Description (leetcode)

You are given an integer array nums (0-indexed). In one operation, you can choose an element of the array and increment it by 1.

  • For example, if nums = [1,2,3], you can choose to increment nums[1] to make nums = [1,**3**,3].

Return the minimum number of operations needed to make nums strictly increasing.

An array nums is strictly increasing if nums[i] < nums[i+1] for all 0 <= i < nums.length - 1. An array of length 1 is trivially strictly increasing.

Example 1:

Input: nums = [1,1,1]

Output: 3

Explanation: You can do the following operations:

  1. Increment nums[2], so nums becomes [1,1,2].

  2. Increment nums[1], so nums becomes [1,2,2].

  3. Increment nums[2], so nums becomes [1,2,3].

Example 2:

Input: nums = [1,5,2,4,1]

Output: 14

Example 3:

Input: nums = [8]

Output: 0

Constraints:

  • 1 <= nums.length <= 5000
  • 1 <= nums[i] <= 10^4

submission

impl Solution {
    pub fn min_operations(nums: Vec<i32>) -> i32 {
        nums.into_iter()
            .fold((0, 0), |(s, mut m), v| {
                m = (m + 1).max(v);
                (s + m - v, m)
            })
            .0
    }
}