Minimum Total Space Wasted With K Resizing Operations

Minimum Total Space Wasted With K Resizing Operations - Problem

Dynamic Array Memory Optimization Challenge

You're building a smart dynamic array system that needs to minimize memory waste! 🚀

Given an array nums where nums[i] represents how many elements your array will contain at time i, and an integer k (maximum allowed resize operations), your goal is to find the minimum total space wasted.

How it works:
• At any time t, your array size must be ≥ nums[t] (enough space for all elements)
• Space wasted at time t = array_size[t] - nums[t]
• You can resize the array at most k times to any size you want
• The initial array size doesn't count as a resize operation

Goal: Return the minimum possible total space wasted across all time periods.

Think of it like managing server capacity - you want to minimize unused resources while handling varying demand!

Input & Output

example_1.py — Basic case

$ Input: nums = [10,20], k = 0

› Output: 10

💡 Note: With 0 resizes allowed, we must use one segment. The optimal size is max(10,20)=20. Total waste = (20-10) + (20-20) = 10.

example_2.py — One resize optimal

$ Input: nums = [10,20,30], k = 1

› Output: 10

💡 Note: With 1 resize: segment [10,20] with size 20 wastes 10, then resize to segment [30] with size 30 wastes 0. Total waste = 10.

example_3.py — Multiple resizes

$ Input: nums = [10,20,15,30,20], k = 2

› Output: 15

💡 Note: Optimal partitioning: [10,20] (size 20, waste 10), [15] (size 15, waste 0), [30,20] (size 30, waste 10). Total = 20.

Constraints

1 ≤ nums.length ≤ 200
1 ≤ nums[i] ≤ 10⁶
0 ≤ k ≤ nums.length - 1
The array size must always be ≥ nums[i] at time i

Visualization

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Understanding the Visualization

Identify Periods

Each number represents employees needed at that time

Choose Renovation Points

Decide when to resize based on demand changes

Calculate Waste

For each period, waste = office_size - actual_employees

Minimize Total Cost

Use DP to find optimal renovation schedule

Key Takeaway

🎯 Key Insight: The optimal array size for any time period is the maximum demand in that period. The challenge becomes finding the best points to resize (partition the timeline) to minimize total waste.

Asked in

G Google 45 a Amazon 32 f Meta 28 ⊞ Microsoft 24

This problem is solved optimally using dynamic programming. The key insight is that for any segment without resizing, the optimal array size is the maximum element in that segment. We can precompute costs for all possible segments in O(n²), then use DP to find the optimal way to partition the array with at most k resizes. Time complexity: O(n²k), space complexity: O(n² + nk).

Common Approaches

Approach	Time	Space	Notes
✓ Dynamic Programming (Bottom-Up)	O(n³)	O(n² + nk)	Build solution table from smaller subproblems to avoid recomputation
Brute Force (Try All Partitions)	O(n^k)	O(k)	Recursively try all possible ways to partition the timeline with k resizes

Dynamic Programming (Bottom-Up) — Algorithm Steps

Precompute waste for all possible segments [i,j] in O(n²) time
Initialize DP table: dp[i][j] = min waste for first i elements with j resizes
Fill DP table: for each position and resize count, try all possible last segments
Return dp[n][k] as the final answer

Visualization

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Step-by-Step Walkthrough

Precompute Costs

Calculate waste for all possible segments [i,j]

Initialize DP

Set dp[0][0] = 0, all others to infinity

Fill Table

For each cell, try all possible previous segments

Find Minimum

Take minimum from last row (all possible resize counts)

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int max(int a, int b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }

int minSpaceWastedKResizing(int* nums, int numsSize, int k) {
    int n = numsSize;
    
    // Allocate waste matrix
    int** waste = malloc(n * sizeof(int*));
    for (int i = 0; i < n; i++) {
        waste[i] = malloc(n * sizeof(int));
    }
    
    // Precompute segment costs
    for (int i = 0; i < n; i++) {
        int maxVal = nums[i];
        for (int j = i; j < n; j++) {
            maxVal = max(maxVal, nums[j]);
            waste[i][j] = (j > i ? waste[i][j-1] : 0) + maxVal - nums[j];
        }
    }
    
    // Allocate DP table
    int** dp = malloc((n + 1) * sizeof(int*));
    for (int i = 0; i <= n; i++) {
        dp[i] = malloc((k + 2) * sizeof(int));
        for (int j = 0; j < k + 2; j++) {
            dp[i][j] = INT_MAX;
        }
    }
    dp[0][0] = 0;
    
    // Fill DP table
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < k + 2; j++) {
            int start = j > 0 ? j - 1 : 0;
            for (int prev = start; prev < i; prev++) {
                int resizeIdx = j > 0 ? j - 1 : 0;
                if (dp[prev][resizeIdx] != INT_MAX) {
                    dp[i][j] = min(dp[i][j], dp[prev][resizeIdx] + waste[prev][i-1]);
                }
            }
        }
    }
    
    int result = INT_MAX;
    for (int j = 0; j < k + 2; j++) {
        result = min(result, dp[n][j]);
    }
    
    // Free memory
    for (int i = 0; i < n; i++) free(waste[i]);
    free(waste);
    for (int i = 0; i <= n; i++) free(dp[i]);
    free(dp);
    
    return result;
}

int main() {
    int nums[] = {10, 20, 30};
    int k = 1;
    printf("%d\n", minSpaceWastedKResizing(nums, 3, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

O(n²) to precompute segment costs + O(n²k) for DP filling = O(n³) overall

⚠ Quadratic Growth

Space Complexity

O(n² + nk)

O(n²) for segment costs matrix + O(nk) for DP table

⚠ Quadratic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Dynamic Programming (Bottom-Up) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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