4 Sum - Find any quadruplet having given sum
Last Updated :
23 Jul, 2025
Given an array arr[] of n integers and an integer target, the task is to find any quadruplet in arr[] such that it's sum is equal to the target.
Note: If there are multiple quadruplets with sum = target, return any one of them.
Examples:
Input: arr[] = {2, 4, 6, 8, 1, 3}, target = 15
Output: {2, 4, 6, 3}
Explanation: The quadruplet {2, 4, 6, 3} has sum = 15.
Input: arr[] = {1, 2, 3, 4, 10}, target = 20
Output: {}
Explanation: No quadruplet adds up to a sum of 20.
[Naive Approach] - Explore all the subsets of size four - O(n^4) Time and O(1) Space
The idea is to generate all possible quadruplets using four nested loops and calculate the sum of each quadruplet. If the sum of a quadruplet is equal to target, return that quadruplet.
C++
// C++ program to find a quadruplet with sum
// equal to target by exploring all quadruplets
#include <iostream>
#include <vector>
using namespace std;
vector<int> findQuadruplet(vector<int>& arr, int target) {
int n = arr.size();
// Generating all possible quadruplets
for (int i = 0; i < n - 3; i++) {
for (int j = i + 1; j < n - 2; j++) {
for (int k = j + 1; k < n - 1; k++) {
for(int l = k + 1; l < n; l++) {
int currSum = arr[i] + arr[j] + arr[k] + arr[l];
// If current sum equals target, we have found desired quadruplet
if (currSum == target) {
return {arr[i], arr[j], arr[k], arr[l]};
}
}
}
}
}
// Return empty vector if no quadruplet found
return {};
}
int main() {
vector<int> arr = {2, 4, 6, 8, 1, 3};
int target = 15;
vector<int> res = findQuadruplet(arr, target);
for (int i = 0; i < res.size(); i++)
cout << res[i] << " ";
return 0;
}
// C program to find a quadruplet with sum
// equal to target by exploring all quadruplets
#include <stdio.h>
int* findQuadruplet(int* arr, int n, int target) {
// Generating all possible quadruplets
for (int i = 0; i < n - 3; i++) {
for (int j = i + 1; j < n - 2; j++) {
for (int k = j + 1; k < n - 1; k++) {
for (int l = k + 1; l < n; l++) {
int currSum = arr[i] + arr[j] + arr[k] + arr[l];
// If current sum equals target, we have found desired quadruplet
if (currSum == target) {
int* res = (int*)malloc(4 * sizeof(int));
res[0] = arr[i];
res[1] = arr[j];
res[2] = arr[k];
res[3] = arr[l];
return res;
}
}
}
}
}
// Return NULL if no quadruplet found
return NULL;
}
int main() {
int arr[] = {2, 4, 6, 8, 1, 3};
int target = 15;
int n = sizeof(arr) / sizeof(arr[0]);
int* res = findQuadruplet(arr, n, target);
if (res != NULL) {
for (int i = 0; i < 4; i++)
printf("%d ", res[i]);
}
return 0;
}
// Java program to find a quadruple with sum
// equal to target by exploring all quadruplets
import java.util.ArrayList;
import java.util.List;
class GfG {
static List<Integer> findQuadruplet(int[] arr, int target) {
int n = arr.length;
// Generating all possible quadruplets
for (int i = 0; i < n - 3; i++) {
for (int j = i + 1; j < n - 2; j++) {
for (int k = j + 1; k < n - 1; k++) {
for (int l = k + 1; l < n; l++) {
int currSum = arr[i] + arr[j] + arr[k] + arr[l];
// If current sum equals target, we have found desired quadruplet
if (currSum == target) {
List<Integer> result = new ArrayList<>();
result.add(arr[i]);
result.add(arr[j]);
result.add(arr[k]);
result.add(arr[l]);
return result;
}
}
}
}
}
// Return empty list if no quadruplet found
return new ArrayList<>();
}
public static void main(String[] args) {
int[] arr = {2, 4, 6, 8, 1, 3};
int target = 15;
List<Integer> res = findQuadruplet(arr, target);
for (int i = 0; i < res.size(); i++) {
System.out.print(res.get(i) + " ");
}
}
}
# Python program to find a quadruple with sum
# equal to target by exploring all quadruplets
def findQuadruplet(arr, target):
n = len(arr)
# Generating all possible quadruplets
for i in range(n - 3):
for j in range(i + 1, n - 2):
for k in range(j + 1, n - 1):
for l in range(k + 1, n):
currSum = arr[i] + arr[j] + arr[k] + arr[l]
# If current sum equals target, we have found desired quadruplet
if currSum == target:
return [arr[i], arr[j], arr[k], arr[l]]
# Return empty list if no quadruplet found
return []
if __name__ == "__main__":
arr = [2, 4, 6, 8, 1, 3]
target = 15
res = findQuadruplet(arr, target)
for i in range(len(res)):
print(res[i], end=" ")
// C# program to find a quadruple with sum
// equal to target by exploring all quadruplets
using System;
using System.Collections.Generic;
class GfG {
static List<int> findQuadruplet(List<int> arr, int target) {
int n = arr.Count;
// Generating all possible quadruplets
for (int i = 0; i < n - 3; i++) {
for (int j = i + 1; j < n - 2; j++) {
for (int k = j + 1; k < n - 1; k++) {
for(int l = k + 1; l < n; l++) {
int currSum = arr[i] + arr[j] + arr[k] + arr[l];
// If current sum equals target, we have found desired quadruplet
if (currSum == target) {
return new List<int> { arr[i], arr[j], arr[k], arr[l] };
}
}
}
}
}
// Return empty list if no quadruplet found
return new List<int>();
}
static void Main() {
List<int> arr = new List<int> { 2, 4, 6, 8, 1, 3 };
int target = 15;
List<int> res = findQuadruplet(arr, target);
for (int i = 0; i < res.Count; i++)
Console.Write(res[i] + " ");
}
}
// JavaScript program to find a quadruplet with sum
// equal to target by exploring all quadruplets
function findQuadruplet(arr, target) {
const n = arr.length;
// Generating all possible quadruplets
for (let i = 0; i < n - 3; i++) {
for (let j = i + 1; j < n - 2; j++) {
for (let k = j + 1; k < n - 1; k++) {
for (let l = k + 1; l < n; l++) {
const currSum = arr[i] + arr[j] + arr[k] + arr[l];
// If current sum equals target, we have found desired quadruplet
if (currSum === target) {
return [arr[i], arr[j], arr[k], arr[l]];
}
}
}
}
}
// Return empty array if no quadruplet found
return [];
}
const arr = [2, 4, 6, 8, 1, 3];
const target = 15;
const res = findQuadruplet(arr, target);
console.log(res.join(" "));
[Expected Approach] Sorting and Two Pointers Technique – O(n^3) time and O(1) space
Initially, we sort the input array so that we can apply two pointers technique. Then, we fix the first two elements of the quadruplet using two nested loops and inside the second nested loop we use two pointers technique to find the remaining two elements. Set one pointer at the beginning (left) and another at the end (right) of the remaining array. We then check the sum of all these four elements and compare it with the given target:
- If sum < target, move left pointer towards right to increase the sum.
- If sum > target, move right pointer towards left to decrease the sum.
- If sum == target, we’ve found the quadruple.
C++
// C++ program to find a quadruplet having
// sum target using two pointers technique
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> findQuadruplet(vector<int> &arr, int target) {
int n = arr.size();
// Sorting the array to use two pointers technique
sort(arr.begin(), arr.end());
for (int i = 0; i < n - 3; i++) {
for (int j = i + 1; j < n - 2; j++) {
int l = j + 1, r = n - 1;
while (l < r) {
int currSum = arr[i] + arr[j] + arr[l] + arr[r];
// If currSum is equal to target,
// we have found the desired quadruplet
if (currSum == target) {
return {arr[i], arr[j], arr[l], arr[r]};
}
// If currSum exceeds the target, move
// the right pointer left to reduce the sum
if (currSum > target)
r--;
// If currSum is less than target, move
// the left pointer right to increase the sum
else
l++;
}
}
}
// Return empty array if no quadruplet found
return {};
}
int main() {
vector<int> arr = {2, 4, 6, 8, 1, 3};
int target = 15;
vector<int> res = findQuadruplet(arr, target);
for (int i = 0; i < res.size(); i++)
cout << res[i] << " ";
return 0;
}
// C program to find a quadruplet having
// sum target using two pointer technique
#include <stdio.h>
int compare(const void* a, const void* b) {
return (*(int*)a - *(int*)b);
}
int* findQuadruplet(int* arr, int n, int target) {
// Sorting the array to use two pointers technique
qsort(arr, n, sizeof(int), compare);
for (int i = 0; i < n - 3; i++) {
for (int j = i + 1; j < n - 2; j++) {
int l = j + 1, r = n - 1;
while (l < r) {
int currSum = arr[i] + arr[j] + arr[l] + arr[r];
// If currSum is equal to target,
// we have found the desired quadruplet
if (currSum == target) {
int* result = (int*)malloc(4 * sizeof(int));
result[0] = arr[i];
result[1] = arr[j];
result[2] = arr[l];
result[3] = arr[r];
return result;
}
// If currSum exceeds the target, move
// the right pointer left to reduce the sum
if (currSum > target)
r--;
// If currSum is less than target, move
// the left pointer right to increase the sum
else
l++;
}
}
}
// Return NULL if no quadruplet found
return NULL;
}
int main() {
int arr[] = {2, 4, 6, 8, 1, 3};
int target = 15;
int n = sizeof(arr) / sizeof(arr[0]);
int* res = findQuadruplet(arr, n, target);
if (res != NULL) {
for (int i = 0; i < 4; i++)
printf("%d ", res[i]);
}
return 0;
}
// Java program to find a quadruplet having
// sum target using two pointer technique
import java.util.Arrays;
import java.util.List;
import java.util.ArrayList;
class GfG {
static List<Integer> findQuadruplet(int[] arr, int target) {
int n = arr.length;
// Sorting the array to use two pointers technique
Arrays.sort(arr);
for (int i = 0; i < n - 3; i++) {
for (int j = i + 1; j < n - 2; j++) {
int l = j + 1, r = n - 1;
while (l < r) {
int currSum = arr[i] + arr[j] + arr[l] + arr[r];
// If currSum is equal to target,
// we have found the desired quadruplet
if (currSum == target)
return Arrays.asList(arr[i], arr[j], arr[l], arr[r]);
// If currSum exceeds the target, move
// the right pointer left to reduce the sum
if (currSum > target)
r--;
// If currSum is less than target, move
// the left pointer right to increase the sum
else
l++;
}
}
}
// Return empty array if no quadruplet found
return new ArrayList<>();
}
public static void main(String[] args) {
int[] arr = {2, 4, 6, 8, 1, 3};
int target = 15;
List<Integer> res = findQuadruplet(arr, target);
for (int i = 0; i < res.size(); i++)
System.out.print(res.get(i) + " ");
}
}
# Python program to find a quadruplet having
# sum target using two pointers technique
def findQuadruplet(arr, target):
n = len(arr)
# Sorting the array to use two pointers technique
arr.sort()
for i in range(n - 3):
for j in range(i + 1, n - 2):
l, r = j + 1, n - 1
while l < r:
currSum = arr[i] + arr[j] + arr[l] + arr[r]
# If currSum is equal to target,
# we have found the desired quadruplet
if currSum == target:
return [arr[i], arr[j], arr[l], arr[r]]
# If currSum exceeds the target, move
# the right pointer left to reduce the sum
if currSum > target:
r -= 1
# If currSum is less than target, move
# the left pointer right to increase the sum
else:
l += 1
# Return empty array if no quadruplet found
return []
if __name__ == "__main__":
arr = [2, 4, 6, 8, 1, 3]
target = 15
res = findQuadruplet(arr, target)
for i in range(len(res)):
print(res[i], end=" ")
// C# program to find a quadruplet having
// sum target using two pointer technique
using System;
using System.Collections.Generic;
class GfG {
static List<int> findQuadruplet(List<int> arr, int target) {
int n = arr.Count;
// Sorting the array to use two pointers technique
arr.Sort();
for (int i = 0; i < n - 3; i++) {
for (int j = i + 1; j < n - 2; j++) {
int l = j + 1, r = n - 1;
while (l < r) {
int currSum = arr[i] + arr[j] + arr[l] + arr[r];
// If currSum is equal to target,
// we have found the desired quadruplet
if (currSum == target) {
return new List<int> { arr[i], arr[j], arr[l], arr[r] };
}
// If currSum exceeds the target, move
// the right pointer left to reduce the sum
if (currSum > target)
r--;
// If currSum is less than target, move
// the left pointer right to increase the sum
else
l++;
}
}
}
// Return empty array if no quadruplet found
return new List<int>();
}
static void Main() {
List<int> arr = new List<int> { 2, 4, 6, 8, 1, 3 };
int target = 15;
List<int> res = findQuadruplet(arr, target);
for (int i = 0; i < res.Count; i++)
Console.Write(res[i] + " ");
}
}
// JavaScript program to find a quadruplet having
// sum target using two pointer technique
function findQuadruplet(arr, target) {
let n = arr.length;
// Sorting the array to use two pointers technique
arr.sort((a, b) => a - b);
for (let i = 0; i < n - 3; i++) {
for (let j = i + 1; j < n - 2; j++) {
let l = j + 1, r = n - 1;
while (l < r) {
let currSum = arr[i] + arr[j] + arr[l] + arr[r];
// If currSum is equal to target,
// we have found the desired quadruplet
if (currSum === target) {
return [arr[i], arr[j], arr[l], arr[r]];
}
// If currSum exceeds the target, move
// the right pointer left to reduce the sum
if (currSum > target) {
r--;
}
// If currSum is less than target, move
// the left pointer right to increase the sum
else {
l++;
}
}
}
}
// Return empty array if no quadruplet found
return [];
}
const arr = [2, 4, 6, 8, 1, 3];
const target = 15;
const res = findQuadruplet(arr, target);
console.log(res.join(" "));
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