import java.util.Arrays; | |
import java.util.Scanner; | |
/** | |
* To find triplet equals to given sum in complexity O(n*log(n)) | |
* | |
* <p>Array must be sorted | |
* | |
* @author Ujjawal Joshi | |
* @date 2020.05.18 | |
* <p>Test Cases: Input: 6 //Length of array 12 3 4 1 6 9 target=24 Output:3 9 12 Explanation: | |
* There is a triplet (12, 3 and 9) present in the array whose sum is 24. | |
*/ | |
class ThreeSum { | |
public static void main(String args[]) { | |
Scanner sc = new Scanner(System.in); | |
int n = sc.nextInt(); // Length of an array | |
int a[] = new int[n]; | |
for (int i = 0; i < n; i++) { | |
a[i] = sc.nextInt(); | |
} | |
System.out.println("Target"); | |
int n_find = sc.nextInt(); | |
Arrays.sort(a); // Sort the array if array is not sorted | |
for (int i = 0; i < n; i++) { | |
int l = i + 1, r = n - 1; | |
while (l < r) { | |
if (a[i] + a[l] + a[r] == n_find) { | |
System.out.println(a[i] + " " + a[l] + " " + a[r]); | |
break; | |
} // if you want all the triplets write l++;r--; insted of break; | |
else if (a[i] + a[l] + a[r] < n_find) l++; | |
else r--; | |
} | |
} | |
sc.close(); | |
} | |
} |