Leetcode #1312: Minimum Insertion Steps to Make a String Palindrome
In this guide, we solve Leetcode #1312 Minimum Insertion Steps to Make a String Palindrome 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
Given a string s. In one step you can insert any character at any index of the string.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: String, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: s = "zzazz"
Output: 0
Explanation: The string "zzazz" is already palindrome we do not need any insertions.
Python Solution
class Solution:
def minInsertions(self, s: str) -> int:
def dfs(i: int, j: int) -> int:
if i >= j:
return 0
if s[i] == s[j]:
return dfs(i + 1, j - 1)
return 1 + min(dfs(i + 1, j), dfs(i, j - 1))
return dfs(0, len(s) - 1)
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
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.