Leetcode #2182: Construct String With Repeat Limit
In this guide, we solve Leetcode #2182 Construct String With Repeat Limit 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 string s and an integer repeatLimit. Construct a new string repeatLimitedString using the characters of s such that no letter appears more than repeatLimit times in a row.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Hash Table, String, Counting, Heap (Priority Queue)
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 = "cczazcc", repeatLimit = 3
Output: "zzcccac"
Explanation: We use all of the characters from s to construct the repeatLimitedString "zzcccac".
The letter 'a' appears at most 1 time in a row.
The letter 'c' appears at most 3 times in a row.
The letter 'z' appears at most 2 times in a row.
Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
The string is the lexicographically largest repeatLimitedString possible so we return "zzcccac".
Note that the string "zzcccca" is lexicographically larger but the letter 'c' appears more than 3 times in a row, so it is not a valid repeatLimitedString.
Python Solution
class Solution:
def repeatLimitedString(self, s: str, repeatLimit: int) -> str:
cnt = [0] * 26
for c in s:
cnt[ord(c) - ord("a")] += 1
ans = []
j = 24
for i in range(25, -1, -1):
j = min(i - 1, j)
while 1:
x = min(repeatLimit, cnt[i])
cnt[i] -= x
ans.append(ascii_lowercase[i] * x)
if cnt[i] == 0:
break
while j >= 0 and cnt[j] == 0:
j -= 1
if j < 0:
break
cnt[j] -= 1
ans.append(ascii_lowercase[j])
return "".join(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.