LeetCode 2439. Minimize Maximum of Array Solution in Java, C++, Python & Go | Explanation + Code

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

Solution 1: Binary Search

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 × log M), where n is the length of the array, and M is the maximum value in the array.

PythonJavaC++Go

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)