Leetcode #2399: Check Distances Between Same Letters
In this guide, we solve Leetcode #2399 Check Distances Between Same Letters 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 consisting of only lowercase English letters, where each letter in s appears exactly twice. You are also given a 0-indexed integer array distance of length 26.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Array, Hash Table, String
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 = "abaccb", distance = [1,3,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: true
Explanation:
- 'a' appears at indices 0 and 2 so it satisfies distance[0] = 1.
- 'b' appears at indices 1 and 5 so it satisfies distance[1] = 3.
- 'c' appears at indices 3 and 4 so it satisfies distance[2] = 0.
Note that distance[3] = 5, but since 'd' does not appear in s, it can be ignored.
Return true because s is a well-spaced string.
Python Solution
class Solution:
def checkDistances(self, s: str, distance: List[int]) -> bool:
d = defaultdict(int)
for i, c in enumerate(map(ord, s), 1):
j = c - ord("a")
if d[j] and i - d[j] - 1 != distance[j]:
return False
d[j] = i
return True
Complexity
The time complexity is , where is the length of the string . The space complexity is , where is the character set, which in this case is the set of lowercase letters.
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.