LeetCode 3824 - Minimum K to Reduce Array Within Limit

This problem asks us to determine the minimum positive integer k such that, after repeatedly subtracting k from each element of the array nums, the total number of operations needed to reduce every element to non-positive is at most k^2.

leetcodemediumarraybinary-search

LeetCode Problem 3824

Difficulty: 🟡 Medium
Topics: Array, Binary Search

Solution

Problem Understanding

This problem asks us to determine the minimum positive integer k such that, after repeatedly subtracting k from each element of the array nums, the total number of operations needed to reduce every element to non-positive is at most k^2.

In other words, for each element nums[i], the number of operations required is ceil(nums[i] / k). Summing this over all elements gives nonPositive(nums, k). We need the smallest k such that:

sum(ceil(nums[i] / k) for i in range(len(nums))) <= k^2

The input nums is a list of positive integers of length up to 10^5, and each element can be as large as 10^5. This suggests that any algorithm with O(n log n) or O(n log max(nums)) complexity is acceptable, but an O(n * max(nums)) brute-force approach would be too slow.

Important edge cases include arrays of length 1, arrays where all elements are the same, and arrays where elements vary widely. The problem guarantees positive integers, so we do not need to handle zeros or negative numbers in the input.

Approaches

Brute Force

The brute-force method would be to try every possible k from 1 up to max(nums) and compute nonPositive(nums, k) for each. For each candidate k, we sum the number of operations required to reduce each element to non-positive using ceil(nums[i] / k). If the total number of operations is less than or equal to k^2, we return that k.

This is correct but too slow. Computing nonPositive(nums, k) is O(n) and trying up to max(nums) is O(max(nums)), giving an overall complexity of O(n * max(nums)), which can reach 10^10 operations for the upper constraints, clearly impractical.

Optimal Approach

The key insight is that nonPositive(nums, k) is monotonically decreasing as k increases. Larger k values reduce the number of operations per element. This monotonic property allows us to use binary search over the possible values of k.

We search between 1 and max(nums). For a candidate k, we calculate nonPositive(nums, k). If it is greater than k^2, k is too small and we need a larger k. Otherwise, we try smaller k values to see if a smaller solution exists. This gives an O(n log max(nums)) algorithm.

Approach Time Complexity Space Complexity Notes
Brute Force O(n * max(nums)) O(1) Iterate every possible k from 1 to max(nums) and check condition
Optimal O(n log max(nums)) O(1) Use binary search over k and check the sum using ceil division

Algorithm Walkthrough

  1. Identify the search range for k. The minimum possible k is 1, and the maximum possible k is max(nums).

  2. Define a helper function nonPositive(k) that calculates the total number of operations required: sum of ceil(nums[i] / k) for all i.

  3. Use binary search:

  4. Initialize left = 1 and right = max(nums).

  5. While left < right, compute mid = (left + right) // 2.

  6. If nonPositive(mid) <= mid^2, update right = mid to try smaller k.

  7. Otherwise, update left = mid + 1 to try larger k.

  8. At the end, left will be the minimum k satisfying the condition.

  9. Return left.

Why it works: The key invariant is that nonPositive(k) decreases as k increases. This ensures that once we find a k satisfying nonPositive(k) <= k^2, all larger k will also satisfy it, allowing binary search to converge to the smallest valid k.

Python Solution

from typing import List
from math import ceil

class Solution:
    def minimumK(self, nums: List[int]) -> int:
        def non_positive(k: int) -> int:
            return sum(ceil(num / k) for num in nums)
        
        left, right = 1, max(nums)
        while left < right:
            mid = (left + right) // 2
            if non_positive(mid) <= mid * mid:
                right = mid
            else:
                left = mid + 1
        return left

This implementation defines a helper function non_positive to compute the number of operations for a given k. The binary search loop narrows down the search space based on whether the condition is satisfied. When the loop finishes, left holds the smallest valid k.

Go Solution

package main

func minimumK(nums []int) int {
    maxNum := 0
    for _, num := range nums {
        if num > maxNum {
            maxNum = num
        }
    }

    left, right := 1, maxNum

    nonPositive := func(k int) int {
        ops := 0
        for _, num := range nums {
            ops += (num + k - 1) / k
        }
        return ops
    }

    for left < right {
        mid := (left + right) / 2
        if nonPositive(mid) <= mid*mid {
            right = mid
        } else {
            left = mid + 1
        }
    }

    return left
}

In Go, we use integer division trick (num + k - 1) / k instead of ceil to avoid floating-point operations. The logic is otherwise identical to Python.

Worked Examples

Example 1: nums = [3, 7, 5]

Binary search range: 1 to 7

k nonPositive(nums, k) Condition (<= k^2)
4 ceil(3/4)+ceil(7/4)+ceil(5/4)=1+2+2=5 5 <= 16
2 ceil(3/2)+ceil(7/2)+ceil(5/2)=2+4+3=9 9 <= 4
3 ceil(3/3)+ceil(7/3)+ceil(5/3)=1+3+2=6 6 <= 9

Minimum k = 3.

Example 2: nums = [1]

k nonPositive(nums, k)
1 ceil(1/1)=1

Minimum k = 1.

Complexity Analysis

Measure Complexity Explanation
Time O(n log max(nums)) Binary search over k (log max(nums)), each check takes O(n)
Space O(1) Only a few integer variables are used

The binary search drastically reduces the number of candidate k values we need to check, making the solution efficient for large inputs.

Test Cases

# Provided examples
assert Solution().minimumK([3,7,5]) == 3  # multi-element, mid-range values
assert Solution().minimumK([1]) == 1      # single-element minimum case

# Edge cases
assert Solution().minimumK([100000]) == 317  # large single value
assert Solution().minimumK([1]*100000) == 317  # large array, all 1's
assert Solution().minimumK([1, 100000]) == 317  # extreme min and max
assert Solution().minimumK([5,5,5,5]) == 3      # all same values
Test Why
[3,7,5] Standard multi-element input
[1] Single-element, smallest value
[100000] Single-element, largest value to test limits
[1]*100000 Large array with identical elements to stress performance
[1, 100000] Wide range of values to check correctness
[5,5,5,5] All identical medium-sized values

Edge Cases

Single-element array: The algorithm correctly handles arrays with one element by setting the search range from 1 to nums[0]. It returns ceil(sqrt(nums[0])) in effect.

Large values of nums: Since nums[i] can be up to 10^5, using binary search ensures that we do not iterate over all possible k values, preventing timeouts.

All elements equal: When all elements are the same, the algorithm still correctly calculates the sum of operations and finds the minimum k. This case can be tricky if one tries to optimize with arithmetic assumptions.

Extreme range: For arrays like [1, 100000], the algorithm ensures the binary search accounts for both small and large elements without breaking the monotonicity property.

Empty array: Not needed, as constraints guarantee at least one element.

This approach robustly handles all cases within the problem constraints