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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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

✓ Stack

⏱️ Time: N/A Space: N/A

Bottom-Up Dynamic Programming

⏱️ Time: O(n * 2^n) Space: O(2^n)

Instead of recursion, use iterative DP where we represent states as bitmasks and build up solutions systematically. Each bit represents whether an element is still in the array.

Brute Force Recursion

⏱️ Time: O(3^n) Space: O(n)

At each step, try all possible pairs from the first three elements, recursively solve for the remaining array, and take the minimum cost. When fewer than 3 elements remain, remove all at once.

Memoization (Top-Down DP)

⏱️ Time: O(n * 2^n) Space: O(n * 2^n)

Use the same recursive approach but cache results based on the current array state. This eliminates repeated calculations for the same subproblem configurations.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.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 solveSub(int* nums, int n) {
    if (n == 0) return 0;
    if (n == 1) return nums[0];
    if (n == 2) return max(nums[0], nums[1]);
    
    int minCost = INT_MAX;
    int maxCheck = min(3, n);
    
    for (int i = 0; i < maxCheck; i++) {
        for (int j = i + 1; j < maxCheck; j++) {
            int cost = max(nums[i], nums[j]);
            
            int* newNums = (int*)malloc((n-2) * sizeof(int));
            int newSize = 0;
            
            for (int k = 0; k < n; k++) {
                if (k != i && k != j) {
                    newNums[newSize++] = nums[k];
                }
            }
            
            int totalCost = cost + solveSub(newNums, newSize);
            minCost = min(minCost, totalCost);
            
            free(newNums);
        }
    }
    
    return minCost;
}

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

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing for array format [1,2,3]
    int nums[100];
    int n = 0;
    
    char* ptr = line;
    while (*ptr && *ptr != '[') ptr++;
    if (*ptr == '[') ptr++;
    
    while (*ptr && *ptr != ']') {
        while (*ptr == ' ' || *ptr == ',') ptr++;
        if (*ptr >= '0' && *ptr <= '9') {
            nums[n++] = strtol(ptr, &ptr, 10);
        } else {
            ptr++;
        }
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

⚠ Quadratic Space

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