Leetcode #2955: Number of Same-End Substrings
In this guide, we solve Leetcode #2955 Number of Same-End Substrings in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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].
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Array, Hash Table, String, Counting, Prefix Sum
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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".
Python Solution
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
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.