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Count of elements in array A left after performing deletion/rotation operation based on given conditions

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  • Last Updated : 29 May, 2022
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Given two binary arrays, A[] and B[] of size N respectively, the task is to find the number of elements in array A[] that will be left after performing the following operation until no elements can be deleted:

  1. If the starting elements of array A[] and B[] are equal, then delete both the elements.
  2. Otherwise, append the starting character of array A[] to the end of the array, A[], after removing it.

Examples:

Input: A[] = {1, 1, 0, 1}, B[] = {1, 0, 1, 1}, N = 4 
Output:
Explanation:
The operations are performed as follows:

  1. A[0]( =1) = B[0]( =1): Delete the elements. Thereafter, the arrays are modified to {1, 0, 1} and {0, 1, 1} respectively.
  2. A[0](=1) != B[0](= 0): Shift the A[0] to the end of the array A[]. Thereafter, the arrays are modified to { 0, 1, 1} and {0, 1, 1} respectively.
  3. A[0]( =0) = B[0]( =0): Delete the elements. Thereafter, the arrays are modified to {1, 1} and {1, 1} respectively.
  4. A[0]( =1) = B[0]( =1): Delete the elements. Thereafter, the arrays are modified to {1} and {1} respectively.
  5. A[0]( =1) = B[0]( =1): Delete the elements. Thereafter, both arrays became empty.

Therefore, no elements are left in the array A[].

Input: A[] = {1, 0, 1, 1, 1, 1}, B[] = {1, 1, 0, 1, 0, 1}, N = 6 
Output:

Approach: The given problem can be solved by removing the common 0s and 1s, and then counting the unique number of 0s and 1s in both arrays. Consider the following observations:

  1. The elements can be deleted as long as there is an element equal to the first element of B[] left in the array A[].
  2. It can also be observed that the order of elements of A[] can be easily changed.
  3. Therefore, the idea is to keep the count of the number of 0s and 1s left in A[] and if an element is encountered in B[] such that the same element is no longer present in A[], then no more operations can be performed.

Follow the steps below to solve the problem:

  • Traverse the array, A[] and count the total number of 0s and 1s in variables and store them in variables, say zero and one respectively.
  • Initialize a variable say count as 0 to store the total number of deletions performed.
  • Traverse the array, B[] using the variable i and do the following:
    • If B[i] is equal to 0 and zero>0, then increment the value of count by 1 and decrement zero by 1.
    • Else, if B[i] is equal to 1 and one>0, then increment the value of count by 1 and decrement one by 1.
    • Otherwise, break out of the loop as no more operations can further be performed.
  • Finally, after completing the above steps, print the difference between N and count as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate minimum size
// of the array A[] after performing
// the given operations
int minimumSizeAfterDeletion(int A[], int B[], int N)
{
    // Stores the count of 0s and 1s
    int zero = 0, one = 0;
 
    // Stores the total deletions performed
    int count = 0;
 
    // Traverse the array A[]
    for (int i = 0; i < N; i++) {
        if (A[i] == 0) {
            zero++;
        }
        else {
            one++;
        }
    }
 
    // Traverse array B[]
    for (int i = 0; i < N; i++) {
 
        // If the B[i] is 0 and zero is
        // greater than 0
        if (B[i] == 0 && zero > 0) {
            // Increment count by 1
            count++;
            // Decrement zero by 1
            zero--;
        }
 
        // Else if the B[i] is 1 and one is
        // greater than 0
        else if (B[i] == 1 && one > 0) {
            // Increment count by 1
            count++;
            // Decrement one by 1
            one--;
        }
 
        // Otherwise
        else {
            break;
        }
    }
 
    // Return the answer
    return N - count;
}
 
// Driver Code
int main()
{
 
    // Given Input
    int A[] = { 1, 0, 1, 1, 1, 1 };
    int B[] = { 1, 1, 0, 1, 0, 1 };
    int N = 6;
 
    // Function Call
    cout << minimumSizeAfterDeletion(A, B, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
 
class GFG{
     
// Function to calculate minimum size
// of the array A[] after performing
// the given operations
static int minimumSizeAfterDeletion(int A[], int B[],
                                    int N)
{
     
    // Stores the count of 0s and 1s
    int zero = 0, one = 0;
 
    // Stores the total deletions performed
    int count = 0;
 
    // Traverse the array A[]
    for(int i = 0; i < N; i++)
    {
        if (A[i] == 0)
        {
            zero++;
        }
        else
        {
            one++;
        }
    }
 
    // Traverse array B[]
    for(int i = 0; i < N; i++)
    {
         
        // If the B[i] is 0 and zero is
        // greater than 0
        if (B[i] == 0 && zero > 0)
        {
             
            // Increment count by 1
            count++;
             
            // Decrement zero by 1
            zero--;
        }
 
        // Else if the B[i] is 1 and one is
        // greater than 0
        else if (B[i] == 1 && one > 0)
        {
             
            // Increment count by 1
            count++;
             
            // Decrement one by 1
            one--;
        }
 
        // Otherwise
        else
        {
            break;
        }
    }
 
    // Return the answer
    return N - count;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Input
    int A[] = { 1, 0, 1, 1, 1, 1 };
    int B[] = { 1, 1, 0, 1, 0, 1 };
    int N = 6;
     
    // Function Call
    minimumSizeAfterDeletion(A, B, N);
    System.out.println(minimumSizeAfterDeletion(A, B, N));
}
}
 
// This code is contributed by Potta Lokesh

Python3




# Python3 program for the above approach
 
# Function to calculate minimum size
# of the array A[] after performing
# the given operations
def minimumSizeAfterDeletion(A, B, N):
     
    # Stores the count of 0s and 1s
    zero = 0
    one = 0
     
    # Stores the total deletions performed
    count = 0
     
    # Traverse the array A[]
    for i in range(N):
        if A[i] == 0:
            zero += 1
        else:
            one += 1
     
    # Traverse array B[]       
    for i in range(N):
         
        # If the B[i] is 0 and zero is
        # greater than 0
        if B[i] == 0 and zero > 0:
             
            # Increment count by 1
            count += 1
             
            # Decrement zero by 1
            zero -= 1
         
        # Else if the B[i] is 1 and one is
        # greater than 0
        elif B[i] == 1 and one > 0:
             
            # Increment count by 1
            count += 1
             
            # Decrement one by 1
            one -= 1
             
        # Otherwise
        else:
            break
     
    # Return the answer   
    return N - count
 
# Driver code
 
# Given input
A = [ 1, 0, 1, 1, 1, 1 ]
B = [ 1, 1, 0, 1, 0, 1 ]
N = 6
 
# Function call
print(minimumSizeAfterDeletion(A, B, N))
 
# This code is contributed by Parth Manchanda

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to calculate minimum size
// of the array A[] after performing
// the given operations
static int minimumSizeAfterDeletion(int []A, int []B, int N)
{
    // Stores the count of 0s and 1s
    int zero = 0, one = 0;
 
    // Stores the total deletions performed
    int count = 0;
 
    // Traverse the array A[]
    for (int i = 0; i < N; i++) {
        if (A[i] == 0) {
            zero++;
        }
        else {
            one++;
        }
    }
 
    // Traverse array B[]
    for (int i = 0; i < N; i++) {
 
        // If the B[i] is 0 and zero is
        // greater than 0
        if (B[i] == 0 && zero > 0) {
            // Increment count by 1
            count++;
            // Decrement zero by 1
            zero--;
        }
 
        // Else if the B[i] is 1 and one is
        // greater than 0
        else if (B[i] == 1 && one > 0) {
            // Increment count by 1
            count++;
            // Decrement one by 1
            one--;
        }
 
        // Otherwise
        else {
            break;
        }
    }
 
    // Return the answer
    return N - count;
}
 
// Driver Code
public static void Main()
{
 
    // Given Input
    int []A = { 1, 0, 1, 1, 1, 1 };
    int []B = { 1, 1, 0, 1, 0, 1 };
    int N = 6;
 
    // Function Call
    Console.Write(minimumSizeAfterDeletion(A, B, N));
}
}
 
// This code is contributed by ipg2016107.

Javascript




<script>
// Javascript program for the above approach
// Function to calculate minimum size
// of the array A[] after performing
// the given operations
function minimumSizeAfterDeletion(A, B, N)
{
     
    // Stores the count of 0s and 1s
    var zero = 0, one = 0;
 
    // Stores the total deletions performed
    var count = 0;
 
    // Traverse the array A[]
    for(var i = 0; i < N; i++)
    {
        if (A[i] == 0)
        {
            zero++;
        }
        else
        {
            one++;
        }
    }
 
    // Traverse array B[]
    for(var i = 0; i < N; i++)
    {
         
        // If the B[i] is 0 and zero is
        // greater than 0
        if (B[i] == 0 && zero > 0)
        {
             
            // Increment count by 1
            count++;
             
            // Decrement zero by 1
            zero--;
        }
 
        // Else if the B[i] is 1 and one is
        // greater than 0
        else if (B[i] == 1 && one > 0)
        {
             
            // Increment count by 1
            count++;
             
            // Decrement one by 1
            one--;
        }
 
        // Otherwise
        else
        {
            break;
        }
    }
 
    // Return the answer
    return N - count;
}
 
// Driver Code
    // Given Input
    var A = [ 1, 0, 1, 1, 1, 1 ];
    var B = [ 1, 1, 0, 1, 0, 1 ];
    var N = 6;
     
    // Function Call
    minimumSizeAfterDeletion(A, B, N);
    document.write(minimumSizeAfterDeletion(A, B, N));
 
// This code is contributed by shivanisinghss2110
</script>

Output

2

Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time.
Auxiliary Space: O(1), as we are not using any extra space.


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