查询以查找字符串S 中 L 到 R 范围内重复字符的总数

给定一个大小为 N的字符串S由小写字母和一个整数 Q组成，该整数 Q 表示对 S 的查询次数。我们的任务是为所有查询 Q 打印子字符串 L 到 R 中重复字符的数量。
注： 1 ≤N ≤ 10 ⁶和 1 ≤ Q≤ 10 ⁶
例子：

Input : 
S = “geeksforgeeks”, Q = 2 
L = 1 R = 5 
L = 4 R = 8 
Output : 
1 
0 
Explanation: 
For the first query ‘e’ is the only duplicate character in S from range 1 to 5. 
For the second query theres is no duplicate character in S.
Input : 
S = “Geekyy”, Q = 1 
L = 1 R = 6 
Output : 
2 
Explanation: 
For the first query ‘e’ and ‘y’ are duplicate characters in S from range 1 to 6. 
 

天真的方法：
最简单的方法是维护一个大小为 26 的频率数组，以存储每个字符的计数。对于每个查询，给定范围 [L, R] 我们将遍历子字符串 S[L] 到 S[R] 并继续计算每个字符的出现次数。现在，如果任何字符的频率大于 1，那么我们将加 1 来回答。
有效的方法：
为了以有效的方式解决上述问题，我们将存储每个字符出现在动态数组中的字符串中的位置。对于每个给定的查询，我们将遍历所有 26 个小写字母。如果当前字母在子串 S[L: R] 中，那么对应向量中大于或等于 L的第一个元素的下一个元素应该存在并且小于或等于 R 。
下图显示了我们如何在动态数组中存储字符：

下面是上述方法的实现：

CPP

// CPP implementation to Find the total
// number of duplicate character in a
// range L to R for Q number of queries in a string S
 
#include 
using namespace std;
 
// Vector of vector to store
// position of all characters
// as they appear in string
vector > v(26);
 
// Function to store position of each character
void calculate(string s)
{
    for (int i = 0; i < s.size(); i++) {
        // Inserting position of each
        // character as they appear
        v[s[i] - 'a'].push_back(i);
    }
}
 
// Function to calculate duplicate
// characters for Q queries
void query(int L, int R)
{
    // Variable to count duplicates
    int duplicates = 0;
 
    // Iterate over all 26 characters
    for (int i = 0; i < 26; i++) {
 
        // Finding the first element which
        // is less than or equal to L
        auto first = lower_bound(v[i].begin(),
                                 v[i].end(), L - 1);
 
        // Check if first pointer exists
        // and is less than R
        if (first != v[i].end() && *first < R) {
            // Incrementing first pointer to check
            // if the next duplicate element exists
            first++;
 
            // Check if the next element exists
            // and is less than R
            if (first != v[i].end() && *first < R)
                duplicates++;
        }
    }
 
    cout << duplicates << endl;
}
 
// Driver Code
int main()
{
    string s = "geeksforgeeks";
 
    int Q = 2;
 
    int l1 = 1, r1 = 5;
    int l2 = 4, r2 = 8;
 
    calculate(s);
 
    query(l1, r1);
    query(l2, r2);
 
    return 0;
}

Python3

# Python implementation to Find the total
# number of duplicate character in a
# range L to R for Q number of queries in a string S
 
import bisect
 
# Vector of vector to store
# position of all characters
# as they appear in string
v = [[] for _ in range(26)]
 
# Function to store position of each character
def calculate(s: str) -> None:
 
    for i in range(len(s)):
        # Inserting position of each
        # character as they appear
        v[ord(s[i]) - ord('a')].append(i)
 
 
# Function to calculate duplicate
# characters for Q queries
def query(L: int, R: int) -> None:
 
    # Variable to count duplicates
    duplicates = 0
 
    # Iterate over all 26 characters
    for i in range(26):
 
        # Finding the first element which
        # is less than or equal to L
        first = bisect.bisect_left(v[i], L - 1)
 
        # Check if first pointer exists
        # and is less than R
        if (first < len(v[i]) and v[i][first] < R):
            # Incrementing first pointer to check
            # if the next duplicate element exists
            first += 1
 
            # Check if the next element exists
            # and is less than R
            if (first < len(v[i]) and v[i][first] < R):
                duplicates += 1
 
    print(duplicates)
 
 
# Driver Code
if __name__ == "__main__":
 
    s = "geeksforgeeks"
 
    Q = 2
 
    l1 = 1
    r1 = 5
    l2 = 4
    r2 = 8
 
    calculate(s)
 
    query(l1, r1)
    query(l2, r2)
 
# This code is contributed by sanjeev2552

输出：

1
0

`
时间复杂度： O( Q * 26 * log N)

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