2020-05-19 01:54:57 +08:00
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package Others;
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import java.util.Scanner;
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import java.util.Arrays;
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/**
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* To find triplet equals to given sum in complexity O(n*log(n))
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*
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*
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* Array must be sorted
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*
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* @author Ujjawal Joshi
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* @date 2020.05.18
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2020-05-29 15:02:24 +08:00
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*
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* Test Cases:
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Input:
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* 6 //Length of array
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12 3 4 1 6 9
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target=24
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* Output:3 9 12
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* Explanation: There is a triplet (12, 3 and 9) present
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in the array whose sum is 24.
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*
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*
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2020-05-19 01:54:57 +08:00
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*/
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class threesum{
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public static void main(String args[])
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{
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Scanner sc =new Scanner(System.in);
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int n=sc.nextInt(); //Length of an array
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int a[]=new int[n];
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for(int i=0;i<n;i++)
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{
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a[i]=sc.nextInt();
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}
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2020-05-29 15:02:24 +08:00
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System.out.println("Target");
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2020-05-19 01:54:57 +08:00
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int n_find=sc.nextInt();
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Arrays.sort(a); // Sort the array if array is not sorted
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for(int i=0;i<n;i++){
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int l=i+1,r=n-1;
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while(l<r){
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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;
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else if(a[i]+a[l]+a[r]<n_find) l++;
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else r--;
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}
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}
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2020-06-30 02:44:04 +08:00
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sc.close();
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2020-05-19 01:54:57 +08:00
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}
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}
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