Longest Common Subsequence (LeetCode 1143) | Full Solution with a natural explanation 

Great explanation. I have seen many dp solution but nobody explanation as clearly as you did

great approach, thanks teacher, the best explanation I have ever seen

Thanks for clearly explaining DP Bottom up approach in normal array traversal instead of reverse traversal. Thanks!

your way of teaching is just great sir , Thanku !!

Best explanation ever!

you are such a nice teacher)

Great explanation sir . Please start a playlist of hard DSA problems.

Thanks a lot sir ..u make hard ques in easy way

Nice explanation bro👏

Could you also do a topdown approach please

This is no less than excellent !

nice explanation

amazing and detail explained

while(true){ print("great explenation") }

Thanks!

Thank you so much brother!

Thanks

class Solution { public int longestCommonSubsequence(String text1, String text2) { // Ensure text1 is the longer string to minimize space usage if (text2.length() > text1.length()) { return longestCommonSubsequence(text2, text1); } // Convert strings to character arrays for easy access char[] s1 = text1.toCharArray(); char[] s2 = text2.toCharArray(); // Create two arrays to store the lengths of longest common subsequences int[] previousRow = new int[s2.length + 1]; int[] currentRow = new int[s2.length + 1]; // Loop through each character in s1 for (int i = 0; i < s1.length; i++) { // Loop through each character in s2 for (int j = 0; j < s2.length; j++) { // If characters match, increment the length from the previous row and previous column if (s1[i] == s2[j]) { currentRow[j + 1] = previousRow[j] + 1; } else { // If characters don't match, take the maximum length by ignoring one character either from s1 or s2 currentRow[j + 1] = Math.max(currentRow[j], previousRow[j + 1]); } } // Swap rows: currentRow becomes previousRow for the next iteration int[] temp = previousRow; previousRow = currentRow; currentRow = temp; } // The length of the longest common subsequence is in the last element of previousRow return previousRow[s2.length]; } public static void main(String[] args) { Solution solution = new Solution(); // Test cases System.out.println(solution.longestCommonSubsequence("abcde", "ace")); // Output: 3 System.out.println(solution.longestCommonSubsequence("abc", "abc")); // Output: 3 System.out.println(solution.longestCommonSubsequence("abc", "def")); // Output: 0 } } this is true dynamic solution

best

No recursive brute force for this??????????????

Sir, in this problem you are iterating from i=1, j=1. but, in the given example the memoization table has dp[0][1]=1. how does it got 1 in program

sir can you explain the backtracking part? To print longest common subsequence

Hello Sir, can you make a good video about iteration and loops ? And I hope you will not give the example " You need to repeat 10 times hello, how would you do it ? " ..

cudo!

sosote rocafuerte