# Given an array A[] and a number x, check for pair in A[] with sum as x

Write a C program that, given an array A[] of n numbers and another number x, determines whether or not there exist two elements in S whose sum is exactly x.

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

METHOD 1 (Use Sorting)

Algorithm :

```hasArrayTwoCandidates (A[], ar_size, sum)
1) Sort the array in non-decreasing order.
2) Initialize two index variables to find the candidate
elements in the sorted array.
(a) Initialize first to the leftmost index: l = 0
(b) Initialize second  the rightmost index:  r = ar_size-1
3) Loop while l < r.
(a) If (A[l] + A[r] == sum)  then return 1
(b) Else if( A[l] + A[r] <  sum )  then l++
(c) Else r--
4) No candidates in whole array - return 0
```

Time Complexity: Depends on what sorting algorithm we use. If we use Merge Sort or Heap Sort then (-)(nlogn) in worst case. If we use Quick Sort then O(n^2) in worst case.
Auxiliary Space : Again, depends on sorting algorithm. For example auxiliary space is O(n) for merge sort and O(1) for Heap Sort.

Example :

Let Array be {1, 4, 45, 6, 10, -8} and sum to find be 16

Sort the array
A = {-8, 1, 4, 6, 10, 45}

Initialize l = 0, r = 5
A[l] + A[r] ( -8 + 45) > 16 => decrement r. Now r = 10
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 are more than one pair having the given sum then this algorithm reports only one. Can be easily extended for this though.

Below is the implementation of the above approach.

## C

```# 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("Array has two elements with given sum");
else
printf("Array doesn't have two elements with given sum");

getchar();
return 0;
}
```

## 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[i] + A[j] > 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 << "Array has two elements with given sum";
else
cout << "Array doesn't have two elements with given sum";

return 0;
}

```

## Java

```// 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("Array has two " +
"elements with given sum");
else
System.out.println("Array doesn't have " +
"two elements with given sum");

}

}

```

## Python

```# Python program to check for the sum condition to be satisified
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 is 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("Array has two elements with the given sum")
else:
print("Array doesn't have two elements with the given sum")

## This code is contributed by __Devesh Agrawal__
```

## C#

```// 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);
}
}

// main function
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("Array has two elements"+
" with given sum");
else
Console.Write("Array doesn't have "+
"two elements with given sum");

}
}

// This code is contributed by Sam007
```

## PHP

```
<?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 "Array has two elements " .
"with given sum";
else
echo "Array doesn't have two " .
"elements with given sum";

// This code is contributed by m_kit
?>

```

Output :

`Array has two elements with the given sum`

METHOD 2 (Use Hashing)
This method works in O(n) time.

```1) Initialize an empty hash table s.
2) Do following for each element A[i] in A[]
(a)    If s[x - A[i]] is set then print the pair (A[i], x - A[i])
(b)    Insert A[i] into s.
```

Below is the implementation of the above approach :

## C

```// Works only if range elements is limited
#include <stdio.h>
#define MAX 100000

void printPairs(int arr[], int arr_size, int sum)
{
int i, temp;
bool s[MAX] = {0}; /*initialize hash set as 0*/

for (i = 0; i < arr_size; i++)
{
temp = sum - arr[i];
if (temp >= 0 && s[temp] == 1)
printf("Pair with given sum %d is (%d, %d) n",
sum, arr[i], temp);
s[arr[i]] = 1;
}
}

/* 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]);

printPairs(A, arr_size, n);

getchar();
return 0;
}
```

## 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 (temp >= 0 && s.find(temp) != s.end())
cout << "Pair with given sum " << sum <<
" is (" << arr[i] << ", " << temp <<
")" << endl;

s.insert(arr[i]);
}
}

/* 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
printPairs(A, arr_size, n);

return 0;
}
```

## Java

```// 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 (temp>=0 && s.contains(temp))
{
System.out.println("Pair with given sum " +
sum + " is (" + arr[i] +
", "+temp+")");
}
s.add(arr[i]);
}
}

// Main to test the above function
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

```# Python program to find if there are
# two elements wtih given sum

# function to check for the given sum
# in the array
def printPairs(arr, arr_size, sum):

# Create an empty hash set
s = set()

for i in range(0,arr_size):
temp = sum-arr[i]
if (temp>=0 and temp in s):
print ("Pair with the given sum is", arr[i], "and", temp)
s.add(arr[i])

# driver program to check the above function
A = [1,4,45,6,10,-8]
n = 16
printPairs(A, len(A), n)

# This code is contributed by __Devesh Agrawal__
```

## C#

```// 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 (temp >= 0 && s.Contains(temp))
{
Console.Write("Pair with given sum " +
sum + " is (" + arr[i] +
", " + temp + ")");
}
s.Add(arr[i]);
}
}

// 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)

```

Output:

`Pair with given sum 16 is (10, 6)`

Time Complexity: O(n)
Auxiliary Space: O(n) where n is size of array.

If range of numbers include negative numbers then also it works. All we have to do for negative numbers is to make everything positive by adding the absolute value of smallest negative integer to all numbers.

### Asked in: Amazon,CarWale,Flipkart,Microsoft

Please write comments if you find any of the above codes/algorithms incorrect, or find other ways to solve the same problem.

My Personal Notes arrow_drop_up

Improved By : jit_t, manishshaw1

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