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Distinct Subsequences LeetCode Solution

Problem – Distinct Subsequences LeetCode Solution

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

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

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

Example 2:

Input: s = "babgbag", t = "bag"
Output: 5
As shown below, there are 5 ways you can generate "bag" from S.


  • 1 <= s.length, t.length <= 1000
  • s and t consist of English letters.

Distinct Subsequences LeetCode Solution in Java

public int numDistinct(String S, String T) {
    // array creation
    int[][] mem = new int[T.length()+1][S.length()+1];

    // filling the first row: with 1s
    for(int j=0; j<=S.length(); j++) {
        mem[0][j] = 1;
    // the first column is 0 by default in every other rows but the first, which we need.
    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()];

Distinct Subsequences LeetCode Solution in C++

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

Distinct Subsequences LeetCode Solution in Python

class Solution:
    def numDistinct(self, s, t):
        def dp(i, j):
            if i == -1: return j == -1
            if j == -1: return j == -1
            return dp(i-1, j) + (s[i] == t[j]) * dp(i-1, j-1)
        return dp(len(s) - 1, len(t) - 1)
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