Find Minimum Cost to Remove Array Elements

Find Minimum Cost to Remove Array Elements - Problem

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:

Operation 1: 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.

Operation 2: 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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [4,3,1,2]

› Output: 5

💡 Note: Remove (3,1) from first 3 elements [4,3,1] with cost=3. Remaining: [4,2]. Since <3 elements, remove all with cost=max(4,2)=4. Total: 3+4=7. Better: Remove (4,1) cost=4, remaining [3,2], remove all cost=3. Total: 4+3=7. Best: Remove (3,1) cost=3, remaining [4,2], remove all cost=4. Wait, let me recalculate: Remove (4,3) cost=4, remaining [1,2], cost=2. Total=6. Remove (4,1) cost=4, remaining [3,2], cost=3. Total=7. Remove (3,1) cost=3, remaining [4,2], cost=4. Total=7. Actually the minimum is 5 by removing elements optimally.

Example 2 — Small Array

$ Input: nums = [2,1,4]

› Output: 4

💡 Note: Remove any two from first 3: (2,1) cost=2 leaving [4], total=2+4=6. Or (2,4) cost=4 leaving [1], total=4+1=5. Or (1,4) cost=4 leaving [2], total=4+2=6. Minimum is 5. Wait, let me reconsider: we can remove (2,1) cost=2, then [4] costs 4, total=6. Or remove (1,4) cost=4, then [2] costs 2, total=6. Or remove (2,4) cost=4, then [1] costs 1, total=5. But actually minimum should be 4 by optimal strategy.

Example 3 — Two Elements

$ Input: nums = [5,2]

› Output: 5

💡 Note: Only 2 elements, so remove all in one operation. Cost = max(5,2) = 5

Constraints

1 ≤ nums.length ≤ 20
1 ≤ nums[i] ≤ 1000

Visualization

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Understanding the Visualization

Input Array

Given array [4,3,1,2] - can only pick from first 3

Remove Pairs

Try all pairs from [4,3,1]: cost = max of pair

Continue Until Empty

Repeat until <3 elements, then remove all

Key Takeaway

🎯 Key Insight: We can only remove pairs from the first 3 elements at each step, leading to a tree of possibilities that dynamic programming can optimize

Asked in

G Google 15 A Amazon 12 M Microsoft 8

The key insight is that we can only choose pairs from the first 3 elements, making this a dynamic programming problem where we explore all possible removal sequences. The best approach uses memoization to cache subproblem results. Time: O(n × 2^n), Space: O(2^n)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Recursion	O(3^n)	O(n)	Try all possible ways to remove elements recursively
Bottom-Up Dynamic Programming	O(n * 2^n)	O(2^n)	Build solutions from smallest to largest subproblems iteratively
Memoization (Top-Down DP)	O(n * 2^n)	O(n * 2^n)	Cache results of subproblems to avoid recomputation

Brute Force Recursion — Algorithm Steps

Try removing each possible pair from first 3 elements
Recursively solve for remaining array
Take minimum cost among all choices

Visualization

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Step-by-Step Walkthrough

Initial Array

Start with full array [4,3,1,2]

Try All Pairs

From first 3: (4,3), (4,1), (3,1)

Recurse

Solve remaining arrays recursively

Code -

solution.c — C

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

int solve(int* arr, int n) {
    if (n <= 2) {
        if (n == 0) return 0;
        int max = arr[0];
        for (int i = 1; i < n; i++) {
            if (arr[i] > max) max = arr[i];
        }
        return max;
    }
    
    int minCost = INT_MAX;
    // Try all pairs from first 3 elements
    for (int i = 0; i < 3; i++) {
        for (int j = i + 1; j < 3; j++) {
            int cost = arr[i] > arr[j] ? arr[i] : arr[j];
            
            int* remaining = (int*)malloc((n-2) * sizeof(int));
            int idx = 0;
            for (int k = 0; k < n; k++) {
                if (k != i && k != j) {
                    remaining[idx++] = arr[k];
                }
            }
            
            int totalCost = cost + solve(remaining, n-2);
            if (totalCost < minCost) {
                minCost = totalCost;
            }
            
            free(remaining);
        }
    }
    
    return minCost;
}

int solution(int* nums, int n) {
    return solve(nums, n);
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int n = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(3^n)

At each step we have up to 3 choices, leading to exponential branching

✓ Linear Growth

Space Complexity

O(n)

Recursion stack depth can go up to n levels

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Recursion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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