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, 19Input : {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.
Code-
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) { for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { for(int k = j + 1; k < n; k++) { if((arr[i]+arr[j]==arr[k]) || (arr[i]+arr[k]==arr[j]) || (arr[j]+arr[k]==arr[i])){ // printing out the first triplet cout << "Numbers are: " << arr[i] << " " << arr[j] << " " << arr[k]; return; } } } } // 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
import java.util.*; public class Main { // Utility function for finding triplet in array public static void findTriplet(int[] arr, int n) { for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { for(int k = j + 1; k < n; k++) { if((arr[i]+arr[j]==arr[k]) || (arr[i]+arr[k]==arr[j]) || (arr[j]+arr[k]==arr[i])) { // printing out the first triplet System.out.println("Numbers are: " + arr[i] + " " + arr[j] + " " + arr[k]); return; } } } } // 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); } } |
Python3
# Python3 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): for i in range(n): for j in range(i + 1, n): for k in range(j + 1, n): if((arr[i]+arr[j] == arr[k]) or (arr[i]+arr[k] == arr[j]) or (arr[j]+arr[k] == arr[i])): # printing out the first triplet print("Numbers are:", arr[i], arr[j], arr[k]) return # No such triplet is found in array print("No such triplet exists") # driver program if __name__ == '__main__': arr = [5, 32, 1, 7, 10, 50, 19, 21, 2] n = len(arr) findTriplet(arr, n) |
C#
// C# program to find three numbers // such that sum of two makes the // third element in array using System; public class MainClass { // Utility function for finding // triplet in array public static void FindTriplet(int[] arr, int n) { for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { for (int k = j + 1; k < n; k++) { if ((arr[i] + arr[j] == arr[k]) || (arr[i] + arr[k] == arr[j]) || (arr[j] + arr[k] == arr[i])) { // printing out the first triplet Console.WriteLine( "Numbers are: " + arr[i] + " " + arr[j] + " " + arr[k]); return; } } } } // No such triplet is found in array Console.WriteLine("No such triplet exists"); } public static void Main() { int[] arr = { 5, 32, 1, 7, 10, 50, 19, 21, 2 }; int n = arr.Length; FindTriplet(arr, n); } } |
Javascript
// Utility function for finding triplet in array function findTriplet(arr) { const n = arr.length; for (let i = 0; i < n; i++) { for (let j = i + 1; j < n; j++) { for (let k = j + 1; k < n; k++) { if ( arr[i] + arr[j] === arr[k] || arr[i] + arr[k] === arr[j] || arr[j] + arr[k] === arr[i] ) { // Printing out the first triplet console.log(`Numbers are: ${arr[i]}, ${arr[j]}, ${arr[k]}`); return; } } } } // No such triplet is found in array console.log("No such triplet exists"); } // Driver program const arr = [5, 32, 1, 7, 10, 50, 19, 21, 2]; findTriplet(arr); |
Output
Numbers are: 5 7 2
Time Complexity: O(N^3)
Auxiliary Space: O(1)
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++
// 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. ?> |
Javascript
<script> // Javascript 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 arr.sort((a,b) => a-b); // for every element in arr // check if a pair exist(in array) whose // sum is equal to arr element for (let i = n - 1; i >= 0; i--) { let j = 0; let 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 document.write("numbers are " + arr[i] + " " + arr[j] + " " + arr[k] + "<br>"); 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 document.write("No such triplet exists"); } // driver program let arr = [ 5, 32, 1, 7, 10, 50, 19, 21, 2 ]; let n = arr.length; findTriplet(arr, n); // This code is contributed by Mayank Tyagi </script> |
Output
numbers are 21 2 19
Time Complexity: O(N^2)
Auxiliary Space: O(1)
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.
Implementation:
C++
// CPP 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 search bool 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 triplets void 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; } 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; } // This code is contributed by Sarthak Delori |
Java
// Java program to find three numbers // such that sum of two makes the // third element in array import java.util.*; class GFG{ // Function to perform binary search static boolean 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 triplets static void findTriplet(int arr[], int n) { // Sorting the array Arrays.sort(arr); // 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 System.out.print("Numbers are: " + arr[i] + " " + arr[j] + " " + (arr[i] + arr[j])); return; } } } // If no such triplets are found System.out.print("No such numbers exist"); } // Driver code 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 target_2 |
Python3
# Python program to find three numbers # such that sum of two makes the # third element in array from functools import cmp_to_key def mycmp(a, b): return a - b def search(sum, start, end, arr): while (start <= end): mid = (start + end) // 2 if (arr[mid] == sum): return True elif (arr[mid] > sum): end = mid - 1 else: start = mid + 1 return False # Utility function for finding # triplet in array def findTriplet(arr, n): # sort the array arr.sort(key = cmp_to_key(mycmp)) # initialising nested loops for i in range(n): for j in range(i + 1,n): if (search((arr[i] + arr[j]), j, n - 1, arr)): print(f"numbers are {arr[i]} {arr[j]} {( arr[i] + arr[j] )}") return # 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 shinjanpatra |
C#
// C# program to find three numbers // such that sum of two makes the // third element in array using System; public class GFG { // function to perform binary search static bool 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; } // 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 = 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)) { // pair found Console.WriteLine("Numbers are " + arr[i] + " " + arr[j] + " " + (arr[i]+arr[j])); return; } } } // 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 Aarti_Rathi |
Javascript
<script> // Javascript program to find three numbers // such that sum of two makes the // third element in array bool search(sum, start, end, arr) { while (start <= end) { let mid = (start + end) / 2; if (arr[mid] == sum) { return true; } else if (arr[mid] > sum) { end = mid - 1; } else { start = mid + 1; } } return false; } // Utility function for finding // triplet in array function findTriplet(arr, n) { // sort the array arr.sort((a,b) => a-b); // initialising nested loops for (let i = 0; i < n; i++) { for (let j = i + 1; j < n; j++) { if (search((arr[i] + arr[j]), j, n - 1, arr)) { document.write("numbers are " + arr[i] + " " + arr[j] + " " + ( arr[i] + arr[j] ) + "<br>"); } } } // No such triplet is found in array document.write("No such triplet exists"); } // driver program let arr = [ 5, 32, 1, 7, 10, 50, 19, 21, 2 ]; let n = arr.length; findTriplet(arr, n); // This code is contributed by Sarthak Delori </script> |
Output
Numbers are: 2 5 7
Time Complexity: O(N^2*log N)
Auxiliary Space: O(1)
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