How do you solve Find Minimum Cost to Remove Array Elements on LeetCode?

Find Minimum Cost to Remove Array Elements (LeetCode #3469) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Removing elements in pairs minimizes costs.

What is the time complexity of Find Minimum Cost to Remove Array Elements?

The optimal solution for Find Minimum Cost to Remove Array Elements runs in O(n) time and uses O(n) space. The linear complexity arises from a single pass through the array, updating the DP array based on previous computations.

What is the brute force solution for Find Minimum Cost to Remove Array Elements?

The brute force approach for Find Minimum Cost to Remove Array Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try all possible pairs of elements to remove from the first three elements, calculating the cost for each combination. This will give us the total cost for removing all elements. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Minimum Cost to Remove Array Elements?

Find Minimum Cost to Remove Array Elements uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Find Minimum Cost to Remove Array Elements?

Explain your thought process clearly. Discuss edge cases upfront. Practice similar DP problems.

What are common mistakes when solving Find Minimum Cost to Remove Array Elements?

Not considering all pairs from the first three elements. Forgetting to handle the last few elements correctly.

#3469

Find Minimum Cost to Remove Array Elements

Medium

ArrayDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can use dynamic programming to keep track of the minimum cost to remove elements from the array. By considering only the first three elements at each step, we can efficiently compute the cost.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP array where dp[i] represents the minimum cost to remove elements from nums[0] to nums[i].
2Step 2: Iterate through the array, calculating costs based on the last three elements.
3Step 3: Return dp[n-1] as the result.

solution.py11 lines

1def minCost(nums):
2    n = len(nums)
3    dp = [0] * n
4    for i in range(n):
5        if i == 0:
6            dp[i] = nums[i]
7        elif i == 1:
8            dp[i] = max(nums[0], nums[1])
9        else:
10            dp[i] = min(dp[i-2] + max(nums[i], nums[i-1]), dp[i-1] + nums[i])
11    return dp[-1]

ℹ

Complexity note: The linear complexity arises from a single pass through the array, updating the DP array based on previous computations.

1Removing elements in pairs minimizes costs.
2Dynamic programming efficiently tracks minimum costs.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.