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2439. Minimize Maximum of Array

Description

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.

 

Constraints:

  • n == nums.length
  • 2 <= n <= 105
  • 0 <= nums[i] <= 109

Solutions

To minimize the maximum value of the array, it is intuitive to use binary search. We binary search for the maximum value $mx$ of the array, and find the smallest $mx$ that satisfies the problem requirements.

The time complexity is $O(n \times \log M)$, where $n$ is the length of the array, and $M$ is the maximum value in the array.

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class Solution:
    def minimizeArrayValue(self, nums: List[int]) -> int:
        def check(mx):
            d = 0
            for x in nums[:0:-1]:
                d = max(0, d + x - mx)
            return nums[0] + d <= mx

        left, right = 0, max(nums)
        while left < right:
            mid = (left + right) >> 1
            if check(mid):
                right = mid
            else:
                left = mid + 1
        return left

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class Solution {
    private int[] nums;

    public int minimizeArrayValue(int[] nums) {
        this.nums = nums;
        int left = 0, right = max(nums);
        while (left < right) {
            int mid = (left + right) >> 1;
            if (check(mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

    private boolean check(int mx) {
        long d = 0;
        for (int i = nums.length - 1; i > 0; --i) {
            d = Math.max(0, d + nums[i] - mx);
        }
        return nums[0] + d <= mx;
    }

    private int max(int[] nums) {
        int v = nums[0];
        for (int x : nums) {
            v = Math.max(v, x);
        }
        return v;
    }
}

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class Solution {
public:
    int minimizeArrayValue(vector<int>& nums) {
        int left = 0, right = *max_element(nums.begin(), nums.end());
        auto check = [&](int mx) {
            long d = 0;
            for (int i = nums.size() - 1; i; --i) {
                d = max(0l, d + nums[i] - mx);
            }
            return nums[0] + d <= mx;
        };
        while (left < right) {
            int mid = (left + right) >> 1;
            if (check(mid))
                right = mid;
            else
                left = mid + 1;
        }
        return left;
    }
};

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func minimizeArrayValue(nums []int) int {
    check := func(mx int) bool {
        d := 0
        for i := len(nums) - 1; i > 0; i-- {
            d = max(0, nums[i]+d-mx)
        }
        return nums[0]+d <= mx
    }

    left, right := 0, slices.Max(nums)
    for left < right {
        mid := (left + right) >> 1
        if check(mid) {
            right = mid
        } else {
            left = mid + 1
        }
    }
    return left
}

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