Find 3 elements of an array with sum equals zero with example in c,c++,python and java

Here are solutions for finding three elements in an array whose sum equals zero in C, C++, Python, and Java:

Solution in C:

#include <stdio.h>

void findElements(int arr[], int n) {
    for (int i = 0; i < n - 2; i++) {
        for (int j = i + 1; j < n - 1; j++) {
            for (int k = j + 1; k < n; k++) {
                if (arr[i] + arr[j] + arr[k] == 0) {
                    printf("Triplet found: %d, %d, %dn", arr[i], arr[j], arr[k]);
                    return;
                }
            }
        }
    }
    printf("No triplet found.n");
}

int main() {
    int arr[] = {2, -5, 7, -3, 1, -1, 0};
    int size = sizeof(arr) / sizeof(arr[0]);

    findElements(arr, size);

    return 0;
}

Solution in C++:

#include <iostream>
#include <unordered_set>

void findElements(int arr[], int n) {
    for (int i = 0; i < n - 2; i++) {
        std::unordered_set<int> elements;

        for (int j = i + 1; j < n; j++) {
            int complement = -(arr[i] + arr[j]);
            if (elements.count(complement)) {
                std::cout << "Triplet found: " << arr[i] << ", " << arr[j] << ", " << complement << std::endl;
                return;
            }
            elements.insert(arr[j]);
        }
    }
    std::cout << "No triplet found." << std::endl;
}

int main() {
    int arr[] = {2, -5, 7, -3, 1, -1, 0};
    int size = sizeof(arr) / sizeof(arr[0]);

    findElements(arr, size);

    return 0;
}

Solution in Python:

def find_elements(arr):
    n = len(arr)
    for i in range(n - 2):
        elements = set()

        for j in range(i + 1, n):
            complement = -(arr[i] + arr[j])
            if complement in elements:
                print("Triplet found:", arr[i], ",", arr[j], ",", complement)
                return
            elements.add(arr[j])

    print("No triplet found.")

arr = [2, -5, 7, -3, 1, -1, 0]
find_elements(arr)

Solution in Java:

import java.util.HashSet;

class Main {
    static void findElements(int[] arr) {
        int n = arr.length;
        for (int i = 0; i < n - 2; i++) {
            HashSet<Integer> elements = new HashSet<>();

            for (int j = i + 1; j < n; j++) {
                int complement = -(arr[i] + arr[j]);
                if (elements.contains(complement)) {
                    System.out.println("Triplet found: " + arr[i] + ", " + arr[j] + ", " + complement);
                    return;
                }
                elements.add(arr[j]);
            }
        }
        System.out.println("No triplet found.");
    }

    public static void main(String[] args) {
        int[] arr = {2, -5, 7, -3, 1, -1, 0};
        findElements(arr);
    }
}

These solutions iterate over the array using nested loops and use different techniques to find three elements whose sum is zero. The C and C++ solutions use three nested loops and an unordered set, while the Python and Java solutions also use two loops and a set or HashSet to store the elements and check for the existence of the complement.