最大化给定二进制字符串中具有相同比例的 0 和 1 的分区

给定一个大小为N的二进制字符串S ，任务是找到字符串S的最大分区数，使得所有分区中0和1的数量之比相同。

例子：

Input: S = “010100001”
Output: 3
Explanation:
The substring [0, 2], [3, 5] and [6, 8] have the same ratio of ‘0′ and ‘1’ is 2:1. Therefore, the maximum possible partition is 3.

Input: str=”001101″
Output: 2

编程需要懂一点英语

朴素方法：朴素方法是生成字符串S 的所有分区，并找到最大频率为0比1的分区。
时间复杂度： O(N*2 ^N )
辅助空间： O(N)

有效的方法：要解决这个问题，首先看一些观察：

Let say there are two prefix strings that obtain the same ratio of 0 and 1. Then it is possible to divide the larger prefix into two with the same ratio. For eg- S = “0101” the prefix [0, 1] has the ratio of 1:1 and the prefix[0, 3] has the ratio of 1:1 then it is possible to partition the string in two with the ratio 1:1. Therefore, just count the number of prefixes that has the same 0 and 1 ratio as that of string S and print the number of prefix with that ratio.

编程需要懂一点英语

请按照以下步骤解决问题：

初始化两个整数变量，比如count_of_0和count_of_1为0 ，分别存储'0'和'1'字符的数量。
初始化一个哈希图，比如M ，它存储 ' 0'到 ' 1'的比率的频率。
使用变量i在[0, N-1]范围内迭代并执行以下步骤：
- 如果 S[i]等于'0'则将count_of_0增加1，否则将count_of_1增加1 。
- 找到count_of_0和count_of_1的 GCD 并将其存储在一个变量中，比如GCD 。
- 如果GCD = 0 ，则将m[{count_of_0, count_of_1}]增加1 。
- 如果GCD != 0 ，则将m[{count_of_0 / GCD, count_of_1 / GCD}]的值增加1 。
打印m[{count_of_0 / GCD, count_of_1 / GCD}]的值作为答案。

以下是上述方法的实现：

C++

// C++ program for the above approach
  
#include 
using namespace std;
  
// Function to find maximum partition such
// that the ratio of 0 and 1 are same
void maximumPartition(string S)
{
  
    // Size of string
    int N = S.size();
  
    // Variable to store the frequency
    // of 0 and 1
    int count_of_0 = 0, count_of_1 = 0;
  
    // Map to store frequency of ratio
    map, int> m;
  
    int ans;
  
    // Traverse the string
    for (int i = 0; i < N; i++) {
        // Increment the frequency
        // of 0 by 1 if s[i] = 0
        if (S[i] == '0')
            count_of_0++;
        // Otherwise increment frequency of 1
        else
            count_of_1++;
  
        int first = count_of_0, second = count_of_1;
  
        // Find GCD of count_of_0 and count_of_1
        int GCD = __gcd(count_of_0, count_of_1);
  
        // Convert the count of 0 and count of 1
        // in the coprime numbers
        if (GCD != 0) {
            first = count_of_0 / GCD;
            second = count_of_1 / GCD;
        }
  
        // Increase the ratio of 0 and 1 by 1
        m[{ first, second }]++;
        ans = m[{ first, second }];
    }
  
    cout << ans << endl;
}
  
// Driver Code
int main()
{
    // Given Input
    string S = "001101";
  
    // Function Call
    maximumPartition(S);
    return 0;
}

输出

时间复杂度： O(N log N)
辅助空间： O(N)