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Minimize count of 0s required to be removed to maximize length of longest substring of 1s
  • Last Updated : 03 Feb, 2021

Given a binary string S of length N, the task is to find the minimum number of 0s required to be removed from the given string S to get the longest substring of 1s.

Examples:

Input: S = “010011”
Output: 2
Explanation:
Removing str[2] and str[3] modifies the string S to “0111”. Therefore, the minimum number of removals required is 2.

Input: S = “011111”
Output: 0

Approach: The idea is to find the leftmost and rightmost index of 1 in the string, then count the number of 0s present between them. Finally, print the value of the count obtained. Follow the steps below to solve the problem:



  • Traverse the string S to find the first and last occurrence of 1 in the string and store its indices in variables, say left and right, respectively.
  • Iterate over the range [left, right] using the variable i, and if the value of str[i] is equal to 0, increment count by 1.
  • After completing the above steps, print the value of count as the result.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the minimum number
// of 0s required to be removed to
// maximize the longest substring of 1s
void minNumZeros(string str)
{
    // Stores leftmost and rightmost
    // indices consisting of 1
    int left = INT_MAX, right = INT_MIN;
 
    // Stores the count of 0s
    // between left and right
    int count = 0;
 
    // Traverse the string str
    for (int i = 0; i < str.length(); i++) {
 
        // If the current character is 1
        if (str[i] == '1') {
 
            // Update leftmost and rightmost
            // index consisting of 1
            left = min(i, left);
            right = max(right, i);
        }
    }
 
    // If string consists only of 0s
    if (left == INT_MAX) {
        cout << "0";
        return;
    }
 
    // Count the number of 0s
    // between left and right
    for (int i = left; i < right; i++) {
 
        if (str[i] == '0') {
            count++;
        }
    }
 
    // Print the result
    cout << count;
}
 
// Driver Code
int main()
{
    string str = "010011";
    minNumZeros(str);
 
    return 0;
}

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Java

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// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to count the minimum number
// of 0s required to be removed to
// maximize the longest substring of 1s
static void minNumZeros(String str)
{
     
    // Stores leftmost and rightmost
    // indices consisting of 1
    int left = Integer.MAX_VALUE;
    int right = Integer.MIN_VALUE;
 
    // Stores the count of 0s
    // between left and right
    int count = 0;
 
    // Traverse the string str
    for(int i = 0; i < str.length(); i++)
    {
         
        // If the current character is 1
        if (str.charAt(i) == '1')
        {
             
            // Update leftmost and rightmost
            // index consisting of 1
            left = Math.min(i, left);
            right = Math.max(right, i);
        }
    }
 
    // If string consists only of 0s
    if (left == Integer.MAX_VALUE)
    {
        System.out.print("0");
        return;
    }
 
    // Count the number of 0s
    // between left and right
    for(int i = left; i < right; i++)
    {
        if (str.charAt(i) == '0')
        {
            count++;
        }
    }
 
    // Print the result
    System.out.print(count);
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "010011";
     
    minNumZeros(str);
}
}
 
// This code is contributed by Dharanendra L V

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Python3

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# Python program for the above approach
import sys
 
# Function to count the minimum number
# of 0s required to be removed to
# maximize the longest substring of 1s
def minNumZeros(st) :
 
    # Stores leftmost and rightmost
    # indices consisting of 1
    left = sys.maxsize
    right = -sys.maxsize - 1
 
    # Stores the count of 0s
    # between left and right
    count = 0
 
    # Traverse the string str
    for i in range(len(st)) :
    #for (int i = 0; i < str.length(); i++) {
 
        # If the current character is 1
        if st[i] == '1' :
 
            # Update leftmost and rightmost
            # index consisting of 1
            left = min(i, left)
            right = max(right, i)
         
    # If string consists only of 0s
    if left == sys.maxsize :
        print("0")
        return
     
    # Count the number of 0s
    # between left and right
    for i in range(left,right):
         
    #for (int i = left; i < right; i++) {
 
        if st[i] == '0' :
            count += 1
         
    # Print the result
    print(count)
 
# Driver Code
if __name__ == "__main__" :    
    st = "010011";
    minNumZeros(st)
     
 # This code is contributed by jana_sayantan.  

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C#

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// C# program for the above approach
using System;
 
class GFG{
 
// Function to count the minimum number
// of 0s required to be removed to
// maximize the longest substring of 1s
static void minNumZeros(string str)
{
     
    // Stores leftmost and rightmost
    // indices consisting of 1
    int left = int.MaxValue;
    int right = int.MinValue;
 
    // Stores the count of 0s
    // between left and right
    int count = 0;
 
    // Traverse the string str
    for(int i = 0; i < str.Length; i++)
    {
         
        // If the current character is 1
        if (str[i] == '1')
        {
             
            // Update leftmost and rightmost
            // index consisting of 1
            left = Math.Min(i, left);
            right = Math.Max(right, i);
        }
    }
 
    // If string consists only of 0s
    if (left == int.MaxValue)
    {
        Console.WriteLine("0");
        return;
    }
 
    // Count the number of 0s
    // between left and right
    for(int i = left; i < right; i++)
    {
        if (str[i] == '0')
        {
            count++;
        }
    }
 
    // Print the result
    Console.WriteLine(count);
}
 
// Driver Code
static public void Main()
{
    string str = "010011";
     
    minNumZeros(str);
}
}
 
// This code is contributed by Dharanendra L V

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Output: 

2

 

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

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