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:
- Traverse the string str and find the position of the first occurrence of 1 and the last occurrence of 0.
- Print 0 if the str has only one type of character.
- Now, count the number of 1 is present prior to the last occurrence of 0 and store in a variable, say cnt1.
- Now, count the number of 0s present after the first occurrence of 1 in a variable, say cnt0.
- 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 codeint 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 approachimport 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 orderfunction 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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