Related Articles

Related Articles

Minimum flips of odd indexed elements from odd length subarrays to make two given arrays equal
  • Last Updated : 07 Dec, 2020

Given two binary arrays X[] and Y[] of size N, the task is to convert array X[] into array Y[] by minimum number of operations of selecting any subarray of odd length and flipping all odd-indexed elements from the subarray.

Examples:

Input: X[] = {1, 0, 0, 0, 0, 1}, Y[] = {1, 1, 0, 1, 1, 1}
Output: 2
Explanation:
Initially, X[] is {1, 0, 0, 0, 0, 1}.
Operation 1: Choose the sub-array {0, 0, 0} from array X[] and change the 2nd and 4th character and convert it to {1, 0, 1}.
Now X becomes {1, 1, 0, 1, 0, 1}.
Operation 2: Choose the sub-array {0} containing only the 5th character and convert it to 1.
Finally, X becomes {1, 1, 0, 1, 1, 1}, which is equal to Y.
Therefore, the count of operations is 2.

Input: X[] = {0, 1, 0}, Y[] = {0, 1, 0}
Output: 0
Explanation:
Since both the arrays X and Y are equal thus the minimum operations would be 0.

Approach: The idea is to count the operations for both even and odd positions individually. Follow the steps below to solve the problem:



  • Initialize a variable C as 0 to store the count of operations.
  • Traverse the array X[] elements over odd positions and take a counter count = 0.
    • Check for consecutive unequal elements between X[i] and Y[i] and increase the counter count by 1 every time.
    • When X[i] and Y[i] are equal increase a global C to increase operation by 1 since in that one operation all odd positions can be made equal to as in Y.
  • Similarly, Traverse the array X[] elements over even positions and again take a counter count = 0.
    • Check for consecutive unequal elements between X[i] and Y[i] and increase the counter count by 1 every time.
    • When X[i] and Y[i] are equal increase a global counter C to increase operation by 1.
  • After the above steps, print the value of C as the resultant count of operations required.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum flip
// of subarrays required at alternate
// index to make binary arrays equals
void minOperation(int X[], int Y[],
                  int n)
{
    // Stores count of total operations
    int C = 0;
 
    // Stores count of consecutive
    // unequal elements
    int count = 0;
 
    // Loop to run on odd positions
    for (int i = 1; i < n; i = i + 2) {
 
        if (X[i] != Y[i]) {
            count++;
        }
        else {
 
            // Incrementing the
            // global counter
            if (count != 0)
                C++;
 
            // Change count to 0
            count = 0;
        }
    }
 
    // If all last elements are equal
    if (count != 0)
        C++;
 
    count = 0;
 
    // Loop to run on even positions
    for (int i = 0; i < n; i = i + 2) {
 
        if (X[i] != Y[i]) {
            count++;
        }
        else {
 
            // Incrementing the
            // global counter
            if (count != 0)
                C++;
 
            // Change count to 0
            count = 0;
        }
    }
 
    if (count != 0)
        C++;
 
    // Print the minimum operations
    cout << C;
}
 
// Driver Code
int main()
{
    int X[] = { 1, 0, 0, 0, 0, 1 };
    int Y[] = { 1, 1, 0, 1, 1, 1 };
    int N = sizeof(X) / sizeof(X[0]);
 
    // Function Call
    minOperation(X, Y, N);
 
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for the above approach
import java.util.*;
    
class GFG{
    
// Function to find the minimum flip
// of subarrays required at alternate
// index to make binary arrays equals
static void minOperation(int X[], int Y[],
                         int n)
{
     
    // Stores count of total operations
    int C = 0;
  
    // Stores count of consecutive
    // unequal elements
    int count = 0;
  
    // Loop to run on odd positions
    for(int i = 1; i < n; i = i + 2)
    {
         
        if (X[i] != Y[i])
        {
            count++;
        }
        else
        {
             
            // Incrementing the
            // global counter
            if (count != 0)
                C++;
  
            // Change count to 0
            count = 0;
        }
    }
  
    // If all last elements are equal
    if (count != 0)
        C++;
  
    count = 0;
  
    // Loop to run on even positions
    for(int i = 0; i < n; i = i + 2)
    {
        if (X[i] != Y[i])
        {
            count++;
        }
        else
        {
             
            // Incrementing the
            // global counter
            if (count != 0)
                C++;
  
            // Change count to 0
            count = 0;
        }
    }
  
    if (count != 0)
        C++;
  
    // Print the minimum operations
    System.out.print(C);
}
    
// Driver Code
public static void main(String[] args)
{
    int X[] = { 1, 0, 0, 0, 0, 1 };
    int Y[] = { 1, 1, 0, 1, 1, 1 };
    int N = X.length;
  
    // Function Call
    minOperation(X, Y, N);
}
}
 
// This code is contributed by susmitakundugoaldanga

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python program for the above approach
 
# Function to find the minimum flip
# of subarrays required at alternate
# index to make binary arrays equals
def minOperation(X, Y, n):
   
    # Stores count of total operations
    C = 0;
 
    # Stores count of consecutive
    # unequal elements
    count = 0;
 
    # Loop to run on odd positions
    for i in range(1, n, 2):
 
        if (X[i] != Y[i]):
            count += 1;
        else:
 
            # Incrementing the
            # global counter
            if (count != 0):
                C += 1;
 
            # Change count to 0
            count = 0;
 
    # If all last elements are equal
    if (count != 0):
        C += 1;
 
    count = 0;
 
    # Loop to run on even positions
    for i in range(0, n, 2):
        if (X[i] != Y[i]):
            count += 1;
        else:
 
            # Incrementing the
            # global counter
            if (count != 0):
                C += 1;
 
            # Change count to 0
            count = 0;
 
    if (count != 0):
        C += 1;
 
    # Prthe minimum operations
    print(C);
 
# Driver Code
if __name__ == '__main__':
    X = [1, 0, 0, 0, 0, 1];
    Y = [1, 1, 0, 1, 1, 1];
    N = len(X);
 
    # Function Call
    minOperation(X, Y, N);
 
    # This code is contributed by 29AjayKumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for the above approach
using System;
 
class GFG{
    
// Function to find the minimum flip
// of subarrays required at alternate
// index to make binary arrays equals
static void minOperation(int []X, int []Y,
                         int n)
{
     
    // Stores count of total operations
    int C = 0;
  
    // Stores count of consecutive
    // unequal elements
    int count = 0;
  
    // Loop to run on odd positions
    for(int i = 1; i < n; i = i + 2)
    {
        if (X[i] != Y[i])
        {
            count++;
        }
        else
        {
             
            // Incrementing the
            // global counter
            if (count != 0)
                C++;
  
            // Change count to 0
            count = 0;
        }
    }
  
    // If all last elements are equal
    if (count != 0)
        C++;
  
    count = 0;
  
    // Loop to run on even positions
    for(int i = 0; i < n; i = i + 2)
    {
        if (X[i] != Y[i])
        {
            count++;
        }
        else
        {
             
            // Incrementing the
            // global counter
            if (count != 0)
                C++;
  
            // Change count to 0
            count = 0;
        }
    }
  
    if (count != 0)
        C++;
  
    // Print the minimum operations
    Console.Write(C);
}
    
// Driver Code
public static void Main(String[] args)
{
    int []X = { 1, 0, 0, 0, 0, 1 };
    int []Y = { 1, 1, 0, 1, 1, 1 };
    int N = X.Length;
     
    // Function Call
    minOperation(X, Y, N);
}
}
 
// This code is contributed by Amit Katiyar

chevron_right


Output: 

2

 

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up
Recommended Articles
Page :