LeetCode #3469 — MEDIUM

Find Minimum Cost to Remove Array Elements

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

The Problem

Problem Statement

You are given an integer array nums. Your task is to remove all elements from the array by performing one of the following operations at each step until nums is empty:

Choose any two elements from the first three elements of nums and remove them. The cost of this operation is the maximum of the two elements removed.
If fewer than three elements remain in nums, remove all the remaining elements in a single operation. The cost of this operation is the maximum of the remaining elements.

Return the minimum cost required to remove all the elements.

Example 1:

Input: nums = [6,2,8,4]

Output: 12

Explanation:

Initially, nums = [6, 2, 8, 4].

In the first operation, remove nums[0] = 6 and nums[2] = 8 with a cost of max(6, 8) = 8. Now, nums = [2, 4].
In the second operation, remove the remaining elements with a cost of max(2, 4) = 4.

The cost to remove all elements is 8 + 4 = 12. This is the minimum cost to remove all elements in nums. Hence, the output is 12.

Example 2:

Input: nums = [2,1,3,3]

Output: 5

Explanation:

Initially, nums = [2, 1, 3, 3].

In the first operation, remove nums[0] = 2 and nums[1] = 1 with a cost of max(2, 1) = 2. Now, nums = [3, 3].
In the second operation remove the remaining elements with a cost of max(3, 3) = 3.

The cost to remove all elements is 2 + 3 = 5. This is the minimum cost to remove all elements in nums. Hence, the output is 5.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. Your task is to remove all elements from the array by performing one of the following operations at each step until nums is empty: Choose any two elements from the first three elements of nums and remove them. The cost of this operation is the maximum of the two elements removed. If fewer than three elements remain in nums, remove all the remaining elements in a single operation. The cost of this operation is the maximum of the remaining elements. Return the minimum cost required to remove all the elements.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

[6,2,8,4]

Example 2

[2,1,3,3]

Core Insight

What unlocks the optimal approach

Can we use dynamic programming here?
Use dynamic programming. The process guarantees that the remaining elements form a prefix of the array with at most one previous element.
Define the state as <code>dp[i][j]</code>, where <code>i</code> represents the last remaining element and <code>j</code> represents the starting index of the current prefix.

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 #3469: Find Minimum Cost to Remove Array Elements
class Solution {
  public int minCost(int[] nums) {
    final int n = nums.length;
    Integer[][] mem = new Integer[n + 1][n + 1];
    return minCost(/*last=*/0, 1, nums, mem);
  }

  private int minCost(int last, int i, int[] nums, Integer[][] mem) {
    final int n = nums.length;
    if (i == n) // Single element left.
      return nums[last];
    if (i == n - 1) // Two elements left.
      return Math.max(nums[last], nums[i]);
    if (mem[i][last] != null)
      return mem[i][last];
    final int a = Math.max(nums[i], nums[i + 1]) + minCost(last, i + 2, nums, mem);
    final int b = Math.max(nums[last], nums[i]) + minCost(i + 1, i + 2, nums, mem);
    final int c = Math.max(nums[last], nums[i + 1]) + minCost(i, i + 2, nums, mem);
    return mem[i][last] = Math.min(a, Math.min(b, c));
  }
}

// Accepted solution for LeetCode #3469: Find Minimum Cost to Remove Array Elements
package main

import "slices"

// https://space.bilibili.com/206214
func minCost(nums []int) int {
	n := len(nums)
	var f []int
	if n%2 == 1 {
		f = slices.Clone(nums)
	} else {
		f = make([]int, n)
		for i, x := range nums {
			f[i] = max(x, nums[n-1])
		}
	}
	for i := n - 3 + n%2; i > 0; i -= 2 {
		b, c := nums[i], nums[i+1]
		for j, a := range nums[:i] {
			f[j] = min(f[j]+max(b, c), f[i]+max(a, c), f[i+1]+max(a, b))
		}
	}
	return f[0]
}

func minCost2(nums []int) int {
	n := len(nums)
	f := make([][]int, n+1)
	f[n] = nums
	f[n-1] = make([]int, n)
	for i, x := range nums {
		f[n-1][i] = max(x, nums[n-1])
	}
	for i := n - 3 + n%2; i > 0; i -= 2 {
		f[i] = make([]int, i)
		b, c := nums[i], nums[i+1]
		for j, a := range nums[:i] {
			f[i][j] = min(f[i+2][j]+max(b, c), f[i+2][i]+max(a, c), f[i+2][i+1]+max(a, b))
		}
	}
	return f[1][0]
}

func minCost1(nums []int) int {
	n := len(nums)
	memo := make([][]int, n-1)
	for i := range memo {
		memo[i] = make([]int, i)
	}
	var dfs func(int, int) int
	dfs = func(i, j int) int {
		if i == n {
			return nums[j]
		}
		if i == n-1 {
			return max(nums[j], nums[i])
		}
		p := &memo[i][j]
		a, b, c := nums[j], nums[i], nums[i+1]
		if *p == 0 {
			*p = min(dfs(i+2, j)+max(b, c), dfs(i+2, i)+max(a, c), dfs(i+2, i+1)+max(a, b))
		}
		return *p
	}
	return dfs(1, 0)
}

# Accepted solution for LeetCode #3469: Find Minimum Cost to Remove Array Elements
class Solution:
  def minCost(self, nums: list[int]) -> int:
    n = len(nums)

    @functools.lru_cache(None)
    def dp(last: int, i: int) -> int:
      if i == n:  # Single element left.
        return nums[last]
      if i == n - 1:  # Two elements left.
        return max(nums[last], nums[i])
      a = max(nums[i], nums[i + 1]) + dp(last, i + 2)
      b = max(nums[last], nums[i]) + dp(i + 1, i + 2)
      c = max(nums[last], nums[i + 1]) + dp(i, i + 2)
      return min(a, b, c)

    return dp(0, 1)

// Accepted solution for LeetCode #3469: Find Minimum Cost to Remove Array Elements
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3469: Find Minimum Cost to Remove Array Elements
// class Solution {
//   public int minCost(int[] nums) {
//     final int n = nums.length;
//     Integer[][] mem = new Integer[n + 1][n + 1];
//     return minCost(/*last=*/0, 1, nums, mem);
//   }
// 
//   private int minCost(int last, int i, int[] nums, Integer[][] mem) {
//     final int n = nums.length;
//     if (i == n) // Single element left.
//       return nums[last];
//     if (i == n - 1) // Two elements left.
//       return Math.max(nums[last], nums[i]);
//     if (mem[i][last] != null)
//       return mem[i][last];
//     final int a = Math.max(nums[i], nums[i + 1]) + minCost(last, i + 2, nums, mem);
//     final int b = Math.max(nums[last], nums[i]) + minCost(i + 1, i + 2, nums, mem);
//     final int c = Math.max(nums[last], nums[i + 1]) + minCost(i, i + 2, nums, mem);
//     return mem[i][last] = Math.min(a, Math.min(b, c));
//   }
// }

// Accepted solution for LeetCode #3469: Find Minimum Cost to Remove Array Elements
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3469: Find Minimum Cost to Remove Array Elements
// class Solution {
//   public int minCost(int[] nums) {
//     final int n = nums.length;
//     Integer[][] mem = new Integer[n + 1][n + 1];
//     return minCost(/*last=*/0, 1, nums, mem);
//   }
// 
//   private int minCost(int last, int i, int[] nums, Integer[][] mem) {
//     final int n = nums.length;
//     if (i == n) // Single element left.
//       return nums[last];
//     if (i == n - 1) // Two elements left.
//       return Math.max(nums[last], nums[i]);
//     if (mem[i][last] != null)
//       return mem[i][last];
//     final int a = Math.max(nums[i], nums[i + 1]) + minCost(last, i + 2, nums, mem);
//     final int b = Math.max(nums[last], nums[i]) + minCost(i + 1, i + 2, nums, mem);
//     final int c = Math.max(nums[last], nums[i + 1]) + minCost(i, i + 2, nums, mem);
//     return mem[i][last] = Math.min(a, Math.min(b, c));
//   }
// }

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 × m)

Space

O(n × m)

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.