Last updated on October 9th, 2024 at 05:56 pm

Here, We see Distinct Subsequences LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Distinct Subsequences LeetCode Solution

Problem Statement

Given two strings s and t, return the number of distinct subsequences of s which equals t.

The test cases are generated so that the answer fits on a 32-bit signed integer.

Example 1:
Input: s = “rabbbit”, t = “rabbit”
Output: 3
Explanation: As shown below, there are 3 ways you can generate “rabbit” from s.
rabbbit
rabbbit
rabbbit

1. Distinct Subsequences LeetCode Solution C++

class Solution {
public:
    int numDistinct(string s, string t) {
        int m = t.length(), n = s.length();
        vector<vector<int>> dp(m + 1, vector<int> (n + 1, 0));
        for (int j = 0; j <= n; j++) dp[0][j] = 1;
        for (int j = 1; j <= n; j++)
            for (int i = 1; i <= m; i++)
                dp[i][j] = dp[i][j - 1] + (t[i - 1] == s[j - 1] ? dp[i - 1][j - 1] : 0);
        return dp[m][n];
    }
};  

2. Distinct Subsequences LeetCode Solution Java

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

3. Distinct Subsequences Solution JavaScript

var numDistinct = function (s, t) {
    if (s.length < t.length) return 0;
    [s, t] = [t, s];
    let m = s.length;
    let n = t.length;
    let aa = Array.from({ length: m + 1 }, () => new Array(n + 1).fill(0));
    for (let j = 0; j <= n; j++) {
        aa[0][j] = 1;
    }
    for (let i = 1; i <= m; i++)
        for (let j = 1; j <= n; j++)
            if (s[i - 1] === t[j - 1]) {
                aa[i][j] = aa[i][j - 1] + aa[i - 1][j - 1];
            } else {
                aa[i][j] = aa[i][j - 1];
            }
    return aa[m][n];
};

4. Distinct Subsequences Solution Python

class Solution(object):
    def numDistinct(self, s, t):
        dp = [[0] * (len(s) + 1) for _ in range(len(t) + 1)]
        for col in range(len(dp[0])):
            dp[0][col] = 1
        for x in range(1, len(s) + 1):
            for y in range(1, len(t) + 1):
                if s[x - 1] == t[y - 1]:
                    dp[y][x] = dp[y - 1][x - 1] + dp[y][x - 1]
                else:
                    dp[y][x] = dp[y][x - 1]
        return dp[-1][-1]