How many times one string appears in another

def num_distinct(s, t):
    ways = [1] + [0] * len(t)            # ways[j]: t[:j] inside the prefix of s seen so far
    for ch in s:
        for j in range(len(t), 0, -1):   # downwards: ways[j - 1] is still the previous row
            if t[j - 1] == ch:
                ways[j] += ways[j - 1]   # use this character of s for t[j - 1]
    return ways[len(t)]

#include <string>
#include <vector>

long long solve(std::string s, std::string t) {
    std::vector<long long> ways(t.size() + 1, 0);
    ways[0] = 1;
    for (char ch : s)
        for (int j = t.size(); j >= 1; --j)          // downwards
            if (t[j - 1] == ch) ways[j] += ways[j - 1];
    return ways[t.size()];
}

class Solution {
    public long numDistinct(String s, String t) {
        long[] ways = new long[t.length() + 1];
        ways[0] = 1;
        for (char ch : s.toCharArray())
            for (int j = t.length(); j >= 1; j--)        // downwards
                if (t.charAt(j - 1) == ch) ways[j] += ways[j - 1];
        return ways[t.length()];
    }
}

function numDistinct(s, t) {
  const ways = [1, ...new Array(t.length).fill(0)];
  for (const ch of s)
    for (let j = t.length; j >= 1; j--) if (t[j - 1] === ch) ways[j] += ways[j - 1];   // downwards
  return ways[t.length];
}

fn solve(s: &str, t: &str) -> i64 {
    let t: Vec<char> = t.chars().collect();
    let mut ways = vec![0i64; t.len() + 1];
    ways[0] = 1;
    for ch in s.chars() {
        for j in (1..=t.len()).rev() {                   // downwards
            if t[j - 1] == ch {
                ways[j] += ways[j - 1];
            }
        }
    }
    ways[t.len()]
}

solve :: String -> String -> Int
solve s t = last (foldl step (1 : map (const 0) t) s)
  where
    -- the new row: ways[0] stays 1; ways[j] adds the old ways[j - 1] when t[j - 1] matches
    step ways ch = head ways : zipWith3 (\c old prev -> if c == ch then old + prev else old) t (tail ways) ways