Minimum Cost to Make Array Equal

Problem Statement

In this problem, you are provided with two arrays: nums and cost, both of size n and filled with positive integers. The array nums represents a sequence of numbers, while cost details the cost associated with modifying each corresponding element in nums.

The available operation allows you to either increment or decrement an element in nums by 1, with the modification cost of manipulating the i-th element given as cost[i]. Your objective is to determine the smallest possible cumulative cost to make all entries in nums uniform (i.e., all entries are the same).

The challenge is to compute this minimum cost efficiently, considering the constraints on the length of the arrays and the values they hold.

Examples

Example 1

Input:

nums = [1,3,5,2], cost = [2,3,1,14]

Output:

8

Explanation:

We can make all the elements equal to 2 in the following way:
- Increase the 0th element one time. The cost is 2.
- Decrease the 1st element one time. The cost is 3.
- Decrease the 2nd element three times. The cost is 1 + 1 + 1 = 3.
The total cost is 2 + 3 + 3 = 8.
It can be shown that we cannot make the array equal with a smaller cost.

Example 2

Input:

nums = [2,2,2,2,2], cost = [4,2,8,1,3]

Output:

0

Explanation:

All the elements are already equal, so no operations are needed.

Constraints

• n == nums.length == cost.length
• 1 <= n <= 105
• 1 <= nums[i], cost[i] <= 106
• Test cases are generated in a way that the output doesn't exceed 253-1

Approach and Intuition

The problem essentially boils down to finding a target value in the nums array that can be attained from all other numbers at the lowest total modification cost, given the cost for changing each element is specific and listed in cost.

1. Understanding the relation between cost and the value adjustments:

   • Since each element can be adjusted by increasing or decreasing its value at a specific cost, you should aim at adjusting all the values in nums to a common target that minimizes the sum of these costs.

2. Analyzing sample cases for patterns:

   • In the first example, adjusting all numbers to 2 was optimal. Meanwhile, in the second example, since all elements were already the same, no cost was necessary.

3. Determining the target value:

   • An effective strategy is to consider changing every number in nums to one common value and calculate the required cost for each potential target.
   • This target could be derived based on the mean or the median of the numbers factored by their respective costs, as these statistical measures often represent central tendencies which could be nearer to the cost-effective target.

4. Calculating the cost for adjusting each potential target:

   • Iterate through each number in nums, calculate the difference from the current target, multiply this by its respective cost in cost, and sum these values.
   • Due to the potentially large size of n, this operation should be as optimized as possible, likely necessitating an algorithm that runs in linear or linearithmic time relative to n.

5. Concluding the minimum total cost:

   • The minimum out of all calculated total costs for possible targets gives the desired answer.

It's essential to identify an efficient way to manage large inputs and outputs, as per the constraints, while calculating the cumulative costs dynamically to avoid recalculating from scratch for each potential target. This means using pre-computed sums or similar techniques to optimize the calculations.

Solutions

class Solution {
public:
    long long calculateTotalCost(vector<int>& elements, vector<int>& adjustments, int value) {
        long long total = 0;
        for (int index = 0; index < elements.size(); ++index)
            total += 1L * abs(elements[index] - value) * adjustments[index];
        return total;
    }
        
    long long findMinCost(vector<int>& elements, vector<int>& adjustments) {
        long long minimalCost = calculateTotalCost(elements, adjustments, elements[0]);
        int lowerBound = *min_element(elements.begin(), elements.end());
        int upperBound = *max_element(elements.begin(), elements.end());
            
        while (lowerBound < upperBound) {
            int middle = (lowerBound + upperBound) / 2;
            long long costLeft = calculateTotalCost(elements, adjustments, middle);
            long long costRight = calculateTotalCost(elements, adjustments, middle + 1);
            minimalCost = min(costLeft, costRight);
            if (costLeft > costRight)
                lowerBound = middle + 1;
            else
                upperBound = middle;
        }
        return minimalCost;
    } 
};