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i want u to calculate for each element what is the total cost of converstion , for example to covert to 1 , what is the total cost , to convert to 3 what is the total cost and so on for all nums

Total Conversion Costs Analysis
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Here are the total costs to convert all elements to each value:
1: 352 units
2: 253 units
3: 186 units
5: 142 units
6: 128 units (minimum)
7: 138 units
8: 160 units
10: 256 units

The optimal value is 6, with minimum total cost of 128 units.

*my understanding* : there is a downward peak when u draw the graph , i did in A4 , i see , its possible there can be multiple downward peak with constant values. so minimum is guarnteed. There can be only one downward peak with same values.

this is a application of finding peak problem.
where x is the nums values and y is the total cost of converting to each nums. we do see a downward parabola/downward peak.

Let me break down the solution and intuition:
Key Insights:

If we plot the total cost against possible target values, it forms a convex function (like a parabola)
The optimal value must lie between the minimum and maximum values in the array
As we move away from the optimal value, the cost increases on both sides

Algorithm:

We use binary search to find the optimal target value:

Search space is from min to max value in nums array
For each mid point, calculate cost for mid and mid+1
If cost(mid) > cost(mid+1), optimal value is on right side
If cost(mid) < cost(mid+1), optimal value is on left side


For calculating cost:

For each number, find absolute difference from target
Multiply difference by cost of operations for that index
Sum up all costs



Time Complexity: O(n * log M) where:

n is the length of arrays
M is the range of values (max - min)

Space Complexity: O(1)
The binary search approach works because:

The cost function is convex (U-shaped)
If cost(x) > cost(x+1), minimum lies to the right
If cost(x) < cost(x+1), minimum lies to the left

This gives us an efficient way to find the optimal value without trying every possible target number.


class Solution {
    /**
     * The key intuition is:
     * 1. The optimal final value will be between min and max of array
     * 2. The cost function forms a parabola - as we move away from optimal value, cost increases
     * 3. We can use binary search to find the value giving minimum cost
     */
    public long minCost(int[] nums, int[] cost) {
        // Find range for binary search
        int low = nums[0], high = nums[0];
        for (int num : nums) {
            low = Math.min(min, num);
            high = Math.max(max, num);
        }
        
        // Binary search for the optimal value
        long answer = getCost(nums, cost, nums[0]);
        while (low <= high) {
            int mid = low + (high - low) / 2;
            long cost1 = getCost(nums, cost, mid);
            long cost2 = getCost(nums, cost, mid + 1);
            
            answer = Math.min(cost1, cost2);
            
            // If cost is decreasing, move right
            if (cost1 > cost2) {
                low = mid + 1;
            }
            // If cost is increasing, move left
            else {
                high = mid-1;
            }
        }
        
        return answer;
    }
    
    /**
     * Calculate total cost to convert all numbers to target
     */
    private long getCost(int[] nums, int[] cost, int mid) {
        long totalCost = 0;
        for (int i = 0; i < nums.length; i++) {
            // Calculate difference and multiply by cost per operation
            long diff = Math.abs(nums[i] - mid);
            totalCost += diff * cost[i];
        }
        return totalCost;
    }
}