2366 Minimum Replacements to Sort the Array.md

2366. Minimum Replacement to Sort the Array

Intuition

This is definitely math question. The two key points are to compute parts and the prev value. As the hint saying, we need to start from the second-to-last since the last one can't be splitted.

Approach

1. Start traversing from the second-to-last element of the array
2. For each element nums[i], if it's greater than the previously processed element prev:
   • Calculate how many parts we need to split the current number into: (num + prev - 1) / prev
   • Update prev to be the minimum value after splitting: num / parts
   • Add the number of operations needed: partss - 1
3. If the current element is less than or equal to prev, no splitting is needed
4. Return the total number of operations

Complexity

• Time complexity: O(n)
• Space complexity: O(1)

Keywords

• Greedy Algorithm
• Mathematics

Code

func minimumReplacement(nums []int) int64 {
    if len(nums) == 1 {
        return 0
    }
    ret := int64(0)
    var split func(num int, prev *int) int
    split = func(num int, prev *int) int {
        if num <= *prev {
            *prev = num
            return 0
        }
        tmp := (num + *prev - 1) / *prev
        *prev = num / tmp
        return tmp - 1
    }
    prev := nums[len(nums) - 1]
    for i := len(nums) - 2; i >= 0; i -= 1 {
        ret += int64(split(nums[i], &prev))
    }
    return ret
}