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📜  二进制字符串中零和一的最大差|设定2(O(n)时间)

📅  最后修改于: 2021-04-25 00:40:20             🧑  作者: Mango

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 Shrikant13


PHP


Javascript


输出:

6 

时间复杂度: O(n)
空间复杂度: O(1)