What is the time complexity of Build Array Where You Can Find The Maximum Exactly K Comparisons?

The optimal solution for Build Array Where You Can Find The Maximum Exactly K Comparisons runs in O(n * m * k) time and uses O(n * m * k) space. The time complexity is O(n * m * k) because we fill a 3D DP table iterating through n, m, and k. The space complexity is also O(n * m * k) due to the storage of the DP table.

What is the brute force solution for Build Array Where You Can Find The Maximum Exactly K Comparisons?

The brute force approach for Build Array Where You Can Find The Maximum Exactly K Comparisons is Brute Force, with O(m^n) time complexity and O(n) space complexity. The brute force approach involves generating all possible arrays of length n with values between 1 and m. We then check each array to see if it meets the search cost condition exactly equal to k. The optimal Optimal Solution approach improves this to O(n * m * k).

What algorithm or data structure is used to solve Build Array Where You Can Find The Maximum Exactly K Comparisons?

Build Array Where You Can Find The Maximum Exactly K Comparisons uses the following concepts: Dynamic Programming, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Backtracking.

What are the interview tips for Build Array Where You Can Find The Maximum Exactly K Comparisons?

Clearly explain your thought process while writing the code. Be prepared to discuss the trade-offs of different approaches. Practice similar problems to build familiarity with dynamic programming.

What are common mistakes when solving Build Array Where You Can Find The Maximum Exactly K Comparisons?

Not considering all possible maximum values when filling the DP table. Overlooking the modulo operation which can lead to incorrect results.

#1420

Build Array Where You Can Find The Maximum Exactly K Comparisons

Hard

Dynamic ProgrammingPrefix SumDynamic ProgrammingBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m^n)	O(n * m * k)
Space	O(n)	O(n * m * k)

💡

Intuition

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

The optimal approach uses dynamic programming to build a table that counts the number of valid arrays based on the current index, maximum value used, and the search cost. This avoids redundant calculations and efficiently computes the result.

⚙️

Algorithm

3 steps

1Step 1: Initialize a 3D DP table where dp[a][b][c] represents the number of ways to build the array starting from index a, with maximum value b, and search cost c.
2Step 2: Fill the DP table iteratively, considering each possible maximum value and updating the search cost based on whether the current maximum is used.
3Step 3: Sum up all valid configurations that result in the desired search cost k.

solution.py13 lines

1def countArrays(n, m, k):
2    MOD = 10**9 + 7
3    dp = [[[0] * (k + 1) for _ in range(m + 1)] for _ in range(n + 1)]
4    dp[0][0][0] = 1
5
6    for i in range(1, n + 1):
7        for max_val in range(1, m + 1):
8            for search_cost in range(k + 1):
9                dp[i][max_val][search_cost] = (dp[i - 1][max_val][search_cost] * max_val) % MOD
10                if search_cost > 0:
11                    dp[i][max_val][search_cost] = (dp[i][max_val][search_cost] + dp[i - 1][max_val][search_cost - 1]) % MOD
12
13    return dp[n][m][k]

ℹ

Complexity note: The time complexity is O(n * m * k) because we fill a 3D DP table iterating through n, m, and k. The space complexity is also O(n * m * k) due to the storage of the DP table.

1Dynamic programming can significantly reduce the time complexity by avoiding redundant calculations.
2Understanding the relationship between the maximum value, index, and search cost is crucial.

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