HackMD
2439.Minimize Maximum of Array
tags: Medium,Array,Binary Search,Greedy,DP,Prefix Sum
2439. 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
isuch that1 <= i < nandnums[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
