# Minimum Operations

This post is part of the Algorithms Problem Solving series.

## Problem description

This is the Minimum Operations problem. The description looks like this:

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.

## Examples

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
``````

## Solution

First solution counting and update the array in place.

``````def min_operations(nums):
previous_num = nums[0]
operations = 0

for index, num in enumerate(nums[1:]):
if num <= previous_num:
new_num = previous_num - num + 1
operations += new_num
nums[index + 1] += new_num

previous_num = nums[index + 1]

return operations
``````

Second and more optimized solution without the need to update the array.

``````def min_operations(nums):
prev, ops = 0, 0

for num in nums:
if num <= prev:
prev += 1
ops += prev - num
else:
prev = num

return ops
``````

## Resources

Have fun, keep learning, and always keep coding!