Minimize Bitwise XOR of array elements with 1 required to make sum of array at least K

Given an array arr[] consisting of N positive integers and a positive integer K, the task is to count minimum Bitwise XOR of array elements with 1 required such that the sum of the array is at least K.

Examples:

Input: arr[] = {0, 1, 1, 0, 1}, K = 4
Output: 1
Explanation: Performing Bitwise XOR of arr[0] and 1 modifies arr[] to {1, 1, 1, 0, 1}. Now, the sum of array = 1 + 1 + 1 + 0 + 1 = 4(= K).

Input: arr[] = {14, 0, 1, 0}, K = 20
Output: -1

Approach: The given problem can be solved using the fact that Bitwise XOR of 1 with an even element increases the element by 1. 
Follow the steps below to solve the problem:

• Find the sum of the given array arr[] and store it in a variable say S.
• Count the number of even elements in the array and store it in a variable, say E.
• If S is greater than or equal to K, then print 0.
• Otherwise, if the value of S + E is less than K, then print -1.
• Otherwise, print the value of (K – S) as the minimum number of Bitwise XOR of an array element with 1 is required.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find minimum number
// of Bitwise XOR of array elements
// with 1 required to make sum of
// the array at least K
int minStepK(int arr[], int N, int K)
{
    // Stores the count of
    // even array elements
    int E = 0;

    // Stores sum of the array
    int S = 0;

    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {

        // Increment sum
        S += arr[i];

        // If array element is even
        if (arr[i] % 2 == 0)

            // Increase count of even
            E += 1;
    }

    // If S is at least K
    if (S >= K)
        return 0;

    // If S + E is less than K
    else if (S + E < K)
        return -1;

    // Otherwise, moves = K - S
    else
        return K - S;
}

// Driver Code
int main()
{
    int arr[] = { 0, 1, 1, 0, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 4;
    cout << minStepK(arr, N, K);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find minimum number
// of Bitwise XOR of array elements
// with 1 required to make sum of
// the array at least K
static int minStepK(int arr[], int N, int K)
{

    // Stores the count of
    // even array elements
    int E = 0;

    // Stores sum of the array
    int S = 0;

    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {

        // Increment sum
        S += arr[i];

        // If array element is even
        if (arr[i] % 2 == 0)

            // Increase count of even
            E += 1;
    }

    // If S is at least K
    if (S >= K)
        return 0;

    // If S + E is less than K
    else if (S + E < K)
        return -1;

    // Otherwise, moves = K - S
    else
        return K - S;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 0, 1, 1, 0, 1 };
    int N = arr.length;
    int K = 4;

    System.out.println(minStepK(arr, N, K));
}
}

// This code is contributed by offbeat

Python3

# Python 3 program for the above approach

# Function to find minimum number
# of Bitwise XOR of array elements
# with 1 required to make sum of
# the array at least K
def minStepK(arr, N, K):

    # Stores the count of
    # even array elements
    E = 0

    # Stores sum of the array
    S = 0

    # Traverse the array arr[]
    for i in range(N):

        # Increment sum
        S += arr[i]

        # If array element is even
        if (arr[i] % 2 == 0):

            # Increase count of even
            E += 1

    # If S is at least K
    if (S >= K):
        return 0

    # If S + E is less than K
    elif (S + E < K):
        return -1

    # Otherwise, moves = K - S
    else:
        return K - S

# Driver Code
if __name__ == "__main__":

    arr = [0, 1, 1, 0, 1]
    N = len(arr)
    K = 4
    print(minStepK(arr, N, K))

    # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;
class GFG
{

// Function to find minimum number
// of Bitwise XOR of array elements
// with 1 required to make sum of
// the array at least K
static int minStepK(int[] arr, int N, int K)
{

    // Stores the count of
    // even array elements
    int E = 0;

    // Stores sum of the array
    int S = 0;

    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {

        // Increment sum
        S += arr[i];

        // If array element is even
        if (arr[i] % 2 == 0)

            // Increase count of even
            E += 1;
    }

    // If S is at least K
    if (S >= K)
        return 0;

    // If S + E is less than K
    else if (S + E < K)
        return -1;

    // Otherwise, moves = K - S
    else
        return K - S;
}

    // Driver Code
    public static void Main()
    {
    int[] arr= { 0, 1, 1, 0, 1 };
    int N = arr.Length;
    int K = 4;

    Console.WriteLine(minStepK(arr, N, K));

    }
}

// This code is contributed by sanjoy_62.

Javascript

<script>

// JavaScript program for the above approach

// Function to find minimum number
// of Bitwise XOR of array elements
// with 1 required to make sum of
// the array at least K
function minStepK(arr, N, K)
{

    // Stores the count of
    // even array elements
    var E = 0;

    // Stores sum of the array
    var S = 0;

    // Traverse the array arr[]
    for(var i = 0; i < N; i++)
    {

        // Increment sum
        S += arr[i];

        // If array element is even
        if (arr[i] % 2 === 0)

            // Increase count of even
            E += 1;
    }

    // If S is at least K
    if (S >= K)
        return 0;

    // If S + E is less than K
    else if (S + E < K)
        return -1;

    // Otherwise, moves = K - S
    else
        return K - S;
}

// Driver Code
var arr = [ 0, 1, 1, 0, 1 ];
var N = arr.length;
var K = 4;

document.write(minStepK(arr, N, K));

// This code is contributed by rdtank

</script>

Output:

1

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