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Find a value X in range [0, K] which can maximize X XOR sum over given array

  • Difficulty Level : Basic
  • Last Updated : 16 Nov, 2021

Given an array a[] of size N and an integer K, the task is to find a value X in the range [0, K] which can maximize the value of the given function

  • Xor-sum(X) = (X XOR A[0]) + (X Xor A[1]) + (X Xor A[2]) + __________+ (X Xor A[N-1]).

Examples:

Input: a[] = {1, 2, 3, 4, 5, 6}, N=6, K=10
Output: 8
Explanation: The value of X is 1 for which the required sum becomes maximum.

Input: a[] = {1, 6}, N=2, K=7
Output: 0

Approach: The idea is to consider each and every value of K and then find the value of X which satisfies the above condition. Follow the steps below to solve the problem:

  • Initialize the variables max and X as 0 and -1 to store the maximum sum defined and the value of X for which it is happening.
  • Iterate over the range [0, K] using the variable j and perform the following tasks:
    • Initialize the variable sum as 0 to store the defined sum.
    • Iterate over the range [0, N) using the variable i and perform the following tasks:
      • Set the value of sum as sum + j^a[i].
    • If sum is greater than max, then set the value of max as sum and X as j.
  • After performing the above steps, print the value of X as the answer.

Below is the implementation of the above approach

C++14




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the value of X for which
// the given sum maximises
void findValue(vector<int>& a, int N, int K)
{
 
    // Variables to store the maximum
    // sum and that particular value
    // of X.
    long long int max = 0, X = -1;
 
    // Check every value of K
    for (int j = 0; j <= K; j++) {
 
        // Variable to store the desired sum
        long long int sum = 0;
        for (int i = 0; i < N; i++) {
 
            (sum = sum + (j ^ a[i]));
        }
 
        // Check the condition
        if (sum > max) {
            max = sum;
            X = j;
        }
    }
 
    // Print the result
    cout << X;
}
 
// Driver Code
int main()
 
{
 
    vector<int> a = { 1, 2, 3, 4, 5, 6 };
    int N = 6, K = 10;
 
    findValue(a, N, K);
 
    return 0;
}

Java




// Java program for the above approach
class GFG {
 
    // Function to find the value of X for which
    // the given sum maximises
    public static void findValue(int[] a, int N, int K) {
 
        // Variables to store the maximum
        // sum and that particular value
        // of X.
        int max = 0, X = -1;
 
        // Check every value of K
        for (int j = 0; j <= K; j++) {
 
            // Variable to store the desired sum
            int sum = 0;
            for (int i = 0; i < N; i++) {
 
                sum = sum + (j ^ a[i]);
            }
 
            // Check the condition
            if (sum > max) {
                max = sum;
                X = j;
            }
        }
 
        // Print the result
        System.out.println(X);
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int[] a = { 1, 2, 3, 4, 5, 6 };
        int N = 6, K = 10;
 
        findValue(a, N, K);
 
    }
 
}
 
// This code is ontributed by saurabh_jaiswal.

Python3




# Python3 program for the above approach
 
# Function to find the value of X for which
# the given sum maximises
def findValue(a, N, K) :
 
    # Variables to store the maximum
    # sum and that particular value
    # of X.
    max = 0; X = -1;
 
    # Check every value of K
    for j in range( K + 1) :
 
        # Variable to store the desired sum
        sum = 0;
        for i in range(N) :
 
            sum = sum + (j ^ a[i]);
 
        # Check the condition
        if (sum > max) :
            max = sum;
            X = j;
      
    # Print the result
    print(X);
 
# Driver Code
if __name__ == "__main__" :
 
    a = [ 1, 2, 3, 4, 5, 6 ];
    N = 6; K = 10;
 
    findValue(a, N, K);
 
    # This code is contributed by AnkThon

C#




// C# program for the above approach
using System;
 
public class GFG {
 
    // Function to find the value of X for which
    // the given sum maximises
    public static void findValue(int[] a, int N, int K) {
 
        // Variables to store the maximum
        // sum and that particular value
        // of X.
        int max = 0, X = -1;
 
        // Check every value of K
        for (int j = 0; j <= K; j++) {
 
            // Variable to store the desired sum
            int sum = 0;
            for (int i = 0; i < N; i++) {
 
                sum = sum + (j ^ a[i]);
            }
 
            // Check the condition
            if (sum > max) {
                max = sum;
                X = j;
            }
        }
 
        // Print the result
        Console.WriteLine(X);
    }
 
    // Driver Code
    public static void Main(string []args)
    {
        int[] a = { 1, 2, 3, 4, 5, 6 };
        int N = 6, K = 10;
 
        findValue(a, N, K);
    }
}
 
// This code is contributed by AnkThon

Javascript




<script>
        // JavaScript Program to implement
        // the above approach
 
        // Function to find the value of X for which
        // the given sum maximises
        function findValue(a, N, K) {
 
            // Variables to store the maximum
            // sum and that particular value
            // of X.
            let max = 0, X = -1;
 
            // Check every value of K
            for (let j = 0; j <= K; j++) {
 
                // Variable to store the desired sum
                let sum = 0;
                for (let i = 0; i < N; i++) {
 
                    (sum = sum + (j ^ a[i]));
                }
 
                // Check the condition
                if (sum > max) {
                    max = sum;
                    X = j;
                }
            }
 
            // Print the result
            document.write(X);
        }
 
        // Driver Code
        let a = [1, 2, 3, 4, 5, 6];
        let N = 6, K = 10;
 
        findValue(a, N, K);
 
    // This code is contributed by Potta Lokesh
    </script>
Output
8

Time Complexity: O(K*N)
Auxiliary Space: O(1)


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