> For the complete documentation index, see [llms.txt](https://zhenchaogan.gitbook.io/leetcode-solution/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://zhenchaogan.gitbook.io/leetcode-solution/leetcode-1827-minimum-operations-to-make-the-array-increasing.md).

# LeetCode 1827. 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`

## Solution

```
class Solution {
public:
    int minOperations(vector<int>& nums) {
        int ans = 0;
        for (int i = 1; i < nums.size(); i++) {
            int previous = nums[i-1];
            int current = nums[i];
            
            if (current <= previous) {
                current = previous + 1;
                ans += (current - nums[i]);
                nums[i] = current;
            }
        }
        
        return ans;
    }
};
```