Given two strings s and t, return the number of distinct subsequences of s which equals t.
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 Explanation: As shown below, there are 3 ways you can generate "rabbit" from s.rabbbitrabbbitrabbbit
Example 2:
Input: s = "babgbag", t = "bag" Output: 5 Explanation: As shown below, there are 5 ways you can generate "bag" from s.babgbagbabgbagbabgbagbabgbagbabgbag
Constraints:
1 <= s.length, t.length <= 1000sandtconsist of English letters.
class Solution:
def numDistinct(self, s: str, t: str) -> int:
m, n = len(s), len(t)
f = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(m + 1):
f[i][0] = 1
for i, a in enumerate(s, 1):
for j, b in enumerate(t, 1):
f[i][j] = f[i - 1][j]
if a == b:
f[i][j] += f[i - 1][j - 1]
return f[m][n]class Solution:
def numDistinct(self, s: str, t: str) -> int:
n = len(t)
f = [1] + [0] * n
for a in s:
for j in range(n, 0, -1):
if a == t[j - 1]:
f[j] += f[j - 1]
return f[n]class Solution {
public int numDistinct(String s, String t) {
int m = s.length(), n = t.length();
int[][] f = new int[m + 1][n + 1];
for (int i = 0; i < m + 1; ++i) {
f[i][0] = 1;
}
for (int i = 1; i < m + 1; ++i) {
for (int j = 1; j < n + 1; ++j) {
f[i][j] = f[i - 1][j];
if (s.charAt(i - 1) == t.charAt(j - 1)) {
f[i][j] += f[i - 1][j - 1];
}
}
}
return f[m][n];
}
}class Solution {
public int numDistinct(String s, String t) {
int n = t.length();
int[] f = new int[n + 1];
f[0] = 1;
for (char a : s.toCharArray()) {
for (int j = n; j > 0; --j) {
char b = t.charAt(j - 1);
if (a == b) {
f[j] += f[j - 1];
}
}
}
return f[n];
}
}func numDistinct(s string, t string) int {
m, n := len(s), len(t)
dp := make([][]int, m+1)
for i := 0; i <= m; i++ {
dp[i] = make([]int, n+1)
dp[i][0] = 1
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
dp[i][j] = dp[i-1][j]
if s[i-1] == t[j-1] {
dp[i][j] += dp[i-1][j-1]
}
}
}
return dp[m][n]
}class Solution {
public:
int numDistinct(string s, string t) {
int m = s.size(), n = t.size();
unsigned long long f[m + 1][n + 1];
memset(f, 0, sizeof(f));
for (int i = 0; i < m + 1; ++i) {
f[i][0] = 1;
}
for (int i = 1; i < m + 1; ++i) {
for (int j = 1; j < n + 1; ++j) {
f[i][j] = f[i - 1][j];
if (s[i - 1] == t[j - 1]) {
f[i][j] += f[i - 1][j - 1];
}
}
}
return f[m][n];
}
};class Solution {
public:
int numDistinct(string s, string t) {
int n = t.size();
unsigned long long f[n + 1];
memset(f, 0, sizeof(f));
f[0] = 1;
for (char& a : s) {
for (int j = n; j; --j) {
char b = t[j - 1];
if (a == b) {
f[j] += f[j - 1];
}
}
}
return f[n];
}
};func numDistinct(s string, t string) int {
m, n := len(s), len(t)
f := make([][]int, m+1)
for i := range f {
f[i] = make([]int, n+1)
}
for i := 0; i <= m; i++ {
f[i][0] = 1
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
f[i][j] = f[i-1][j]
if s[i-1] == t[j-1] {
f[i][j] += f[i-1][j-1]
}
}
}
return f[m][n]
}func numDistinct(s string, t string) int {
n := len(t)
f := make([]int, n+1)
f[0] = 1
for _, a := range s {
for j := n; j > 0; j-- {
if b := t[j-1]; byte(a) == b {
f[j] += f[j-1]
}
}
}
return f[n]
}function numDistinct(s: string, t: string): number {
const m = s.length;
const n = t.length;
const f: number[][] = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
for (let i = 0; i <= m; ++i) {
f[i][0] = 1;
}
for (let i = 1; i <= m; ++i) {
for (let j = 1; j <= n; ++j) {
f[i][j] = f[i - 1][j];
if (s[i - 1] === t[j - 1]) {
f[i][j] += f[i - 1][j - 1];
}
}
}
return f[m][n];
}function numDistinct(s: string, t: string): number {
const n = t.length;
const f: number[] = new Array(n + 1).fill(0);
f[0] = 1;
for (const a of s) {
for (let j = n; j; --j) {
const b = t[j - 1];
if (a === b) {
f[j] += f[j - 1];
}
}
}
return f[n];
}impl Solution {
#[allow(dead_code)]
pub fn num_distinct(s: String, t: String) -> i32 {
let n = s.len();
let m = t.len();
let mut dp: Vec<Vec<u64>> = vec![vec![0; m + 1]; n + 1];
// Initialize the dp vector
for i in 0..=n {
dp[i][0] = 1;
}
// Begin the actual dp process
for i in 1..=n {
for j in 1..=m {
dp[i][j] = if s.as_bytes()[i - 1] == t.as_bytes()[j - 1] {
dp[i - 1][j] + dp[i - 1][j - 1]
} else {
dp[i - 1][j]
}
}
}
dp[n][m] as i32
}
}