2439.Minimize Maximum of Array
===

###### tags: `Medium`,`Array`,`Binary Search`,`Greedy`,`DP`,`Prefix Sum`

[2439. Minimize Maximum of Array](https://leetcode.com/problems/minimize-maximum-of-array/)

### 題目描述

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.*

### 範例

**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.

**Constraints**:

* `n` == `nums.length`
* 2 <= `n` <= 10^5^
* 0 <= `nums[i]` <= 10^9^

### 解答
#### C++

class Solution {
public:
    int minimizeArrayValue(vector<int>& nums) {
        long long ans = 0, sum = 0;
        for (int i = 0; i < nums.size(); i++) {
            sum += nums[i];
            ans = max(ans, (sum + i) / (i + 1));
        }
        return ans;
    }
};

> [name=Yen-Chi Chen][time=Wed, Apr 5, 2023]

#### Python

class Solution:
    def minimizeArrayValue(self, nums: List[int]) -> int:
        ans, total = 0, 0
        for i, num in enumerate(nums):
            total += num
            ans = max(ans, (total + i) // (i + 1))
        return ans

> [name=Yen-Chi Chen][time=Wed, Apr 5, 2023]

#### Javascript

function minimizeArrayValue(nums) {
  let min = nums[0];
  let max = Math.max(...nums);

  while (max > min) {
    const mid = Math.floor((max + min) / 2);
    let sum = 0;
    for (const num of nums) {
      sum += mid - num;
      if (sum < 0) break;
    }

    if (sum < 0) {
      min = mid + 1;
    } else {
      max = mid;
    }
  }

  return min;
}

> binary search版本
> [name=Marsgoat][time=Apr 11, 2023]

function minimizeArrayValue(nums) {
  let result = 0;
  let sum = 0;
  for (let i = 0; i < nums.length; i++) {
    sum += nums[i];
    result = Math.max(result, Math.ceil(sum / (i + 1)));
  }

  return result;
}

> 跟吉神學習了，真的好神...
> [name=Marsgoat][time=Apr 11, 2023]

### Reference

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