Given a binary string of 0s and 1s. The task is to find the maximum difference between the number of 0s and number of 1s in any sub-string of the given binary string. That is maximize ( number of 0s – number of 1s ) for any sub-string in the given binary string.
Examples:
Input : S = "11000010001"
Output : 6
From index 2 to index 9, there are 7
0s and 1 1s, so number of 0s - number
of 1s is 6.
Input : S = "1111"
Output : -1
We have discussed Dynamic Programming approach in below post :
Maximum difference of zeros and ones in binary string | Set 1.
In the post we seen an efficient method that work in O(n) time and in O(1) extra space. Idea behind that if we convert all zeros into 1 and all ones into -1.now our problem reduces to find out the maximum sum sub_array Using Kadane’s Algorithm.
Input : S = "11000010001"
After converting '0' into 1 and
'1' into -1 our S Look Like
S = -1 -1 1 1 1 1 -1 1 1 1 -1
Now we have to find out Maximum Sum sub_array
that is : 6 is that case
Output : 6
Below is the implementation of above idea.
C++
#include <iostream>
using namespace std;
int findLength(string str, int n)
{
int current_sum = 0;
int max_sum = 0;
for ( int i = 0; i < n; i++) {
current_sum += (str[i] == '0' ? 1 : -1);
if (current_sum < 0)
current_sum = 0;
max_sum = max(current_sum, max_sum);
}
return max_sum == 0 ? -1 : max_sum;
}
int main()
{
string s = "11000010001" ;
int n = 11;
cout << findLength(s, n) << endl;
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG {
public static int findLength(String str, int n)
{
int current_sum = 0 ;
int max_sum = 0 ;
for ( int i = 0 ; i < n; i++) {
current_sum += (str.charAt(i) == '0' ? 1 : - 1 );
if (current_sum < 0 )
current_sum = 0 ;
max_sum = Math.max(current_sum, max_sum);
}
return max_sum == 0 ? - 1 : max_sum;
}
public static void main(String[] args)
{
String str = "11000010001" ;
int n = str.length();
System.out.println(findLength(str, n));
}
}
|
Python3
def findLength(string, n):
current_sum = 0
max_sum = 0
for i in range (n):
current_sum + = ( 1 if string[i] = = '0' else - 1 )
if current_sum < 0 :
current_sum = 0
max_sum = max (current_sum, max_sum)
return max_sum if max_sum else 0
s = "11000010001"
n = 11
print (findLength(s, n))
|
C#
using System;
class GFG
{
public static int findLength( string str,
int n)
{
int current_sum = 0;
int max_sum = 0;
for ( int i = 0; i < n; i++)
{
current_sum += (str[i] == '0' ? 1 : -1);
if (current_sum < 0)
{
current_sum = 0;
}
max_sum = Math.Max(current_sum, max_sum);
}
return max_sum == 0 ? -1 : max_sum;
}
public static void Main( string [] args)
{
string str = "11000010001" ;
int n = str.Length;
Console.WriteLine(findLength(str, n));
}
}
|
PHP
<?php
function findLength( $str , $n )
{
$current_sum = 0;
$max_sum = 0;
for ( $i = 0; $i < $n ; $i ++)
{
$current_sum += ( $str [ $i ] == '0' ? 1 : -1);
if ( $current_sum < 0)
$current_sum = 0;
$max_sum = max( $current_sum , $max_sum );
}
return $max_sum == 0 ? -1 : $max_sum ;
}
$s = "11000010001" ;
$n = 11;
echo findLength( $s , $n ), "\n" ;
?>
|
Javascript
<script>
function findLength(str, n)
{
let current_sum = 0;
let max_sum = 0;
for (let i = 0; i < n; i++)
{
current_sum += (str[i] == '0' ? 1 : -1);
if (current_sum < 0)
{
current_sum = 0;
}
max_sum = Math.max(current_sum, max_sum);
}
return max_sum == 0 ? -1 : max_sum;
}
let str = "11000010001" ;
let n = str.length;
document.write(findLength(str, n));
</script>
|
Time Complexity: O(n)
Space complexity: O(n), since string gets copied when we pass it to a function.