2439.Minimize Maximum of Array

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

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 <= 10⁵
0 <= nums[i] <= 10⁹

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;
    }
};

Yen-Chi ChenWed, 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

Yen-Chi ChenWed, 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
MarsgoatApr 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…
MarsgoatApr 11, 2023

Reference

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