What is the brute force solution for Number of Ways to Form a Target String Given a Dictionary?

The brute force approach for Number of Ways to Form a Target String Given a Dictionary is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible combination of characters from the given words to form the target string. It checks all possible ways to select characters while adhering to the constraints of the problem. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Number of Ways to Form a Target String Given a Dictionary?

Number of Ways to Form a Target String Given a Dictionary uses the following concepts: Array, String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Character Frequency Counting.

What are the interview tips for Number of Ways to Form a Target String Given a Dictionary?

Always clarify the constraints and requirements of the problem before jumping into coding. Think about how you can use dynamic programming to break down the problem into smaller subproblems. Practice explaining your thought process as you code — it helps in interviews.

What are common mistakes when solving Number of Ways to Form a Target String Given a Dictionary?

Failing to account for character frequency correctly, leading to undercounting or overcounting combinations. Not properly implementing the constraints of character selection, which can lead to invalid combinations.

#1639

Number of Ways to Form a Target String Given a Dictionary

Hard

ArrayStringDynamic ProgrammingDynamic ProgrammingCharacter Frequency Counting

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n)

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Intuition

Time O(n * m)Space O(n)

The optimal solution uses dynamic programming to efficiently count the ways to form the target string by leveraging character frequency counts from the words. This avoids the need to generate all combinations explicitly.

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Algorithm

3 steps

1Step 1: Create a frequency array to count occurrences of each character in each position of the words.
2Step 2: Initialize a DP array where dp[i] represents the number of ways to form the first i characters of the target.
3Step 3: Iterate over each character in the target and update the DP array based on the frequency of matching characters from the words.

solution.py23 lines

1# Full working Python code
2def countWays(words, target):
3    MOD = 10**9 + 7
4    m, n = len(words[0]), len(target)
5    dp = [0] * (n + 1)
6    dp[0] = 1
7    freq = [[0] * 26 for _ in range(m)]
8
9    for word in words:
10        for j in range(m):
11            freq[j][ord(word[j]) - ord('a')] += 1
12
13    for i in range(1, n + 1):
14        for j in range(m):
15            if target[i - 1] == words[j][i - 1]:
16                dp[i] = (dp[i] + dp[i - 1] * freq[i - 1][ord(target[i - 1]) - ord('a')]) % MOD) % MOD
17
18    return dp[n]
19
20# Example usage
21words = ["acca", "bbbb", "caca"]
22target = "aba"
23print(countWays(words, target))

ℹ

Complexity note: The time complexity is O(n * m) because we iterate through the target string and for each character, we check the frequency of characters in the words, which takes linear time relative to the length of the target and the number of words.

1Understanding how to use dynamic programming can greatly reduce the complexity of combinatorial problems.
2Character frequency counts can help optimize the selection process in string formation problems.

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