How do you solve Minimum Cost to Make Array Equal on LeetCode?

Minimum Cost to Make Array Equal (LeetCode #2448) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log m) time complexity and O(1) space complexity. Key insight: Changing elements to a target value that already exists in nums is optimal.

What is the brute force solution for Minimum Cost to Make Array Equal?

The brute force approach for Minimum Cost to Make Array Equal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calculating the total cost to make all elements equal to each possible target value in the nums array. This is straightforward but inefficient, as it checks every possible target. The optimal Optimal Solution approach improves this to O(n log m).

What algorithm or data structure is used to solve Minimum Cost to Make Array Equal?

Minimum Cost to Make Array Equal uses the following concepts: Array, Binary Search, Greedy, Sorting, Prefix Sum. The recommended patterns to study are: Binary Search, Greedy Algorithms, Convex Functions.

What are the interview tips for Minimum Cost to Make Array Equal?

Always clarify the problem constraints and edge cases before diving into coding. Explain your thought process clearly while coding to demonstrate your understanding. Practice calculating costs for small examples to build intuition before implementing.

What are common mistakes when solving Minimum Cost to Make Array Equal?

Not considering the cost associated with each operation correctly. Failing to optimize the search space when calculating costs.

#2448

Minimum Cost to Make Array Equal

Hard

ArrayBinary SearchGreedySortingPrefix SumBinary SearchGreedy AlgorithmsConvex Functions

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log m)Space O(1)

The optimal solution leverages binary search to find the best target value that minimizes the total cost. By calculating the cost for a range of potential targets, we can efficiently hone in on the minimum cost.

⚙️

Algorithm

4 steps

1Step 1: Define a function to calculate the total cost for a given target.
2Step 2: Use binary search to find the optimal target value between the minimum and maximum values of nums.
3Step 3: For each midpoint in the binary search, calculate the cost and adjust the search range based on whether the cost increases or decreases.
4Step 4: Return the minimum cost found.

solution.py13 lines

1def minCost(nums, cost):
2    def calculate_cost(target):
3        return sum(abs(num - target) * c for num, c in zip(nums, cost))
4    left, right = min(nums), max(nums)
5    while left < right:
6        mid = (left + right) // 2
7        cost_mid = calculate_cost(mid)
8        cost_mid_plus_one = calculate_cost(mid + 1)
9        if cost_mid < cost_mid_plus_one:
10            right = mid
11        else:
12            left = mid + 1
13    return calculate_cost(left)

ℹ

Complexity note: The time complexity is O(n log m) where n is the number of elements in nums and m is the range of values in nums, due to the binary search over possible target values and the cost calculation for each target.

1Changing elements to a target value that already exists in nums is optimal.
2The cost function is convex, allowing for efficient search methods like binary search.

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