1143. Longest Common Subsequence

# Medium

Very similar to #72, two-sequence DP problem

Subsequence: 子序列，可以跳着来

Substring: 子串，必须元素连续着

Solution:

1. Initialize maxCommon[i][j], which means longest common substring between text1[:i] and text2[:j]. Empty element is necessarily needed

2. Three cases:

   1. maxCommon[i-1][j]. eg, abc+d and ace

   2. maxCommon[i][j-1]. eg, abc and ac+d

   3. If text1[i] = text2[j], then maxCommon[i][j] + 1 common char. eg, abc+d and ace+d

   Compare to assign biggest number to maxCommon[i][j]

Python

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        # initilize maxCommon
        maxCommon = [[-1]*(len(text2)+1) for i in range(len(text1)+1)]
        for i in range(len(text1)+1):
            maxCommon[i][0] = 0
        for j in range(len(text2)+1):
            maxCommon[0][j] = 0
            
        # compute
        for i in range(1, len(text1)+1):
            for j in range(1, len(text2)+1):
                maxCommon[i][j] = max(maxCommon[i-1][j], maxCommon[i][j-1])
                if text1[i-1] == text2[j-1]:
                    maxCommon[i][j] = max(maxCommon[i][j], maxCommon[i-1][j-1]+1)
                    
        return maxCommon[len(text1)][len(text2)]

Java

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int[][] longest = new int[text1.length()+1][text2.length()+1];
        for(int i = 0; i <= text1.length(); i ++)
            Arrays.fill(longest[i], 0);
        
        for(int i = 1; i <= text1.length(); i ++)
            for(int j = 1; j <= text2.length(); j ++) {
                if(text1.charAt(i-1) == text2.charAt(j-1))
                    longest[i][j] = longest[i-1][j-1] + 1;
                else
                    longest[i][j] = Math.max(longest[i-1][j], longest[i][j-1]);
            }
        return longest[text1.length()][text2.length()];
    }
}