给定长度为N的字符串S和{l，r，y}形式的查询数组Q [] [] 。对于每个查询，任务是打印出现在[l，r]范围内的字符y的数量。
例子：
Input: S = “aabv”, Q[][] = {{0, 3, ‘a’}, {1, 2, ‘b’}}
Output: 2 1
Explanation:
Query 1: Number of character ‘a’ present in the range [0, 3] is 2.
Query 2: Number of character ‘b’ present in the range [1, 2] is 1.
Input: S = “abcd”, Q[][] = {{1, 3, ‘c’}, {1, 1, ‘b’}}
Output: 1 1
Explanation:
Query 1: Number of character ‘c’ present in the range [1, 3] is 1.
Query 2: Number of character ‘b’ present in the range [1, 1] is 1.
天真的方法：最简单的方法是在每个查询{l，r，y}的索引i处的字符等于y的情况下遍历[l，r]范围内的字符串并将计数器增加1 。遍历后，为每个查询打印计数器。
时间复杂度： O(N * Q)
辅助空间： O(N)
高效方法：我们的想法是在范围为1〜预先计算本字符数为i对于其中1≤I≤N使用前缀和技术从“a”到“Z”的每个字符。请按照以下步骤解决问题：
- 初始化数组dp [N + 1] [26] ，其中dp [i] [j]存储范围为[0，i]的字符数(i +’a’) 。
- 现在，预先计算每个字符存在于该范围中的数目[1，I]，其中1≤I≤N乘递增DP [i] [j]由DP [I – 1] [j]的其中0≤Ĵ<26和增量dp [i] [S [j] –’a’]减1 。
- 对于每个查询{l，r，y} ，将dp [r] [y –’a’] – dp [l – 1] [y –’a’]的值打印为范围内存在的字符y的数量[l，r]
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to print count of char
// y present in the range [l, r]
void noOfChars(string s, char queries[][3], int q)
{
// Length of the string
int n = s.length();
// Stores the precomputed results
int dp[n + 1][26];
memset(dp, 0, sizeof(dp));
// Iterate the given string
for(int i = 0; i < n; i++)
{
// Increment dp[i][y-'a'] by 1
dp[i + 1][s[i]-'a']++;
// Pre-compute
for(int j = 0; j < 26; j++)
{
dp[i + 1][j] += dp[i][j];
}
}
// Traverse each query
for(int i = 0; i < q; i++)
{
int l = (int)queries[i][0];
int r = (int)queries[i][1];
int c = queries[i][2] - 'a';
// Print the result for
// each query
cout << dp[r] - dp[l - 1] << " ";
}
}
// Driver Code
int main()
{
// Given string
string S = "aabv";
// Given Queries
char queries[2][3] = {{ 1, 2, 'a' },{ 2, 3, 'b' }};
// Function Call
noOfChars(S, queries, 2);
return 0;
}
// This code is contributed by avanitrachhadiya2155 Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to print count of char
// y present in the range [l, r]
static void noOfChars(String s,
char[][] queries)
{
// Length of the string
int n = s.length();
// Stores the precomputed results
int dp[][] = new int[n + 1][26];
// Iterate the given string
for (int i = 0; i < n; i++) {
// Increment dp[i][y-'a'] by 1
dp[i + 1][s.charAt(i) - 'a']++;
// Pre-compute
for (int j = 0; j < 26; j++) {
dp[i + 1][j] += dp[i][j];
}
}
// Number of queries
int q = queries.length;
// Traverse each query
for (int i = 0; i < q; i++) {
int l = (int)queries[i][0];
int r = (int)queries[i][1];
int c = queries[i][2] - 'a';
// Print the result for
// each query
System.out.print(
dp[r] - dp[l - 1]
+ " ");
}
}
// Driver Code
public static void main(String[] args)
{
// Given string
String S = "aabv";
// Given Queries
char queries[][]
= new char[][] { { 1, 2, 'a' },
{ 2, 3, 'b' } };
// Function Call
noOfChars(S, queries);
}
}Python3
# Python3 program for the above approach
# Function to prcount of char
# y present in the range [l, r]
def noOfChars(s, queries):
# Length of the string
n = len(s)
# Stores the precomputed results
dp = [[0 for i in range(26)]
for i in range(n + 1)]
# Iterate the given string
for i in range(n):
# Increment dp[i][y-'a'] by 1
dp[i + 1][ord(s[i]) - ord('a')] += 1
# Pre-compute
for j in range(26):
dp[i + 1][j] += dp[i][j]
# Number of queries
q = len(queries)
# Traverse each query
for i in range(q):
l = queries[i][0]
r = queries[i][1]
c = ord(queries[i][2]) - ord('a')
# Print the result for
# each query
print(dp[r] - dp[l - 1], end = " ")
# Driver Code
if __name__ == '__main__':
# Given string
S = "aabv"
# Given Queries
queries = [ [ 1, 2, 'a' ],
[ 2, 3, 'b' ] ]
# Function Call
noOfChars(S, queries)
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
public class GFG {
// Function to print count of char
// y present in the range [l, r]
static void noOfChars(String s,
char[,] queries)
{
// Length of the string
int n = s.Length;
// Stores the precomputed results
int [,]dp = new int[n + 1, 26];
// Iterate the given string
for (int i = 0; i < n; i++) {
// Increment dp[i,y-'a'] by 1
dp[i + 1, s[i] - 'a']++;
// Pre-compute
for (int j = 0; j < 26; j++) {
dp[i + 1, j] += dp[i, j];
}
}
// Number of queries
int q = queries.GetLength(0);
// Traverse each query
for (int i = 0; i < q; i++) {
int l = (int)queries[i, 0];
int r = (int)queries[i, 1];
int c = queries[i, 2] - 'a';
// Print the result for
// each query
Console.Write(
dp[r, c] - dp[l - 1, c]
+ " ");
}
}
// Driver Code
public static void Main(String[] args)
{
// Given string
String S = "aabv";
// Given Queries
char [,]queries
= new char[,] { { (char)1, (char)2, 'a' },
{ (char)2, (char)3, 'b' } };
// Function Call
noOfChars(S, queries);
}
}
// This code is contributed by 29AjayKumar输出：
2 1时间复杂度： O(Q + N * 26)
辅助空间： O(N)