What is the brute force solution for Minimum Total Space Wasted With K Resizing Operations?

The brute force approach for Minimum Total Space Wasted With K Resizing Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible way to resize the array at each time step, calculating the total wasted space for each configuration. This method is simple but inefficient, as it explores all combinations of resizing operations. The optimal Optimal Solution approach improves this to O(n * k).

What algorithm or data structure is used to solve Minimum Total Space Wasted With K Resizing Operations?

Minimum Total Space Wasted With K Resizing Operations uses the following concepts: Array, Dynamic Programming, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Prefix Sum.

What are the interview tips for Minimum Total Space Wasted With K Resizing Operations?

Always clarify the problem constraints and edge cases before diving into coding. Explain your thought process and approach before writing code to ensure clarity. Practice writing both brute force and optimal solutions to understand the trade-offs.

What are common mistakes when solving Minimum Total Space Wasted With K Resizing Operations?

Failing to initialize the DP array correctly, leading to incorrect results. Not considering edge cases, such as when k is 0 or when nums has only one element.

#1959

Minimum Total Space Wasted With K Resizing Operations

Medium

ArrayDynamic ProgrammingPrefix SumDynamic ProgrammingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal approach uses dynamic programming to efficiently calculate the minimum wasted space by storing intermediate results. This avoids redundant calculations and allows us to build solutions incrementally.

⚙️

Algorithm

3 steps

1Step 1: Create a DP array where dp[i][j] represents the minimum wasted space using the first i elements with j resizes.
2Step 2: Iterate through each element and each possible number of resizes.
3Step 3: For each state, calculate the total wasted space based on previous states and update the DP array.

solution.py11 lines

1def minSpaceWasted(nums, k):
2    n = len(nums)
3    dp = [[float('inf')] * (k + 1) for _ in range(n + 1)]
4    dp[0][0] = 0
5    for i in range(1, n + 1):
6        total = 0
7        for j in range(i, 0, -1):
8            total += nums[j - 1]
9            for resizes in range(k + 1):
10                dp[i][resizes] = min(dp[i][resizes], dp[j - 1][resizes - 1] + total - nums[j - 1])
11    return min(dp[n])

ℹ

Complexity note: This complexity is due to the nested loops iterating over the number of elements and the number of resizing operations, leading to a manageable growth in states compared to the brute force approach.

1Dynamic programming can significantly reduce the number of calculations by storing intermediate results.
2Understanding how to break down the problem into smaller subproblems is crucial for applying DP effectively.

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