115. Distinct Subsequences
Last updated
Last updated
/**
* 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];
}
}