Find a triplet such that sum of two equals to third element
Given an array of integers you have to find three numbers such that sum of two elements equals the third element.
Examples:
Input : {5, 32, 1, 7, 10, 50, 19, 21, 2}
Output : 21, 2, 19
Input : {5, 32, 1, 7, 10, 50, 19, 21, 0}
Output : no such triplet exist
Question source : Arcesium Interview Experience | Set 7 (On campus for Internship)
Simple approach: Run three loops and check if there exists a triplet such that sum of two elements equals the third element.
Time complexity: O(n^3)
Efficient approach : The idea is similar to Find a triplet that sum to a given value.
- Sort the given array first.
- Start fixing the greatest element of three from back and traverse the array to find other two numbers which sum upto the third element.
- Take two pointers j(from front) and k(initially i-1) to find the smallest of the two number and from i-1 to find the largest of the two remaining numbers
- If the addition of both the numbers is still less than A[i], then we need to increase the value of the summation of two numbers, thereby increase the j pointer, so as to increase the value of A[j] + A[k].
- If the addition of both the numbers is more than A[i], then we need to decrease the value of the summation of two numbers, thereby decrease the k pointer so as to decrease the overall value of A[j] + A[k].
Below image is a dry run of the above approach:
Below is the implementation of the above approach:
C++
// CPP program to find three numbers // such that sum of two makes the // third element in array #include <bits/stdc++.h> using namespace std; // Utility function for finding // triplet in array void findTriplet(int arr[], int n) { // sort the array sort(arr, arr + n); // for every element in arr // check if a pair exist(in array) whose // sum is equal to arr element for (int i = n - 1; i >= 0; i--) { int j = 0; int k = i - 1; // Iterate forward and backward to find // the other two elements while (j < k) { // If the two elements sum is // equal to the third element if (arr[i] == arr[j] + arr[k]) { // pair found cout << "numbers are " << arr[i] << " " << arr[j] << " " << arr[k] << endl; return; } // If the element is greater than // sum of both the elements, then try // adding a smaller number to reach the // equality else if (arr[i] > arr[j] + arr[k]) j += 1; // If the element is smaller, then // try with a smaller number // to reach equality, so decrease K else k -= 1; } } // No such triplet is found in array cout << "No such triplet exists"; } // driver program int main() { int arr[] = { 5, 32, 1, 7, 10, 50, 19, 21, 2 }; int n = sizeof(arr) / sizeof(arr[0]); findTriplet(arr, n); return 0; } |
Java
// Java program to find three numbers // such that sum of two makes the // third element in array import java.util.Arrays; public class GFG { // utility function for finding // triplet in array static void findTriplet(int arr[], int n) { // sort the array Arrays.sort(arr); // for every element in arr // check if a pair exist(in array) whose // sum is equal to arr element for (int i = n - 1; i >= 0; i--) { int j = 0; int k = i - 1; while (j < k) { if (arr[i] == arr[j] + arr[k]) { // pair found System.out.println("numbers are " + arr[i] + " " + arr[j] + " " + arr[k]); return; } else if (arr[i] > arr[j] + arr[k]) j += 1; else k -= 1; } } // no such triplet is found in array System.out.println("No such triplet exists"); } // driver program public static void main(String args[]) { int arr[] = { 5, 32, 1, 7, 10, 50, 19, 21, 2 }; int n = arr.length; findTriplet(arr, n); } } // This code is contributed by Sumit Ghosh |
Python
# Python program to find three numbers # such that sum of two makes the # third element in array # utility function for finding # triplet in array def findTriplet(arr, n): # sort the array arr.sort() # for every element in arr # check if a pair exist(in array) whose # sum is equal to arr element i = n - 1 while(i >= 0): j = 0 k = i - 1 while (j < k): if (arr[i] == arr[j] + arr[k]): # pair found print "numbers are ", arr[i], arr[j], arr[k] return elif (arr[i] > arr[j] + arr[k]): j += 1 else: k -= 1 i -= 1 # no such triplet is found in array print "No such triplet exists" # driver program arr = [ 5, 32, 1, 7, 10, 50, 19, 21, 2 ] n = len(arr) findTriplet(arr, n) # This code is contributed by Sachin Bisht |
C#
// C# program to find three numbers // such that sum of two makes the // third element in array using System; public class GFG { // utility function for finding // triplet in array static void findTriplet(int[] arr, int n) { // sort the array Array.Sort(arr); // for every element in arr // check if a pair exist(in // array) whose sum is equal // to arr element for (int i = n - 1; i >= 0; i--) { int j = 0; int k = i - 1; while (j < k) { if (arr[i] == arr[j] + arr[k]) { // pair found Console.WriteLine("numbers are " + arr[i] + " " + arr[j] + " " + arr[k]); return; } else if (arr[i] > arr[j] + arr[k]) j += 1; else k -= 1; } } // no such triplet is found in array Console.WriteLine("No such triplet exists"); } // driver program public static void Main() { int[] arr = { 5, 32, 1, 7, 10, 50, 19, 21, 2 }; int n = arr.Length; findTriplet(arr, n); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to find three // numbers such that sum of // two makes the third // element in array // utility function for // finding triplet in array function findTriplet($arr, $n) { // sort the array sort($arr); // for every element in // arr check if a pair // exist(in array) whose // sum is equal to arr element for ($i = $n - 1; $i >= 0; $i--) { $j = 0; $k = $i - 1; while ($j < $k) { if ($arr[$i] == $arr[$j] + $arr[$k]) { // pair found echo "numbers are ", $arr[$i], " ", $arr[$j], " ", $arr[$k]; return; } else if ($arr[$i] > $arr[$j] + $arr[$k]) $j += 1; else $k -= 1; } } // no such triplet // is found in array echo "No such triplet exists"; } // Driver Code $arr = array(5, 32, 1, 7, 10, 50, 19, 21, 2 ); $n = count($arr); findTriplet($arr, $n); // This code is contributed by anuj_67. ?> |
Output:
numbers are 21 2 19
Time complexity: O(N^2)
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