计算所有子串，重量atmostķ字符

给定一个由小英文字母组成的字符串P和一个由英文字母表中所有字符的权重组成的字符串Q ，使得对于所有 ‘i’，0 ≤ Q[i] ≤ 9。 任务是找到唯一子串的总数权重之和至多K 。
例子：

Input: P = “ababab”, Q = “12345678912345678912345678”, K = 5 
Output: 7 
Explanation: 
The substrings with the sum of weights of individual characters ≤ 5 are: 
“a”, “ab”, “b”, “bc”, “c”, “d”, “e”
Input: P = “acbacbacaa”, Q = “12300045600078900012345000”, K = 2 
Output: 3 
Explanation: 
The substrings with the sum of weights of individual characters ≤ 2 are: 
“a”, “b”, “aa” 
 

方法：想法是使用无序集合来存储唯一值。按照以下步骤计算答案：

• 使用嵌套循环遍历所有子字符串并保持到目前为止遇到的所有字符的权重总和。
• 如果字符总和不大于 K，则将其插入到哈希映射中。
• 最后，输出hashmap的大小。

下面是上述方法的实现：

C++

// C++ program to find the count of
// all the sub-strings with weight of
// characters atmost K
#include 
using namespace std;
 
// Function to find the count of
// all the substrings with weight
// of characters atmost K
int distinctSubstring(string& P, string& Q,
                      int K, int N)
{
 
    // Hashmap to store all substrings
    unordered_set S;
 
    // Iterate over all substrings
    for (int i = 0; i < N; ++i) {
 
        // Maintain the sum of all characters
        // encountered so far
        int sum = 0;
 
        // Maintain the substring till the
        // current position
        string s;
 
        for (int j = i; j < N; ++j) {
 
            // Get the position of the
            // character in string Q
            int pos = P[j] - 'a';
 
            // Add weight to current sum
            sum += Q[pos] - '0';
 
            // Add current character to substring
            s += P[j];
 
            // If sum of characters is <=K
            // then insert in into the set
            if (sum <= K) {
                S.insert(s);
            }
 
            else {
                break;
            }
        }
    }
 
    // Finding the size of the set
    return S.size();
}
 
// Driver code
int main()
{
    string P = "abcde";
    string Q = "12345678912345678912345678";
    int K = 5;
    int N = P.length();
 
    cout << distinctSubstring(P, Q, K, N);
 
    return 0;
}

Java

// Java program to find the count of
// all the sub-Strings with weight of
// characters atmost K
import java.util.*;
 
class GFG{
  
// Function to find the count of
// all the subStrings with weight
// of characters atmost K
static int distinctSubString(String P, String Q,
                      int K, int N)
{
  
    // Hashmap to store all subStrings
    HashSet S = new HashSet();
  
    // Iterate over all subStrings
    for (int i = 0; i < N; ++i) {
  
        // Maintain the sum of all characters
        // encountered so far
        int sum = 0;
  
        // Maintain the subString till the
        // current position
        String s = "";
  
        for (int j = i; j < N; ++j) {
  
            // Get the position of the
            // character in String Q
            int pos = P.charAt(j) - 'a';
  
            // Add weight to current sum
            sum += Q.charAt(pos) - '0';
  
            // Add current character to subString
            s += P.charAt(j);
  
            // If sum of characters is <=K
            // then insert in into the set
            if (sum <= K) {
                S.add(s);
            }
  
            else {
                break;
            }
        }
    }
  
    // Finding the size of the set
    return S.size();
}
  
// Driver code
public static void main(String[] args)
{
    String P = "abcde";
    String Q = "12345678912345678912345678";
    int K = 5;
    int N = P.length();
  
    System.out.print(distinctSubString(P, Q, K, N));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python program to find the count of
# all the sub-strings with weight of
# characters atmost K
 
# Function to find the count of
# all the substrings with weight
# of characters atmost K
def distinctSubstring(P, Q, K, N):
 
    # Hashmap to store all substrings
    S = set()
 
    # Iterate over all substrings
    for i in range(0,N):
 
        # Maintain the sum of all characters
        # encountered so far
        sum = 0;
 
        # Maintain the substring till the
        # current position
        s = ''
 
        for j in range(i,N):
 
            # Get the position of the
            # character in string Q
            pos = ord(P[j]) - 97
 
            # Add weight to current sum
            sum = sum + ord(Q[pos]) - 48
 
            # Add current character to substring
            s += P[j]
 
            # If sum of characters is <=K
            # then insert in into the set
            if (sum <= K):
                S.add(s)
            else:
                break
             
    # Finding the size of the set
    return len(S)
 
# Driver code
P = "abcde"
Q = "12345678912345678912345678"
K = 5
N = len(P)
 
print(distinctSubstring(P, Q, K, N))
 
# This code is contributed by Sanjit_Prasad

C#

// C# program to find the count of
// all the sub-Strings with weight of
// characters atmost K
using System;
using System.Collections.Generic;
 
class GFG{
   
// Function to find the count of
// all the subStrings with weight
// of characters atmost K
static int distinctSubString(String P, String Q,
                      int K, int N)
{
   
    // Hashmap to store all subStrings
    HashSet S = new HashSet();
   
    // Iterate over all subStrings
    for (int i = 0; i < N; ++i) {
   
        // Maintain the sum of all characters
        // encountered so far
        int sum = 0;
   
        // Maintain the subString till the
        // current position
        String s = "";
   
        for (int j = i; j < N; ++j) {
   
            // Get the position of the
            // character in String Q
            int pos = P[j] - 'a';
   
            // Add weight to current sum
            sum += Q[pos] - '0';
   
            // Add current character to subString
            s += P[j];
   
            // If sum of characters is <=K
            // then insert in into the set
            if (sum <= K) {
                S.Add(s);
            }
   
            else {
                break;
            }
        }
    }
   
    // Finding the size of the set
    return S.Count;
}
   
// Driver code
public static void Main(String[] args)
{
    String P = "abcde";
    String Q = "12345678912345678912345678";
    int K = 5;
    int N = P.Length;
   
    Console.Write(distinctSubString(P, Q, K, N));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript program to find the count of
// all the sub-Strings with weight of
// characters atmost K
 
// Function to find the count of
// all the subStrings with weight
// of characters atmost K
function distinctSubString(P, Q, K, N)
{
    
    // Hashmap to store all subStrings
    let S = new Set();
    
    // Iterate over all subStrings
    for (let i = 0; i < N; ++i) {
    
        // Maletain the sum of all characters
        // encountered so far
        let sum = 0;
    
        // Maletain the subString till the
        // current position
        let s = "";
    
        for (let j = i; j < N; ++j) {
    
            // Get the position of the
            // character in String Q
            let pos = P[j].charCodeAt() - 'a'.charCodeAt();
    
            // Add weight to current sum
            sum += Q[pos].charCodeAt() - '0'.charCodeAt();
    
            // Add current character to subString
            s += P[j];
    
            // If sum of characters is <=K
            // then insert in leto the set
            if (sum <= K) {
                S.add(s);
            }
    
            else {
                break;
            }
        }
    }
    
    // Finding the size of the set
    return S.size;
}
  
// Driver code
     
      let P = "abcde";
    let Q = "12345678912345678912345678";
    let K = 5;
    let N = P.length;
    
    document.write(distinctSubString(P, Q, K, N));
                                                                                          

输出：

7

时间复杂度： O(N ² )

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