Minimum operations require to balance Binary String ❤️

Given a binary string S of length N, the task is to find the minimum number of operations required to balance (equal number of ‘0’ and ‘1’) the given binary string, you can perform two types of operations one at a time, which is given below.

Remove a character from the front of the given string.
Remove a character from the end of the given string.

Examples:

Input: S = “0111000”
Output: 1
Explanation: Remove 1 character from the front of the given string.

Input: S = “11100100”
Output: 0

Approach (Using prefix Sum Technique):

If we assign value -1 to ‘0’ characters and the value of +1 to ‘1’ characters, then any valid substring in which the count of ‘0’s and ‘1’s are equal will result in zero sum. Now, the only task is to determine the length of the longest substring whose sum would be zero. The remaining character n – length of the longest substring with 0 sum would be the result.

Steps to solve this problem:

Create a map unmap to store the prefix sums with their first occurrence indices.
Iterate through the characters in the string s:
- For each character, update the prefixSum by +1 if the character is ‘1’, or -1 if the character is ‘0’.
- If the prefixSum becomes zero at any point, update result by (i + 1), this ensures that this is the longest valid substring encountered so far.
- Check if the current prefixSum value exists in the unmap.
  - If true, update result again with the maximum of its current value and i – unmap[prefixSum], which calculates the length of the current valid substring.
  - Otherwise, store current prefixSum in unmap with the index i.
Finally, return the remaining length of the string, which is n – result.

Below is the implementation of the above approach:

C++

// C++ code to implement the approach.
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to solve the problem
int minOperation(string& s)
{
    // Storing prefixSum with index
    // of its first occurance
    unordered_map<int, int> unmap;
    int n = s.size();
 
    // For storing the prefix Sum
    // ending at ith index
    int prefixSum = 0;
 
    // For keeping the length of
    // longest binary string where
    // number of zero and ones are equal
    int result = 0;
 
    // Iterate over the string
    for (int i = 0; i < n; i++) {
 
        prefixSum += ((s[i] == '1') ? 1 : -1);
 
        if (prefixSum == 0) {
            result = max(result, i + 1);
        }
 
        // Check if prefixSum have
        // previously occured or not
        if (unmap.count(prefixSum)) {
 
            // Update the result with
            // this valid substring
            result = max(result, i - unmap[prefixSum]);
        }
        else {
 
            // Store this prefixSum has
            // occur at ith index
            // in the map.
            unmap[prefixSum] = i;
        }
    }
 
    // Return the remaining length
    // other than the longest
    // valid substring.
    return n - result;
}
 
// Driver code
int main()
{
    string S = "0111000";
 
    // Function call
    int result = minOperation(S);
    cout << result << endl;
    return 0;
}

Java

import java.util.HashMap;
 
public class Main {
    // Function to solve the problem
    public static int minOperation(String s)
    {
        // Storing prefixSum with the index of its first
        // occurrence
        HashMap<Integer, Integer> unmap = new HashMap<>();
        int n = s.length();
 
        // For storing the prefix Sum ending at the ith
        // index
        int prefixSum = 0;
 
        // For keeping the length of the longest binary
        // string where the number of zeros and ones are
        // equal
        int result = 0;
 
        // Iterate over the string
        for (int i = 0; i < n; i++) {
            prefixSum += (s.charAt(i) == '1' ? 1 : -1);
 
            if (prefixSum == 0) {
                result = Math.max(result, i + 1);
            }
 
            // Check if prefixSum has previously occurred or
            // not
            if (unmap.containsKey(prefixSum)) {
                // Update the result with this valid
                // substring
                result = Math.max(result,
                                  i - unmap.get(prefixSum));
            }
            else {
                // Store this prefixSum has occurred at the
                // ith index in the map.
                unmap.put(prefixSum, i);
            }
        }
 
        // Return the remaining length other than the
        // longest valid substring.
        return n - result;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String S = "0111000";
 
        // Function call
        int result = minOperation(S);
        System.out.println(result);
    }
}

Python3

def minOperation(s):
    # Storing prefixSum with index
    # of its first occurrence
    unmap = {}
    n = len(s)
 
    # For storing the prefix Sum
    # ending at ith index
    prefixSum = 0
 
    # For keeping the length of
    # the longest binary string where
    # the number of zeros and ones are equal
    result = 0
 
    # Iterate over the string
    for i in range(n):
        if s[i] == '1':
            prefixSum += 1
        else:
            prefixSum -= 1
 
        if prefixSum == 0:
            result = max(result, i + 1)
 
        # Check if prefixSum has
        # previously occurred or not
        if prefixSum in unmap:
            # Update the result with
            # this valid substring
            result = max(result, i - unmap[prefixSum])
        else:
            # Store this prefixSum has
            # occurred at ith index
            # in the map.
            unmap[prefixSum] = i
 
    # Return the remaining length
    # other than the longest
    # valid substring.
    return n - result
 
# Driver code
if __name__ == "__main__":
    S = "0111000"
 
    # Function call
    result = minOperation(S)
    print(result)

C#

using System;
using System.Collections.Generic;
 
class Program
{
    // Function to solve the problem
    static int MinOperation(string s)
    {
        // Storing prefixSum with index
        // of its first occurrence
        Dictionary<int, int> unmap = new Dictionary<int, int>();
        int n = s.Length;
 
        // For storing the prefix Sum
        // ending at ith index
        int prefixSum = 0;
 
        // For keeping the length of
        // longest binary string where
        // number of zero and ones are equal
        int result = 0;
 
        // Iterate over the string
        for (int i = 0; i < n; i++)
        {
            prefixSum += (s[i] == '1') ? 1 : -1;
 
            if (prefixSum == 0)
            {
                result = Math.Max(result, i + 1);
            }
 
            // Check if prefixSum has
            // previously occurred or not
            if (unmap.ContainsKey(prefixSum))
            {
                // Update the result with
                // this valid substring
                result = Math.Max(result, i - unmap[prefixSum]);
            }
            else
            {
                // Store this prefixSum has
                // occurred at ith index
                // in the map.
                unmap[prefixSum] = i;
            }
        }
 
        // Return the remaining length
        // other than the longest
        // valid substring.
        return n - result;
    }
 
    // Driver code
    static void Main()
    {
        string S = "0111000";
 
        // Function call
        int result = MinOperation(S);
        Console.WriteLine(result);
    }
}

Javascript

function MinOperation(s) {
    // Storing prefixSum with index
    // of its first occurrence
    let unmap = new Map();
    let n = s.length;
 
    // For storing the prefix Sum
    // ending at ith index
    let prefixSum = 0;
 
    // For keeping the length of
    // longest binary string where
    // number of zeros and ones are equal
    let result = 0;
 
    // Iterate over the string
    for (let i = 0; i < n; i++) {
        prefixSum += (s[i] === '1') ? 1 : -1;
 
        if (prefixSum === 0) {
            result = Math.max(result, i + 1);
        }
 
        // Check if prefixSum has
        // previously occurred or not
        if (unmap.has(prefixSum)) {
            // Update the result with
            // this valid substring
            result = Math.max(result, i - unmap.get(prefixSum));
        } else {
            // Store this prefixSum has
            // occurred at ith index
            // in the map.
            unmap.set(prefixSum, i);
        }
    }
 
    // Return the remaining length
    // other than the longest
    // valid substring.
    return n - result;
}
 
// Driver code
let S = "0111000";
 
// Function call
let result = MinOperation(S);
console.log(result);

Output

Time Complexity: O(N), where N is the length of the given binary string.
Auxiliary Space: O(N)

Minimum operations require to balance Binary String

Approach (Using prefix Sum Technique):

C++

Java

Python3

C#

Javascript

Contact Us