二进制字符串的K大小子字符串中设置位的最大数目

给定大小为N的二进制字符串S和整数K。任务是找到出现在大小为K的子字符串中的最大置位位数。

例子：

Input: S = “100111010”, K = 3
Output: 3
Explanation:
The substring “111” contains 3 set bits.

Input:S = “0000000”, K = 4
Output: 0
Explanation: S doesn’t have any set bits in it, so ans is 0.

天真的方法：

1. 生成所有大小为K的子字符串。
2. 在所有子串中找到最大置位数。

时间复杂度： O(N ² )。
辅助空间： O(1)。

高效的方法：可以使用滑动窗口技术解决该问题。

1. 以maxcount变量存储设置位的最大计数，以Count变量存储当前窗口的计数设置位。
2. 从1横移字符串至K并计算组位的计数和存储作为MAXCOUNT。
3. 从K +1遍历字符串到字符串的长度。
4. 在每次迭代中，如果设置了^第(K – i)位，则减少计数。如果设置了^第i位，则增加计数。比较并更新maxcount 。
5. 完成数组遍历后，最后返回maxcount。

下面是上述方法的实现：

C++

// C++ program to find the maximum
// set bits in a substring of size K
#include
using namespace std;
  
// Function that find Maximum number
// of set bit appears in a substring
// of size K.
int maxSetBitCount(string s, int k)
{
    int maxCount = 0, n = s.length();
    int count = 0;
  
    // Traverse string 1 to k
    for(int i = 0; i < k; i++) 
    {
          
       // Incerement count if
       // character is set bit
       if (s[i] == '1')
           count++;
    }
    maxCount = count;
  
    // Traverse string k+1
    // to length of string
    for(int i = k; i < n; i++)
    {
         
       // Remove the contribution of the
       // (i - k)th character which is no
       // longer in the window
       if (s[i - k] == '1')
           count--;
         
       // Add the contribution of
       // the current character
       if (s[i] == '1')
           count++;
             
       // Update maxCount at for
       // each window of size k
       maxCount = max(maxCount, count);
    }
      
    // Return maxCount
    return maxCount;
}
  
// Driver code
int main()
{
    string s = "100111010";
    int k = 3;
  
    cout << (maxSetBitCount(s, k));
    return 0;
}
  
// This code is contributed by Rajput-Ji

Java

// Java program to find the maximum
// set bits in a substring of size K
import java.util.*;
  
class GFG {
  
    // Function that find Maximum number
    // of set bit appears in a substring
    // of size K.
    static int maxSetBitCount(String s, int k)
    {
  
        int maxCount = 0, n = s.length();
        int count = 0;
  
        // Traverse string 1 to k
        for (int i = 0; i < k; i++) {
            // Incerement count if
            // character is set bit
            if (s.charAt(i) == '1')
                count++;
        }
  
        maxCount = count;
  
        // Traverse string k+1
        // to length of string
        for (int i = k; i < n; i++) {
  
            // remove the contribution of the
            // (i - k)th character which is no
            // longer in the window
            if (s.charAt(i - k) == '1')
                count--;
  
            // add the contribution of
            // the current character
            if (s.charAt(i) == '1')
                count++;
  
            // update maxCount at for
            // each window of size k
            maxCount = Math.max(maxCount, count);
        }
  
        // return maxCount
        return maxCount;
    }
    // Driver Program
    public static void main(String[] args)
    {
        String s = "100111010";
        int k = 3;
  
        System.out.println(maxSetBitCount(s, k));
    }
}

Python3

# Python3 program to find the maximum
# set bits in a substring of size K
  
# Function that find Maximum number
# of set bit appears in a substring
# of size K.
def maxSetBitCount(s, k):
  
    maxCount = 0
    n = len(s)
    count = 0
  
    # Traverse string 1 to k
    for i in range(k):
          
        # Incerement count if
        # character is set bit
        if (s[i] == '1'):
            count += 1
  
    maxCount = count
  
    # Traverse string k+1
    # to length of string
    for i in range(k, n):
  
        # Remove the contribution of the
        # (i - k)th character which is no
        # longer in the window
        if (s[i - k] == '1'):
            count -= 1
  
        # Add the contribution of
        # the current character
        if (s[i] == '1'):
            count += 1
  
        # Update maxCount at for
        # each window of size k
        maxCount = max(maxCount, count)
  
    # Return maxCount
    return maxCount
  
  
# Driver code
if __name__ == '__main__':
      
    s = "100111010"
    k = 3
  
    print(maxSetBitCount(s, k))
      
# This code is contributed by mohit kumar 29

C#

// C# program to find the maximum
// set bits in a substring of size K
using System;
class GFG {
  
// Function that find Maximum number
// of set bit appears in a substring
// of size K.
static int maxSetBitCount(string s, int k)
{
  
    int maxCount = 0, n = s.Length;
    int count = 0;
  
    // Traverse string 1 to k
    for (int i = 0; i < k; i++) 
    {
        // Incerement count if
        // character is set bit
        if (s[i] == '1')
            count++;
    }
  
    maxCount = count;
  
    // Traverse string k+1
    // to length of string
    for (int i = k; i < n; i++) 
    {
  
        // remove the contribution of the
        // (i - k)th character which is no
        // longer in the window
        if (s[i - k] == '1')
            count--;
  
        // add the contribution of
        // the current character
        if (s[i] == '1')
            count++;
  
        // update maxCount at for
        // each window of size k
        maxCount = Math.Max(maxCount, count);
    }
  
    // return maxCount
    return maxCount;
}
// Driver Program
public static void Main()
{
    string s = "100111010";
    int k = 3;
  
    Console.Write(maxSetBitCount(s, k));
}
}
  
// This code is contributed by Code_Mech

输出：

3

时间复杂度： O(N)。
辅助空间： O(1)。