2439. Minimize Maximum of Array

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# 題目敘述

You are given a 0-indexed array nums comprising of n non-negative integers.

In one operation, you must:

• Choose an integer i such that 1 <= i < n and nums[i] > 0 .
• Decrease nums[i] by 1.
• Increase nums[i - 1] by 1.

Return the minimum possible value of the maximum integer of nums after performing any number of operations.

# Example 1:

Input: nums = [3,7,1,6]
Output: 5
Explanation:
One set of optimal operations is as follows:

1. Choose i = 1, and nums becomes [4,6,1,6].
2. Choose i = 3, and nums becomes [4,6,2,5].
3. Choose i = 1, and nums becomes [5,5,2,5].
   The maximum integer of nums is 5. It can be shown that the maximum number cannot be less than 5.
   Therefore, we return 5.

# Example 2:

Input: nums = [10,1]
Output: 10
Explanation:
It is optimal to leave nums as is, and since 10 is the maximum value, we return 10.

# 解題思路

在題目敘述中我們可以知道， nums 這個陣列無法將前面較大的數值向後移，只能將後面的數值向前移動。

因此我們可以藉由 prefix sum 來計算到目前的總值，除以現在的陣列數目，得到平均值利用高斯取頂 (ceiling function) $\Rightarrow$ 得到當前 subArray 數目的最大值

並與前面的最大值比較:

• 如果比較大 $\Rightarrow$ 就代表後面的值可以再向前移動。
• 如果比較小 $\Rightarrow$ 就表示雖然平均最大值可能有更小的，但是前面的值不可以往後移，因此不採用。

# Algorithm

1. Initialize ans = 0 and prefixSum = 0 .
2. Iterate over nums , for each index i :
   • Update the prefix sum as prefixSum += nums[i] .
   • Check the maximum value we can obtain by averaging prefixSum into i + 1 evenly using ceiling division.
   • Take the larger one from ans and the result from the previous integer division.
3. Return ans

# Solution

class Solution {

    public int minimizeArrayValue(int[] nums) {

        long prefixSum = 0;

        int ans = 0;

        for(int i = 0; i < nums.length; i++){

            prefixSum += nums[i];

            ans = Math.max(ans, (int)Math.ceil(prefixSum * 1.0 / (i + 1)));

        }

        return ans;

    }

}