Palindrome Removal — LeetCode 1246 Python Solution
HardLeetCode PremiumArrayDynamic Programming
- Problem
- #1246
- Pattern
- Dynamic Programming
- Reading time
- 3 min
- Source
- leetcode.com
The problem
You are given an integer array arr. In one move, you can select a palindromic subarray arr[i], arr[i + 1], ..., arr[j] where i <= j, and remove that subarray from the given array.
Example
- Input
- arr = [1,2]
- Output
- 2
Python solution
Python
class Solution:
def minimumMoves(self, arr: List[int]) -> int:
n = len(arr)
f = [[0] * n for _ in range(n)]
for i in range(n):
f[i][i] = 1
for i in range(n - 2, -1, -1):
for j in range(i + 1, n):
if i + 1 == j:
f[i][j] = 1 if arr[i] == arr[j] else 2
else:
t = f[i + 1][j - 1] if arr[i] == arr[j] else inf
for k in range(i, j):
t = min(t, f[i][k] + f[k + 1][j])
f[i][j] = t
return f[0][n - 1]Complexity
| Measure | Complexity |
|---|---|
| Time | O(n^3) |
| Space | O(n^2) auxiliary |
Pattern: Dynamic Programming
Define a state, write the transition, and stop recomputing the same subproblem. LeetCode 1246. Palindrome Removal is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Dynamic Programming.
The dynamic programming guide has the Python template for the pattern and the 481 LeetCode problems that use it.
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Frequently asked questions
- How hard is LeetCode 1246. Palindrome Removal?
- LeetCode 1246. Palindrome Removal is rated Hard on LeetCode.
- What is the time complexity of LeetCode 1246. Palindrome Removal?
- The Python solution on this page runs in O(n^3).
- What is the space complexity of LeetCode 1246. Palindrome Removal?
- The Python solution on this page uses O(n^2) auxiliary space.
- What topics does LeetCode 1246. Palindrome Removal cover?
- LeetCode 1246. Palindrome Removal is tagged Array and Dynamic Programming on LeetCode.
- Is LeetCode 1246. Palindrome Removal a premium problem?
- Yes. LeetCode 1246. Palindrome Removal is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.