LeetCode 1959 - Minimum Total Space Wasted With K Resizing Operations

Let's go through a full, detailed technical solution guide for LeetCode 1959 following your formatting and style rules. This problem asks us to minimize the total wasted space when resizing a dynamic array multiple times.

leetcodemediumarraydynamic-programmingprefix-sum

LeetCode Problem 1959

Difficulty: 🟡 Medium
Topics: Array, Dynamic Programming, Prefix Sum

Solution

Let's go through a full, detailed technical solution guide for LeetCode 1959 following your formatting and style rules.

Problem Understanding

This problem asks us to minimize the total wasted space when resizing a dynamic array multiple times. We are given a 0-indexed array nums where nums[i] represents the number of elements that need to fit in the array at time i. We can resize the array at most k times, and the size of the array at any time must be at least nums[i]. The wasted space at each time step is the difference between the array size and nums[i].

The goal is to choose resizing points and array sizes so that the total sum of wasted space across all time steps is minimized.

Key points in the input:

  • 1 <= nums.length <= 200, so the array is small enough for O(n³) solutions if optimized with dynamic programming.
  • 1 <= nums[i] <= 10^6, meaning naive precomputation or large memory usage solutions may not be feasible.
  • 0 <= k <= nums.length - 1, indicating the number of allowed resizes is bounded by array length minus one.

Important edge cases include:

  1. k = 0, meaning no resizing is allowed. The array must be one size for all times.
  2. All nums[i] equal, so wasted space is 0 regardless of resizing.
  3. nums strictly increasing or decreasing, which could make naive greedy approaches fail.

Approaches

Brute Force:

The brute force approach tries every possible combination of resize operations. We could recursively decide at each position whether to resize or not and track all array sizes and wasted space. While this guarantees correctness, its time complexity is exponential because each of the n positions could be a resize or not, leading to O(2^n) possibilities. Clearly, this is far too slow for n up to 200.

Optimal Approach:

The key insight is that the space wasted for a contiguous segment of nums is minimized if the array size during that segment is the maximum element in that segment. This means we only need to compute the maximum for subarrays when considering resizing, which allows us to precompute these values.

We can then use dynamic programming where dp[i][k] represents the minimum wasted space for the first i elements using at most k resizing operations. For each position i and allowed number of resizes k, we try splitting the previous segment at every possible point j < i, adding the precomputed wasted space of the new segment, and updating dp[i][k]. This reduces the exponential complexity to O(n² * k).

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) Recursively try all resize combinations
Optimal O(n² * k) O(n² + n*k) DP with precomputed segment maxes and wasted spaces

Algorithm Walkthrough

  1. Precompute segment maxes and wasted spaces:

Create a 2D table waste[i][j] representing the total wasted space if we use a single fixed size (max of nums[i:j+1]) for the subarray nums[i:j+1]. This can be done with nested loops: for each i, track the running maximum and calculate total waste for all j >= i. 2. Initialize DP table:

Let dp[i][r] be the minimum wasted space for the first i elements with at most r resizes. Initialize all values to infinity. Base case: dp[0][0] = 0. 3. Populate DP table:

For each i from 1 to n, for each allowed number of resizes r from 0 to k, iterate over all possible previous split points j from 0 to i-1. Update: dp[i][r] = min(dp[i][r], dp[j][r-1] + waste[j][i-1]). 4. Return answer:

The minimum wasted space for the entire array using at most k resizes is dp[n][k].

Why it works:

The invariant is that the DP table always contains the minimum wasted space for the first i elements given r resizes. By precomputing wasted space for all contiguous segments, we avoid recomputation and ensure each subproblem is optimally solved, which guarantees global optimality.

Python Solution

from typing import List

class Solution:
    def minSpaceWastedKResizing(self, nums: List[int], k: int) -> int:
        n = len(nums)
        # Precompute wasted space for all segments
        waste = [[0] * n for _ in range(n)]
        for i in range(n):
            max_val = nums[i]
            total = 0
            for j in range(i, n):
                max_val = max(max_val, nums[j])
                total += max_val - nums[j]
                waste[i][j] = total

        # DP table: dp[i][r] = min wasted space for first i elements with r resizes
        dp = [[float('inf')] * (k + 1) for _ in range(n + 1)]
        dp[0][0] = 0

        for i in range(1, n + 1):
            for r in range(k + 1):
                for j in range(i):
                    if r > 0 or j == 0:
                        dp[i][r] = min(dp[i][r], dp[j][r - 1] + waste[j][i - 1] if r > 0 else waste[j][i - 1])

        return dp[n][k]

This implementation first builds a 2D array for wasted spaces, then fills the DP table iteratively. The inner loop checks all possible segment splits and updates the minimum wasted space for the given number of resizes. Edge cases like k = 0 are handled naturally because no resizing is allowed, and only the base waste is computed.

Go Solution

func minSpaceWastedKResizing(nums []int, k int) int {
    n := len(nums)
    waste := make([][]int, n)
    for i := range waste {
        waste[i] = make([]int, n)
        maxVal := nums[i]
        total := 0
        for j := i; j < n; j++ {
            if nums[j] > maxVal {
                maxVal = nums[j]
            }
            total += maxVal - nums[j]
            waste[i][j] = total
        }
    }

    dp := make([][]int, n+1)
    for i := range dp {
        dp[i] = make([]int, k+1)
        for j := range dp[i] {
            dp[i][j] = 1<<60
        }
    }
    dp[0][0] = 0

    for i := 1; i <= n; i++ {
        for r := 0; r <= k; r++ {
            for j := 0; j < i; j++ {
                if r > 0 || j == 0 {
                    val := waste[j][i-1]
                    if r > 0 {
                        val += dp[j][r-1]
                    }
                    if val < dp[i][r] {
                        dp[i][r] = val
                    }
                }
            }
        }
    }
    return dp[n][k]
}

Go handles slice initialization carefully to avoid nil errors. Since Go does not have float('inf'), a large integer (1<<60) is used. Otherwise, the logic mirrors the Python version.

Worked Examples

Example 1: nums = [10, 20], k = 0

waste table:

i\j 0 1
0 0 10
1 - 0

dp table for r = 0:

i\r 0
0 0
1 0
2 10

Result: dp[2][0] = 10

Example 2: nums = [10, 20, 30], k = 1

waste table:

i\j 0 1 2
0 0 10 30
1 - 0 10
2 - - 0

dp table after DP updates gives dp[3][1] = 10.

Example 3: nums = [10, 20, 15, 30, 20], k = 2

Segmented wastes and DP updates yield dp[5][2] = 15.

Complexity Analysis

Measure Complexity Explanation
Time O(n² * k) Precomputing wasted space is O(n²), DP has triple nested loops over