logseq_notes/pages/OJ notes/pages/Leetcode Maximum-Difference-Between-Increasing-Elements.md

# Leetcode Maximum-Difference-Between-Increasing-Elements

#### 2022-06-27 11:09

> ##### Algorithms:
>
> #algorithm #Kadane_s_algorithm
>
> ##### Difficulty:
>
> #coding_problems #difficulty_easy
>
> ##### Additional tags:
>
> #leetcode
>
> ##### Revisions:
>
> N/A

##### Related topics:

##### Links:

- [Link to problem](https://leetcode.com/problems/maximum-difference-between-increasing-elements/)

  ***

### Problem

Given a **0-indexed** integer array `nums` of size `n`, find the **maximum difference** between `nums[i]` and `nums[j]` (i.e., `nums[j] - nums[i]`), such that `0 <= i < j < n` and `nums[i] < nums[j]`.

Return _the **maximum difference**._ If no such `i` and `j` exists, return `-1`.

#### Examples

Example 1:

```
Input: nums = [7,1,5,4]
Output: 4
Explanation:
The maximum difference occurs with i = 1 and j = 2, nums[j] - nums[i] = 5 - 1 = 4.
Note that with i = 1 and j = 0, the difference nums[j] - nums[i] = 7 - 1 = 6, but i > j, so it is not valid.
```

Example 2:

```
Input: nums = [9,4,3,2]
Output: -1
Explanation:
There is no i and j such that i < j and nums[i] < nums[j].
```

Example 3:

```
Input: nums = [1,5,2,10]
Output: 9
Explanation:
The maximum difference occurs with i = 0 and j = 3, nums[j] - nums[i] = 10 - 1 = 9.
```

#### Constraints

- n == nums.length
- 2 <= n <= 1000
- 1 <= nums[i] <= 109

### Thoughts

Since 0 <= i < j <=n, this can be completed using kadane's algo in one pass.

> [!summary]
> This is a #Kadane_s_algorithm

### Solution

```cpp
class Solution {
public:
int maximumDifference(vector<int>& nums) {
  // Kadane's algorithm, since 0 <= i < j < n, and it can be done using one loop.
  int minNum = nums[0];
  int maxNum = 0;

  for (int i : nums) {
    minNum = min(minNum, i);
    maxNum = max(maxNum, i - minNum);
  }

  if (maxNum == 0) {
    return -1;
  } else {
    return maxNum;
  }
}
};
```