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Leetcode #1959: Minimum Total Space Wasted With K Resizing Operations

In this guide, we solve Leetcode #1959 Minimum Total Space Wasted With K Resizing Operations in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You are currently designing a dynamic array. You are given a 0-indexed integer array nums, where nums[i] is the number of elements that will be in the array at time i.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Array, Dynamic Programming

Intuition

The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.

A carefully chosen DP state captures exactly what we need to build the final answer.

Approach

Define the DP state and recurrence, then compute states in the correct order.

Optionally compress space once the recurrence is clear.

Steps:

  • Choose a DP state definition.
  • Write the recurrence and base cases.
  • Compute states in the correct order.

Example

Input: nums = [10,20], k = 0 Output: 10 Explanation: size = [20,20]. We can set the initial size to be 20. The total wasted space is (20 - 10) + (20 - 20) = 10.

Python Solution

class Solution: def minSpaceWastedKResizing(self, nums: List[int], k: int) -> int: k += 1 n = len(nums) g = [[0] * n for _ in range(n)] for i in range(n): s = mx = 0 for j in range(i, n): s += nums[j] mx = max(mx, nums[j]) g[i][j] = mx * (j - i + 1) - s f = [[inf] * (k + 1) for _ in range(n + 1)] f[0][0] = 0 for i in range(1, n + 1): for j in range(1, k + 1): for h in range(i): f[i][j] = min(f[i][j], f[h][j - 1] + g[h][i - 1]) return f[-1][-1]

Complexity

The time complexity is O(n2×k)O(n^2 \times k)O(n2×k), and the space complexity is O(n×(n+k))O(n \times (n + k))O(n×(n+k)). The space complexity is O(n×(n+k))O(n \times (n + k))O(n×(n+k)).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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