C++ Program to Find a triplet such that sum of two equals to third element - GeeksforGeeks

11 Jan, 2022
Given an array of integers, you have to find three numbers such that the 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 the back and traverse the array to find the other two numbers which sum up to 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 increasing 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++
// C++ 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 arrayvoid 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 codeint main(){    int arr[] = {5, 32, 1, 7, 10,                  50, 19, 21, 2};    int n = sizeof(arr) / sizeof(arr[0]);    findTriplet(arr, n);    return 0;}
Output:  
numbers are 21 2 19
Time complexity: O(N^2)
Another Approach: The idea is similar to previous approach.
Sort the given array.Start a nested loop, fixing the first element i(from 0 to n-1) and moving the other one j (from i+1 to n-1).Take the sum of both the elements and search it in the remaining array using Binary Search.
Sort the given array.
Start a nested loop, fixing the first element i(from 0 to n-1) and moving the other one j (from i+1 to n-1).
Take the sum of both the elements and search it in the remaining array using Binary Search.
C++
// C++ program to find three numbers// such that sum of two makes the// third element in array#include <bits/stdc++.h>#include <iostream>using namespace std;  // Function to perform binary searchbool search(int sum, int start,             int end, int arr[]){    while (start <= end)     {        int mid = (start + end) / 2;        if (arr[mid] == sum)         {            return true;        }        else if (arr[mid] > sum)         {            end = mid - 1;        }        else         {            start = mid + 1;        }    }    return false;}  // Function to find the tripletsvoid findTriplet(int arr[], int n){    // Sorting the array    sort(arr, arr + n);      // Initialising nested loops    for (int i = 0; i < n; i++)     {        for (int j = i + 1; j < n; j++)         {            // Finding the sum of the numbers            if (search((arr[i] + arr[j]),                 j, n - 1, arr))             {                // Printing out the first triplet                cout << "Numbers are: " << arr[i] <<                         " " << arr[j] << " " <<                         (arr[i] + arr[j]);                return;            }        }    }    // If no such triplets are found    cout << "No such numbers exist" << endl;}  // Driver codeint main(){    int arr[] = {5, 32, 1, 7, 10,                  50, 19, 21, 2};    int n = sizeof(arr) / sizeof(arr[0]);    findTriplet(arr, n);    return 0;}// This code is contributed by Sarthak Delori
Time Complexity: O(N^2*log N)
Space Complexity:  O(1)
Please refer complete article on Find a triplet such that sum of two equals to third element for more details!
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