The least operation of leetcode simple problem makes the array increment

subject

Give you an array of integers nums （ Subscript from 0 Start ）. In every operation , You can select an element in the array , And add it 1 .
For example , If nums = [1,2,3] , You can choose to increase it nums[1] obtain nums = [1,3,3] .
Please return to make nums Strictly increasing Of least Operating frequency .
We call it an array nums yes Strictly increasing , When it satisfies for all 0 <= i < nums.length - 1 There are nums[i] < nums[i+1] . A length of 1 Is a special case of strictly incrementing .
Example 1：
Input ：nums = [1,1,1]
Output ：3
explain ： You can do the following ：

1. increase nums[2] , The array becomes [1,1,2] .
2. increase nums[1] , The array becomes [1,2,2] .
3. increase nums[2] , The array becomes [1,2,3] .
   Example 2：
   Input ：nums = [1,5,2,4,1]
   Output ：14
   Example 3：
   Input ：nums = [8]
   Output ：0
   Tips ：
   1 <= nums.length <= 5000
   1 <= nums[i] <= 10^4
   source ： Power button （LeetCode）

Their thinking

   It can be like an example 1 That way, iterate over the array several times to change the array , You can also correct the currently unqualified value once during forward traversal , In this way, multiple backtracking is avoided .

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        count=0
        for i in range(1,len(nums)):
            if nums[i]<=nums[i-1]:
                count+=nums[i-1]+1-nums[i]
                nums[i]=nums[i-1]+1
        return count