Leetcode #1177: Can Make Palindrome from Substring
In this guide, we solve Leetcode #1177 Can Make Palindrome from Substring 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 array queries where queries[i] = [lefti, righti, ki]. We may rearrange the substring s[lefti...righti] for each query and then choose up to ki of them to replace with any lowercase English letter.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Bit Manipulation, Array, Hash Table, String, 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 = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]
Output: [true,false,false,true,true]
Explanation:
queries[0]: substring = "d", is palidrome.
queries[1]: substring = "bc", is not palidrome.
queries[2]: substring = "abcd", is not palidrome after replacing only 1 character.
queries[3]: substring = "abcd", could be changed to "abba" which is palidrome. Also this can be changed to "baab" first rearrange it "bacd" then replace "cd" with "ab".
queries[4]: substring = "abcda", could be changed to "abcba" which is palidrome.
Python Solution
class Solution:
def canMakePaliQueries(self, s: str, queries: List[List[int]]) -> List[bool]:
n = len(s)
ss = [[0] * 26 for _ in range(n + 1)]
for i, c in enumerate(s, 1):
ss[i] = ss[i - 1][:]
ss[i][ord(c) - ord("a")] += 1
ans = []
for l, r, k in queries:
cnt = sum((ss[r + 1][j] - ss[l][j]) & 1 for j in range(26))
ans.append(cnt // 2 <= k)
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.