Pair with given Sum (Two Sum)
Last Updated :
28 May, 2024
Given an array A[] of n numbers and another number x, the task is to check whether or not there exist two elements in A[] whose sum is exactly x.
Examples:
Input: arr[] = {0, -1, 2, -3, 1}, x= -2
Output: Yes
Explanation: If we calculate the sum of the output,1 + (-3) = -2
Input: arr[] = {1, -2, 1, 0, 5}, x = 0
Output: No
Two Sum using Naive Approach:
The basic approach to solve this problem is by nested traversal.
- Traverse the array using a loop
- For each element:
- Check if there exists another in the array with sum as x
- Return true if yes, else continue
- If no such pair is found, return false.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find and print pair
bool chkPair(int A[], int size, int x)
{
for (int i = 0; i < (size - 1); i++) {
for (int j = (i + 1); j < size; j++) {
if (A[i] + A[j] == x) {
return 1;
}
}
}
return 0;
}
// Driver code
int main()
{
int A[] = { 0, -1, 2, -3, 1 };
int x = -2;
int size = sizeof(A) / sizeof(A[0]);
if (chkPair(A, size, x)) {
cout << "Yes" << endl;
}
else {
cout << "No" << x << endl;
}
return 0;
}
// This code is contributed by Samim Hossain Mondal.
/*
* This C program tells if there exists a pair in array
* whose sum results in x.
*/
#include <stdio.h>
// Function to find and print pair
int chkPair(int A[], int size, int x)
{
for (int i = 0; i < (size - 1); i++) {
for (int j = (i + 1); j < size; j++) {
if (A[i] + A[j] == x) {
return 1;
}
}
}
return 0;
}
int main(void)
{
int A[] = { 0, -1, 2, -3, 1 };
int x = -2;
int size = sizeof(A) / sizeof(A[0]);
if (chkPair(A, size, x)) {
printf("Yes\n");
}
else {
printf("No\n");
}
return 0;
}
// This code is contributed by Manish Kumar (mkumar2789)
// Java program to check if there exists a pair
// in array whose sum results in x.
import java.io.*;
class GFG {
// Function to find and print pair
static boolean chkPair(int A[], int size, int x)
{
for (int i = 0; i < (size - 1); i++) {
for (int j = (i + 1); j < size; j++) {
if (A[i] + A[j] == x) {
return true;
}
}
}
return false;
}
public static void main(String[] args)
{
int A[] = { 0, -1, 2, -3, 1 };
int x = -2;
int size = A.length;
if (chkPair(A, size, x)) {
System.out.println("Yes");
}
else {
System.out.println("No");
}
}
}
// This code is contributed by umadevi9616
# This python program tells if there exists a pair in array whose sum results in x.
# Function to find and print pair
def chkPair(A, size, x):
for i in range(0, size - 1):
for j in range(i + 1, size):
if (A[i] + A[j] == x):
return 1
return 0
if __name__ == "__main__":
A = [0, -1, 2, -3, 1]
x = -2
size = len(A)
if (chkPair(A, size, x)):
print("Yes")
else:
print("No")
# This code is contributed by rakeshsahni
// C# program to check if there exists a pair
// in array whose sum results in x.
using System;
class GFG {
// Function to find and print pair
static bool chkPair(int[] A, int size, int x)
{
for (int i = 0; i < (size - 1); i++) {
for (int j = (i + 1); j < size; j++) {
if (A[i] + A[j] == x) {
return true;
}
}
}
return false;
}
public static void Main()
{
int[] A = { 0, -1, 2, -3, 1 };
int x = -2;
int size = A.Length;
if (chkPair(A, size, x)) {
Console.WriteLine("Yes");
}
else {
Console.WriteLine("No");
}
}
}
// This code is contributed by Samim Hossain Mondal.
<script>
// Javascript program to check if there exists a pair
// in array whose sum results in x.
// Function to find and print pair
function chkPair(A , size , x) {
for (i = 0; i < (size - 1); i++) {
for (j = (i + 1); j < size; j++) {
if (A[i] + A[j] == x) {
document.write("Pair with a given sum " + x + " is (" + A[i] + ", " + A[j] + ")");
return true;
}
}
}
return false;
}
let A = [ 0, -1, 2, -3, 1 ];
let x = -2;
let size = A.length;
if (chkPair(A, size, x)) {
document.write("<br/>Valid pair exists");
}
else {
document.write("<br/>No valid pair exists for " + x);
}
// This code is contributed by Samim Hossain Mondal.
</script>
Time Complexity: O(N2), Finding pair for every element in the array of size N.
Auxiliary Space: O(1)
The idea is to use the two-pointer technique. But for using the two-pointer technique, the array must be sorted. Once the array is sorted the two pointers can be taken which mark the beginning and end of the array respectively. If the sum is greater than the sum of those two elements, shift the right pointer to decrease the value of the required sum and if the sum is lesser than the required value, shift the left pointer to increase the value of the required sum.
Illustration:
Let an array be {1, 4, 45, 6, 10, -8} and sum to find be 16
After sorting the array
A = {-8, 1, 4, 6, 10, 45}
Now, increment ‘l’ when the sum of the pair is less than the required sum and decrement ‘r’ when the sum of the pair is more than the required sum.
This is because when the sum is less than the required sum then to get the number which could increase the sum of pair, start moving from left to right(also sort the array) thus “l++” and vice versa.
Initialize l = 0, r = 5
A[l] + A[r] ( -8 + 45) > 16 => decrement r. Now r = 4
A[l] + A[r] ( -8 + 10) increment l. Now l = 1
A[l] + A[r] ( 1 + 10) increment l. Now l = 2
A[l] + A[r] ( 4 + 10) increment l. Now l = 3
A[l] + A[r] ( 6 + 10) == 16 => Found candidates (return 1)
Note: If there is more than one pair having the given sum then this algorithm reports only one. Can be easily extended for this though.
Follow the steps below to solve the problem:
- hasArrayTwoCandidates (A[], ar_size, sum)
- Sort the array in non-decreasing order.
- Initialize two index variables to find the candidate
elements in the sorted array.
- Initialize first to the leftmost index: l = 0
- Initialize second the rightmost index: r = ar_size-1
- Loop while l < r.
- If (A[l] + A[r] == sum) then return 1
- Else if( A[l] + A[r] < sum ) then l++
- Else r–
- No candidates in the whole array – return 0
Below is the implementation of the above approach:
C++
// C++ program to check if given array
// has 2 elements whose sum is equal
// to the given value
#include <bits/stdc++.h>
using namespace std;
// Function to check if array has 2 elements
// whose sum is equal to the given value
bool hasArrayTwoCandidates(int A[], int arr_size, int sum)
{
int l, r;
/* Sort the elements */
sort(A, A + arr_size);
/* Now look for the two candidates in
the sorted array*/
l = 0;
r = arr_size - 1;
while (l < r) {
if (A[l] + A[r] == sum)
return 1;
else if (A[l] + A[r] < sum)
l++;
else // A[l] + A[r] > sum
r--;
}
return 0;
}
/* Driver program to test above function */
int main()
{
int A[] = { 1, 4, 45, 6, 10, -8 };
int n = 16;
int arr_size = sizeof(A) / sizeof(A[0]);
// Function calling
if (hasArrayTwoCandidates(A, arr_size, n))
cout << "Yes";
else
cout << "No";
return 0;
}
// C program to check if given array
// has 2 elements whose sum is equal
// to the given value
#include <stdio.h>
#define bool int
void quickSort(int*, int, int);
bool hasArrayTwoCandidates(int A[], int arr_size, int sum)
{
int l, r;
/* Sort the elements */
quickSort(A, 0, arr_size - 1);
/* Now look for the two candidates in the sorted
array*/
l = 0;
r = arr_size - 1;
while (l < r) {
if (A[l] + A[r] == sum)
return 1;
else if (A[l] + A[r] < sum)
l++;
else // A[i] + A[j] > sum
r--;
}
return 0;
}
/* FOLLOWING FUNCTIONS ARE ONLY FOR SORTING
PURPOSE */
void exchange(int* a, int* b)
{
int temp;
temp = *a;
*a = *b;
*b = temp;
}
int partition(int A[], int si, int ei)
{
int x = A[ei];
int i = (si - 1);
int j;
for (j = si; j <= ei - 1; j++) {
if (A[j] <= x) {
i++;
exchange(&A[i], &A[j]);
}
}
exchange(&A[i + 1], &A[ei]);
return (i + 1);
}
/* Implementation of Quick Sort
A[] --> Array to be sorted
si --> Starting index
ei --> Ending index
*/
void quickSort(int A[], int si, int ei)
{
int pi; /* Partitioning index */
if (si < ei) {
pi = partition(A, si, ei);
quickSort(A, si, pi - 1);
quickSort(A, pi + 1, ei);
}
}
/* Driver program to test above function */
int main()
{
int A[] = { 1, 4, 45, 6, 10, -8 };
int n = 16;
int arr_size = 6;
if (hasArrayTwoCandidates(A, arr_size, n))
printf("Yes");
else
printf("No");
getchar();
return 0;
}
// Java program to check if given array
// has 2 elements whose sum is equal
// to the given value
import java.util.*;
class GFG {
// Function to check if array has 2 elements
// whose sum is equal to the given value
static boolean
hasArrayTwoCandidates(int A[], int arr_size, int sum)
{
int l, r;
/* Sort the elements */
Arrays.sort(A);
/* Now look for the two candidates
in the sorted array*/
l = 0;
r = arr_size - 1;
while (l < r) {
if (A[l] + A[r] == sum)
return true;
else if (A[l] + A[r] < sum)
l++;
else // A[i] + A[j] > sum
r--;
}
return false;
}
// Driver code
public static void main(String args[])
{
int A[] = { 1, 4, 45, 6, 10, -8 };
int n = 16;
int arr_size = A.length;
// Function calling
if (hasArrayTwoCandidates(A, arr_size, n))
System.out.println("Yes");
else
System.out.println("No");
}
}
# Python program to check for the sum condition to be satisfied
def hasArrayTwoCandidates(A, arr_size, sum):
# sort the array
quickSort(A, 0, arr_size-1)
l = 0
r = arr_size-1
# traverse the array for the two elements
while l < r:
if (A[l] + A[r] == sum):
return 1
elif (A[l] + A[r] < sum):
l += 1
else:
r -= 1
return 0
# Implementation of Quick Sort
# A[] --> Array to be sorted
# si --> Starting index
# ei --> Ending index
def quickSort(A, si, ei):
if si < ei:
pi = partition(A, si, ei)
quickSort(A, si, pi-1)
quickSort(A, pi + 1, ei)
# Utility function for partitioning
# the array(used in quick sort)
def partition(A, si, ei):
x = A[ei]
i = (si - 1)
for j in range(si, ei):
if A[j] <= x:
i += 1
# This operation is used to swap two variables in Python
A[i], A[j] = A[j], A[i]
A[i + 1], A[ei] = A[ei], A[i + 1]
return i + 1
# Driver program to test the functions
A = [1, 4, 45, 6, 10, -8]
n = 16
if (hasArrayTwoCandidates(A, len(A), n)):
print("Yes")
else:
print("No")
# This code is contributed by __Devesh Agrawal__
// C# program to check for pair
// in A[] with sum as x
using System;
class GFG {
static bool hasArrayTwoCandidates(int[] A, int arr_size,
int sum)
{
int l, r;
/* Sort the elements */
sort(A, 0, arr_size - 1);
/* Now look for the two candidates
in the sorted array*/
l = 0;
r = arr_size - 1;
while (l < r) {
if (A[l] + A[r] == sum)
return true;
else if (A[l] + A[r] < sum)
l++;
else // A[i] + A[j] > sum
r--;
}
return false;
}
/* Below functions are only to sort the
array using QuickSort */
/* This function takes last element as pivot,
places the pivot element at its correct
position in sorted array, and places all
smaller (smaller than pivot) to left of
pivot and all greater elements to right
of pivot */
static int partition(int[] arr, int low, int high)
{
int pivot = arr[high];
// index of smaller element
int i = (low - 1);
for (int j = low; j <= high - 1; j++) {
// If current element is smaller
// than or equal to pivot
if (arr[j] <= pivot) {
i++;
// swap arr[i] and arr[j]
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
// swap arr[i+1] and arr[high] (or pivot)
int temp1 = arr[i + 1];
arr[i + 1] = arr[high];
arr[high] = temp1;
return i + 1;
}
/* The main function that
implements QuickSort()
arr[] --> Array to be sorted,
low --> Starting index,
high --> Ending index */
static void sort(int[] arr, int low, int high)
{
if (low < high) {
/* pi is partitioning index, arr[pi]
is now at right place */
int pi = partition(arr, low, high);
// Recursively sort elements before
// partition and after partition
sort(arr, low, pi - 1);
sort(arr, pi + 1, high);
}
}
// Driver code
public static void Main()
{
int[] A = { 1, 4, 45, 6, 10, -8 };
int n = 16;
int arr_size = 6;
if (hasArrayTwoCandidates(A, arr_size, n))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Sam007
<script>
// Javascript program to check if given array
// has 2 elements whose sum is equal
// to the given value
// Function to check if array has 2 elements
// whose sum is equal to the given value
function hasArrayTwoCandidates(A, arr_size, sum)
{
var l, r;
/* Sort the elements */
A.sort();
/* Now look for the two candidates in
the sorted array*/
l = 0;
r = arr_size - 1;
while (l < r) {
if (A[l] + A[r] == sum)
return 1;
else if (A[l] + A[r] < sum)
l++;
else // A[i] + A[j] > sum
r--;
}
return 0;
}
/* Driver program to test above function */
var A = [ 1, 4, 45, 6, 10, -8 ]
var n = 16;
var arr_size = A.length;
// Function calling
if (hasArrayTwoCandidates(A, arr_size, n))
document.write("Array has two elements" +
" with the given sum");
else
document.write("Array doesn't have two" +
" elements with the given sum");
</script>
<?php
// PHP program to check if given
// array has 2 elements whose sum
// is equal to the given value
// Function to check if array has
// 2 elements whose sum is equal
// to the given value
function hasArrayTwoCandidates($A, $arr_size,
$sum)
{
$l; $r;
/* Sort the elements */
//sort($A, A + arr_size);
sort($A);
/* Now look for the two candidates
in the sorted array*/
$l = 0;
$r = $arr_size - 1;
while ($l < $r)
{
if($A[$l] + $A[$r] == $sum)
return 1;
else if($A[$l] + $A[$r] < $sum)
$l++;
else // A[i] + A[j] > sum
$r--;
}
return 0;
}
// Driver Code
$A = array (1, 4, 45, 6, 10, -8);
$n = 16;
$arr_size = sizeof($A);
// Function calling
if(hasArrayTwoCandidates($A, $arr_size, $n))
echo "Yes";
else
echo "No";
// This code is contributed by m_kit
?>
Time Complexity: O(NlogN), Time complexity for sorting the array
Auxiliary Space: O(1)
Sort the array, then traverse the array elements and perform binary search for (target – a[i]) on the remaining part
Follow the below steps to solve the problem:
- Sort the array in non-decreasing order.
- Traverse from 0 to N-1
- Initialize searchKey = sum – A[i]
- If(binarySearch(searchKey, A, i + 1, N) == True
- Return False
Below is the implementation of the above approach:
C++
// C++ program to check if given array
// has 2 elements whose sum is equal
// to the given value
#include <bits/stdc++.h>
using namespace std;
bool binarySearch(int A[], int low, int high, int searchKey)
{
while (low <= high) {
int m = low + (high - low) / 2;
// Check if searchKey is present at mid
if (A[m] == searchKey)
return true;
// If searchKey greater, ignore left half
if (A[m] < searchKey)
low = m + 1;
// If searchKey is smaller, ignore right half
else
high = m - 1;
}
// if we reach here, then element was
// not present
return false;
}
bool checkTwoSum(int A[], int arr_size, int sum)
{
int l, r;
/* Sort the elements */
sort(A, A + arr_size);
// Traversing all element in an array search for
// searchKey
for (int i = 0; i < arr_size - 1; i++) {
int searchKey = sum - A[i];
// calling binarySearch function
if (binarySearch(A, i + 1, arr_size - 1, searchKey)
== true) {
return true;
}
}
return false;
}
/* Driver program to test above function */
int main()
{
int A[] = { 1, 4, 45, 6, 10, -8 };
int n = 14;
int arr_size = sizeof(A) / sizeof(A[0]);
// Function calling
if (checkTwoSum(A, arr_size, n))
cout << "Yes";
else
cout << "No";
return 0;
}
// Java program to check if given array
// has 2 elements whose sum is equal
// to the given value
import java.util.*;
class GFG {
static boolean binarySearch(int A[], int low, int high,
int searchKey)
{
while (low <= high) {
int m = low + (high - low) / 2;
// Check if searchKey is present at mid
if (A[m] == searchKey)
return true;
// If searchKey greater, ignore left half
if (A[m] < searchKey)
low = m + 1;
// If searchKey is smaller, ignore right half
else
high = m - 1;
}
// if we reach here, then element was
// not present
return false;
}
static boolean checkTwoSum(int A[], int arr_size,
int sum)
{
int l, r;
/* Sort the elements */
Arrays.sort(A);
// Traversing all element in an array search for
// searchKey
for (int i = 0; i < arr_size - 1; i++) {
int searchKey = sum - A[i];
if (binarySearch(A, i + 1, arr_size - 1,
searchKey)
== true) {
return true;
}
}
return false;
}
// Driver code
public static void main(String args[])
{
int A[] = { 1, 4, 45, 6, 10, -8 };
int n = 14;
int arr_size = A.length;
// Function calling
if (checkTwoSum(A, arr_size, n))
System.out.println("Yes");
else
System.out.println("No");
}
}
# Python program to check for the sum
# condition to be satisfied
def binarySearch(A, low, high, searchKey):
m = 0
while (low <= high):
m = (high + low) // 2
# Check if searchKey is present at mid
if (A[m] == searchKey):
return 1
# If searchKey greater, ignore left half
if (A[m] < searchKey):
low = m + 1
# If searchKey is smaller, ignore right half
else:
high = m - 1
# if we reach here, then element was
# not present
return 0
def checkTwoSum(A, arr_size, sum):
# sort the array
A.sort()
l = 0
r = arr_size-1
# Traversing all element in an array search for searchKey
i = 0
while i < arr_size-1:
searchKey = sum-A[i]
# calling binarySearch function
if(binarySearch(A, i+1, r, searchKey) == 1):
return 1
i = i+1
return 0
# Driver program to test the functions
A = [1, 4, 45, 6, 10, -8]
n = 14
if (checkTwoSum(A, len(A), n)):
print("Yes")
else:
print("No")
// C# program to check for pair
// in A[] with sum as x
using System;
using System.Collections.Generic;
class GFG {
static bool binarySearch(int[] A, int low, int high,
int searchKey)
{
while (low <= high) {
int m = low + (high - low) / 2;
// Check if searchKey is present at mid
if (A[m] == searchKey)
return true;
// If searchKey greater, ignore left half
if (A[m] < searchKey)
low = m + 1;
// If searchKey is smaller, ignore right half
else
high = m - 1;
}
// if we reach here, then element was
// not present
return false;
}
static bool checkTwoSum(int[] A, int arr_size, int sum)
{
Array.Sort(A);
// Traversing all element in an array search for
// searchKey
for (int i = 0; i < arr_size - 1; i++) {
int searchKey = sum - A[i];
// calling binarySearch function
if (binarySearch(A, i + 1, arr_size - 1,
searchKey)
== true) {
return true;
}
}
return false;
}
// Driver code
public static void Main()
{
int[] A = { 1, 4, 45, 6, 10, -8 };
int n = 16;
int arr_size = 6;
if (checkTwoSum(A, arr_size, n))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by garg28harsh.
function binarySearch( A, low, high, searchKey)
{
while (low <= high) {
let m = low + (high - low) / 2;
// Check if searchKey is present at mid
if (A[m] == searchKey)
return true;
// If searchKey greater, ignore left half
if (A[m] < searchKey)
low = m + 1;
// If searchKey is smaller, ignore right half
else
high = m - 1;
}
// if we reach here, then element was
// not present
return false;
}
function checkTwoSum( A, arr_size, sum)
{
/* Sort the elements */
A.sort();
// Traversing all element in an array search for
// searchKey
for (let i = 0; i < arr_size - 1; i++) {
let searchKey = sum - A[i];
// calling binarySearch function
if (binarySearch(A, i + 1, arr_size - 1, searchKey) == true) {
return true;
}
}
return false;
}
/* Driver program to test above function */
let A = [ 1, 4, 45, 6, 10, -8 ];
let n = 14;
let arr_size = 6;
// Function calling
if (checkTwoSum(A, arr_size, n))
console.log( "Yes");
else
console.log("No");
// This code is contributed by garg28harsh.
Time Complexity: O(NlogN)
Auxiliary Space: O(1)
Two Sum using Hashing:
This problem can be solved efficiently by using the technique of hashing. Use a hash_map to check for the current array value x(let), if there exists a value target_sum-x which on adding to the former gives target_sum. This can be done in constant time.
Illustration:
arr[] = {0, -1, 2, -3, 1}
sum = -2
Now start traversing:
Step 1: For ‘0’ there is no valid number ‘-2’ so store ‘0’ in hash_map.
Step 2: For ‘-1’ there is no valid number ‘-1’ so store ‘-1’ in hash_map.
Step 3: For ‘2’ there is no valid number ‘-4’ so store ‘2’ in hash_map.
Step 4: For ‘-3’ there is no valid number ‘1’ so store ‘-3’ in hash_map.
Step 5: For ‘1’ there is a valid number ‘-3’ so answer is 1, -3
unordered_set s
for(i=0 to end)
if(s.find(target_sum – arr[i]) == s.end)
insert(arr[i] into s)
else
print arr[i], target-arr[i]
Follow the steps below to solve the problem:
- Initialize an empty hash table s.
- Do the following for each element A[i] in A[]
- If s[x – A[i]] is set then print the pair (A[i], x – A[i])
- Insert A[i] into s.
Below is the implementation of the above approach:
C++
// C++ program to check if given array
// has 2 elements whose sum is equal
// to the given value
#include <bits/stdc++.h>
using namespace std;
void printPairs(int arr[], int arr_size, int sum)
{
unordered_set<int> s;
for (int i = 0; i < arr_size; i++) {
int temp = sum - arr[i];
if (s.find(temp) != s.end()) {
cout << "Yes" << endl;
return;
}
s.insert(arr[i]);
}
cout << "No" << endl;
}
/* Driver Code */
int main()
{
int A[] = { 1, 4, 45, 6, 10, 8 };
int n = 16;
int arr_size = sizeof(A) / sizeof(A[0]);
// Function calling
printPairs(A, arr_size, n);
return 0;
}
// C program to check if given array
// has 2 elements whose sum is equal
// to the given value
#include <stdio.h>
#define MAX 100000
// NOTE: Works only if range elements is limited
// target - arr[i] >= 0 && target - arr[i] < MAX
void printPairs(int arr[], int arr_size, int target)
{
int i, temp;
/*initialize hash set as 0*/
int s[MAX] = { 0 };
for (i = 0; i < arr_size; i++) {
temp = target - arr[i];
if (s[temp] == 1) {
printf("Yes");
return;
}
s[arr[i]] = 1;
}
printf("No");
}
/* Driver Code */
int main()
{
int A[] = { 1, 4, 45, 6, 10, 8 };
int target = 16;
int arr_size = sizeof(A) / sizeof(A[0]);
printPairs(A, arr_size, target);
getchar();
return 0;
}
// Java implementation using Hashing
import java.io.*;
import java.util.HashSet;
class PairSum {
static void printpairs(int arr[], int sum)
{
HashSet<Integer> s = new HashSet<Integer>();
for (int i = 0; i < arr.length; ++i) {
int temp = sum - arr[i];
// checking for condition
if (s.contains(temp)) {
System.out.println("Yes");
return;
}
s.add(arr[i]);
}
System.out.println("No");
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 1, 4, 45, 6, 10, 8 };
int n = 16;
printpairs(A, n);
}
}
// This article is contributed by Aakash Hasija
# Python program to find if there are
# two elements with given sum
# function to check for the given sum
# in the array
def printPairs(arr, arr_size, sum):
# Create an empty hash map
# using an hashmap allows us to store the indices
hashmap = {}
for i in range(0, arr_size):
temp = sum-arr[i]
if (temp in hashmap):
print('Yes')
return
hashmap[arr[i]] = i
print("No")
# driver code
A = [1, 4, 45, 6, 10, 8]
n = 16
printPairs(A, len(A), n)
# This code will also work in case the array has the same number twice
# and target is the sum of those numbers
# Eg: Array = [4,6,4] Target = 8
# This code is contributed by __Achyut Upadhyay__
// C# implementation using Hashing
using System;
using System.Collections.Generic;
class GFG {
static void printpairs(int[] arr, int sum)
{
HashSet<int> s = new HashSet<int>();
for (int i = 0; i < arr.Length; ++i) {
int temp = sum - arr[i];
// checking for condition
if (s.Contains(temp)) {
Console.Write("Yes");
return;
}
s.Add(arr[i]);
}
Console.Write("No");
}
// Driver Code
static void Main()
{
int[] A = new int[] { 1, 4, 45, 6, 10, 8 };
int n = 16;
printpairs(A, n);
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
<script>
// JavaScript program to check if given array
// has 2 elements whose sum is equal
// to the given value
// Javascript implementation using Hashing
function printpairs(arr, sum)
{
let s = new Set();
for (let i = 0; i < arr.length; ++i)
{
let temp = sum - arr[i];
// checking for condition
if (s.has(temp)) {
document.write(
"Pair with given sum "
+ sum + " is (" + arr[i]
+ ", " + temp + ")");
}
s.add(arr[i]);
}
}
// Driver Code
let A = [ 1, 4, 45, 6, 10, 8 ];
let n = 16;
printpairs(A, n);
</script>
Time Complexity: O(N), As the whole array is needed to be traversed only once.
Auxiliary Space: O(N), A hash map has been used to store array elements.
Note: The solution will work even if the range of numbers includes negative numbers + if the pair is formed by numbers recurring twice in array eg: array = [3,4,3]; pair = (3,3); target sum = 6.
Two Sum Using remainders of the elements less than x:
The idea is to count the elements with remainders when divided by x, i.e 0 to x-1, each remainder separately. Suppose we have x as 6, then the numbers which are less than 6 and have remainders which add up to 6 gives sum as 6 when added. For example, we have elements, 2,4 in the array and 2%6 = 2 and 4%6 =4, and these remainders add up to give 6. Like that we have to check for pairs with remainders (1,5),(2,4),(3,3). if we have one or more elements with remainder 1 and one or more elements with remainder 5, then surely we get a sum as 6. Here we do not consider (0,6) as the elements for the resultant pair should be less than 6. when it comes to (3,3) we have to check if we have two elements with remainder 3, then we can say that “There exists a pair whose sum is x”.
Follow the steps below to solve the problem:
- 1. Create an array with size x.
- 2. Initialize all rem elements to zero.
- 3. Traverse the given array
- Do the following if arr[i] is less than x:
- r=arr[i]%x which is done to get the remainder.
- rem[r]=rem[r]+1 i.e. increasing the count of elements that have remainder r when divided with x.
- 4. Now, traverse the rem array from 1 to x/2.
- If(rem[i]> 0 and rem[x-i]>0) then print “YES” and come out of the loop. This means that we have a pair that results in x upon doing.
- 5. Now when we reach at x/2 in the above loop
- If x is even, for getting a pair we should have two elements with remainder x/2.
- If rem[x/2]>1 then print “YES” else print “NO”
- If it is not satisfied that is x is odd, it will have a separate pair with x-x/2.
- If rem[x/2]>0 and rem[x-x/2]>0 , then print “Yes” else, print”No”;
Below is the implementation of the above approach:
C++
// Code in cpp to tell if there exists a pair in array whose
// sum results in x.
#include <iostream>
using namespace std;
// Function to print pairs
void printPairs(int a[], int n, int x)
{
int i;
int rem[x];
// initializing the rem values with 0's.
for (i = 0; i < x; i++)
rem[i] = 0;
// Perform the remainder operation only if the element
// is x, as numbers greater than x can't be used to get
// a sum x. Updating the count of remainders.
for (i = 0; i < n; i++)
if (a[i] < x)
rem[a[i] % x]++;
// Traversing the remainder list from start to middle to
// find pairs
for (i = 1; i < x / 2; i++) {
if (rem[i] > 0 && rem[x - i] > 0) {
// The elements with remainders i and x-i will
// result to a sum of x. Once we get two
// elements which add up to x , we print x and
// break.
cout << "Yes\n";
break;
}
}
// Once we reach middle of remainder array, we have to
// do operations based on x.
if (i >= x / 2) {
if (x % 2 == 0) {
// if x is even and we have more than 1 elements
// with remainder x/2, then we will have two
// distinct elements which add up to x. if we
// dont have more than 1 element, print "No".
if (rem[x / 2] > 1)
cout << "Yes\n";
else
cout << "No\n";
}
else {
// When x is odd we continue the same process
// which we did in previous loop.
if (rem[x / 2] > 0 && rem[x - x / 2] > 0)
cout << "Yes\n";
else
cout << "No\n";
}
}
}
/* Driver Code */
int main()
{
int A[] = { 1, 4, 45, 6, 10, 8 };
int n = 16;
int arr_size = sizeof(A) / sizeof(A[0]);
// Function calling
printPairs(A, arr_size, n);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// Code in c to tell if there exists a pair in array whose
// sum results in x.
#include <stdio.h>
// Function to print pairs
void printPairs(int a[], int n, int x)
{
int i;
int rem[x];
// initializing the rem values with 0's.
for (i = 0; i < x; i++)
rem[i] = 0;
// Perform the remainder operation only if the element
// is x, as numbers greater than x can't be used to get
// a sum x. Updating the count of remainders.
for (i = 0; i < n; i++)
if (a[i] < x)
rem[a[i] % x]++;
// Traversing the remainder list from start to middle to
// find pairs
for (i = 1; i < x / 2; i++) {
if (rem[i] > 0 && rem[x - i] > 0) {
// The elements with remainders i and x-i will
// result to a sum of x. Once we get two
// elements which add up to x , we print x and
// break.
printf("Yes\n");
break;
}
}
// Once we reach middle of remainder array, we have to
// do operations based on x.
if (i >= x / 2) {
if (x % 2 == 0) {
// if x is even and we have more than 1 elements
// with remainder x/2, then we will have two
// distinct elements which add up to x. if we
// dont have more than 1 element, print "No".
if (rem[x / 2] > 1)
printf("Yes\n");
else
printf("No\n");
}
else {
// When x is odd we continue the same process
// which we did in previous loop.
if (rem[x / 2] > 0 && rem[x - x / 2] > 0)
printf("Yes\n");
else
printf("No\n");
}
}
}
/* Driver Code */
int main()
{
int A[] = { 1, 4, 45, 6, 10, 8 };
int n = 16;
int arr_size = sizeof(A) / sizeof(A[0]);
// Function calling
printPairs(A, arr_size, n);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// Code in Java to tell if there exists a pair in array
// whose sum results in x.
import java.util.*;
class GFG {
// Function to print pairs
static void printPairs(int a[], int n, int x)
{
int i;
int[] rem = new int[x];
// initializing the rem values with 0's.
for (i = 0; i < x; i++)
rem[i] = 0;
// Perform the remainder operation only if
// the element is x, as numbers greater than
// x can't be used to get a sum x. Updating
// the count of remainders.
for (i = 0; i < n; i++)
if (a[i] < x)
rem[a[i] % x]++;
// Traversing the remainder list from start to
// middle to find pairs
for (i = 1; i < x / 2; i++) {
if (rem[i] > 0 && rem[x - i] > 0) {
// The elements with remainders i and x-i
// will result to a sum of x. Once we get
// two elements which add up to x , we print
// x and break.
System.out.println("Yes");
break;
}
}
// Once we reach middle of remainder array, we have
// to do operations based on x.
if (i >= x / 2) {
if (x % 2 == 0) {
// if x is even and we have more than 1
// elements with remainder x/2, then we
// will have two distinct elements which
// add up to x. if we dont have more
// than 1 element, print "No".
if (rem[x / 2] > 1)
System.out.println("Yes");
else
System.out.println("No");
}
else {
// When x is odd we continue the same
// process which we did in previous loop.
if (rem[x / 2] > 0 && rem[x - x / 2] > 0)
System.out.println("Yes");
else
System.out.println("No");
}
}
}
/* Driver Code */
public static void main(String[] args)
{
int A[] = { 1, 4, 45, 6, 10, 8 };
int n = 16;
int arr_size = A.length;
// Function calling
printPairs(A, arr_size, n);
}
}
// This code is contributed by Aditya Kumar (adityakumar129)
# Code in Python3 to tell if there
# exists a pair in array whose
# sum results in x.
# Function to print pairs
def printPairs(a, n, x):
rem = []
for i in range(x):
# Initializing the rem
# values with 0's.
rem.append(0)
for i in range(n):
if (a[i] < x):
# Perform the remainder operation
# only if the element is x, as
# numbers greater than x can't
# be used to get a sum x.Updating
# the count of remainders.
rem[a[i] % x] += 1
# Traversing the remainder list from
# start to middle to find pairs
for i in range(1, x // 2):
if (rem[i] > 0 and rem[x - i] > 0):
# The elements with remainders
# i and x-i will result to a
# sum of x. Once we get two
# elements which add up to x,
# we print x and break.
print("Yes")
break
# Once we reach middle of
# remainder array, we have to
# do operations based on x.
if (i >= x // 2):
if (x % 2 == 0):
if (rem[x // 2] > 1):
# If x is even and we have more
# than 1 elements with remainder
# x/2, then we will have two
# distinct elements which add up
# to x. if we dont have than 1
# element, print "No".
print("Yes")
else:
print("No")
else:
# When x is odd we continue
# the same process which we
# did in previous loop.
if (rem[x // 2] > 0 and
rem[x - x // 2] > 0):
print("Yes")
else:
print("No")
# Driver Code
A = [1, 4, 45, 6, 10, 8]
n = 16
arr_size = len(A)
# Function calling
printPairs(A, arr_size, n)
# This code is contributed by subhammahato348
// C# Code in C# to tell if there
// exists a pair in array whose
// sum results in x.
using System;
class GFG {
// Function to print pairs
static void printPairs(int[] a, int n, int x)
{
int i;
int[] rem = new int[x];
for (i = 0; i < x; i++) {
// initializing the rem
// values with 0's.
rem[i] = 0;
}
for (i = 0; i < n; i++) {
if (a[i] < x) {
// Perform the remainder
// operation only if the
// element is x, as numbers
// greater than x can't
// be used to get a sum x.
// Updating the count of remainders.
rem[a[i] % x]++;
}
}
// Traversing the remainder list
// from start to middle to
// find pairs
for (i = 1; i < x / 2; i++) {
if (rem[i] > 0 && rem[x - i] > 0) {
// The elements with remainders
// i and x-i will
// result to a sum of x.
// Once we get two
// elements which add up to x ,
// we print x and
// break.
Console.Write("Yes"
+ "\n");
break;
}
}
// Once we reach middle of
// remainder array, we have to
// do operations based on x.
if (i >= x / 2) {
if (x % 2 == 0) {
if (rem[x / 2] > 1) {
// if x is even and
// we have more than 1
// elements with remainder
// x/2, then we will
// have two distinct elements
// which add up
// to x. if we dont have
// more than 1
// element, print "No".
Console.Write("Yes"
+ "\n");
}
else {
Console.Write("No"
+ "\n");
}
}
else {
// When x is odd we continue
// the same process
// which we did in previous loop.
if (rem[x / 2] > 0 && rem[x - x / 2] > 0) {
Console.Write("Yes"
+ "\n");
}
else {
Console.WriteLine("No"
+ "\n");
}
}
}
}
/* Driver Code */
public static void Main(string[] args)
{
int[] A = { 1, 4, 45, 6, 10, 8 };
int n = 16;
int arr_size = A.Length;
// Function calling
printPairs(A, arr_size, n);
}
}
// This code is contributed by SoumikMondal
<script>
// Code in Javascript to tell if there
// exists a pair in array whose
// sum results in x.
// Function to print pairs
function printPairs(a, n, x)
{
let i;
let rem = new Array(x);
for(i = 0; i < x; i++)
{
// Initializing the rem
// values with 0's.
rem[i] = 0;
}
for(i = 0; i < n; i++)
{
if (a[i] < x)
{
// Perform the remainder
// operation only if the
// element is x, as numbers
// greater than x can't
// be used to get a sum x.
// Updating the count of remainders.
rem[a[i] % x]++;
}
}
// Traversing the remainder list
// from start to middle to
// find pairs
for(i = 1; i < x / 2; i++)
{
if (rem[i] > 0 && rem[x - i] > 0)
{
// The elements with remainders
// i and x-i will
// result to a sum of x.
// Once we get two
// elements which add up to x ,
// we print x and
// break.
document.write("Yes" + "</br>");
break;
}
}
// Once we reach middle of
// remainder array, we have to
// do operations based on x.
if (i >= x / 2)
{
if (x % 2 == 0)
{
if (rem[x / 2] > 1)
{
// If x is even and
// we have more than 1
// elements with remainder
// x/2, then we will
// have two distinct elements
// which add up
// to x. if we dont have
//more than 1
// element, print "No".
document.write("Yes" + "</br>");
}
else
{
document.write("No" + "</br>");
}
}
else
{
// When x is odd we continue
// the same process
// which we did in previous loop.
if (rem[x / 2] > 0 &&
rem[x - x / 2] > 0)
{
document.write("Yes" + "</br>");
}
else
{
document.write("No" + "</br>");
}
}
}
}
// Driver code
let A = [ 1, 4, 45, 6, 10, 8 ];
let n = 16;
let arr_size = A.length;
// Function calling
printPairs(A, arr_size, n);
// This code is contributed by suresh07
</script>
Time Complexity: O(N+X), Traversing over the array of size N and Checking for remainders till X
Auxiliary Space: O(X), Space for storing remainders
Related Problems:
Please write comments if you find any of the above codes/algorithms incorrect, or find other ways to solve the same problem.
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