给定二进制字符串0s和1s。任务是在给定二进制字符串的任何子字符串中找到0的数量和1的数量之间的最大差值。对于给定的二进制字符串中的任何子字符串，这都是最大值(0的数量– 1的数量)。
例子：
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我们在下面的文章中讨论了动态编程方法：
二进制字符串中零和一的最大差|设置1。
在帖子中，我们看到了一种有效的方法，该方法可以在O(n)时间和O(1)额外空间中工作。如果我们将所有零转换为1，将所有零转换为-1，则背后的想法是，现在我们的问题将减少，可以使用Kadane算法找到最大和sub_array。
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以下是上述想法的实现。
C++
// CPP Program to find the length of
// substring with maximum difference of
// zeros and ones in binary string.
#include
using namespace std;
// Returns the length of substring with
// maximum difference of zeroes and ones
// in binary string
int findLength(string str, int n)
{
int current_sum = 0;
int max_sum = 0;
// traverse a binary string from left
// to right
for (int i = 0; i < n; i++) {
// add current value to the current_sum
// according to the Character
// if it's '0' add 1 else -1
current_sum += (str[i] == '0' ? 1 : -1);
if (current_sum < 0)
current_sum = 0;
// update maximum sum
max_sum = max(current_sum, max_sum);
}
// return -1 if string does not contain
// any zero that means all ones
// otherwise max_sum
return max_sum == 0 ? -1 : max_sum;
}
// Driven Program
int main()
{
string s = "11000010001";
int n = 11;
cout << findLength(s, n) << endl;
return 0;
} Java
// Java Program to find the length of
// substring with maximum difference of
// zeroes and ones in binary string.
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG {
// Find the length of substring with maximum
// difference of zeros and ones in binary
// string
public static int findLength(String str, int n)
{
int current_sum = 0;
int max_sum = 0;
// traverse a binary string from left to right
for (int i = 0; i < n; i++) {
// add current value to the current_sum
// according to the Character
// if it's '0' add 1 else -1
current_sum += (str.charAt(i) == '0' ? 1 : -1);
if (current_sum < 0)
current_sum = 0;
// update maxium sum
max_sum = Math.max(current_sum, max_sum);
}
// return -1 if string does not contain any zero
// that means string contains all ones otherwise 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
# Python Program to find the length of
# substring with maximum difference of
# zeros and ones in binary string.
# Returns the length of substring with
# maximum difference of zeroes and ones
# in binary string
def findLength(string, n):
current_sum = 0
max_sum = 0
# traverse a binary string from left
# to right
for i in range(n):
# add current value to the current_sum
# according to the Character
# if it's '0' add 1 else -1
current_sum += (1 if string[i] == '0' else -1)
if current_sum < 0:
current_sum = 0
# update maximum sum
max_sum = max(current_sum, max_sum)
# return -1 if string does not contain
# any zero that means all ones
# otherwise max_sum
return max_sum if max_sum else 0
# Driven Program
s = "11000010001"
n = 11
print(findLength(s, n))
# This code is contributed by Ansu Kumari.C#
// C# Program to find the length of
// substring with maximum difference of
// zeroes and ones in binary string.
using System;
class GFG
{
// Find the length of substring with
// maximum difference of zeros and
// ones in binary string
public static int findLength(string str,
int n)
{
int current_sum = 0;
int max_sum = 0;
// traverse a binary string
// from left to right
for (int i = 0; i < n; i++)
{
// add current value to the current_sum
// according to the Character
// if it's '0' add 1 else -1
current_sum += (str[i] == '0' ? 1 : -1);
if (current_sum < 0)
{
current_sum = 0;
}
// update maxium sum
max_sum = Math.Max(current_sum, max_sum);
}
// return -1 if string does not contain
// any zero that means string contains
// all ones otherwise max_sum
return max_sum == 0 ? -1 : max_sum;
}
// Driver Code
public static void Main(string[] args)
{
string str = "11000010001";
int n = str.Length;
Console.WriteLine(findLength(str, n));
}
}
// This code is contributed by Shrikant13PHP
Javascript
输出：
6 时间复杂度： O(n)
空间复杂度： O(1)