Leetcode #2983: Palindrome Rearrangement Queries
In this guide, we solve Leetcode #2983 Palindrome Rearrangement Queries 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 having an even length n. You are also given a 0-indexed 2D integer array, queries, where queries[i] = [ai, bi, ci, di].
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: 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 = "abcabc", queries = [[1,1,3,5],[0,2,5,5]]
Output: [true,true]
Explanation: In this example, there are two queries:
In the first query:
- a0 = 1, b0 = 1, c0 = 3, d0 = 5.
- So, you are allowed to rearrange s[1:1] => abcabc and s[3:5] => abcabc.
- To make s a palindrome, s[3:5] can be rearranged to become => abccba.
- Now, s is a palindrome. So, answer[0] = true.
In the second query:
- a1 = 0, b1 = 2, c1 = 5, d1 = 5.
- So, you are allowed to rearrange s[0:2] => abcabc and s[5:5] => abcabc.
- To make s a palindrome, s[0:2] can be rearranged to become => cbaabc.
- Now, s is a palindrome. So, answer[1] = true.
Python Solution
class Solution:
def canMakePalindromeQueries(self, s: str, queries: List[List[int]]) -> List[bool]:
def count(pre: List[List[int]], i: int, j: int) -> List[int]:
return [x - y for x, y in zip(pre[j + 1], pre[i])]
def sub(cnt1: List[int], cnt2: List[int]) -> List[int]:
res = []
for x, y in zip(cnt1, cnt2):
if x - y < 0:
return []
res.append(x - y)
return res
def check(
pre1: List[List[int]], pre2: List[List[int]], a: int, b: int, c: int, d: int
) -> bool:
if diff[a] > 0 or diff[m] - diff[max(b, d) + 1] > 0:
return False
if d <= b:
return count(pre1, a, b) == count(pre2, a, b)
if b < c:
return (
diff[c] - diff[b + 1] == 0
and count(pre1, a, b) == count(pre2, a, b)
and count(pre1, c, d) == count(pre2, c, d)
)
cnt1 = sub(count(pre1, a, b), count(pre2, a, c - 1))
cnt2 = sub(count(pre2, c, d), count(pre1, b + 1, d))
return bool(cnt1) and bool(cnt2) and cnt1 == cnt2
n = len(s)
m = n // 2
t = s[m:][::-1]
s = s[:m]
pre1 = [[0] * 26 for _ in range(m + 1)]
pre2 = [[0] * 26 for _ in range(m + 1)]
diff = [0] * (m + 1)
for i, (c1, c2) in enumerate(zip(s, t), 1):
pre1[i] = pre1[i - 1][:]
pre2[i] = pre2[i - 1][:]
pre1[i][ord(c1) - ord("a")] += 1
pre2[i][ord(c2) - ord("a")] += 1
diff[i] = diff[i - 1] + int(c1 != c2)
ans = []
for a, b, c, d in queries:
c, d = n - 1 - d, n - 1 - c
ok = (
check(pre1, pre2, a, b, c, d)
if a <= c
else check(pre2, pre1, c, d, a, b)
)
ans.append(ok)
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.