LCSOfThreeStrings.java

Given 3 strings A, B and C, the task is to find the longest common sub-sequence in all three given sequences.

Example 1:

Input:
A = "geeks", B = "geeksfor", 
C = "geeksforgeeks"
Output: 5
Explanation: "geeks"is the longest common
subsequence with length 5.
​Example 2:

Input: 
A = "abcd", B = "efgh", C = "ijkl"
Output: 0
Explanation: There's no common subsequence
in all the strings.

Your Task:
You don't need to read input or print anything. Your task is to complete the function LCSof3() which takes the strings A, B, C and their lengths n1, n2, n3 as input and returns the length of the longest common subsequence in all the 3 strings.


Expected Time Complexity: O(n1*n2*n3).
Expected Auxiliary Space: O(n1*n2*n3).


Constraints:
1<=n1, n2, n3<=20


class Solution 
{ 
    int LCSof3(String a, String b, String c, int n1, int n2, int n3) 
    { 
        // code here
        // we just expand lcs of 2 string dp approach to 3 string
        // there we uses 2d dp, here we expand it into 3d dp
        int [][][] dp = new int [n1 + 1][n2 + 1][n3 + 1];
        for (int i=0; i<=n1; i++) {
            for (int j=0; j<=n2; j++) {
                for (int k=0; k<=n3; k++) {
                    if (i == 0 || j == 0 || k == 0) {
                        dp[i][j][k] = 0;
                    }
                    else if (a.charAt(i - 1) == b.charAt(j - 1) && b.charAt(j - 1) == c.charAt(k - 1)) {
                        dp[i][j][k] = 1 + dp[i - 1][j - 1][k - 1];
                    }
                    else {
                        dp[i][j][k] = Math.max(dp[i - 1][j][k], Math.max(dp[i][j - 1][k], dp[i][j][k - 1]));
                    }
                }
            }
        }
        return dp[n1][n2][n3];
    }
}