n-1);
}
}
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
ArrayList<String> arr = new ArrayList<>();
arr.add("()");
arr = premutaionsB(arr, n);
for(int i=0;i<arr.size();i++)
System.out.println(arr.get(i));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an index K, return the K-th row of <a href = "https://en.wikipedia.org/wiki/Pascal%27s_triangle">Pascal’s triangle</a>.
You must print the K-th row modulo 10<sup>9</sup> + 7.
<b>The rows of Pascal's triangle are conventionally enumerated starting with row K=0 at the top (the 0th row) </b>The only line of input contains the input K.
<b>Constraints:-</b>
0 ≤ K ≤ 3000
Print k-th row of Pascal's Triangle containing k+1 integers modulo 10^9+7.Sample Input :
3
Sample Output:
1 3 3 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int k = sc.nextInt();
int digit = 1;
int m = 1000000007;;
int arr[] = new int[k+1];
arr[0] = 1;
arr[1] = 1;
for(int i=2; i<=k; i++){
for(int j=i;j>=1;j--){
arr[j] += arr[j-1];
arr[j] = arr[j]%m;
}
}
for(int i=0;i<=k;i++){
System.out.print(arr[i] + " ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an index K, return the K-th row of <a href = "https://en.wikipedia.org/wiki/Pascal%27s_triangle">Pascal’s triangle</a>.
You must print the K-th row modulo 10<sup>9</sup> + 7.
<b>The rows of Pascal's triangle are conventionally enumerated starting with row K=0 at the top (the 0th row) </b>The only line of input contains the input K.
<b>Constraints:-</b>
0 ≤ K ≤ 3000
Print k-th row of Pascal's Triangle containing k+1 integers modulo 10^9+7.Sample Input :
3
Sample Output:
1 3 3 1, I have written this Solution Code: def generateNthRow (N):
prev = 1
print(prev, end = '')
for i in range(1, N + 1):
curr = (prev * (N - i + 1)) // i
print("", curr%1000000007, end = '')
prev = curr
N = int(input())
generateNthRow(N), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an index K, return the K-th row of <a href = "https://en.wikipedia.org/wiki/Pascal%27s_triangle">Pascal’s triangle</a>.
You must print the K-th row modulo 10<sup>9</sup> + 7.
<b>The rows of Pascal's triangle are conventionally enumerated starting with row K=0 at the top (the 0th row) </b>The only line of input contains the input K.
<b>Constraints:-</b>
0 ≤ K ≤ 3000
Print k-th row of Pascal's Triangle containing k+1 integers modulo 10^9+7.Sample Input :
3
Sample Output:
1 3 3 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
signed main()
{
int t,n;
t=1;
vector<vector<long long> > v(3002);
for(int i=0; i<=3001; i++)
{
for (int j=0; j<=i; j++)
{
if (j==0 || j==i) v[i].push_back(1);
else v[i].push_back((v[i-1][j]%mod + v[i-1][j-1]%mod)%mod);
}
}
while (t--)
{
cin>>n;
n=n+1;
for (int i=0; i<v[n-1].size(); i++) cout<<v[n-1][i]<<" ";
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita likes a number if it is stored in an integer while Doraemon likes it when it is stored in a String. Your task is to write a code so that they can easily convert an integer to a string or a string to an integer whenever they want.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the following functions:-
<b>StringToInt()</b> that takes String S as parameter.
<b>IntToString()</b> that takes the integer N as parameter.
Constraints:-
1 <= (Given Number) <= 100Return an integer in <b>StringToInt()</b> while return a integer integer in <b>IntToString()</b>. The driver code will print "<b>Nice Job</b>" if your code is correct otherwise "<b>Wrong answer</b>".Sample Input:-
5
Sample Output:-
Nice Job
Sample Input:-
12
Sample Output:-
Nice Job, I have written this Solution Code: def StringToInt(a):
return int(a)
def IntToString(a):
return str(a)
if __name__ == "__main__":
n = input()
s = StringToInt(n)
if n == str(s):
a=1
# print("Nice Job")
else:
print("Wrong answer")
quit()
p = IntToString(s)
if s == int(p):
print("Nice Job")
else:
print("Wrong answer"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita likes a number if it is stored in an integer while Doraemon likes it when it is stored in a String. Your task is to write a code so that they can easily convert an integer to a string or a string to an integer whenever they want.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the following functions:-
<b>StringToInt()</b> that takes String S as parameter.
<b>IntToString()</b> that takes the integer N as parameter.
Constraints:-
1 <= (Given Number) <= 100Return an integer in <b>StringToInt()</b> while return a integer integer in <b>IntToString()</b>. The driver code will print "<b>Nice Job</b>" if your code is correct otherwise "<b>Wrong answer</b>".Sample Input:-
5
Sample Output:-
Nice Job
Sample Input:-
12
Sample Output:-
Nice Job, I have written this Solution Code: static int StringToInt(String S)
{
return Integer.parseInt(S);
}
static String IntToString(int N){
return String.valueOf(N);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: def DivisorProblem(N):
ans=0
while N>1:
cnt=2
while N%cnt!=0:
cnt=cnt+1
N = N//cnt
ans=ans+1
return ans
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: static int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: def Print_Digit(n):
dc = {1: "one", 2: "two", 3: "three", 4: "four",
5: "five", 6: "six", 7: "seven", 8: "eight", 9: "nine", 0: "zero"}
final_list = []
while (n > 0):
final_list.append(dc[int(n%10)])
n = int(n / 10)
for val in final_list[::-1]:
print(val, end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: class Solution {
public static void Print_Digits(int N){
if(N==0){return;}
Print_Digits(N/10);
int x=N%10;
if(x==1){System.out.print("one ");}
else if(x==2){System.out.print("two ");}
else if(x==3){System.out.print("three ");}
else if(x==4){System.out.print("four ");}
else if(x==5){System.out.print("five ");}
else if(x==6){System.out.print("six ");}
else if(x==7){System.out.print("seven ");}
else if(x==8){System.out.print("eight ");}
else if(x==9){System.out.print("nine ");}
else if(x==0){System.out.print("zero ");}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: public static void For_Loop(int n){
for(int i=1;i<=n;i++){
if(i%2==1){System.out.print("odd ");}
else{
System.out.print("even ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: n = int(input())
for i in range(1, n+1):
if(i%2)==0:
print("even ",end="")
else:
print("odd ",end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan is a college student. Nutan arrived at college and listened to his friends discussing 'loyal' numbers in a matrix. But Nutan had no idea what 'loyal' numbers were, so he asked his friends. A 'loyal' number, according to one of his friends, is a matrix element that is the smallest in its row and the largest in its column.
Now it's time to test Nutan's understanding. He is given a m x n matrix of distinct numbers and is asked to return all ‘loyal’ numbers in any order.
As a friend of Nutan, help him to solve the problem.The first line contains the numbers of rows r and number of columns c as space separated.
For the next r lines, contains the c elements of the matrix each line as a space separated.
<b>Constraints</b>
m == mat. length
n == mat[i]. length
1 <= n, m <= 10
-100 <= matrix[i][j] <= 100.
All elements in the matrix are distinct.Print the elements of the array containing the loyal numbers in a single line space separated, if it's not present then print -1.Sample Input
2 2
7 8
1 2
Sample Output
7
Explanation
7 is the only loyal number since it is the minimum in its row and the maximum in its column., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
int m=sc.nextInt();
int n=sc.nextInt();
int[][]mat=new int[m][n];
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
mat[i][j]=sc.nextInt();
}
}
Vector<Integer> ans = new Vector<Integer>();
ans = minmaxNumbers(mat,ans);
if(ans.size()==0){
System.out.println("-1");
}
for(int i=0;i<ans.size();i++){
System.out.println(ans.get(i));
}
}
public static Vector<Integer> minmaxNumbers(int[][] matrix, Vector<Integer> res)
{
Set<Integer> set = new HashSet<Integer>();
for(int i=0;i<matrix.length;i++){
int minr = Integer.MAX_VALUE;
for(int j=0;j<matrix[i].length;j++){
minr = Math.min(minr,matrix[i][j]);
}
set.add(minr);
}
for(int j=0;j<matrix[0].length;j++){
int maxc =Integer.MIN_VALUE;
for(int i=0;i<matrix.length;i++){
maxc = Math.max(maxc,matrix[i][j]);
}
if(set.contains(maxc)){
res.add(maxc);
}
}
return res;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
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