Skip to content
Programming101
Programming101

Learn everything about programming

  • Home
  • CS Subjects
    • IoT – Internet of Things
    • Digital Communication
    • Human Values
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
  • HackerRank Solutions
    • HackerRank Algorithms Solutions
    • HackerRank C problems solutions
    • HackerRank C++ problems solutions
    • HackerRank Java problems solutions
    • HackerRank Python problems solutions
Programming101
Programming101

Learn everything about programming

Leetcode Distinct Subsequences problem solution

YASH PAL, 31 July 2024

In this Leetcode Distinct Subsequences problem solution we have 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). It is guaranteed the answer fits on a 32-bit signed integer.

Leetcode Distinct Subsequences problem solution

Problem solution in Python.

class Solution:
    def numDistinct(self, s: str, t: str) -> int:
        dp = [1] * (len(s) + 1)
        for r, t_char in enumerate(t, 1):
            prev, dp[0] = dp[0], 0
            for c, s_char in enumerate(s, 1):
                curr = dp[c]
                dp[c] = dp[c-1] + (t_char == s_char and prev)
                prev = curr
        return dp[-1]

Problem solution in Java.

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

Problem solution in C++.

class Solution {
public:
int numDistinct(string s, string t)
{
int n = t.size(), m = s.size();

    if(n>m)return 0;
    
    vector<vector<unsigned int>> dp(n,vector<unsigned int>(m,0));
    
    for(int x=0; x<m; x++)
    {
        if(t[0] == s[x])
        {
            dp[0][x] = dp[0][max(x-1,0)]+1;
        }
        else
        {
            dp[0][x] = dp[0][max(x-1,0)];
        }
    }
    
    for(int x=1; x<n; x++)
    {
        for(int y=x; y<m; y++)
        {
            if(t[x] == s[y])
            {
           
                    
                dp[x][y] = dp[x-1][y-1]+dp[x][y-1];
            }
            else
            {
                dp[x][y] = dp[x][y-1];
            }
        }
    }

    
    return dp[n-1][m-1];
    
}
};

Problem solution in C.

int numDistinct(char* s, char* t) {
	int len1, len2;
	if((len1 = strlen(s)) == 0 || (len2 = strlen(t)) == 0){
		return 0;
	}

	int dp[len2], i, j;
	int cur = 0;
	memset(dp, 0, sizeof(int) * len2);

	for(i = 0; i < len1; ++i){
		if(cur != len2){
			if(cur == 0){
				if(s[i] == t[cur]){
					dp[cur++] = 1;
				}
				continue;
			}

			for(j = cur; j > -1; --j){
				if(t[j] == s[i]){
				    if(j == 0){
				        dp[j] = dp[j] + 1;
				        continue;
				    }
					dp[j] = dp[j] + 1 * dp[j-1];
					if(j == cur){
						cur++;
					}
				}
			}
		}
		else{
			for(j = cur - 1; j > -1; --j){
				if(t[j] == s[i]){
				    if(j == 0){
				        dp[j] = dp[j] + 1;
				        continue;
				    }
					dp[j] = dp[j] + 1 * dp[j-1];
				}
			}
		}
	}

	return dp[len2-1];
}

coding problems

Post navigation

Previous post
Next post
  • HackerRank Separate the Numbers solution
  • How AI Is Revolutionizing Personalized Learning in Schools
  • GTA 5 is the Game of the Year for 2024 and 2025
  • Hackerrank Day 5 loops 30 days of code solution
  • Hackerrank Day 6 Lets Review 30 days of code solution
©2025 Programming101 | WordPress Theme by SuperbThemes