LeetCode 1639: Number of Ways to Form a Target String Given a Dictionary Solution

LeetCode 1639 Solution Explanation

Explanation:

To solve this problem, we can use dynamic programming. We will iterate over the target string and for each character in the target, we will maintain a count of ways to form that character. We will update this count based on the characters in the words list.

For each character in the target string, we will iterate over the words list and update the count based on the count of ways to form the previous character. This is because each character in the target can be formed based on the characters formed in the previous positions. We will keep track of the count modulo 10^9 + 7 to avoid overflow.

After iterating over all characters in the target string, the final count will represent the total number of ways to form the target string.

Time Complexity:

The time complexity of this approach is O(n * m * l), where n is the length of the target string, m is the length of the words list, and l is the length of each word.

Space Complexity:

The space complexity of this approach is O(n), where n is the length of the target string.

LeetCode 1639 Solutions in Java, C++, Python

class Solution {
    public int numWays(String[] words, String target) {
        int mod = 1000000007;
        int n = target.length();
        int m = words[0].length();

        long[][] dp = new long[m + 1][n + 1];
        dp[0][0] = 1;

        for (int i = 1; i <= m; i++) {
            for (int j = 0; j <= n; j++) {
                dp[i][j] = dp[i - 1][j];
                if (j > 0) {
                    dp[i][j] += dp[i - 1][j - 1];
                }
                dp[i][j] %= mod;
            }
        }

        long[] count = new long[n + 1];
        for (String word : words) {
            for (int i = 0; i < m; i++) {
                for (int j = n - 1; j >= 0; j--) {
                    if (target.charAt(j) == word.charAt(i)) {
                        count[j + 1] += count[j] + dp[i][j];
                        count[j + 1] %= mod;
                    }
                }
            }
        }

        return (int) count[n];
    }
}

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