What is the time complexity of Minimum Insertion Steps to Make a String Palindrome?

The optimal solution has a time complexity of O(n^2).

What is the space complexity of Minimum Insertion Steps to Make a String Palindrome?

The space complexity is O(n^2).

What data structures are used to solve Minimum Insertion Steps to Make a String Palindrome?

This problem uses String, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Insertion Steps to Make a String Palindrome?

Insight 1: The key insight is realizing that the problem can be reframed as finding the Longest Palindromic Subsequence (LPS) of the given string. The number of insertions needed is simply the difference between the length of the original string and the length of its LPS. Insight 2: Dynamic Programming is the appropriate technique to efficiently compute the length of the LPS. A recursive approach with memoization avoids recomputation of overlapping subproblems, leading to an optimized solution.

Which companies ask LeetCode 1312. Minimum Insertion Steps to Make a String Palindrome in interviews?

This problem is frequently asked at Arcesium.

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Minimum Insertion Steps to Make a String Palindrome - LeetCode 1312 Solution

Minimum Insertion Steps to Make a String Palindrome - Complete Solution Guide

Minimum Insertion Steps to Make a String Palindrome is LeetCode problem 1312, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a string s . In one step you can insert any character at any index of the string. Return the minimum number of steps to make s palindrome. A Palindrome String is one that reads the same backward as well as forward. Example 1: Input: s = "zzazz" Output: 0 Explanation: The string "zzazz" is already palindrome we do not need any insertions. Example 2: Input: s = "mbadm" Output: 2 Explanation: String can be "mbdadbm" or "mdbabdm". Example 3: Input: s = "leetcode" Output: 5 Explanation: Inserti

Detailed Explanation

The problem asks us to find the minimum number of character insertions needed to transform a given string `s` into a palindrome. A palindrome is a string that reads the same forwards and backward. We are allowed to insert characters at any index within the string. The goal is to determine the fewest such insertions to achieve palindromicity.

Solution Approach

The provided code calculates the minimum insertions needed by first finding the length of the Longest Palindromic Subsequence (LPS) using dynamic programming. The difference between the original string's length and the LPS length gives the minimum number of insertions required to make the string a palindrome. The dynamic programming approach builds a solution from smaller subproblems: determining the LPS of smaller substrings and then using these results to solve larger substrings until the entire string is considered. Memoization (using `lru_cache` in Python, `HashMap` in Java, `vector<vector<int>>` in C++, and `int**` in C) avoids recomputing already solved subproblems, improving efficiency.

Step-by-Step Algorithm

Step 1: Initialize a memoization table to store the results of subproblems, preventing redundant calculations.
Step 2: Define a recursive function `lps(i, j)` to compute the length of the longest palindromic subsequence of the substring `s[i...j]`.
Step 3: Base cases: If `i > j`, the substring is empty, so the LPS length is 0. If `i == j`, the substring contains a single character, so the LPS length is 1.
Step 4: Recursive step: If `s[i]` and `s[j]` are equal, the LPS includes these characters, so `lps(i, j) = 2 + lps(i + 1, j - 1)`. If `s[i]` and `s[j]` are not equal, the LPS excludes either `s[i]` or `s[j]`, so `lps(i, j) = max(lps(i + 1, j), lps(i, j - 1))`.
Step 5: Memoize the result of `lps(i, j)` before returning it.
Step 6: Call the `lps` function with `i = 0` and `j = n - 1` to compute the LPS of the entire string.
Step 7: Return `n - lps(0, n - 1)`, which is the minimum number of insertions needed.

Key Insights

Insight 1: The key insight is realizing that the problem can be reframed as finding the Longest Palindromic Subsequence (LPS) of the given string. The number of insertions needed is simply the difference between the length of the original string and the length of its LPS.
Insight 2: Dynamic Programming is the appropriate technique to efficiently compute the length of the LPS. A recursive approach with memoization avoids recomputation of overlapping subproblems, leading to an optimized solution.
Insight 3: The base cases for the dynamic programming recurrence are crucial. An empty string or a single-character string are inherently palindromic, defining the initial conditions for the LPS calculation.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: Arcesium.

Minimum Insertion Steps to Make a String Palindrome | LeetCode 1312

Hard

String Dynamic Programming

Time: O(n^2)Space: O(n^2)

Solution Code

Available in:

import java.util.HashMap;

class Solution {
    public int minInsertions(String s) {
        int n = s.length();

        // The problem is equivalent to finding the difference between
        // the string length and the length of its longest palindromic subsequence (LPS).
        // n - lps(s)
        // We can find the length of the LPS using dynamic programming.
        // Let lps(i, j) be the length of the LPS of s[i...j].
        //
        // Recurrence for LPS:
        // if s[i] == s[j]:
        //   lps(i, j) = 2 + lps(i + 1, j - 1)
        // else:
        //   lps(i, j) = max(lps(i + 1, j), lps(i, j - 1))
        //
        // Base cases:
        // lps(i, i) = 1
        // lps(i, j) = 0 if i > j

        HashMap<String, Integer> memo = new HashMap<>();

        class Helper {
            public int lps(int i, int j, String s) {
                if (i > j) {
                    return 0;
                }
                if (i == j) {
                    return 1;
                }

                String key = i + "," + j;
                if (memo.containsKey(key)) {
                    return memo.get(key);
                }
                
                if (s.charAt(i) == s.charAt(j)) {
                    int result = 2 + lps(i + 1, j - 1, s);
                    memo.put(key, result);
                    return result;
                } else {
                    int result = Math.max(lps(i + 1, j, s), lps(i, j - 1, s));
                    memo.put(key, result);
                    return result;
                }
            }
        }
        
        Helper helper = new Helper();

        return n - helper.lps(0, n - 1, s);
    }
}

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Problem Info

Description

Problem 1312: Minimum Insertion Steps to Make a String Palindrome - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

Arcesium

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1312

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize a memoization table to store the results of subproblems, preventing redundant calculations.
2
Step 2: Define a recursive function `lps(i, j)` to compute the length of the longest palindromic subsequence of the substring `s[i...j]`.
3
Step 3: Base cases: If `i > j`, the substring is empty, so the LPS length is 0. If `i == j`, the substring contains a single character, so the LPS length is 1.
4
Step 4: Recursive step: If `s[i]` and `s[j]` are equal, the LPS includes these characters, so `lps(i, j) = 2 + lps(i + 1, j - 1)`. If `s[i]` and `s[j]` are not equal, the LPS excludes either `s[i]` or `s[j]`, so `lps(i, j) = max(lps(i + 1, j), lps(i, j - 1))`.
5
Step 5: Memoize the result of `lps(i, j)` before returning it.
6
Step 6: Call the `lps` function with `i = 0` and `j = n - 1` to compute the LPS of the entire string.
7
Step 7: Return `n - lps(0, n - 1)`, which is the minimum number of insertions needed.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is derived from the number of subproblems solved by the dynamic programming algorithm. Since there are O(n^2) subproblems and each subproblem is solved in O(1) time due to memoization, the total time complexity is O(n^2). The space complexity is also O(n^2) because we are storing the solutions to O(n^2) subproblems in the memoization table.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Iterative Dynamic Programming

Instead of using a recursive approach with memoization, we can use an iterative dynamic programming approach with a bottom-up tabular method. We can create a 2D array `dp` of size `n x n` where `dp[i][j]` stores the length of the LPS of the substring `s[i...j]`. We iterate through the array filling the `dp` values based on the same recurrence relation.

Time: O(n^2)Space: O(n^2)

Trade-offs: Iterative DP can sometimes be slightly faster than recursive DP because it avoids the overhead of function calls. However, the core logic and complexity remain the same. It can also be slightly more complex to implement and understand compared to the recursive approach with memoization.

✨

Key Insights

1Insight 1: The key insight is realizing that the problem can be reframed as finding the Longest Palindromic Subsequence (LPS) of the given string. The number of insertions needed is simply the difference between the length of the original string and the length of its LPS.
2Insight 2: Dynamic Programming is the appropriate technique to efficiently compute the length of the LPS. A recursive approach with memoization avoids recomputation of overlapping subproblems, leading to an optimized solution.
3Insight 3: The base cases for the dynamic programming recurrence are crucial. An empty string or a single-character string are inherently palindromic, defining the initial conditions for the LPS calculation.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string. Although the problem statement specifies `1 <= s.length <= 500`, an empty string should ideally be handled correctly, though not explicitly tested. The code implicitly handles it in the base case where i > j, returning 0, which leads to a correct answer of 0 since n - 0 = 0.
Edge case 2: String is already a palindrome. The algorithm correctly calculates the LPS to be the length of the string, resulting in 0 insertions needed.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to memoize the results of the `lps` function. This leads to exponential time complexity as the same subproblems are recomputed multiple times.

✗Common mistake 2: Incorrectly defining the base cases for the recursion. Ensure that the base cases accurately reflect the LPS length for empty or single-character substrings.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: DNA sequencing: Finding the longest common subsequence between two DNA strands to identify similarities or evolutionary relationships. The LPS is a specific instance when the two sequences are a string and its reverse.

🏦Real-world scenario 2: Text editing: Implementing a 'diff' utility that highlights the differences between two versions of a document. The longest common subsequence (related to LPS) can be used to identify the unchanged portions of the text.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 23, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Insertion Steps to Make a String Palindrome"?

For this hard-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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