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.
Below is the implementation of the above approach:
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); }} |
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.
Step-by-step approach:
- 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:
Java
// Java program to find three numbers// such that sum of two makes the// third element in arrayimport 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 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 Sumit Ghosh |
Output
numbers are 21 2 19
Time complexity: O(N^2)
Auxiliary Space: O(1) as no extra space has been used.
Java Program to Find a triplet such that sum of two equals to third element 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.
Below is the implementation of the above approach:
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 |
Output
Numbers are: 2 5 7
Time Complexity: O(N^2*log N)
Auxiliary Space: 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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