post_href
stringlengths 57
213
| python_solutions
stringlengths 71
22.3k
| slug
stringlengths 3
77
| post_title
stringlengths 1
100
| user
stringlengths 3
29
| upvotes
int64 -20
1.2k
| views
int64 0
60.9k
| problem_title
stringlengths 3
77
| number
int64 1
2.48k
| acceptance
float64 0.14
0.91
| difficulty
stringclasses 3
values | __index_level_0__
int64 0
34k
|
|---|---|---|---|---|---|---|---|---|---|---|---|
https://leetcode.com/problems/a-number-after-a-double-reversal/discuss/1798270/1-Line-Python-Solution-oror-90-Faster-oror-Memory-less-than-90 | class Solution:
def isSameAfterReversals(self, num: int) -> bool:
return True if str(num)[-1]!='0' or num==0 else False | a-number-after-a-double-reversal | 1-Line Python Solution || 90% Faster || Memory less than 90% | Taha-C | 0 | 31 | a number after a double reversal | 2,119 | 0.758 | Easy | 29,300 |
https://leetcode.com/problems/a-number-after-a-double-reversal/discuss/1791770/Python3-accepted-one-liner-solution | class Solution:
def isSameAfterReversals(self, num: int) -> bool:
return num == int(str(int(str(num)[::-1]))[::-1]) | a-number-after-a-double-reversal | Python3 accepted one-liner solution | sreeleetcode19 | 0 | 29 | a number after a double reversal | 2,119 | 0.758 | Easy | 29,301 |
https://leetcode.com/problems/a-number-after-a-double-reversal/discuss/1695969/Python-3-1-line-O(1)-time-O(1)-space | class Solution:
def isSameAfterReversals(self, num: int) -> bool:
return num % 10 != 0 or num == 0 | a-number-after-a-double-reversal | Python 3, 1 line, O(1) time, O(1) space | dereky4 | 0 | 56 | a number after a double reversal | 2,119 | 0.758 | Easy | 29,302 |
https://leetcode.com/problems/a-number-after-a-double-reversal/discuss/1690405/**-Python-code%3A-1-Linear | class Solution:
def isSameAfterReversals(self, num: int) -> bool:
return num==int(str(int(str(num)[::-1]))[::-1]) | a-number-after-a-double-reversal | ** Python code: 1-Linear | Anilchouhan181 | 0 | 19 | a number after a double reversal | 2,119 | 0.758 | Easy | 29,303 |
https://leetcode.com/problems/a-number-after-a-double-reversal/discuss/1683461/return-num-10-!-0-or-num-0-lessEOPgreater | class Solution:
def isSameAfterReversals(self, num: int) -> bool:
return num % 10 != 0 or num == 0 | a-number-after-a-double-reversal | return num % 10 != 0 or num == 0 <EOP> | snagsbybalin | 0 | 14 | a number after a double reversal | 2,119 | 0.758 | Easy | 29,304 |
https://leetcode.com/problems/a-number-after-a-double-reversal/discuss/1658533/Python-1-Liner-Simple | class Solution:
def isSameAfterReversals(self, num: int) -> bool:
return num < 10 or num % 10 | a-number-after-a-double-reversal | Python 1 Liner Simple | dhnam2234 | 0 | 46 | a number after a double reversal | 2,119 | 0.758 | Easy | 29,305 |
https://leetcode.com/problems/a-number-after-a-double-reversal/discuss/1647456/Python3-1-line | class Solution:
def isSameAfterReversals(self, num: int) -> bool:
return num == 0 or num % 10 | a-number-after-a-double-reversal | [Python3] 1-line | ye15 | 0 | 50 | a number after a double reversal | 2,119 | 0.758 | Easy | 29,306 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/2838969/Python-Straight-forward-O(n-2)-solution-(still-fast-85) | class Solution:
def executeInstructions(self, n: int, startPos: list[int], s: str) -> list[int]:
def num_of_valid_instructions(s, pos, start, end):
row, colon = pos
k = 0
for i in range(start, end):
cur = s[i]
row += (cur == 'D') - (cur == 'U')
colon += (cur == 'R') - (cur == 'L')
if not (0 <= row < n and 0 <= colon < n):
return k
k += 1
return k
ans = []
for i in range(len(s)):
ans.append(num_of_valid_instructions(s, startPos, i, len(s)))
return ans | execution-of-all-suffix-instructions-staying-in-a-grid | [Python] Straight-forward O(n ^ 2) solution (still fast - 85%) | Mark_computer | 1 | 4 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,307 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/1696798/python-brute-force | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
d = {'U':(-1, 0),'D':(1, 0), 'R':(0, 1), 'L':(0, -1)}
m = len(s)
ans = [0]*m
def isBounded(start, end):
return (0<= start < n and 0<= end < n)
for i in range(m):
x, y = startPos
for j in range(i, m):
dx, dy = d[s[j]]
x += dx; y += dy
if not isBounded(x, y):
break
ans[i] += 1
return ans | execution-of-all-suffix-instructions-staying-in-a-grid | python brute force | abkc1221 | 1 | 82 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,308 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/1647470/Python3-brute-force | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
ans = []
for k in range(len(s)):
i, j = startPos
val = 0
for kk in range(k, len(s)):
if s[kk] == 'L': j -= 1
elif s[kk] == 'R': j += 1
elif s[kk] == 'U': i -= 1
else: i += 1
if 0 <= i < n and 0 <= j < n: val += 1
else: break
ans.append(val)
return ans | execution-of-all-suffix-instructions-staying-in-a-grid | [Python3] brute-force | ye15 | 1 | 61 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,309 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/1647470/Python3-brute-force | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
@cache
def fn(i, j, k):
"""Return valid number of instructions at (i, j) starting from kth."""
if k == len(s): return 0
if s[k] == 'L': j -= 1
elif s[k] == 'R': j += 1
elif s[k] == 'U': i -= 1
else: i += 1
if 0 <= i < n and 0 <= j < n: return 1 + fn(i, j, k+1)
return 0
return [fn(*startPos, k) for k in range(len(s))] | execution-of-all-suffix-instructions-staying-in-a-grid | [Python3] brute-force | ye15 | 1 | 61 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,310 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/2792292/HOLY-COW!-Explained-trivial-python3-solution | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
ans = []
for i in range(len(s)):
# create short version of instructions
# for current iteration
instructions = s[i:]
cnt = 0
x, y = startPos[0], startPos[1]
# for every instruction in this iteration
for inst in instructions:
# move
if inst == "R":
y += 1
elif inst == "L":
y -= 1
elif inst == "D":
x += 1
else:
x -= 1
# check conditions
if x < 0 or y < 0 or x >= n or y >= n:
break
# counting steps
cnt += 1
ans.append(cnt)
return ans | execution-of-all-suffix-instructions-staying-in-a-grid | HOLY COW! Explained trivial python3 solution | jestpunk | 0 | 4 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,311 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/2778967/Easy-Python-Code | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
lenn=len(s)
res=[0]*lenn
def rec(i,j,steps,k):
if i<0 or j<0 or i==n or j==n:
return steps-1
if k==lenn:
return steps
if s[k]=='R':
return rec(i,j+1,steps+1,k+1)
if s[k]=='L':
return rec(i,j-1,steps+1,k+1)
if s[k]=='U':
return rec(i-1,j,steps+1,k+1)
if s[k]=='D':
return rec(i+1,j,steps+1,k+1)
for i in range(lenn):
res[i]=rec(startPos[0],startPos[1],0,i)
return res | execution-of-all-suffix-instructions-staying-in-a-grid | Easy Python Code | Chetan_007 | 0 | 1 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,312 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/2755985/python-easy-solutionoror-Bruteforce-approach | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
ans = []
r = startPos[0]
c = startPos[1]
count = 0
while s!='':
for j in range(len(s)):
if s[j]=="R":
if c<n-1:
c+=1
count+=1
else:
ans.append(count)
count = 0
break
elif s[j]=="L":
if c>0:
c-=1
count+=1
else:
ans.append(count)
count = 0
break
elif s[j]=="D":
if r<n-1:
r+=1
count+=1
else:
ans.append(count)
count = 0
break
elif s[j]=="U":
if r>0:
r-=1
count+=1
else:
ans.append(count)
count = 0
break
if count==len(s):
ans.append(count)
count = 0
r = startPos[0]
c = startPos[1]
s = s[1:]
return ans | execution-of-all-suffix-instructions-staying-in-a-grid | ✅✅python easy solution|| Bruteforce approach | chessman_1 | 0 | 3 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,313 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/2752939/Python-bruteforce | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
directions = {"R":(0,1),"L":(0,-1),"U":(-1,0),"D":(1,0)}
res = []
for startingInstruction in range(len(s)):
x,y = startPos
successful_instructions = 0
for instruction in s[startingInstruction:]:
dx,dy = directions[instruction]
x,y = x+dx,y+dy
if x >= 0 and x < n and y >= 0 and y < n:
successful_instructions += 1
else: break
res.append(successful_instructions)
return res | execution-of-all-suffix-instructions-staying-in-a-grid | Python bruteforce | shriyansnaik | 0 | 2 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,314 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/2700326/Python3-Store-FIRST-time-at-X-and-Y-positions-in-dictionary | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
fx, fy, m = {startPos[1] : 0}, {startPos[0]: 0}, {'U': (0, -1) , 'D': (0, 1), 'L': (-1, 0) , 'R': (1, 0)}
ret = []
for i in reversed(range(len(s))):
offx, offy = m[s[i]]
fx = {k + offx: v + 1 for k, v in fx.items()}
fy = {k + offy: v + 1 for k, v in fy.items()}
fx.update({startPos[1] : 0})
fy.update({startPos[0] : 0})
mo = len(s) + 1 - i
for d in [fx, fy]:
for k, v in d.items():
if (k < 0 or k >= n) and v<mo:
mo = v
ret.insert(0, mo-1)
return ret | execution-of-all-suffix-instructions-staying-in-a-grid | Python3 Store FIRST time at X and Y positions in dictionary | godshiva | 0 | 6 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,315 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/2699569/Python3-Solution-nested-loops-beats-90 | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
dict_ = {'R':1, 'D':1, 'L':-1, 'U':-1, }
ans = []
for i in range(len(s)):
m = 0
x, y = startPos
for c in s[i:]:
if c in ('R', 'L'):
y += dict_[c]
if y == n or y < 0:
break
else:
m += 1
else:
x += dict_[c]
if x == n or x < 0:
print('break')
break
else:
m += 1
ans.append(m)
return ans | execution-of-all-suffix-instructions-staying-in-a-grid | Python3 Solution - nested loops, beats 90% | sipi09 | 0 | 4 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,316 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/2698507/Python3-Simple-Solution | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
res = []
for i in range(len(s)):
row, col = startPos
tmp = 0
for inst in s[i:]:
if inst == 'R' and col + 1 < n:
tmp, col = tmp + 1, col + 1
elif inst == 'L' and col - 1 >= 0:
tmp, col = tmp + 1, col - 1
elif inst == 'U' and row - 1 >= 0:
tmp, row = tmp + 1, row - 1
elif inst == 'D' and row + 1 < n:
tmp, row = tmp + 1, row + 1
else: break
res.append(tmp)
return res | execution-of-all-suffix-instructions-staying-in-a-grid | Python3 Simple Solution | mediocre-coder | 0 | 3 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,317 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/2688270/Python | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
move = {'L': [0,-1], 'D': [1,0], 'U': [-1,0], 'R': [0,1]}
res = []
for i in range(len(s)):
counter = 0
x = startPos[0]
y = startPos[1]
for j in range(i, len(s)):
if 0 <= x+ move[s[j]][0] < n and 0 <= y+move[s[j]][1] < n :
x = x+ move[s[j]][0]
y = y+ move[s[j]][1]
counter +=1
else:
break
res.append(counter)
return res | execution-of-all-suffix-instructions-staying-in-a-grid | Python | naveenraiit | 0 | 3 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,318 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/2206750/C%2B%2B-Python-Kotlin-oror-Easy-Explained-Solution-oror-O(n2) | class Solution(object):
def executeInstructions(self, n, startPos, s):
li = []
count = 0;
ini = startPos[0]
inj = startPos[1]
for i in range(len(s)):
for j in range(i, len(s)):
if s[j] == 'R' : startPos[1] += 1
elif s[j] == 'L' : startPos[1] -= 1
elif s[j] == 'U' : startPos[0] -= 1
elif s[j] == 'D' : startPos[0] += 1
if (startPos[1] > -1 and startPos[1] < n and startPos[0] > -1 and startPos[0] < n) : count += 1
else : startPos = [-1,-1]
startPos = [ini, inj]
li.append(count)
count = 0
return li | execution-of-all-suffix-instructions-staying-in-a-grid | C++, Python, Kotlin || Easy Explained Solution || O(n^2) | r_o_xx_ | 0 | 24 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,319 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/2044450/Python-Easy-Solution | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
result = []
for i in range(len(s)):
moves = s[i:]
number_of_moves = 0
position = list(startPos)
for move in moves:
if move == "L":
position[1] -= 1
if move == "R":
position[1] += 1
if move == "U":
position[0] -= 1
if move == "D":
position[0] += 1
if position[0] >= n or position[1] >= n or position[0] < 0 or position[1] < 0:
break
number_of_moves += 1
result.append(number_of_moves)
return result | execution-of-all-suffix-instructions-staying-in-a-grid | Python - Easy Solution | TrueJacobG | 0 | 49 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,320 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/1939495/Python3-Brute-Force-Solution | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
output = []
ns = len(s)
for i in range(ns):
count = 0
Pos = [startPos[0], startPos[1]]
for j in range(i, ns):
if s[j] == 'L':
Pos[1] -= 1
elif s[j] == 'R':
Pos[1] += 1
elif s[j] == 'U':
Pos[0] -= 1
elif s[j] == 'D':
Pos[0] += 1
if Pos[0] < 0 or Pos[0] >= n:
break
if Pos[1] < 0 or Pos[1] >= n:
break
count += 1
output.append(count)
return output | execution-of-all-suffix-instructions-staying-in-a-grid | [Python3] Brute Force Solution | terrencetang | 0 | 22 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,321 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/1794625/Python-Simple-Brute-Force-Solution | class Solution:
def executeInstructions(self, n: int, sp: List[int], s: str) -> List[int]:
def move_up(pos):
pos[0] -= 1
return pos
def move_down(pos):
pos[0] += 1
return pos
def move_right(pos):
pos[1] += 1
return pos
def move_left(pos):
pos[1] -= 1
return pos
ans = []
funcs = {
'R': move_right,
'L': move_left,
'U': move_up,
'D': move_down
}
for i in range(len(s)):
count = 0
temp = sp.copy()
for j in range(i, len(s)):
temp = funcs[s[j]](temp)
if temp[0] >= 0 and temp[1] >= 0 and temp[0] < n and temp[1] < n:
count += 1
else:
break
ans.append(count)
return ans | execution-of-all-suffix-instructions-staying-in-a-grid | Python Simple Brute-Force Solution | dos_77 | 0 | 46 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,322 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/1695189/Python-O(m2)-using-imaginary-numbers | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
d = dict(U=-1, D=1, L=-1j, R=1j)
start = startPos[0] + 1j*startPos[1]
m = len(s)
l = [0]*m
for i in range(m):
pos = start
for j in range(i, m):
pos += d[s[j]]
if not (0 <= pos.real < n and 0 <= pos.imag < n):
break
l[i] += 1
return l | execution-of-all-suffix-instructions-staying-in-a-grid | Python O(m^2), using imaginary numbers | emwalker | 0 | 49 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,323 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/1659660/Python-Super-Easy-but-Slow-. | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
arr=[]
if(n==1):
for i in range(len(s)):
arr.append(0)
return arr
else:
for a in range(len(s)):
count=0
i=startPos[0]
j=startPos[1]
for k in range(a,len(s)):
if s[k]=='R':
j+=1
elif s[k]=='L':
j-=1
elif s[k]=='D':
i+=1
elif s[k]=='U':
i-=1
if(j>=0) and (j<=n-1):
if(i>=0) and (i<=n-1):
count+=1
else:
break
else:
break
arr.append(count)
return arr | execution-of-all-suffix-instructions-staying-in-a-grid | Python Super Easy but Slow . | alimlalani | 0 | 54 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,324 |
https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/discuss/1647557/Python3-Brute-Force | class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
grid = [[0]*n for _ in range(n)]
directions = {'R':(0,1),'L':(0,-1),'U':(-1,0),'D':(1,0)}
l = len(s)
ans = [0]*l
temp = startPos.copy()
for i in range(l):
step = 0
startPos = temp.copy()
for j in range(i,l):
d = directions[s[j]]
new_start = [startPos[0]+d[0],startPos[1]+d[1]]
if 0<=new_start[0]<n and 0<=new_start[1]<n :
step += 1
startPos = new_start
else:
break
ans[i] = step
return ans | execution-of-all-suffix-instructions-staying-in-a-grid | Python3 Brute-Force | AndrewHou | 0 | 34 | execution of all suffix instructions staying in a grid | 2,120 | 0.836 | Medium | 29,325 |
https://leetcode.com/problems/intervals-between-identical-elements/discuss/1647480/Python3-prefix-sum | class Solution:
def getDistances(self, arr: List[int]) -> List[int]:
loc = defaultdict(list)
for i, x in enumerate(arr): loc[x].append(i)
for k, idx in loc.items():
prefix = list(accumulate(idx, initial=0))
vals = []
for i, x in enumerate(idx):
vals.append(prefix[-1] - prefix[i] - prefix[i+1] - (len(idx)-2*i-1)*x)
loc[k] = deque(vals)
return [loc[x].popleft() for x in arr] | intervals-between-identical-elements | [Python3] prefix sum | ye15 | 9 | 1,100 | intervals between identical elements | 2,121 | 0.433 | Medium | 29,326 |
https://leetcode.com/problems/intervals-between-identical-elements/discuss/1706167/Python3-concise-code-with-visual-notes. | class Solution:
def getDistances(self, arr: List[int]) -> List[int]:
n = len(arr)
d = defaultdict(list)
for i, v in enumerate(arr): d[v].append(i)
res = defaultdict(list)
for v, idx in d.items():
ps = list(accumulate(idx, initial=0))
vals = []
idn = len(idx)
for i, x in enumerate(idx):
vals.append(i*x-ps[i] + ps[-1]-ps[i+1]-(idn-i-1)*x)
res[v] = deque(vals)
return [res[v].popleft() for v in arr] | intervals-between-identical-elements | Python3 concise code with visual notes. | danielxuforever | 1 | 191 | intervals between identical elements | 2,121 | 0.433 | Medium | 29,327 |
https://leetcode.com/problems/intervals-between-identical-elements/discuss/1647662/Clear-Prefix-Sum-O(n) | class Solution:
def getDistances(self, arr: List[int]) -> List[int]:
"""
The key fact is that
result[i] = sum(i - indices below i) + sum(indices above i - i)
This implies
results[i] = sum(indices above i) - sum(indices below i) +
i * (number of indices above i - number of indices below i)
Fortunately, you can update the sums in constant time.
"""
indicesAbove = {}
indicesBelow = {}
runningSumAbove = {}
runningSumBelow = {}
result = [0] * len(arr)
for i, n in enumerate(arr):
if n not in indicesAbove:
indicesAbove[n] = 1
indicesBelow[n] = 0
runningSumAbove[n] = i
runningSumBelow[n] = 0
else:
indicesAbove[n] += 1
runningSumAbove[n] += i
# result = sum of numbers above - sum of #s below + pivot * (Nb - Na)
for i, n in enumerate(arr):
runningSumAbove[n] -= i
indicesAbove[n] -= 1
result[i] = runningSumAbove[n] - runningSumBelow[n] + i * (indicesBelow[n] - indicesAbove[n])
indicesBelow[n] += 1
runningSumBelow[n] += i
return result
``` | intervals-between-identical-elements | Clear Prefix Sum O(n) | ealster2004 | 1 | 63 | intervals between identical elements | 2,121 | 0.433 | Medium | 29,328 |
https://leetcode.com/problems/intervals-between-identical-elements/discuss/2596930/Python-Explained-Preifx-Sum | class Solution:
def getDistances(self, arr: List[int]) -> List[int]:
# make a dict to save indices
idces = collections.defaultdict(list)
# search for all common numbers and make the computations
for idx, num in enumerate(arr):
# check whether we already found one
idces[num].append(idx)
# reset the array
arr = [0 for _ in arr]
# go through all of the numbers and calculate the sums
for ls in idces.values():
# go through all other elements
for idx, num in enumerate(ls):
for idx2, num2 in enumerate(ls):
if idx != idx2:
arr[num] += abs(num-num2)
return arr | intervals-between-identical-elements | [Python] - Explained Preifx-Sum | Lucew | 0 | 35 | intervals between identical elements | 2,121 | 0.433 | Medium | 29,329 |
https://leetcode.com/problems/intervals-between-identical-elements/discuss/2596930/Python-Explained-Preifx-Sum | class Solution:
def getDistances(self, arr: List[int]) -> List[int]:
# make a dict to save indices
idces = collections.defaultdict(list)
# search for all common numbers and make the computations
for idx, num in enumerate(arr):
# check whether we already found one
idces[num].append(idx)
# go through all of the numbers and calculate the sums
for ls in idces.values():
# make the prefix sum
prefix = [ls[0] for _ in ls]
n = len(ls)
for idx, num in enumerate(ls[1:], 1):
prefix[idx] = prefix[idx-1] + num
# calculate the result
for idx, num in enumerate(ls):
# get the left part (current index - sum before the index)
left = 0
if idx > 0:
left = idx*num - prefix[idx-1]
# get the right part (sum after the index, minus current index)
right = 0
if idx < n-1:
right = prefix[-1] - prefix[idx] - (n-1-idx)*num
arr[num] = left + right
return arr | intervals-between-identical-elements | [Python] - Explained Preifx-Sum | Lucew | 0 | 35 | intervals between identical elements | 2,121 | 0.433 | Medium | 29,330 |
https://leetcode.com/problems/intervals-between-identical-elements/discuss/2404854/Why-my-solution-doesn't-work | class Solution:
def getDistances(self, arr: List[int]) -> List[int]:
idx = {}
for i,n in enumerate(arr):
idx[n] = idx.get(n, [])+[i]
ans = [0]*len(arr)
for l in idx.values():
tot = sum(l)
length = len(l)
for i in range(len(l)):
ans[l[i]] = tot+l[i]*i-l[i]*(length-i)
tot-=l[i]*2
return ans | intervals-between-identical-elements | Why my solution doesn't work? | li87o | 0 | 27 | intervals between identical elements | 2,121 | 0.433 | Medium | 29,331 |
https://leetcode.com/problems/intervals-between-identical-elements/discuss/1651838/Python-Simple-Prefix-Sum-Solution | class Solution:
def getDistances(self, arr: List[int]) -> List[int]:
pre, hmap = [(0,0)]*len(arr), {}
for i, val in enumerate(arr):
if val not in hmap: hmap[val] = (0,0)
//hmap[num] stores : [sum of indexes where num is present, frequency of occurence of num]
hmap[val] = [hmap.get(val, 0)[0] + i, hmap.get(val, 0)[1] + 1]
pre[i] = hmap[val]
res = [0] * len(arr)
for i, val in enumerate(arr):
//right = (sum of indexes where arr[i] is present towards right of i) - ( (current index i) * (frequency of arr[i] from i+1 to n
//left = (frequency of arr[i] before i ( excluding i, so -1)) * (current index i)- (sum of indexes where arr[i] is present from 0 to i, excluding i)
right = (hmap[val][0] - pre[i][0]) - (hmap[val][1] - pre[i][1]) * i
left = (pre[i][1]-1) * i - (pre[i][0] - i)
res[i] = right + left
return res | intervals-between-identical-elements | [Python] Simple Prefix Sum Solution | Saksham003 | 0 | 136 | intervals between identical elements | 2,121 | 0.433 | Medium | 29,332 |
https://leetcode.com/problems/intervals-between-identical-elements/discuss/1650782/Linear-solution-100-speed | class Solution:
def getDistances(self, arr: List[int]) -> List[int]:
val_idx = defaultdict(list)
for i, n in enumerate(arr):
val_idx[n].append(i)
ans = [0] * len(arr)
for lst_idx in val_idx.values():
len_lst_idx = len(lst_idx)
if len_lst_idx == 1:
continue
sum_l, sum_r = 0, sum(lst_idx)
len_lst_idx1 = len_lst_idx - 1
for i, n in enumerate(lst_idx):
sum_r -= n
ans[lst_idx[i]] = sum_r + n * (i * 2 - len_lst_idx1) - sum_l
sum_l += n
return ans | intervals-between-identical-elements | Linear solution, 100% speed | EvgenySH | 0 | 97 | intervals between identical elements | 2,121 | 0.433 | Medium | 29,333 |
https://leetcode.com/problems/intervals-between-identical-elements/discuss/1648226/Simple-approach%3A-code-%2B-comments | class Solution:
# computes for each element of the list the sum of intervals with other elements when their values are equal; returns computed sums in a list
def getDistances(self, arr: List[int]) -> List[int]:
# e1 e2 e3 -> e4 e5 e6 e7
# |--L---|-di-|----R----|
# soi += di*(L-R)
n=len(arr)
ans=[None]*n
D={} # [soi,ei,L,R]
# inits the first soi for each value
for i,v in enumerate(arr):
if v not in D:
D[v]=[0,i,0,1]
else:
soi,first_i,_,k=D[v]
D[v]=soi+(i-first_i),first_i,0,k+1
# fills the list ans
for i,v in enumerate(arr):
soi,prev_i,L,R=D[v]
soi+=(i-prev_i)*(L-R)
D[v]=soi,i,L+1,R-1
ans[i]=soi
return ans | intervals-between-identical-elements | Simple approach: code + comments | volokh0x | 0 | 45 | intervals between identical elements | 2,121 | 0.433 | Medium | 29,334 |
https://leetcode.com/problems/intervals-between-identical-elements/discuss/1647619/Python3-Index-hashmap-%2B-prefix-sum-%2B-binary-search | class Solution:
def getDistances(self, arr: List[int]) -> List[int]:
indices = defaultdict(list)
prefsum = {}
for i, a in enumerate(arr):
indices[a].append(i)
if a not in prefsum:
prefsum[a] = [i]
else:
prefsum[a].append(i + prefsum[a][-1])
res = []
for i, a in enumerate(arr):
idx = bisect_right(indices[a], i) - 1
# count of smaller, count of larger
sc, lc = idx, len(indices[a]) - idx - 1
ss = prefsum[a][idx - 1] if idx > 0 else 0 # sum of smaller
ls = prefsum[a][-1] - prefsum[a][idx] # sum of larger
res.append(i * sc - ss + ls - i * lc)
return res | intervals-between-identical-elements | [Python3] Index hashmap + prefix sum + binary search | FanchenBao | 0 | 49 | intervals between identical elements | 2,121 | 0.433 | Medium | 29,335 |
https://leetcode.com/problems/recover-the-original-array/discuss/1647488/Python3-brute-force | class Solution:
def recoverArray(self, nums: List[int]) -> List[int]:
nums.sort()
cnt = Counter(nums)
for i in range(1, len(nums)):
diff = nums[i] - nums[0]
if diff and diff&1 == 0:
ans = []
freq = cnt.copy()
for k, v in freq.items():
if v:
if freq[k+diff] < v: break
ans.extend([k+diff//2]*v)
freq[k+diff] -= v
else: return ans | recover-the-original-array | [Python3] brute-force | ye15 | 11 | 503 | recover the original array | 2,122 | 0.381 | Hard | 29,336 |
https://leetcode.com/problems/recover-the-original-array/discuss/1647692/Python3-Try-All-Valid-K-Values | class Solution:
def recoverArray(self, nums):
self.n, self.nums = len(nums) // 2, sorted(nums)
small = self.nums[0]
for x in self.nums[1:]:
k = x - small # this represents 2 * k from the problem statement
if k % 2 or not k: continue
temp = self.valid(k)
if temp: return temp
return []
def valid(self, k):
counts = defaultdict(list)
for i in range(len(self.nums) - 1, -1, -1):
counts[self.nums[i]].append(i)
# go through each value from smallest to largest
# at each value check for corresponding (val + k)
# keep track of which values are used when checking
# ahead for the (val + k)
# finally add (val + k / 2) if we find the corresponding
# (val + k) as it is the value from the original array
ans, used = [], [False] * len(self.nums)
for i, v in enumerate(self.nums):
if used[i]: continue
if not counts[v + k]: return []
used[counts[v + k].pop()] = True
ans.append(v + k // 2)
return ans | recover-the-original-array | [Python3] Try All Valid K Values | simonesestili | 1 | 63 | recover the original array | 2,122 | 0.381 | Hard | 29,337 |
https://leetcode.com/problems/recover-the-original-array/discuss/2491703/Clean-and-well-structured-Python3-implementation-(Top-93.2)-oror-Very-simple | class Solution(object):
def recoverArray(self, nums):
nums.sort()
mid = len(nums) // 2
# All possible k are (nums[j] - nums[0]) // 2, otherwise there is no num that satisfies nums[0] + k = num - k.
# For nums is sorted, so that any 2 elements (x, y) in nums[1:j] cannot satisfy x + k = y - k.
# In other words, for any x in nums[1:j], it needs to find y from nums[j + 1:] to satisfy x + k = y - k, but
# unfortunately if j > mid, then len(nums[j + 1:]) < mid <= len(nums[1:j]), nums[j + 1:] are not enough.
# The conclusion is j <= mid.
# If you think it’s not easy to understand why mid is enough, len(nums) can also work well
# for j in range(1, len(nums)):
for j in range(1, mid + 1): # O(N)
if nums[j] - nums[0] > 0 and (nums[j] - nums[0]) % 2 == 0: # Note the problem described k is positive.
k, counter, ans = (nums[j] - nums[0]) // 2, collections.Counter(nums), []
# For each number in lower, we try to find the corresponding number from higher list.
# Because nums is sorted, current n is always the current lowest num which can only come from lower
# list, so we search the corresponding number of n which equals to n + 2 * k in the left
# if it can not be found, change another k and continue to try.
for n in nums: # check if n + 2 * k available as corresponding number in higher list of n
if counter[n] == 0: # removed by previous num as its corresponding number in higher list
continue
if counter[n + 2 * k] == 0: # not found corresponding number in higher list
break
ans.append(n + k)
counter[n] -= 1 # remove n
counter[n + 2 * k] -= 1 # remove the corresponding number in higher list
if len(ans) == mid:
return ans | recover-the-original-array | ✔️ Clean and well structured Python3 implementation (Top 93.2%) || Very simple | Kagoot | 0 | 35 | recover the original array | 2,122 | 0.381 | Hard | 29,338 |
https://leetcode.com/problems/recover-the-original-array/discuss/1713646/Using-Counter-and-sorted-keys-82-speed | class Solution:
def recoverArray(self, nums: List[int]) -> List[int]:
half_len_nums = len(nums) // 2
cnt = Counter(nums)
keys = sorted(cnt.keys())
min_key = keys[0]
diffs = [diff for i in range(1, len(keys))
if not (diff := keys[i] - min_key) % 2]
for d in diffs:
ans = []
cnt_copy = cnt.copy()
for k in keys:
if cnt_copy[k]:
if k + d in cnt_copy and cnt_copy[k] <= cnt_copy[k + d]:
ans.extend([k + d // 2] * cnt_copy[k])
cnt_copy[k + d] -= cnt_copy[k]
else:
break
if len(ans) == half_len_nums:
return ans
return [] | recover-the-original-array | Using Counter and sorted keys, 82% speed | EvgenySH | 0 | 83 | recover the original array | 2,122 | 0.381 | Hard | 29,339 |
https://leetcode.com/problems/recover-the-original-array/discuss/1648740/Python-3-Hint-solution | class Solution:
def recoverArray(self, nums: List[int]) -> List[int]:
nums.sort()
n = len(nums)
def check(k):
cnt = defaultdict(int)
ans = []
for x in nums:
if cnt[x - 2*k] > 0:
cnt[x - 2*k] -= 1
ans.append(x - k)
else:
cnt[x] += 1
if len(ans) == n // 2: return ans
return []
# maximum k should not exceed half of the array
cand_k = sorted(set((nums[i] - nums[0]) // 2 for i in range(1, n // 2 + 1) if (nums[i] - nums[0]) % 2 == 0))
for k in cand_k:
if k == 0: continue
ans = check(k)
if ans: return ans | recover-the-original-array | [Python 3] Hint solution | chestnut890123 | 0 | 51 | recover the original array | 2,122 | 0.381 | Hard | 29,340 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1662586/Python-O(NlogN)-sort-solution-and-O(N)-linear-scan-solution | class Solution:
def checkString(self, s: str) -> bool:
return ''.join(sorted(s)) == s | check-if-all-as-appears-before-all-bs | [Python] O(NlogN) sort solution and O(N) linear scan solution | kryuki | 4 | 193 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,341 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1662586/Python-O(NlogN)-sort-solution-and-O(N)-linear-scan-solution | class Solution:
def checkString(self, s: str) -> bool:
appeared_b = False
for char in s:
if char == 'b':
appeared_b = True
else:
if appeared_b:
return False
return True | check-if-all-as-appears-before-all-bs | [Python] O(NlogN) sort solution and O(N) linear scan solution | kryuki | 4 | 193 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,342 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1661317/Python | class Solution:
def checkString(self, s: str) -> bool:
found = False
for c in s:
if c == 'b':
found = True
elif found:
return False
return True | check-if-all-as-appears-before-all-bs | Python | blue_sky5 | 2 | 108 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,343 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2725389/Python3ororO(n)oror-easy | class Solution:
def checkString(self, s: str) -> bool:
c=s[0]
if s[0]=='b' and 'a' in s:
return False
n=len(s)
for i in range(n):
if c==s[i]:
continue
elif c in s[i+1:]:
return False
return True | check-if-all-as-appears-before-all-bs | Python3||O(n)|| easy | Sneh713 | 1 | 153 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,344 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1816145/Python-solution-in-one-line | class Solution:
def checkString(self, s: str) -> bool:
return not 'ba' in s | check-if-all-as-appears-before-all-bs | Python solution in one line | iamamirhossein | 1 | 62 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,345 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1666354/Easy-To-Understand-Python-Simple-Solution | class Solution:
def checkString(self, s: str) -> bool:
if "ba" in s:
return False
return True
``` | check-if-all-as-appears-before-all-bs | Easy To Understand Python Simple Solution | prasadghadage | 1 | 32 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,346 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2825276/Simple-and-Fast-Python-Solution | class Solution:
def checkString(self, s: str) -> bool:
if "b" in s:
b_index = s.index("b")
new_s = s[b_index: len(s)]
return not "a" in new_s
else:
return True | check-if-all-as-appears-before-all-bs | Simple and Fast Python Solution | PranavBhatt | 0 | 2 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,347 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2810706/Very-very-easy-solution-using-Python | class Solution:
def checkString(self, s: str) -> bool:
c = s.count('a')
if c == 0:
return True
if c > 0 and s[:c] == "a" * c:
return True
return False | check-if-all-as-appears-before-all-bs | Very very easy solution using Python | ankurbhambri | 0 | 7 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,348 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2803509/Simple-Python-Solution | class Solution:
def checkString(self, s: str) -> bool:
seen = set()
for c in s:
if c == 'a' and 'b' in seen:
return False
seen.add(c)
return True | check-if-all-as-appears-before-all-bs | Simple Python Solution | mansoorafzal | 0 | 2 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,349 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2788366/python-solution | class Solution:
def checkString(self, s: str) -> bool:
a = 0
b = 0
for i in s:
if i == 'a' and b == 0:
a += 1
elif i == 'b':
b += 1
else:
return False
return True | check-if-all-as-appears-before-all-bs | python solution | shingnapure_shilpa17 | 0 | 3 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,350 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2786357/Simplest-solution. | class Solution:
def checkString(self, s: str) -> bool:
start, end = 0, len(s)
while start < end and s[start] == 'a':
start += 1
while start < end and s[start] == 'b':
start += 1
return start == end | check-if-all-as-appears-before-all-bs | Simplest solution. | ahmedsamara | 0 | 2 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,351 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2781351/python-solution-very-easy | class Solution:
def checkString(self, s: str) -> bool:
if (sorted(s)) == list(s):
return True;
else:
return False; | check-if-all-as-appears-before-all-bs | python solution very easy | seifsoliman | 0 | 1 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,352 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2758719/python-code | class Solution:
def checkString(self, s: str) -> bool:
return ''.join(sorted(s)) == s | check-if-all-as-appears-before-all-bs | python code | ayushigupta2409 | 0 | 6 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,353 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2758719/python-code | class Solution:
def checkString(self, s: str) -> bool:
s=list(s)
while 'a' in s and 'b' in s:
if s.index('a')<s.index('b'):
s.remove('a')
else:
return False
return True | check-if-all-as-appears-before-all-bs | python code | ayushigupta2409 | 0 | 6 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,354 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2662250/Python-One-liner-Solution | class Solution:
def checkString(self, s: str) -> bool:
return "ba" not in s | check-if-all-as-appears-before-all-bs | Python One-liner Solution | kcstar | 0 | 5 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,355 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2629475/Python3-Simple-One-Liner-with-not-any() | class Solution:
def checkString(self, s: str) -> bool:
return not any('a' in s[i:] for i, char in enumerate(s) if char == 'b') | check-if-all-as-appears-before-all-bs | [Python3] Simple One-Liner with not any() | ivnvalex | 0 | 7 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,356 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2576200/Python-Simple-Python-Solution | class Solution:
def checkString(self, s: str) -> bool:
last_index_a = -1
last_index_b = -1
for i in range(len(s)):
if s[i] == 'a':
last_index_a = i
if s[i] == 'b' and last_index_b == -1:
last_index_b = i
if last_index_a == -1 or last_index_b == -1:
return True
else:
if last_index_a < last_index_b:
return True
else:
return False | check-if-all-as-appears-before-all-bs | [ Python ] ✅✅ Simple Python Solution 🥳✌👍 | ASHOK_KUMAR_MEGHVANSHI | 0 | 36 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,357 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2564972/Solution | class Solution:
def checkString(self, s: str) -> bool:
for i in range(len(s)-1):
if s[i] == "b" and s[i+1] == "a":
return False
return True | check-if-all-as-appears-before-all-bs | Solution | fiqbal997 | 0 | 10 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,358 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2535891/Simple-Python-solution | class Solution:
def checkString(self, s: str) -> bool:
for i in range(len(s)):
if s[i] == "a":
continue
else:
break
if i == len(s)-1:
return True
while i < len(s):
if s[i] == "a":
return False
i += 1
return True | check-if-all-as-appears-before-all-bs | Simple Python solution | aruj900 | 0 | 21 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,359 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2466249/Iterative-Solution-with-slicing | class Solution:
def checkString(self, s: str) -> bool:
# iterate the string s
# if when you locate a b, verify that no a's suceed it.
# use a flag to verify when you see an a
# if it does return False, otherwise return True after iteration
# Time O(N) Space: O(1)
for i in range(1, len(s)):
if s[i] == "a" and "b" in s[:i]:
return False
return True | check-if-all-as-appears-before-all-bs | Iterative Solution with slicing | andrewnerdimo | 0 | 18 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,360 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2385082/Super-easy-python | class Solution:
def checkString(self, s: str) -> bool:
l=len(s)
for i in range(l-1):
if s[i]=='b' and s[i+1]=='a':
return False
return True | check-if-all-as-appears-before-all-bs | Super easy python | sunakshi132 | 0 | 34 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,361 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2183214/faster-than-93.65-of-Python3 | class Solution:
def checkString(self, s: str) -> bool:
return (''.join(sorted(list(s)))) == s | check-if-all-as-appears-before-all-bs | faster than 93.65% of Python3 | writemeom | 0 | 41 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,362 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2166818/Super-simple-python-one-liner | class Solution:
def checkString(self, s: str) -> bool:
return list(s) == sorted(list(s)) | check-if-all-as-appears-before-all-bs | Super simple python one-liner | pro6igy | 0 | 31 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,363 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/2069192/Python3-Easy-Understanding-(Faster-than-98.47) | class Solution:
def checkString(self, s: str) -> bool:
a_count = s.count("a") ## Count the occurances of a
# b_count = s.count("b")
for i in range(a_count): ## Check if the first of a_count in the string are "a"
if s[i] != 'a':
return False
return True | check-if-all-as-appears-before-all-bs | Python3- Easy Understanding (Faster than 98.47%) | prithuls | 0 | 50 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,364 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1988717/Python3-100-faster-with-explanation | class Solution:
def checkString(self, s: str) -> bool:
mapper = {}
for x in s:
if 'b' in mapper and x == 'a':
return False
else:
mapper[x] = 1
return True | check-if-all-as-appears-before-all-bs | Python3, 100% faster with explanation | cvelazquez322 | 0 | 76 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,365 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1940863/Python-dollarolution | class Solution:
def checkString(self, s: str) -> bool:
flag = 0
for i in s:
if i == 'a' and flag == 1:
return False
elif i == 'b':
flag = 1
return True | check-if-all-as-appears-before-all-bs | Python $olution | AakRay | 0 | 41 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,366 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1926995/Python3-simple-solution | class Solution:
def checkString(self, s: str) -> bool:
flag = True
for i in s:
if i == 'b' and flag:
flag = False
if i == 'a' and not flag:
return False
return True | check-if-all-as-appears-before-all-bs | Python3 simple solution | EklavyaJoshi | 0 | 30 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,367 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1911761/python-3-oror-very-simple-solution | class Solution:
def checkString(self, s: str) -> bool:
flag = False
for c in s:
if c == 'b':
flag = True
elif flag:
return False
return True | check-if-all-as-appears-before-all-bs | python 3 || very simple solution | dereky4 | 0 | 34 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,368 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1899274/Python-clean-and-simple!-%2B-One-Liner | class Solution:
def checkString(self, s):
sawB = False
for c in s:
if c=='a' and sawB: return False
elif c=='b': sawB = True
return True | check-if-all-as-appears-before-all-bs | Python - clean and simple! + One Liner | domthedeveloper | 0 | 28 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,369 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1899274/Python-clean-and-simple!-%2B-One-Liner | class Solution:
def checkString(self, s):
sawB = False
for c in s:
match c:
case 'a':
if sawB: return False
case 'b':
sawB = True
return True | check-if-all-as-appears-before-all-bs | Python - clean and simple! + One Liner | domthedeveloper | 0 | 28 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,370 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1899274/Python-clean-and-simple!-%2B-One-Liner | class Solution:
def checkString(self, s):
return "ba" not in s | check-if-all-as-appears-before-all-bs | Python - clean and simple! + One Liner | domthedeveloper | 0 | 28 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,371 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1846953/4-Line-Python3-Easy-Solution-oror-100-Faster-oror-Easy-to-understand-Python-code | class Solution:
def checkString(self, s: str) -> bool:
ct=1
cm=1
for i in range(0,len(s)-1):
if s[i]=='b':
if s[i+1]=='a':
ct=-3
return ct==cm | check-if-all-as-appears-before-all-bs | 4-Line Python3 Easy Solution || 100% Faster || Easy to understand Python code | RatnaPriya | 0 | 53 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,372 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1823504/easy-python-solution | class Solution:
def checkString(self, s: str) -> bool:
flag = 0
if "a" not in s:
return True
elif "b" not in s:
return True
else:
for i in s:
if i == "b":
flag = 1
elif i == "a" and flag == 1:
return False
return True | check-if-all-as-appears-before-all-bs | easy python solution | dakash682 | 0 | 40 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,373 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1791880/Python-easy-solution | class Solution:
def checkString(self, s: str) -> bool:
n = len(s)
# initial values take care if s has only a's or only b's
lastA = -1
firstB = n
for i in range(n):
if s[i] == 'a':
lastA = max(lastA, i)
else:
firstB = min(firstB, i)
return lastA < firstB # check if last occurrence of 'a' is before first occurrence of 'b' | check-if-all-as-appears-before-all-bs | Python easy solution | abkc1221 | 0 | 50 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,374 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1673948/Python3Java-Simple-clean-and-concise-O(n)-time-O(1)-space-100-Java | class Solution:
def checkString(self, s: str) -> bool:
max_a_index = float("-inf")
min_b_index = float("inf")
for i, c in enumerate(s):
if c == "a":
max_a_index = max(max_a_index, i)
elif c == "b":
min_b_index = min(min_b_index, i)
return max_a_index < min_b_index | check-if-all-as-appears-before-all-bs | [Python3/Java] Simple, clean and concise - O(n) time, O(1) space - 100% Java | hanelios | 0 | 42 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,375 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1670879/python-easy-O(n)-time-O(1)-space-solution | class Solution:
def checkString(self, s: str) -> bool:
n = len(s)
if n == 1:
return True
if s.count('a') == 0 or s.count('a') == n:
return True
i, j = 0, n-1
while i <= n-1 and s[i] == 'a':
i += 1
while j >=0 and s[j] =='b':
j -= 1
return i == j+1 | check-if-all-as-appears-before-all-bs | python easy O(n) time, O(1) space solution | byuns9334 | 0 | 39 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,376 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1661534/Python3-Easy-Solution-(Accepted)-(Commented)-(Clean) | class Solution:
def checkString(self, s: str) -> bool:
if 'a' not in s: //Base case is if there are no a's in the string
return True
for i in range(1, len(s)): //Iterate from the 2nd element of string to last
if s[i] == 'a' and s[i-1] == 'b': //If there is a 'b' before 'a' then return False
return False
return True //There was no 'b' before 'a' | check-if-all-as-appears-before-all-bs | Python3 Easy Solution (Accepted) (Commented) (Clean) | sdasstriver9 | 0 | 36 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,377 |
https://leetcode.com/problems/check-if-all-as-appears-before-all-bs/discuss/1660977/Python3-1-line | class Solution:
def checkString(self, s: str) -> bool:
return "ba" not in s | check-if-all-as-appears-before-all-bs | [Python3] 1-line | ye15 | 0 | 47 | check if all as appears before all bs | 2,124 | 0.716 | Easy | 29,378 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2272999/PYTHON-3-99.13-LESS-MEMORY-or-94.93-FASTER-or-EXPLANATION | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
a, s = [x.count("1") for x in bank if x.count("1")], 0
# ex: bank is [[00101], [01001], [00000], [11011]]
# a would return [2, 2, 4]
for c in range(len(a)-1):
s += (a[c]*a[c+1])
# basic math to find the total amount of lasers
# for the first iteration: s += 2*2
# for the second iteration: s += 2*4
# returns s = 12
return s | number-of-laser-beams-in-a-bank | [PYTHON 3] 99.13% LESS MEMORY | 94.93% FASTER | EXPLANATION | omkarxpatel | 7 | 130 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,379 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/1660994/Python3-simulation | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
ans = prev = 0
for row in bank:
curr = row.count('1')
if curr:
ans += prev * curr
prev = curr
return ans | number-of-laser-beams-in-a-bank | [Python3] simulation | ye15 | 7 | 417 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,380 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2677735/Simple-Python-Solution | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
ans = prev = 0
for s in bank:
c = s.count('1')
if c:
ans += prev * c
prev = c
return ans | number-of-laser-beams-in-a-bank | Simple Python Solution | Sneh713 | 1 | 70 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,381 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2633920/Python3-Simple-Solution | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
pre = 0
nn = 0
ans = 0
for i in bank:
nn= 0
for j in i:
if j == '1':
nn+=1
if nn:
ans+=nn*pre
pre= nn
return ans
## PLease upvote if you like the Solution | number-of-laser-beams-in-a-bank | Python3 Simple Solution | hoo__mann | 1 | 11 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,382 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/1938928/Python3-Simple-Solution | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
beams = 0
prev = 0
for b in bank:
last = b.count('1')
if last:
beams += prev * last
prev = last
return beams | number-of-laser-beams-in-a-bank | [Python3] Simple Solution | terrencetang | 1 | 70 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,383 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/1832792/Python-Solution-or-95-lesser-memory-or-5-Lines-of-code | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
laser, n = [], len(bank[0])
bank = list(filter(("0"*n).__ne__, bank))
for i in range(len(bank)-1):
laser.append(bank[i].count("1")*bank[i+1].count("1"))
return sum(laser) | number-of-laser-beams-in-a-bank | ✔Python Solution | 95% lesser memory | 5 Lines of code | Coding_Tan3 | 1 | 59 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,384 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2840449/Join-method | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
if len(bank) > 1:
last_value = len(''.join(bank[0].split('0')))
v_sum = 0
for floor in bank[1:]:
value = len(''.join(floor.split('0')))
if value != 0:
v_sum += (last_value*value)
last_value = value
return v_sum
else:
return 0 | number-of-laser-beams-in-a-bank | Join method | Robert1914 | 0 | 1 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,385 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2785334/Python-Recursion-solution | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
def recu(bank,i,prev,res):
if i == len(bank):
return res
count = bank[i].count("1")
if count:
res += count*prev
prev = count
return recu(bank,i+1,prev,res)
return recu(bank,0,0,0) | number-of-laser-beams-in-a-bank | Python Recursion solution | anshsharma17 | 0 | 3 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,386 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2785334/Python-Recursion-solution | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
res = []
for b1 in range(len(bank)):
count = 0
for b2 in range(len(bank[b1])):
if bank[b1][b2] == "1":
count+=1
if count > 0:
res.append(count)
if len(res) == 0 or len(res) ==1:
return 0
return (sum([res[i]*res[i+1] for i in range(len(res)-1)])) | number-of-laser-beams-in-a-bank | Python Recursion solution | anshsharma17 | 0 | 3 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,387 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2784963/Golang-Rust-Python-solution | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
def recu(bank,i,prev,res):
if i == len(bank):
return res
count = bank[i].count("1")
if count:
res += count*prev
prev = count
return recu(bank,i+1,prev,res)
return recu(bank,0,0,0) | number-of-laser-beams-in-a-bank | Golang-Rust-Python solution | anshsharma17 | 0 | 1 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,388 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2784963/Golang-Rust-Python-solution | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
res = []
for b1 in range(len(bank)):
count = 0
for b2 in range(len(bank[b1])):
if bank[b1][b2] == "1":
count+=1
if count > 0:
res.append(count)
if len(res) == 0 or len(res) ==1:
return 0
return (sum([res[i]*res[i+1] for i in range(len(res)-1)])) | number-of-laser-beams-in-a-bank | Golang-Rust-Python solution | anshsharma17 | 0 | 1 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,389 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2759791/easy-python-solutionorortime-complexity-O(n) | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
ans= 0
i = 0
j = 1
while i<len(bank) and j<len(bank):
a = bank[i].count("1")
b = bank[j].count("1")
if a!=0 and b!=0:
ans+=a*b
i = j
if j<len(bank):
j +=1
elif a==0 and b==0:
i+=1
j+=1
elif a==0 and b!=0:
i+=1
j+=1
elif b==0 and a!=0:
j+=1
return ans | number-of-laser-beams-in-a-bank | ✅✅easy python solution||time complexity O(n) | chessman_1 | 0 | 5 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,390 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2710420/Python3 | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
count_l = [i.count('1') for i in bank if i.count('1')>0]
ans = 0
i = 0
while i < len(count_l) - 1:
ans = ans + count_l[i] * count_l[i + 1]
i += 1
return ans | number-of-laser-beams-in-a-bank | Python3 | sipi09 | 0 | 3 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,391 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2709062/Python3-One-Liner-Using-List-Comprehension | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
return [sum(map(lambda a: a[0]*a[1], zip(x, x[1:]))) for x in [[g for g in [sum([int(c) for c in b]) for b in bank] if g!=0]]][0] | number-of-laser-beams-in-a-bank | Python3 One Liner Using List Comprehension | godshiva | 0 | 2 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,392 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2698714/Python3-Simple-Solution | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
res = 0
for i, row1 in enumerate(bank):
c1 = row1.count('1')
if c1 == 0: continue
for row2 in bank[i + 1:]:
c2 = row2.count('1')
if c2 > 0:
res += c1 * c2
break
return res | number-of-laser-beams-in-a-bank | Python3 Simple Solution | mediocre-coder | 0 | 3 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,393 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2688168/Python-or-93 | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
res=0
last = 0
for i in range(len(bank)):
new = bank[i].count('1')
if new >0:
res = res + last*new
last = new
return res | number-of-laser-beams-in-a-bank | Python | 93% | naveenraiit | 0 | 7 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,394 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2665920/python3-Faster-than-95 | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
upper = 0
lower = 0
total = 0
for i in bank:
beam = i.count('1')
if upper == 0:
upper = beam
continue
else:
if beam != 0:
lower = beam
total = total + lower * upper
upper = lower
else:
continue
return total | number-of-laser-beams-in-a-bank | python3 Faster than 95% | Noisy47 | 0 | 3 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,395 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2629077/Python-solution | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
ans = 0
arr = []
length = 0
for row in bank:
if set(row) == {'0'}:
continue
arr.append(row.count('1'))
length += 1
for i in range(length -1):
ans += (arr[i] * arr[i + 1])
return ans | number-of-laser-beams-in-a-bank | Python solution | Guild_Arts | 0 | 2 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,396 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2539313/Python-or-looping-or-O(n)-Time-or-O(1)-Space | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
ans = 0
prev = 0
for floor in bank:
count = floor.count("1")
ans += count*prev
if count>0:
prev = count
return ans | number-of-laser-beams-in-a-bank | Python | looping | O(n) Time | O(1) Space | coolakash10 | 0 | 5 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,397 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2499957/Python-or-Basic-Combinatorics | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
laserCount = 0; prevDeviceCount = 0
for row in bank:
currentDeviceCount = row.count("1")
if currentDeviceCount != 0: # this skips lines that have no devices
laserCount += prevDeviceCount * currentDeviceCount
prevDeviceCount = currentDeviceCount
return laserCount | number-of-laser-beams-in-a-bank | Python | Basic Combinatorics | sr_vrd | 0 | 8 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,398 |
https://leetcode.com/problems/number-of-laser-beams-in-a-bank/discuss/2272241/Simple-Python-Solution-96-faster-than-other | class Solution:
def numberOfBeams(self, bank: List[str]) -> int:
output = 0
curr = 0
currPlus = 1
if len(bank)<2:
return 0
while currPlus < len(bank):
curr1count = bank[curr].count("1")
currPlus1count = bank[currPlus].count("1")
if curr1count > 0:
if currPlus1count > 0:
output += curr1count*currPlus1count
curr = currPlus
currPlus+=1
else:
currPlus+=1
else:
curr+=1
currPlus+=1
return output | number-of-laser-beams-in-a-bank | Simple Python Solution 96% faster than other | sunnysharma03 | 0 | 21 | number of laser beams in a bank | 2,125 | 0.826 | Medium | 29,399 |
Subsets and Splits