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
YouTube
Examples
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)
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
/**
* 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];
}
}
/**
* 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];
}
}
