查找指定子字符串中的字符串频率的查询

给定一个字符串S和一个查询矩阵Q ，分别指定任务S的子字符串的起始索引L(= Q [i] [0])和R(= Q [i] [0]) 。是在子字符串[L，R]中找到字符串K的频率。

注意：范围遵循从1开始的索引。

例子：

Input: S = “GFGFFGFG”, K = “GFG”, Q = {{1, 8}, {3, 5}, {5, 8}} 
Output: 
2 
0 
1 
Explanation: For query 1, there are 2 (“GFG”) substrings from index 1 to index 8. One is from index 1 to 3 and the other is from index 6 to 8. 
For query 2, there are 0 (“GFG”) substrings from index 3 to 5. 
For query 3, there are 1 (“GFG”) substrings from index 5 to index 8. The one and only substring are from index 6 to 8.

Input: S = “ABCABCABABC”, K = “ABC”, Q = {{1, 6}, {5, 11}} 
Output: 
2 
1 

天真的方法：
对所有查询从L到R运行循环。计算字符串K的出现次数并返回计数。

时间复杂度： O(Q * | S |的长度)。

高效方法：
预计算并存储每个索引的K频率。现在，为了计算字符串在[L，R]范围内的频率，我们只需要计算索引为(R-1)和(L-1)的K的频率之间的差即可。

下面是上述方法的实现：

C++

// C++ Program to find
// frequency of a string K
// in a substring [L, R] in S
 
#include 
#define max_len 100005
using namespace std;
 
// Store the frequency of
// string for each index
int cnt[max_len];
 
// Compute and store frequencies
// for every index
void precompute(string s, string K)
{
 
    int n = s.size();
    for (int i = 0; i < n - 1; i++) {
        cnt[i + 1]
            = cnt[i]
              + (s.substr(i, K.size()) == K);
    }
}
 
// Driver Code
int main()
{
    string s = "ABCABCABABC";
    string K = "ABC";
    precompute(s, K);
 
    vector > Q
        = { { 1, 6 }, { 5, 11 } };
 
    for (auto it : Q) {
        cout << cnt[it.second - 1]
                    - cnt[it.first - 1]
             << endl;
    }
 
    return 0;
}

Java

// Java program to find
// frequency of a string K
// in a substring [L, R] in S
class GFG{
     
static int max_len = 100005;
 
// Store the frequency of
// string for each index
static int cnt[] = new int[max_len];
 
// Compute and store frequencies
// for every index
public static void precompute(String s,
                              String K)
{
    int n = s.length();
     
    for(int i = 0; i < n - 2; i++)
    {
        cnt[i + 1] = cnt[i];
        if (s.substring(
            i, i + K.length()).equals(K))
        {
            cnt[i + 1] += 1;
        }
    }
    cnt[n - 2 + 1] = cnt[n - 2];
}
 
// Driver code
public static void main(String[] args)
{
    String s = "ABCABCABABC";
    String K = "ABC";
    precompute(s, K);
   
    int Q[][] = { { 1, 6 }, { 5, 11 } };
     
    for(int it = 0; it < Q.length; it++)
    {
        System.out.println(cnt[Q[it][1] - 1] -
                           cnt[Q[it][0] - 1]);
    }
}
}
 
// This code is contributed by divyesh072019

Python3

# Python3 Program to find
# frequency of a string K
# in a substring [L, R] in S
max_len = 100005
 
# Store the frequency of
# string for each index
cnt = [0] * max_len
 
# Compute and store frequencies
# for every index
def precompute(s, K):
 
    n = len(s)
    for i in range(n - 1):
        cnt[i + 1] = cnt[i]
        if s[i : len(K) + i] == K:
            cnt[i + 1] += 1
 
# Driver Code
if __name__ == "__main__":
 
    s = "ABCABCABABC"
    K = "ABC"
    precompute(s, K)
    Q = [[1, 6], [5, 11]]
 
    for it in Q:
        print(cnt[it[1] - 1] -
              cnt[it[0] - 1])
 
# This code is contributed by Chitranayal

C#

// C# program to find frequency of
// a string K in a substring [L, R] in S
using System.IO;
using System;
 
class GFG{
     
static int max_len = 100005;
 
// Store the frequency of
// string for each index
static int[] cnt = new int[max_len];
 
// Compute and store frequencies
// for every index
static void precompute(string s,string K)
{
    int n = s.Length;
     
    for(int i = 0; i < n - 2; i++)
    {
        cnt[i + 1] = cnt[i];
         
        if (s.Substring(i, K.Length).Equals(K))
        {
            cnt[i + 1] += 1;
        }
    }
    cnt[n - 2 + 1] = cnt[n - 2];
}
 
// Driver code
static void Main()
{
    string s = "ABCABCABABC";
    string K = "ABC";
    precompute(s, K);
    int[,] Q = { { 1, 6 }, { 5, 11 } };
     
    for(int it = 0; it < Q.GetLength(0); it++)
    {  
        Console.WriteLine(cnt[Q[it, 1] - 1] -
                          cnt[Q[it, 0] - 1]);
    }
}
}
 
// This code is contributed by rag2127

输出：

2
1

时间复杂度： O(| S | + Q的长度) ，因为每个查询都在O(1)中回答。
辅助空间： O(| S |)