Home - News ( United States | United Kingdom | Italy | Germany ) - Football scores

Showing content from https://www.geeksforgeeks.org/javascript-program-for-two-pointers-technique/ below:

Javascript Program for Two Pointers Technique

Last Updated : 16 Sep, 2024

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.

// Naive solution to find if there is a
// pair in A[0..N-1] with given sum.

function isPairSum(A, N, X) {
    for (let i = 0; i < N - 1; i++) {
        for (let j = i + 1; j < N; j++) {
            // as equal i and j means same element
            if (i == j)
                continue;

            // pair exists
            if (A[i] + A[j] == X)
                return 1;

            // as the array is sorted
            if (A[i] + A[j] > X)
                break;
        }
    }

    // No pair found with given sum.
    return 0;
}

// Driver code

// array declaration
let arr = [2, 3, 5, 8, 9, 10, 11];

// value to search
let val = 17;

// size of the array
let arrSize = 7;

// Function call
console.log(isPairSum(arr, arrSize, val));

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. 

// Two pointer technique based solution to find
// if there is a pair in A[0..N-1] with a given sum.
function isPairSum(A, N, X) {
    // represents first pointer
    let i = 0;

    // represents second pointer
    let j = N - 1;

    while (i < j) {
        // If we find a pair
        if (A[i] + A[j] == X)
            return true;

        // 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 false;
}

// Driver code

let arr = [2, 3, 5, 8, 9, 10, 11];
let val = 17;
let arrSize = 7;

let result = isPairSum(arr, arrSize, val);

console.log(result);

Illustration : 

• Time Complexity: O(n)
• Auxiliary Space: O(1) since using constant space

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.

• 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!

RetroSearch is an open source project built by @garambo | Open a GitHub Issue

Search and Browse the WWW like it's 1997 | Search results from DuckDuckGo

HTML: 3.2 | Encoding: UTF-8 | Version: 0.7.4