How do you solve Construct String with Minimum Cost on LeetCode?

Construct String with Minimum Cost (LeetCode #3213) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n) space complexity. Key insight: Dynamic programming can significantly reduce the number of calculations needed by storing intermediate results.

What is the time complexity of Construct String with Minimum Cost?

The optimal solution for Construct String with Minimum Cost runs in O(n * m) time and uses O(n) space. The time complexity is O(n * m) where n is the length of the target and m is the number of words, as we check each word for each position in the target.

What is the brute force solution for Construct String with Minimum Cost?

The brute force approach for Construct String with Minimum Cost is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying all possible combinations of the words to see if we can construct the target string. This method is straightforward but inefficient, as it checks every possible way to form the target. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Construct String with Minimum Cost?

Construct String with Minimum Cost uses the following concepts: Array, String, Dynamic Programming, Suffix Array. The recommended patterns to study are: Dynamic Programming, String Matching.

What are the interview tips for Construct String with Minimum Cost?

Always explain your thought process clearly; it shows your understanding of the problem. Try to identify patterns in the problem that can lead to a more efficient solution. Practice explaining your code as you write it; this helps in interviews.

What are common mistakes when solving Construct String with Minimum Cost?

Forgetting to initialize the DP array correctly, which can lead to incorrect results. Not considering all possible combinations of words when constructing the target.

#3213

Construct String with Minimum Cost

Hard

ArrayStringDynamic ProgrammingSuffix ArrayDynamic ProgrammingString Matching

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal solution uses dynamic programming to build the target string incrementally while tracking the minimum cost at each step. This approach is efficient and avoids redundant calculations by storing results.

⚙️

Algorithm

3 steps

1Step 1: Create a DP array where dp[i] represents the minimum cost to construct the first i characters of the target.
2Step 2: Initialize dp[0] to 0 (cost of constructing an empty string) and all other values to infinity.
3Step 3: For each word, check if it can be appended to the current target substring, and update the DP array accordingly.

solution.py11 lines

1# Full working Python code
2
3def min_cost_to_construct(target, words, costs):
4    n = len(target)
5    dp = [float('inf')] * (n + 1)
6    dp[0] = 0
7    for i in range(n):
8        for j in range(len(words)):
9            if target.startswith(words[j], i):
10                dp[i + len(words[j])] = min(dp[i + len(words[j])], dp[i] + costs[j])
11    return dp[n] if dp[n] != float('inf') else -1

ℹ

Complexity note: The time complexity is O(n * m) where n is the length of the target and m is the number of words, as we check each word for each position in the target.

1Dynamic programming can significantly reduce the number of calculations needed by storing intermediate results.
2Understanding how to break down the problem into smaller subproblems is crucial in dynamic programming.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.