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++ 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 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 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# 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 |
<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> |
2
Time Complexity: O(N)
Auxiliary Space: O(1)