Javascript Program to Find array sum using Bitwise OR after splitting given array in two halves after K circular shifts

Last Updated : 31 Mar, 2022

Given an array A[] of length N, where N is an even number, the task is to answer Q independent queries where each query consists of a positive integer K representing the number of circular shifts performed on the array and find the sum of elements by performing Bitwise OR operation on the divided array.
Note: Each query begins with the original array.
Examples:

Input: A[] = {12, 23, 4, 21, 22, 76}, Q = 1, K = 2
Output: 117
Explanation:
Since K is 2, modified array A[]={22, 76, 12, 23, 4, 21}.
Bitwise OR of first half of array = (22 | 76 | 12) = 94
Bitwise OR of second half of array = (21 | 23 | 4) = 23
Sum of OR values is 94 + 23 = 117
Input: A[] = {7, 44, 19, 86, 65, 39, 75, 101}, Q = 1, K = 4
Output: 238
Since K is 4, modified array A[]={65, 39, 75, 101, 7, 44, 19, 86}.
Bitwise OR of first half of array = 111
Bitwise OR of second half of array = 127
Sum of OR values is 111 + 127 = 238

Naive Approach:
To solve the problem mentioned above the simplest approach is to shift each element of the array by K % (N / 2) and then traverse the array to calculate the OR of the two halves for every query. But this method is not efficient and hence can be optimized further.
Efficient Approach:
To optimize the above mentioned approach we can take the help of Segment Tree data structure.

Observation:

We can observe that after exactly N / 2 right circular shifts the two halves of the array become the same as in the original array. This effectively reduces the number of rotations to K % (N / 2).
Performing a right circular shift is basically shifting the last element of the array to the front. So for any positive integer X performing X right circular shifts is equal to shifting the last X elements of the array to the front.

Following are the steps to solve the problem :

Construct a segment tree for the original array A[] and assign a variable let's say i = K % (N / 2).
Then for each query we use the segment tree of find the bitwise OR; that is Bitwise OR of i elements from the end OR bitwise OR of the first (N / 2) - i - 1 elements.
Then calculate the bitwise OR of elements in range [(N / 2) - i, N - i - 1].
Add the two results to get the answer for the ith query.

Below is the implementation of the above approach:

JavaScript

<script>
// javascript program to find Bitwise OR of two
// equal halves of an array after performing
// K right circular shifts
    const MAX = 100005;

    // Array for storing
    // the segment tree
    var seg = Array(4 * MAX).fill(0);

    // Function to build the segment tree
    function build(node , l , r , a) {
        if (l == r)
            seg[node] = a[l];

        else {
            var mid = parseInt((l + r) / 2);

            build(2 * node, l, mid, a);
            build(2 * node + 1, mid + 1, r, a);

            seg[node] = (seg[2 * node] | seg[2 * node + 1]);
        }
    }

    // Function to return the OR
    // of elements in the range [l, r]
    function query(node , l , r , start , end , a) {

        // Check for out of bound condition
        if (l > end || r < start)
            return 0;

        if (start <= l && r <= end)
            return seg[node];

        // Find middle of the range
        var mid = parseInt((l + r) / 2);

        // Recurse for all the elements in array
        return ((query(2 * node, l, mid, start, end, a)) | (query(2 * node + 1, mid + 1, r, start, end, a)));
    }

    // Function to find the OR sum
    function orsum(a , n , q , k) {

        // Function to build the segment Tree
        build(1, 0, n - 1, a);

        // Loop to handle q queries
        for (j = 0; j < q; j++) {

            // Effective number of
            // right circular shifts
            var i = k[j] % (n / 2);

            // Calculating the OR of
            // the two halves of the
            // array from the segment tree

            // OR of second half of the
            // array [n/2-i, n-1-i]
            var sec = query(1, 0, n - 1, n / 2 - i, n - i - 1, a);

            // OR of first half of the array
            // [n-i, n-1]OR[0, n/2-1-i]
            var first = (query(1, 0, n - 1, 0, n / 2 - 1 - i, a) | query(1, 0, n - 1, n - i, n - 1, a));

            var temp = sec + first;

            // Print final answer to the query
            document.write(temp + "<br/>");
        }
    }

    // Driver Code
        var a = [ 7, 44, 19, 86, 65, 39, 75, 101 ];
        var n = a.length;
        var q = 2;

        var k = [ 4, 2 ];

        orsum(a, n, q, k);

// This code is contributed by Rajput-Ji. 
</script>