Length of Longest Common Subsequence (LCS) and Longest common subsequence string leetcode
Tags: java, dp, algorithms, datastructure
Published: 2022-07-18 06:16:00

Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.

A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

For example, "ace" is a subsequence of "abcde".
A common subsequence of two strings is a subsequence that is common to both strings.

 

**Example 1**:

```
Input: text1 = "abcde", text2 = "ace" 
Output: 3  
Explanation: The longest common subsequence is "ace" and its length is 3.
```
**Solution**:
Bottom up Approach(Memoization) : 
Time complexity : `O(m*n)` where is `m` and `n` is the length of two strings `a` and `b`
space complexity : o(m*n) for using `2d dp` and `O(m+n)` for auxiliary stack space because in worst case we will make `m+n` recursive calls.
```java
class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
         // we will start with the last index if its a match then we will decrement both the index
        //else we will decrement text1 index keeping text2 index same and in second method call we will decrement text2 index keeping the text1 index same 
        // by this we will cover all the possibility
        // and we will be  able to get substring with the largest length common in both the Strings
        // lets optimize with dp
        int dp[][] = new int[text1.length()][text2.length()];
        for(int row[]: dp) Arrays.fill(row,-1);
        return findLcsLength(text1,text2,text1.length()-1,text2.length()-1,dp);
    }
    public int findLcsLength(String a, String b, int i,int j,int dp[][]){
        if(i<0 || j<0) return 0;
        if(dp[i][j]!=-1) return dp[i][j];
        if(a.charAt(i) ==b.charAt(j)){
            return dp[i][j] =  1 + findLcsLength(a,b,i-1,j-1,dp);
        }
        else {
            return dp[i][j]= 0+Integer.max(findLcsLength(a,b,i-1,j,dp),findLcsLength(a,b,i,j-1,dp));
        }
        
    }
}
```

**Tabulation**

```java
class Solution {
    public int longestCommonSubsequence(String str1, String str2) {
	 int dp[][] = new int[str1.length()+1][str2.length()+1];
	 for(int i=0;i<=str1.length();i++){
				dp[i][0] =0;
			}
			for( int i =0;i<=str2.length();i++){
				dp[0][i] = 0;
			}
			
			for( int i =1;i<=str1.length();i++){
				for(int j =1;j<=str2.length();j++){
					if(str1.charAt(i-1)==str2.charAt(j-1)){
						dp[i][j] = 1 + dp[i-1][j-1];
					}
					else dp[i][j] = Integer.max(dp[i][j-1],dp[i-1][j]);
				}
			}
	return dp[str1.length()][str2.length()];
	}
}
```
**If we have to print the longest common subsequence string, just add the following in the above code.**
```java
int p = str1.length(), q = str2.length();
        while(p>0 && q>0){
            if(str1.charAt(p-1) == str2.charAt(q-1)){
                str = str1.charAt(p-1)+ str;
                p--;
                q--;
                
            }
            else if(dp[p][q-1] > dp[p-1][q]){
                q--;
            }
            else p--;
        }
        System.out.println(str);
```