Number of Ways to Form a Target String Given a Dictionary — C++

// Time:  O(l * (w + n)), l is the length of a word, w is the number of words, n is the length of target
// Space: O(n)

// optimized from Solution2
class Solution {
public:
    int numWays(vector<string>& words, string target) {
        static const int MOD = 1e9 + 7;

        vector<uint64_t> dp(size(target) + 1);  // use uint64_t to avoid overflow
        dp[0] = 1;
        for (int i = 0; i < size(words[0]); ++i) {
            vector<int> count(26);
            for (const auto& w : words) {
                ++count[w[i] - 'a'];
            }
            for (int j = size(target) - 1; j >= 0; --j) {
                dp[j + 1] += dp[j] * count[target[j] - 'a'] % MOD;
            }
        }
        return dp.back() % MOD;
    }
};

// Time:  O(l * (w + n)), l is the length of a word, w is the number of words, n is the length of target
// Space: O(n)
class Solution2 {
public:
    int numWays(vector<string>& words, string target) {
        static const int MOD = 1e9 + 7;
        
        // dp[i+1][j+1]: number of ways of target[0..j] using count[0..i].
        vector<vector<uint64_t>> dp(2, vector<uint64_t>(size(target) + 1));  // use uint64_t to avoid overflow
        dp[0][0] = dp[1][0] = 1;
        for (int i = 0; i < size(words[0]); ++i) {
            vector<int> count(26);
            for (const auto& w : words) {
                ++count[w[i] - 'a'];
            }
            for (int j = 0; j < size(target); ++j) {
                dp[(i + 1) % 2][j + 1] = dp[i % 2][j + 1] + dp[i % 2][j] * count[target[j] - 'a'] % MOD;
            }
        }
        return dp[size(words[0]) % 2].back() % MOD;
    }
};