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Java-based LeetCode algorithm problem solutions, regularly updated
package g0101_0200.s0115_distinct_subsequences;
// #Hard #String #Dynamic_Programming #2022_06_23_Time_2_ms_(100.00%)_Space_40.2_MB_(99.14%)
public class Solution {
public int numDistinct(String text, String text2) {
if (text.length() < text2.length()) {
return 0;
}
if (text.length() == text2.length()) {
return (text.equals(text2) ? 1 : 0);
}
int move = text.length() - text2.length() + 2;
// Only finite number of character in s can occupy first position in T. Same applies for
// every character in T.
int[] dp = new int[move];
int j = 1;
int k = 1;
for (int i = 0; i < text2.length(); i++) {
boolean firstMatch = true;
for (; j < move; j++) {
if (text2.charAt(i) == text.charAt(i + j - 1)) {
if (firstMatch) {
// Keep track of first match. To avoid useless comparisons on next
// iteration.
k = j;
firstMatch = false;
}
if (i == 0) {
dp[j] = 1;
}
dp[j] += dp[j - 1];
} else {
dp[j] = dp[j - 1];
}
}
// No match found for current character of t in s. No point in checking others.
if (dp[move - 1] == 0) {
return 0;
}
j = k;
}
return dp[move - 1];
}
}