LeetCode #1959 — MEDIUM

Minimum Total Space Wasted With K Resizing Operations

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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. In addition, you are given an integer k, the maximum number of times you can resize the array (to any size).

The size of the array at time t, size_t, must be at least nums[t] because there needs to be enough space in the array to hold all the elements. The space wasted at time t is defined as size_t - nums[t], and the total space wasted is the sum of the space wasted across every time t where 0 <= t < nums.length.

Return the minimum total space wasted if you can resize the array at most k times.

Note: The array can have any size at the start and does not count towards the number of resizing operations.

Example 1:

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.

Example 2:

Input: nums = [10,20,30], k = 1
Output: 10
Explanation: size = [20,20,30].
We can set the initial size to be 20 and resize to 30 at time 2. 
The total wasted space is (20 - 10) + (20 - 20) + (30 - 30) = 10.

Example 3:

Input: nums = [10,20,15,30,20], k = 2
Output: 15
Explanation: size = [10,20,20,30,30].
We can set the initial size to 10, resize to 20 at time 1, and resize to 30 at time 3.
The total wasted space is (10 - 10) + (20 - 20) + (20 - 15) + (30 - 30) + (30 - 20) = 15.

Constraints:

1 <= nums.length <= 200
1 <= nums[i] <= 10⁶
0 <= k <= nums.length - 1

Step 01

Brute Force Baseline

Problem summary: 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. In addition, you are given an integer k, the maximum number of times you can resize the array (to any size). The size of the array at time t, sizet, must be at least nums[t] because there needs to be enough space in the array to hold all the elements. The space wasted at time t is defined as sizet - nums[t], and the total space wasted is the sum of the space wasted across every time t where 0 <= t < nums.length. Return the minimum total space wasted if you can resize the array at most k times. Note: The array can have any size at the start and does not count towards the number of resizing operations.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[10,20]
0

Example 2

[10,20,30]
1

Example 3

[10,20,15,30,20]
2

Step 02

Core Insight

What unlocks the optimal approach

Given a range, how can you find the minimum waste if you can't perform any resize operations?
Can we build our solution using dynamic programming using the current index and the number of resizing operations performed as the states?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1959: Minimum Total Space Wasted With K Resizing Operations
class Solution {
    public int minSpaceWastedKResizing(int[] nums, int k) {
        ++k;
        int n = nums.length;
        int[][] g = new int[n][n];
        for (int i = 0; i < n; ++i) {
            int s = 0, mx = 0;
            for (int j = i; j < n; ++j) {
                s += nums[j];
                mx = Math.max(mx, nums[j]);
                g[i][j] = mx * (j - i + 1) - s;
            }
        }
        int[][] f = new int[n + 1][k + 1];
        int inf = 0x3f3f3f3f;
        for (int i = 0; i < f.length; ++i) {
            Arrays.fill(f[i], inf);
        }
        f[0][0] = 0;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= k; ++j) {
                for (int h = 0; h < i; ++h) {
                    f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h][i - 1]);
                }
            }
        }
        return f[n][k];
    }
}

// Accepted solution for LeetCode #1959: Minimum Total Space Wasted With K Resizing Operations
func minSpaceWastedKResizing(nums []int, k int) int {
	k++
	n := len(nums)
	g := make([][]int, n)
	for i := range g {
		g[i] = make([]int, n)
	}
	for i := 0; i < n; i++ {
		s, mx := 0, 0
		for j := i; j < n; j++ {
			s += nums[j]
			mx = max(mx, nums[j])
			g[i][j] = mx*(j-i+1) - s
		}
	}
	f := make([][]int, n+1)
	inf := 0x3f3f3f3f
	for i := range f {
		f[i] = make([]int, k+1)
		for j := range f[i] {
			f[i][j] = inf
		}
	}
	f[0][0] = 0
	for i := 1; i <= n; i++ {
		for j := 1; j <= k; j++ {
			for h := 0; h < i; h++ {
				f[i][j] = min(f[i][j], f[h][j-1]+g[h][i-1])
			}
		}
	}
	return f[n][k]
}

# Accepted solution for LeetCode #1959: Minimum Total Space Wasted With K Resizing Operations
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]

// Accepted solution for LeetCode #1959: Minimum Total Space Wasted With K Resizing Operations
impl Solution {
    pub fn min_space_wasted_k_resizing(nums: Vec<i32>, k: i32) -> i32 {
        let mut k = k + 1;
        let n = nums.len();
        let mut g = vec![vec![0; n]; n];

        for i in 0..n {
            let (mut s, mut mx) = (0, 0);
            for j in i..n {
                s += nums[j];
                mx = mx.max(nums[j]);
                g[i][j] = mx * (j as i32 - i as i32 + 1) - s;
            }
        }

        let inf = 0x3f3f3f3f;
        let mut f = vec![vec![inf; (k + 1) as usize]; n + 1];
        f[0][0] = 0;

        for i in 1..=n {
            for j in 1..=k as usize {
                for h in 0..i {
                    f[i][j] = f[i][j].min(f[h][j - 1] + g[h][i - 1]);
                }
            }
        }

        f[n][k as usize]
    }
}

// Accepted solution for LeetCode #1959: Minimum Total Space Wasted With K Resizing Operations
function minSpaceWastedKResizing(nums: number[], k: number): number {
    k += 1;
    const n = nums.length;
    const g: number[][] = Array.from({ length: n }, () => Array(n).fill(0));

    for (let i = 0; i < n; i++) {
        let s = 0,
            mx = 0;
        for (let j = i; j < n; j++) {
            s += nums[j];
            mx = Math.max(mx, nums[j]);
            g[i][j] = mx * (j - i + 1) - s;
        }
    }

    const inf = Number.POSITIVE_INFINITY;
    const f: number[][] = Array.from({ length: n + 1 }, () => Array(k + 1).fill(inf));
    f[0][0] = 0;

    for (let i = 1; i <= n; i++) {
        for (let j = 1; j <= k; j++) {
            for (let h = 0; h < i; h++) {
                f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h][i - 1]);
            }
        }
    }

    return f[n][k];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2 × k)

Space

O(n × (n + k)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.