Introduction

In this blog post, we will explore how to solve the LeetCode problem "1639" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a list of strings of the same length words and a string target. Your task is to form target using the given words under the following rules: target should be formed from left to right. To form the ith character (0-indexed) of target, you can choose the kth character of the jth string in words if target[i] = words[j][k]. Once you use the kth character of the jth string of words, you can no longer use the xth character of any string in words where x <= k. In other words, all characters to the left of or at index k become unusuable for every string. Repeat the process until you form the string target. Notice that you can use multiple characters from the same string in words provided the conditions above are met. Return the number of ways to form target from words. Since the answer may be too large, return it modulo 109 + 7. Example 1: Input: words = ["acca","bbbb","caca"], target = "aba" Output: 6 Explanation: There are 6 ways to form target. "aba" -> index 0 ("acca"), index 1 ("bbbb"), index 3 ("caca") "aba" -> index 0 ("acca"), index 2 ("bbbb"), index 3 ("caca") "aba" -> index 0 ("acca"), index 1 ("bbbb"), index 3 ("acca") "aba" -> index 0 ("acca"), index 2 ("bbbb"), index 3 ("acca") "aba" -> index 1 ("caca"), index 2 ("bbbb"), index 3 ("acca") "aba" -> index 1 ("caca"), index 2 ("bbbb"), index 3 ("caca") Example 2: Input: words = ["abba","baab"], target = "bab" Output: 4 Explanation: There are 4 ways to form target. "bab" -> index 0 ("baab"), index 1 ("baab"), index 2 ("abba") "bab" -> index 0 ("baab"), index 1 ("baab"), index 3 ("baab") "bab" -> index 0 ("baab"), index 2 ("baab"), index 3 ("baab") "bab" -> index 1 ("abba"), index 2 ("baab"), index 3 ("baab") Constraints: 1 <= words.length <= 1000 1 <= words[i].length <= 1000 All strings in words have the same length. 1 <= target.length <= 1000 words[i] and target contain only lowercase English letters.

Explanation

Here's the solution to the problem:

• High-Level Approach:

  • Use dynamic programming to store the number of ways to form a prefix of the target string using a prefix of the words' character indices.
  • Iterate through the target string and the possible character indices in the words. If a character in the words matches the current character in the target, add the number of ways to form the previous prefix of the target.
  • Use a frequency count of characters at each index of words to optimize.

• Complexity:

  • Runtime Complexity: O(M*N), where M is the length of the target string and N is the length of each word in words.
  • Storage Complexity: O(M*N) due to the DP table.

Code

    def num_ways(words, target):
    MOD = 10**9 + 7
    m = len(target)
    n = len(words[0])
    word_count = len(words)

    # char_counts[i][char_code] is the count of char 'char_code' at index i in all words
    char_counts = [[0] * 26 for _ in range(n)]
    for word in words:
        for i in range(n):
            char_counts[i][ord(word[i]) - ord('a')] += 1

    dp = [[0] * (n + 1) for _ in range(m + 1)]
    dp[0][0] = 1

    for i in range(m + 1):
        for j in range(n + 1):
            if i == 0 and j == 0:
                continue

            if i > 0:
                dp[i][j] = dp[i][j] % MOD
                dp[i][j] += dp[i][j-1] % MOD
                dp[i][j] = dp[i][j] % MOD
            elif j > 0:
                dp[i][j] = dp[i][j] % MOD
                dp[i][j] += dp[i][j-1] % MOD
                dp[i][j] = dp[i][j] % MOD

            if i > 0 and j > 0:
                target_char_index = ord(target[i-1]) - ord('a')
                count = char_counts[j-1][target_char_index]
                dp[i][j] = dp[i][j] % MOD
                dp[i][j] += (dp[i-1][j-1] * count) % MOD
                dp[i][j] = dp[i][j] % MOD

    return dp[m][n] % MOD