115. Distinct Subsequences

Description

Given a string S and a string T, count the number of distinct subsequences of S which equals T.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).

It's guaranteed the answer fits on a 32-bit signed integer.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input: S = "rabbbit", T = "rabbit"

Output: 3

Explanation: As shown below, there are 3 ways you can generate "rabbit" from S.

(The caret symbol ^ means the chosen letters)

Example 2

Input: S = "babgbag", T = "bag"

Output: 5

Explanation: As shown below, there are 5 ways you can generate "bag" from S.

(The caret symbol ^ means the chosen letters)

Solutions

Solution 1

/**
 * Time complexity : O(M*N) where M is the length of S and N is the length of T.
 * Space complexity : O(M*N) which is occupied by the 2D dp array that we create.
 */

class Solution {
    public int numDistinct(String s, String t) {
        int m = s.length(), n = t.length();
        int[][] dp = new int[m+1][n+1];
        
        for(int i = 0; i <= m; i++) {
            dp[i][0] = 1;
        }
        
        for(int i = 1; i <= m; i++) {
            for(int j = 1; j <= n; j++) {
                if(s.charAt(i-1) == t.charAt(j-1)) {
                    dp[i][j] = dp[i-1][j] + dp[i-1][j-1];
                } else {
                    dp[i][j] = dp[i-1][j];
                }
            }
        }
        
        return dp[m][n];
    }
}

Solution 2

/**
 * Time complexity : O(M*N) where M is the length of S and N is the length of T.
 * Space complexity : O(N) since we are using a single array which is the 
 *    size of the string T.
 */

class Solution {
    public int numDistinct(String s, String t) {
        int m = s.length(), n = t.length();
        int[] dp = new int[n+1];
        dp[0] = 1;
        
        for(int i = 1; i <= m; i++) {
            for(int j = n; j > 0; j--) {
                if(s.charAt(i-1) == t.charAt(j-1)) {
                    dp[j] += dp[j-1];
                }
            }
        }
        
        return dp[n];
    }
}

Follow up