# LeetCode 1827. Minimum Operations to Make the Array Increasing

## Description

https://leetcode.com/problems/minimum-operations-to-make-the-array-increasing/

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] <= 104`

## Explanation

Iterate the list and whenever find a number is no greater than the previous one, increase that number and count the operations.

## Python Solution

``````class Solution:
def minOperations(self, nums: List[int]) -> int:

count = 0

if len(nums) < 2:
return count

prev = None
for i, num in enumerate(nums):
if prev and num <= prev:
count += prev - num + 1
nums[i] = prev + 1
prev = nums[i]

return count
``````
• Time Complexity: O(N).
• Space Complexity: O(1).