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

Last Updated : 18 May, 2021

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:

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

Minimize steps required to make all array elements same by adding 1, 2 or 5

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

C++

Java

Python3

C#

Javascript

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