Last updated on January 18th, 2025 at 03:28 am

Here, we see a Distinct Subsequences LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.

List of all LeetCode Solution

Topics

Dynamic Programming, String

Companies

Yelp

Level of Question

Hard

Distinct Subsequences LeetCode Solution

1. 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

2. Coding Pattern Used in Solution

The provided code in all four languages (C++, Java, JavaScript, and Python) follows the “Dynamic Programming” pattern. Specifically, it solves a “2D DP Table Problem” where the solution is built incrementally using a 2D table (dp or mem or aa) to store intermediate results.

3. Code Implementation in Different Languages

3.1 Distinct Subsequences 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];
    }
};  

3.2 Distinct Subsequences 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()];
}
}

3.3 Distinct Subsequences 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];
};

3.4 Distinct Subsequences 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]

4. Time and Space Complexity

	Time Complexity	Space Complexity
C++	O(m * n)	O(m * n)
Java	O(m * n)	O(m * n)
JavaScript	O(m * n)	O(m * n)
Python	O(m * n)	O(m * n)

• The code uses a Dynamic Programming approach to solve the problem of counting distinct subsequences.
• It builds a 2D DP table to store intermediate results and iteratively fills it based on character matches between s and t.
• The time complexity is O(m * n), and the space complexity is O(m * n) for all four implementations.