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Minimum elements to be removed to make count of 0s and 1s equal
  • Last Updated : 29 Sep, 2020

Given two binary arrays A and B of equal length, the task is to find the minimum number of elements to be removed, from the front, to make the count of 0s and 1s equal in the given two Binary Arrays.

Examples: 

Input: A = {0, 1, 0, 1, 1, 0}, B = {1, 0, 0, 1, 1, 1} 
Output:
Explanation: 
-> Remove first 5 elements from array A 
-> Remove the first element from B. 
-> Therefore at least 6 elements are needed to be removed in order to make the total number of 1’s and 0’s equal.

Input: a = {1, 0}, b = {0, 1} 
Output:
Explanation: 
The count of 1s and 0s are already equal. Therefore no element needs to be removed. 

Approach: The idea is to find the difference between the total number of 1’s and the total number of 0’s separately and minimising this difference. Removing 1 will decrease the difference by 1 and removing 0 will increase diff by 1. 
Therefore, the steps needed to make the difference equal to 0 are: 



  • Count initial difference in total number of 1’s and 0’s. Let’s call it initial_diff.
  • Remove first l elements from the 1st array and store the difference in a variable left_diff.
  • Remove first r elements from the 2nd array and store the difference in a variable right_diff.
  • Now find such values of l and r such that:

initial_diff – left_diff – right_diff = 0.

  • In order to find l and r optimally, for each unique value of right_diff, save the smallest r to reach this value in an unordered_map. Then finally, iterate over l to find the minimum answer. 

Below is the implementation of the above approach

C++

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// C++ program to find minimum elements
// to be removed to make count
// of 0s and 1s equal
 
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns minimum elements
// need to be removed
int MinRemovedElements(int a[], int b[],
                    int n)
{
 
    int no_of_ones = 0;
    int no_of_zeroes = 0;
 
    // Counting total number of
    // zeroes and ones
    for (int i = 0; i < n; i++) {
        if (a[i] == 1)
            no_of_ones++;
        else
            no_of_zeroes++;
 
        if (b[i] == 1)
            no_of_ones++;
        else
            no_of_zeroes++;
    }
 
    // Computing difference in ones and
    // zeroes
    int diff = no_of_ones - no_of_zeroes;
 
    unordered_map<int, int> mp;
    mp[0] = 0;
    int curr = 0;
 
    // Filling the difference of number
    // of 1's and 0's of 1st array in
    // unoredered_map
    for (int i = 0; i < n; i++) {
        if (a[i] == 1)
            curr++;
        else
            curr--;
 
        // Condition if current difference
        // not found in map
        if (mp.find(curr) == mp.end())
            mp[curr] = i + 1;
    }
 
    curr = 0;
    int answer = 2 * n;
 
    for (int i = 0; i < n; i++) {
        if (b[i] == 1)
            curr++;
        else
            curr--;
 
        // Condition if current difference is
        // present in map then compute the
        // minimum one
        if (mp.find(diff - curr) != mp.end())
            answer
                = min(answer,
                    i + 1 + mp);
    }
 
    // Condition if the total difference
    // is present in 1st array
    if (mp.find(diff) != mp.end())
        answer = min(answer, mp);
 
    return answer;
}
 
// Driver Code
int main()
{
    int a[] = { 0, 1, 0, 1, 1, 0 };
    int b[] = { 1, 0, 0, 1, 1, 1 };
    int size = sizeof(a) / sizeof(a[0]);
 
    cout << MinRemovedElements(a, b, size)
        << "\n";
 
    return 0;
}

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Java

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// Java program to find minimum elements
// to be removed to make count
// of 0s and 1s equal
import java.util.*;
 
class GFG{
 
// Function that returns minimum elements
// need to be removed
static int MinRemovedElements(int a[], int b[],
                    int n)
{
 
    int no_of_ones = 0;
    int no_of_zeroes = 0;
 
    // Counting total number of
    // zeroes and ones
    for (int i = 0; i < n; i++) {
        if (a[i] == 1)
            no_of_ones++;
        else
            no_of_zeroes++;
 
        if (b[i] == 1)
            no_of_ones++;
        else
            no_of_zeroes++;
    }
 
    // Computing difference in ones and
    // zeroes
    int diff = no_of_ones - no_of_zeroes;
 
    HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();
    mp.put(0, 0);
    int curr = 0;
 
    // Filling the difference of number
    // of 1's and 0's of 1st array in
    // unoredered_map
    for (int i = 0; i < n; i++) {
        if (a[i] == 1)
            curr++;
        else
            curr--;
 
        // Condition if current difference
        // not found in map
        if (!mp.containsKey(curr))
            mp.put(curr, i + 1);
    }
 
    curr = 0;
    int answer = 2 * n;
 
    for (int i = 0; i < n; i++) {
        if (b[i] == 1)
            curr++;
        else
            curr--;
 
        // Condition if current difference is
        // present in map then compute the
        // minimum one
        if (mp.containsKey(diff - curr))
            answer
                = Math.min(answer,mp.get(diff - curr) + 1 + i);
    }
 
    // Condition if the total difference
    // is present in 1st array
    if (mp.containsKey(diff))
        answer = Math.min(answer, mp.get(diff));
 
    return answer;
}
 
// Driver Code
public static void main(String[] args)
{
    int a[] = { 0, 1, 0, 1, 1, 0 };
    int b[] = { 1, 0, 0, 1, 1, 1 };
    int size = a.length;
 
    System.out.print(MinRemovedElements(a, b, size)
        + "\n");
}
}
 
// This code is contributed by Rajput-Ji

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Python3

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# Python3 program to find minimum
# elements to be removed to make
# count of 0s and 1s equal
 
# Function that returns minimum
# elements need to be removed
def MinRemovedElements(a, b, n):
 
    no_of_ones = 0;
    no_of_zeroes = 0;
 
    # Counting total number of
    # zeroes and ones
    for i in range(n):
        if (a[i] == 1):
            no_of_ones += 1;
        else:
            no_of_zeroes += 1;
 
        if (b[i] == 1):
            no_of_ones += 1;
        else:
            no_of_zeroes += 1;
 
    # Computing difference in ones and
    # zeroes
    diff1 = no_of_ones - no_of_zeroes;
 
    mp = {};
    mp[0] = 0;
    curr = 0;
 
    # Filling the difference of number
    # of 1's and 0's of 1st array in
    # unoredered_map
    for i in range(n):
        if (a[i] == 1):
            curr += 1;
        else :
            curr -= 1;
 
        # Condition if current difference
        # not found in map
        if curr not in mp:
            mp[curr] = i + 1;
     
    curr = 0;
    answer = 2 * n;
 
    for i in range(n):
        if (b[i] == 1):
            curr += 1;
        else:
            curr -= 1;
 
        # Condition if current difference is
        # present in map then compute the
        # minimum one
        if (diff1 - curr) in mp :
            answer = min(answer, i + 1 + mp[diff1 -
                                            curr]);
 
    # Condition if the total difference
    # is present in 1st array
    if diff1 in mp:
        answer = min(answer, mp[diff1]);
 
    return answer;
 
# Driver Code
if __name__ == "__main__" :
 
    a = [ 0, 1, 0, 1, 1, 0 ];
    b = [ 1, 0, 0, 1, 1, 1 ];
     
    size = len(a);
 
    print(MinRemovedElements(a, b, size));
     
# This code is contributed by Yash_R

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C#

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// C# program to find minimum elements
// to be removed to make count
// of 0s and 1s equal
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function that returns minimum elements
// need to be removed
static int MinRemovedElements(int[] a, int[] b,
                              int n)
{
    int no_of_ones = 0;
    int no_of_zeroes = 0;
   
    // Counting total number of
    // zeroes and ones
    for(int i = 0; i < n; i++)
    {
        if (a[i] == 1)
            no_of_ones++;
        else
            no_of_zeroes++;
   
        if (b[i] == 1)
            no_of_ones++;
        else
            no_of_zeroes++;
    }
   
    // Computing difference in ones and
    // zeroes
    int diff = no_of_ones - no_of_zeroes;
     
    Dictionary<int,
               int> mp = new Dictionary<int,
                                        int>();
                                         
    mp.Add(0, 0);
    int curr = 0;
   
    // Filling the difference of number
    // of 1's and 0's of 1st array in
    // unoredered_map
    for(int i = 0; i < n; i++)
    {
        if (a[i] == 1)
            curr++;
        else
            curr--;
   
        // Condition if current difference
        // not found in map
        if (!mp.ContainsKey(curr))
            mp.Add(curr, i + 1);
    }
   
    curr = 0;
    int answer = 2 * n;
   
    for(int i = 0; i < n; i++)
    {
        if (b[i] == 1)
            curr++;
        else
            curr--;
   
        // Condition if current difference is
        // present in map then compute the
        // minimum one
        if (mp.ContainsKey(diff - curr))
            answer = Math.Min(answer,
             i + 1 + mp);
    }
   
    // Condition if the total difference
    // is present in 1st array
    if (mp.ContainsKey(diff))
        answer = Math.Min(answer, mp);
   
    return answer;
}
 
// Driver code
static void Main()
{
    int[] a = { 0, 1, 0, 1, 1, 0 };
    int[] b = { 1, 0, 0, 1, 1, 1 };
    int size = a.Length;
      
    Console.WriteLine(MinRemovedElements(
        a, b, size));
}
}
 
// This code is contributed by divyeshrabadiya07

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Output:

6

Time complexity: O(N)

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