2 Sum - Find All Pairs With Given Sum
Last Updated :
11 Oct, 2024
Given an array arr[] and a target value, the task is to find all possible indices (i, j) of pairs (arr[i], arr[j]) whose sum is equal to target and i != j. We can return pairs in any order, but all the returned pairs should be internally sorted, that is for any pair(i, j), i should be less than j.
Examples:
Input: arr[] = {10, 20, 30, 20, 10, 30}, target = 50
Output: {{1, 2}, {1, 5}, {2, 3}, {3, 5}}
Explanation: All pairs with sum = 50 are:
- arr[1] + arr[2] = 20 + 30 = 50
- arr[1] + arr[5] = 20 + 30 = 50
- arr[2] + arr[3] = 30 + 20 = 50
- arr[3] + arr[5] = 20 + 30 = 50
Input: arr[] = {10, 20, 30, 20, 10, 30}, target = 80
Output: { }
Explanation: No pairs have sum = 80.
[Naive Approach] Using Nested Loops - O(n^2) Time and O(1) Space
The simplest approach is to generate all possible pairs from the given array arr[] and if the sum of elements of the pairs is equal to target, then add it to the result.
C++
// C++ Code to find all pairs using nested loops
#include <iostream>
#include <vector>
using namespace std;
// function to find all pairs
vector<vector<int>> findAllPairs(vector<int> &arr, int target) {
int n = arr.size();
vector<vector<int>> res;
// Two nested loops to generate all pairs
for(int i = 0; i < n; i++) {
for(int j = i + 1; j < n; j++) {
// If sum of pair equals target, add it to result
if(arr[i] + arr[j] == target) {
res.push_back({i, j});
}
}
}
return res;
}
int main() {
vector<int> arr = {10, 20, 30, 20, 10, 30};
int target = 50;
vector<vector<int>> res = findAllPairs(arr, target);
for(auto pair: res) {
cout << pair[0] << " " << pair[1] << "\n";
}
return 0;
}
// C Code to find all pairs using nested loops
#include <stdio.h>
// function to find all pairs
void findAllPairs(int *arr, int n, int target) {
// Two nested loops to generate all pairs
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
// If sum of pair equals target, print it
if (arr[i] + arr[j] == target) {
printf("%d %d\n", i, j);
}
}
}
}
int main() {
int arr[] = {10, 20, 30, 20, 10, 30};
int target = 50;
int n = sizeof(arr) / sizeof(arr[0]);
findAllPairs(arr, n, target);
return 0;
}
// Java Code to find all pairs using nested loops
import java.util.ArrayList;
import java.util.List;
class GfG {
// function to find all pairs
static List<List<Integer>> findAllPairs(int[] arr, int target) {
int n = arr.length;
List<List<Integer>> res = new ArrayList<>();
// Two nested loops to generate all pairs
for(int i = 0; i < n; i++) {
for(int j = i + 1; j < n; j++) {
// If sum of pair equals target, add it to result
if(arr[i] + arr[j] == target) {
List<Integer> pair = new ArrayList<>();
pair.add(i);
pair.add(j);
res.add(pair);
}
}
}
return res;
}
public static void main(String[] args) {
int[] arr = {10, 20, 30, 20, 10, 30};
int target = 50;
List<List<Integer>> res = findAllPairs(arr, target);
for(List<Integer> pair: res) {
System.out.println(pair.get(0) + " " + pair.get(1));
}
}
}
# Python Code to find all pairs using nested loops
def findAllPairs(arr, target):
n = len(arr)
res = []
# Two nested loops to generate all pairs
for i in range(n):
for j in range(i + 1, n):
# If sum of pair equals target, add it to result
if arr[i] + arr[j] == target:
res.append([i, j])
return res
if __name__ == "__main__":
arr = [10, 20, 30, 20, 10, 30]
target = 50
res = findAllPairs(arr, target)
for pair in res:
print(pair[0], pair[1])
// C# Code to find all pairs using nested loops
using System;
using System.Collections.Generic;
class GfG {
// function to find all pairs
static List<List<int>> findAllPairs(List<int> arr, int target) {
int n = arr.Count;
List<List<int>> res = new List<List<int>>();
// Two nested loops to generate all pairs
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
// If sum of pair equals target, add it to result
if (arr[i] + arr[j] == target) {
res.Add(new List<int> { i, j });
}
}
}
return res;
}
static void Main(string[] args) {
List<int> arr = new List<int> { 10, 20, 30, 20, 10, 30 };
int target = 50;
List<List<int>> res = findAllPairs(arr, target);
foreach (var pair in res) {
Console.WriteLine(pair[0] + " " + pair[1]);
}
}
}
// JavaScript Code to find all pairs using nested loops
// function to find all pairs
function findAllPairs(arr, target) {
let n = arr.length;
let res = [];
// Two nested loops to generate all pairs
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
// If sum of pair equals target, add it to result
if (arr[i] + arr[j] === target) {
res.push([i, j]);
}
}
}
return res;
}
const arr = [10, 20, 30, 20, 10, 30];
const target = 50;
const res = findAllPairs(arr, target);
for (const pair of res) {
console.log(pair[0] + " " + pair[1]);
}
[Expected Approach] Using Hashing - O(n^2) Time and O(n) Space
The idea is to maintain a hash map with element as key and its indices as values. Iterate over the array and for each element arr[i], if (target - arr[i]) exists in the hash map, then pair i with all indices of (target - arr[i]) and store them in result.
In the worst case, this approach also takes O(n^2) time but in the average case, it is much faster than Naive approach as we are iterating over only those pairs whose sum is equal to target.
C++
// C++ Code to find all pairs using hashing
#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
// function to find all pairs
vector<vector<int>> findAllPairs(vector<int> &arr, int target) {
int n = arr.size();
vector<vector<int>> res;
unordered_map<int, vector<int>> mp;
for (int i = 0; i < n; i++) {
// Check if there exists an element that can pair with arr[i]
if (mp.find(target - arr[i]) != mp.end()) {
// If such an element exists, iterate through its indices
for (int idx: mp[target - arr[i]]) {
res.push_back({idx, i});
}
}
// Store the index of the current element in the map
mp[arr[i]].push_back(i);
}
return res;
}
int main() {
vector<int> arr = {10, 20, 30, 20, 10, 30};
int target = 50;
vector<vector<int>> res = findAllPairs(arr, target);
for (auto pair : res) {
cout << pair[0] << " " << pair[1] << "\n";
}
return 0;
}
// Java Code to find all pairs using hashing
import java.util.*;
class GfG {
// function to find all pairs
static List<List<Integer>> findAllPairs(int[] arr, int target) {
int n = arr.length;
List<List<Integer>> res = new ArrayList<>();
Map<Integer, List<Integer>> mp = new HashMap<>();
for (int i = 0; i < n; i++) {
// Check if there exists an element that can pair with arr[i]
if (mp.containsKey(target - arr[i])) {
// If such an element exists, iterate through its indices
for (int idx : mp.get(target - arr[i])) {
res.add(Arrays.asList(idx, i));
}
}
// Store the index of the current element in the map
if (!mp.containsKey(arr[i])) {
mp.put(arr[i], new ArrayList<>());
}
mp.get(arr[i]).add(i);
}
return res;
}
public static void main(String[] args) {
int[] arr = {10, 20, 30, 20, 10, 30};
int target = 50;
List<List<Integer>> res = findAllPairs(arr, target);
for (List<Integer> pair : res) {
System.out.println(pair.get(0) + " " + pair.get(1));
}
}
}
# Python Code to find all pairs using hashing
# function to find all pairs
def findAllPairs(arr, target):
n = len(arr)
res = []
mp = {}
for i in range(n):
# Check if there exists an element that can pair with arr[i]
if target - arr[i] in mp:
# If such an element exists, iterate through its indices
for idx in mp[target - arr[i]]:
res.append([idx, i])
# Store the index of the current element in the map
if arr[i] not in mp:
mp[arr[i]] = []
mp[arr[i]].append(i)
return res
if __name__ == "__main__":
arr = [10, 20, 30, 20, 10, 30]
target = 50
res = findAllPairs(arr, target)
for pair in res:
print(pair[0], pair[1])
// C# Code to find all pairs using hashing
using System;
using System.Collections.Generic;
class GfG {
// function to find all pairs
static List<List<int>> findAllPairs(int[] arr, int target) {
int n = arr.Length;
List<List<int>> res = new List<List<int>>();
Dictionary<int, List<int>> mp = new Dictionary<int, List<int>>();
for (int i = 0; i < n; i++) {
// Check if there exists an element that can pair with arr[i]
if (mp.ContainsKey(target - arr[i])) {
// If such an element exists, iterate through its indices
foreach (int idx in mp[target - arr[i]]) {
res.Add(new List<int> { idx, i });
}
}
// Store the index of the current element in the map
if (!mp.ContainsKey(arr[i])) {
mp[arr[i]] = new List<int>();
}
mp[arr[i]].Add(i);
}
return res;
}
static void Main(string[] args) {
int[] arr = { 10, 20, 30, 20, 10, 30 };
int target = 50;
List<List<int>> res = findAllPairs(arr, target);
foreach (var pair in res) {
Console.WriteLine(pair[0] + " " + pair[1]);
}
}
}
// JavaScript Code to find all pairs using hashing
// function to find all pairs
function findAllPairs(arr, target) {
const n = arr.length;
const res = [];
const mp = new Map();
for (let i = 0; i < n; i++) {
// Check if there exists an element that can pair with arr[i]
if (mp.has(target - arr[i])) {
// If such an element exists, iterate through its indices
for (const idx of mp.get(target - arr[i])) {
res.push([idx, i]);
}
}
// Store the index of the current element in the map
if (!mp.has(arr[i])) {
mp.set(arr[i], []);
}
mp.get(arr[i]).push(i);
}
return res;
}
const arr = [10, 20, 30, 20, 10, 30];
const target = 50;
const res = findAllPairs(arr, target);
for (const pair of res) {
console.log(pair[0], pair[1]);
}
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