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2955. Number of Same-End Substrings
Description
You are given a 0-indexed string s, and a 2D array of integers queries, where queries[i] = [li, ri] indicates a substring of s starting from the index li and ending at the index ri (both inclusive), i.e. s[li..ri].
Return an array ans where ans[i] is the number of same-end substrings of queries[i].
A 0-indexed string t of length n is called same-end if it has the same character at both of its ends, i.e., t[0] == t[n - 1].
A substring is a contiguous non-empty sequence of characters within a string.
Example 1:
Input: s = "abcaab", queries = [[0,0],[1,4],[2,5],[0,5]] Output: [1,5,5,10] Explanation: Here is the same-end substrings of each query: 1st query: s[0..0] is "a" which has 1 same-end substring: "a". 2nd query: s[1..4] is "bcaa" which has 5 same-end substrings: "bcaa", "bcaa", "bcaa", "bcaa", "bcaa". 3rd query: s[2..5] is "caab" which has 5 same-end substrings: "caab", "caab", "caab", "caab", "caab". 4th query: s[0..5] is "abcaab" which has 10 same-end substrings: "abcaab", "abcaab", "abcaab", "abcaab", "abcaab", "abcaab", "abcaab", "abcaab", "abcaab", "abcaab".
Example 2:
Input: s = "abcd", queries = [[0,3]] Output: [4] Explanation: The only query is s[0..3] which is "abcd". It has 4 same-end substrings: "abcd", "abcd", "abcd", "abcd".
Constraints:
2 <= s.length <= 3 * 104sconsists only of lowercase English letters.1 <= queries.length <= 3 * 104queries[i] = [li, ri]0 <= li <= ri < s.length
Solutions
Solution 1: Prefix Sum + Enumeration
We can preprocess the prefix sum for each letter and record it in the array $cnt$, where $cnt[i][j]$ represents the number of times the $i$-th letter appears in the first $j$ characters. In this way, for each interval $[l, r]$, we can enumerate each letter $c$ in the interval, quickly calculate the number of times $c$ appears in the interval $x$ using the prefix sum array. We can arbitrarily choose two of them to form a tail-equal substring, the number of substrings is $C_x^2=\frac{x(x-1)}{2}$, plus the situation where each letter in the interval can form a tail-equal substring alone, there are $r - l + 1$ letters in total. Therefore, for each query $[l, r]$, the number of tail-equal substrings that meet the conditions is $r - l + 1 + \sum_{c \in \Sigma} \frac{x_c(x_c-1)}{2}$, where $x_c$ represents the number of times the letter $c$ appears in the interval $[l, r]$.
The time complexity is $O((n + m) \times |\Sigma|)$, and the space complexity is $O(n \times |\Sigma|)$. Here, $n$ and $m$ are the lengths of the string $s$ and the number of queries, respectively, and $\Sigma$ represents the set of letters appearing in the string $s$, in this problem $|\Sigma|=26$.
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class Solution { public int[] sameEndSubstringCount(String s, int[][] queries) { int n = s.length(); int[][] cnt = new int[26][n + 1]; for (int j = 1; j <= n; ++j) { for (int i = 0; i < 26; ++i) { cnt[i][j] = cnt[i][j - 1]; } cnt[s.charAt(j - 1) - 'a'][j]++; } int m = queries.length; int[] ans = new int[m]; for (int k = 0; k < m; ++k) { int l = queries[k][0], r = queries[k][1]; ans[k] = r - l + 1; for (int i = 0; i < 26; ++i) { int x = cnt[i][r + 1] - cnt[i][l]; ans[k] += x * (x - 1) / 2; } } return ans; } } -
class Solution { public: vector<int> sameEndSubstringCount(string s, vector<vector<int>>& queries) { int n = s.size(); int cnt[26][n + 1]; memset(cnt, 0, sizeof(cnt)); for (int j = 1; j <= n; ++j) { for (int i = 0; i < 26; ++i) { cnt[i][j] = cnt[i][j - 1]; } cnt[s[j - 1] - 'a'][j]++; } vector<int> ans; for (auto& q : queries) { int l = q[0], r = q[1]; ans.push_back(r - l + 1); for (int i = 0; i < 26; ++i) { int x = cnt[i][r + 1] - cnt[i][l]; ans.back() += x * (x - 1) / 2; } } return ans; } }; -
class Solution: def sameEndSubstringCount(self, s: str, queries: List[List[int]]) -> List[int]: n = len(s) cs = set(s) cnt = {c: [0] * (n + 1) for c in cs} for i, a in enumerate(s, 1): for c in cs: cnt[c][i] = cnt[c][i - 1] cnt[a][i] += 1 ans = [] for l, r in queries: t = r - l + 1 for c in cs: x = cnt[c][r + 1] - cnt[c][l] t += x * (x - 1) // 2 ans.append(t) return ans -
func sameEndSubstringCount(s string, queries [][]int) []int { n := len(s) cnt := make([][]int, 26) for i := 0; i < 26; i++ { cnt[i] = make([]int, n+1) } for j := 1; j <= n; j++ { for i := 0; i < 26; i++ { cnt[i][j] = cnt[i][j-1] } cnt[s[j-1]-'a'][j]++ } var ans []int for _, q := range queries { l, r := q[0], q[1] ans = append(ans, r-l+1) for i := 0; i < 26; i++ { x := cnt[i][r+1] - cnt[i][l] ans[len(ans)-1] += x * (x - 1) / 2 } } return ans } -
function sameEndSubstringCount(s: string, queries: number[][]): number[] { const n: number = s.length; const cnt: number[][] = Array.from({ length: 26 }, () => Array(n + 1).fill(0)); for (let j = 1; j <= n; j++) { for (let i = 0; i < 26; i++) { cnt[i][j] = cnt[i][j - 1]; } cnt[s.charCodeAt(j - 1) - 'a'.charCodeAt(0)][j]++; } const ans: number[] = []; for (const [l, r] of queries) { ans.push(r - l + 1); for (let i = 0; i < 26; i++) { const x: number = cnt[i][r + 1] - cnt[i][l]; ans[ans.length - 1] += (x * (x - 1)) / 2; } } return ans; } -
impl Solution { pub fn same_end_substring_count(s: String, queries: Vec<Vec<i32>>) -> Vec<i32> { let n = s.len(); let mut cnt: Vec<Vec<i32>> = vec![vec![0; n + 1]; 26]; for j in 1..=n { for i in 0..26 { cnt[i][j] = cnt[i][j - 1]; } cnt[(s.as_bytes()[j - 1] as usize) - (b'a' as usize)][j] += 1; } let mut ans: Vec<i32> = Vec::new(); for q in queries.iter() { let l = q[0] as usize; let r = q[1] as usize; let mut t = (r - l + 1) as i32; for i in 0..26 { let x = cnt[i][r + 1] - cnt[i][l]; t += (x * (x - 1)) / 2; } ans.push(t); } ans } }