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Pair with sum 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.

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) < 2 => increment l. Now l = 1
A[l] + A[r] ( 1 + 10) < 16 => increment l. Now l = 2
A[l] + A[r] ( 4 + 10) < 14 => 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.

Implementation:

# 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;
}
 
/* 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 sum 16");
    else
        printf("Array doesn't have two elements with sum 16 ");
 
    getchar();
    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);
    }
}



METHOD 2 (Use Hash Map)
Thanks to Bindu for suggesting this method and thanks to Shekhu for providing code.
This method works in O(n) time if range of numbers is known.
Let sum be the given sum and A[] be the array in which we need to find pair.

1) Initialize Binary Hash Map M[] = {0, 0, …}
2) Do following for each element A[i] in A[]
   (a)	If M[x  - A[i]] is set then print the pair (A[i], x – A[i])
   (b)	Set M[A[i]]

Implementation:

#include <stdio.h>
#define MAX 100000
 
void printPairs(int arr[], int arr_size, int sum)
{
  int i, temp;
  bool binMap[MAX] = {0}; /*initialize hash map as 0*/
 
  for(i = 0; i < arr_size; i++)
  {
    temp = sum - arr[i];
    if(temp >= 0 && binMap[temp] == 1)
    {
      printf("Pair with given sum %d is (%d, %d) \n", sum, arr[i], temp);
    }
    binMap[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 = 6;
 
    printPairs(A, arr_size, n);
 
    getchar();
    return 0;
}

Time Complexity: O(n)
Auxiliary Space: O(R) where R is range of integers.

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.

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