Minimum Insertion Steps to Make a String Palindrome | Blue Print | DP On Strings | Leetcode 1312
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This is the 26th Video of our Playlist "DP Concepts & Qns" by codestorywithMIK
Minimum Insertion Steps to Make a String Palindrome : Recursion + Memo - • Minimum Insertion Steps to Make a Str...
This is the eighth video of the "DP On Strings" series in this playlist.
In this video we will try to solve a good DP on strings problem : Minimum Insertion Steps to Make a String Palindrome | Using Blue Print | DP On Strings | Leetcode 1312 | DP Concepts & Qns-26 | codestorywithMIK
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Minimum Insertion Steps to Make a String Palindrome | Using Blue Print | DP On Strings | Leetcode 1312 | DP Concepts & Qns-26 | codestorywithMIK
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Approaches Summary :
Initialization:
Initialize a 2D vector dp where dp[i][j] represents the minimum number of insertions required to make the substring s[i..j] a palindrome. Initialize all values to 0.
Base Cases:
If the substring length L is 2 or more, iterate over all possible substring lengths L from 2 to n.
For each substring starting index i, calculate the end index j as i + L - 1.
Palindrome Check:
If s[i] equals s[j], then s[i..j] is already a palindrome and no insertions are needed beyond what's already considered for s[i+1..j-1]. Therefore, set dp[i][j] = dp[i + 1][j - 1].
Insertion Calculation:
If s[i] does not equal s[j], then to make s[i..j] a palindrome, calculate the minimum insertions required. This involves either:
Inserting the character at j before i (i.e., making s[i..j-1] a palindrome), or
Inserting the character at i after j (i.e., making s[i+1..j] a palindrome).
Update dp[i][j] to 1 + min(dp[i + 1][j], dp[i][j - 1]), representing the minimum of the two insertion scenarios plus one.
Result:
The result for the entire string s is stored in dp[0][n - 1], where n is the length of s. This represents the minimum number of insertions needed to make the entire string s a palindrome.
Time Complexity:
The nested loops iterate over all substrings of s, resulting in a time complexity of O(n^2), where n is the length of the string s.
Each substring length is processed in constant time operations, ensuring efficient computation.
Space Complexity:
The space complexity is also O(n^2) due to the use of the 2D vector dp to store results for all substrings of s.
✨ Timelines✨
00:00 - Introduction
00:10 - Motivation
01:23 - Problem Explanation
04:00 - Example Tree Diagram
11:45 - Bottom Up (Blue Print)
23:52 - Coding it up
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