C Program for Two Pointers Technique - GeeksforGeeks

13 Jan, 2022
Two pointers is really an easy and effective technique which is typically used for searching pairs in a sorted array.Given a sorted array A (sorted in ascending order), having N integers, find if there exists any pair of elements (A[i], A[j]) such that their sum is equal to X.
Let’s see the naive solution.  
C
// Naive solution to find if there is a// pair in A[0..N-1] with given sum.#include <stdio.h>  int isPairSum(int A[],int  N,int X){    for (int i = 0; i < N; i++) {        for (int j = 0; j < N; j++)         {            // as equal i and j means same element            if (i == j)                 continue;                        // pair exists            if (A[i] + A[j] == X)                return true;               // as the array is sorted            if (A[i] + A[j] > X)                break;         }    }      // No pair found with given sum.    return 0;}  // Driver Codeint main(){    int arr[]={3,5,9,2,8,10,11};    int val=17;    int arrSize = sizeof(arr)/sizeof(arr[0]);        // Function call    printf("%d",isPairSum(arr,arrSize,val));      return 0;}
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Time Complexity:  O(n2).
Now let’s see how the two-pointer technique works. We take two pointers, one representing the first element and other representing the last element of the array, and then we add the values kept at both the pointers. If their sum is smaller than X then we shift the left pointer to right or if their sum is greater than X then we shift the right pointer to left, in order to get closer to the sum. We keep moving the pointers until we get the sum as X. 
C
#include <stdio.h>  // Two pointer technique based solution to find// if there is a pair in A[0..N-1] with a given sum.int isPairSum(int A[], int N, int X){    // represents first pointer    int i = 0;      // represents second pointer    int j = N - 1;      while (i < j)    {        // If we find a pair        if (A[i] + A[j] == X)            return 1;          // If sum of elements at current        // pointers is less, we move towards        // higher values by doing i++        else if (A[i] + A[j] < X)            i++;          // If sum of elements at current        // pointers is more, we move towards        // lower values by doing j--        else            j--;    }    return 0;}  // Driver codeint main(){    // array declaration    int arr[] = { 3, 5, 9, 2, 8, 10, 11 };          // value to search    int val = 17;          // size of the array    int arrSize = sizeof(arr) / sizeof(arr[0]);        // Function call    printf("%d", isPairSum(arr, arrSize, val));      return 0;}
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Illustration :  
Time Complexity:  O(n)
How does this work? The algorithm basically uses the fact that the input array is sorted. We start the sum of extreme values (smallest and largest) and conditionally move both pointers. We move left pointer i when the sum of A[i] and A[j] is less than X. We do not miss any pair because the sum is already smaller than X. Same logic applies for right pointer j.
More problems based on two pointer technique. 
Find the closest pair from two sorted arrays
Find the pair in array whose sum is closest to x
Find all triplets with zero sum
Find a triplet that sum to a given value
Find a triplet such that sum of two equals to third element
Find four elements that sum to a given value
Please refer complete article on Two Pointers Technique for more details!
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