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Minimum characters required to be removed to sort binary string in ascending order
  • Difficulty Level : Medium
  • Last Updated : 14 Apr, 2021

Given a binary string str, the task is to remove the minimum number of characters from the given binary string such that the characters in the remaining string form a sorted order.

Examples:

Input: str = “1000101”
Output: 2
Explanation: 
Removal of the first two occurrences of ‘1’ modifies the string to “00001”, which is a sorted order.
Therefore, the minimum count of characters to be removed is 2.

Input: str = “001111”
Output: 0
Explanation:
The string is already sorted.
Therefore, the minimum count of character to be removed is 0.

Approach: The idea is to count the number of 1s before the last occurrence of 0 and the number of 0s after the first occurrence of 1. The minimum of the two counts is the required number of characters to be removed. Below are the steps:



  1. Traverse the string str and find the position of the first occurrence of 1 and the last occurrence of 0.
  2. Print 0 if the str has only one type of character.
  3. Now, count the number of 1 is present prior to the last occurrence of 0 and store in a variable, say cnt1.
  4. Now, count the number of 0s present after the first occurrence of 1 in a variable, say cnt0.
  5. Print the minimum of cnt0 and cnt1 as the minimum count of character required to be removed.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum count
// of characters to be removed to make
// the string sorted in ascending order
int minDeletion(string str)
{
     
    // Length of given string
    int n = str.length();
 
    // Stores the first
    // occurrence of '1'
    int firstIdx1 = -1;
 
    // Stores the last
    // occurrence of '0'
    int lastIdx0 = -1;
 
    // Traverse the string to find
    // the first occurrence of '1'
    for(int i = 0; i < n; i++)
    {
        if (str[i] == '1')
        {
            firstIdx1 = i;
            break;
        }
    }
 
    // Traverse the string to find
    // the last occurrence of '0'
    for(int i = n - 1; i >= 0; i--)
    {
        if (str[i] == '0')
        {
            lastIdx0 = i;
            break;
        }
    }
 
    // Return 0 if the str have
    // only one type of character
    if (firstIdx1 == -1 ||
         lastIdx0 == -1)
        return 0;
 
    // Initialize count1 and count0 to
    // count '1's before lastIdx0
    // and '0's after firstIdx1
    int count1 = 0, count0 = 0;
 
    // Traverse the string to count0
    for(int i = 0; i < lastIdx0; i++)
    {
        if (str[i] == '1')
        {
            count1++;
        }
    }
 
    // Traverse the string to count1
    for(int i = firstIdx1 + 1; i < n; i++)
    {
        if (str[i] == '1')
        {
            count0++;
        }
    }
 
    // Return the minimum of
    // count0 and count1
    return min(count0, count1);
}
     
// Driver code
int main()
{
     
    // Given string str
    string str = "1000101";
     
    // Function call
    cout << minDeletion(str);
     
    return 0;
}
 
// This code is contributed by bikram2001jha

Java




// Java program for the above approach
 
import java.util.*;
import java.lang.*;
 
class GFG {
 
    // Function to find the minimum count
    // of characters to be removed to make
    // the string sorted in ascending order
    static int minDeletion(String str)
    {
 
        // Length of given string
        int n = str.length();
 
        // Stores the first
        // occurrence of '1'
        int firstIdx1 = -1;
 
        // Stores the last
        // occurrence of '0'
        int lastIdx0 = -1;
 
        // Traverse the string to find
        // the first occurrence of '1'
        for (int i = 0; i < n; i++) {
 
            if (str.charAt(i) == '1') {
 
                firstIdx1 = i;
                break;
            }
        }
 
        // Traverse the string to find
        // the last occurrence of '0'
        for (int i = n - 1; i >= 0; i--) {
 
            if (str.charAt(i) == '0') {
 
                lastIdx0 = i;
                break;
            }
        }
 
        // Return 0 if the str have
        // only one type of character
        if (firstIdx1 == -1
            || lastIdx0 == -1)
            return 0;
 
        // Initialize count1 and count0 to
        // count '1's before lastIdx0
        // and '0's after firstIdx1
        int count1 = 0, count0 = 0;
 
        // Traverse the string to count0
        for (int i = 0; i < lastIdx0; i++) {
 
            if (str.charAt(i) == '1') {
 
                count1++;
            }
        }
 
        // Traverse the string to count1
        for (int i = firstIdx1 + 1; i < n; i++) {
 
            if (str.charAt(i) == '1') {
 
                count0++;
            }
        }
 
        // Return the minimum of
        // count0 and count1
        return Math.min(count0, count1);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given string str
        String str = "1000101";
 
        // Function Call
        System.out.println(minDeletion(str));
    }
}

Python3




# Python3 program for the above approach
 
# Function to find the minimum count
# of characters to be removed to make
# the string sorted in ascending order
def minDeletion(s):
     
    # Length of given string
    n = len(s)
 
    # Stores the first
    # occurrence of '1'
    firstIdx1 = -1
 
    # Stores the last
    # occurrence of '0'
    lastIdx0 = -1
 
    # Traverse the string to find
    # the first occurrence of '1'
    for i in range(0, n):
        if (str[i] == '1'):
            firstIdx1 = i
            break
 
    # Traverse the string to find
    # the last occurrence of '0'
    for i in range(n - 1, -1, -1):
        if (str[i] == '0'):
            lastIdx0 = i
            break
 
    # Return 0 if the str have
    # only one type of character
    if (firstIdx1 == -1 or
         lastIdx0 == -1):
        return 0
 
    # Initialize count1 and count0 to
    # count '1's before lastIdx0
    # and '0's after firstIdx1
    count1 = 0
    count0 = 0
 
    # Traverse the string to count0
    for i in range(0, lastIdx0):
        if (str[i] == '1'):
            count1 += 1
 
    # Traverse the string to count1
    for i in range(firstIdx1 + 1, n):
        if (str[i] == '1'):
            count0 += 1
 
    # Return the minimum of
    # count0 and count1
    return min(count0, count1)
     
# Driver code
 
# Given string str
str = "1000101"
     
# Function call
print(minDeletion(str))
 
# This code is contributed by Stream_Cipher

C#




// C# program for the above approach
using System.Collections.Generic;
using System;
 
class GFG{
 
// Function to find the minimum count
// of characters to be removed to make
// the string sorted in ascending order
static int minDeletion(string str)
{
 
    // Length of given string
    int n = str.Length;
 
    // Stores the first
    // occurrence of '1'
    int firstIdx1 = -1;
 
    // Stores the last
    // occurrence of '0'
    int lastIdx0 = -1;
 
    // Traverse the string to find
    // the first occurrence of '1'
    for(int i = 0; i < n; i++)
    {
        if (str[i] == '1')
        {
            firstIdx1 = i;
            break;
        }
    }
 
    // Traverse the string to find
    // the last occurrence of '0'
    for(int i = n - 1; i >= 0; i--)
    {
        if (str[i] == '0')
        {
            lastIdx0 = i;
            break;
        }
    }
 
    // Return 0 if the str have
    // only one type of character
    if (firstIdx1 == -1 ||
         lastIdx0 == -1)
        return 0;
 
    // Initialize count1 and count0 to
    // count '1's before lastIdx0
    // and '0's after firstIdx1
    int count1 = 0, count0 = 0;
 
    // Traverse the string to count0
    for(int i = 0; i < lastIdx0; i++)
    {
        if (str[i] == '1')
        {
            count1++;
        }
    }
 
    // Traverse the string to count1
    for(int i = firstIdx1 + 1; i < n; i++)
    {
        if (str[i] == '1')
        {
            count0++;
        }
    }
 
    // Return the minimum of
    // count0 and count1
    return Math.Min(count0, count1);
}
 
// Driver Code
public static void Main()
{
     
    // Given string str
    string str = "1000101";
 
    // Function call
    Console.WriteLine(minDeletion(str));
}
}
 
// This code is contributed by Stream_Cipher

Javascript




<script>
 
// JavaScript program to implement
// the above approach
 
// Function to find the minimum count
// of characters to be removed to make
// the string sorted in ascending order
function minDeletion(str)
{
  
    // Length of given string
    let n = str.length;
  
    // Stores the first
    // occurrence of '1'
    let firstIdx1 = -1;
  
    // Stores the last
    // occurrence of '0'
    let lastIdx0 = -1;
  
    // Traverse the string to find
    // the first occurrence of '1'
    for(let i = 0; i < n; i++)
    {
        if (str[i] == '1')
        {
            firstIdx1 = i;
            break;
        }
    }
  
    // Traverse the string to find
    // the last occurrence of '0'
    for(let i = n - 1; i >= 0; i--)
    {
        if (str[i] == '0')
        {
            lastIdx0 = i;
            break;
        }
    }
  
    // Return 0 if the str have
    // only one type of character
    if (firstIdx1 == -1 ||
         lastIdx0 == -1)
        return 0;
  
    // Initialize count1 and count0 to
    // count '1's before lastIdx0
    // and '0's after firstIdx1
    let count1 = 0, count0 = 0;
  
    // Traverse the string to count0
    for(let i = 0; i < lastIdx0; i++)
    {
        if (str[i] == '1')
        {
            count1++;
        }
    }
  
    // Traverse the string to count1
    for(let i = firstIdx1 + 1; i < n; i++)
    {
        if (str[i] == '1')
        {
            count0++;
        }
    }
  
    // Return the minimum of
    // count0 and count1
    return Math.min(count0, count1);
}
 
// Driver code
 
    // Given string str
    let str = "1000101";
  
    // Function call
    document.write(minDeletion(str));
 
// This code is contributed by target_2.
</script>
Output
2

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

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