给定一个包含小英文字母的字符串P和一个包含英文字母所有字符权重的字符串Q ，使得对于所有’i’，0≤Q [i]≤9。任务是找到唯一子字符串的总数权重之和不超过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。遵循以下步骤来计算答案：
- 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之和。
- 如果字符总和不大于K，则将其插入哈希图中。
- 最后，输出哈希图的大小。
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_PrasadC#
// 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 输出：
7
时间复杂度： O(N 2 )