Maximum Difference Between Increasing Elements

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.

For example:

nums = [7,1,5,4] should return 4 nums = [9,4,3,2] should return -1 nums = [1,5,2,10] should return 9

## Maximum Difference Between Increasing Elements

### Problem Description

Given a **0-indexed** integer array `nums` of size `n`, the goal is to 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]`. If no such `i` and `j` exist, return `-1`.

### Naive Solution

A brute-force approach involves iterating through all possible pairs `(i, j)` such that `i < j`, checking if `nums[i] < nums[j]`, and updating the maximum difference accordingly. This approach has a time complexity of O(n^2).

```python
def maximumDifference_naive(nums):
    max_diff = -1
    n = len(nums)
    for i in range(n):
        for j in range(i + 1, n):
            if nums[i] < nums[j]:
                max_diff = max(max_diff, nums[j] - nums[i])
    return max_diff

Optimal Solution

A more efficient approach is to iterate through the array once, keeping track of the minimum element seen so far. For each element nums[j], we check if it's greater than the minimum element seen before nums[i]. If it is, we update the maximum difference. This approach has a time complexity of O(n).

def maximumDifference_optimal(nums):
    max_diff = -1
    min_element = nums[0]
    for j in range(1, len(nums)):
        if nums[j] > min_element:
            max_diff = max(max_diff, nums[j] - min_element)
        else:
            min_element = nums[j]  # Update min_element if a smaller number is found
    return max_diff

Example

For nums = [7, 1, 5, 4], the optimal solution works as follows:

1. Initialize max_diff = -1 and min_element = 7.
2. Iterate through the array:
   • j = 1, nums[j] = 1. Since 1 < 7, update min_element = 1.
   • j = 2, nums[j] = 5. Since 5 > 1, update max_diff = max(-1, 5 - 1) = 4.
   • j = 3, nums[j] = 4. Since 4 > 1, update max_diff = max(4, 4 - 1) = 4.
3. Return max_diff = 4.

Big(O) Run-time Analysis

The optimal solution iterates through the array once. Therefore, the time complexity is O(n), where n is the number of elements in the array. The naive solution has nested loops, resulting in O(n^2) time complexity.

Big(O) Space Usage Analysis

The optimal solution uses a constant amount of extra space to store max_diff and min_element. Therefore, the space complexity is O(1), which means it's constant and doesn't depend on the size of the input array.

Edge Cases

1. Empty Array: If the input array is empty, the algorithm should return -1.
2. Array with One Element: If the input array has only one element, it doesn't satisfy the condition i < j, so the algorithm should return -1. Note that based on the constraints in the problem statement, the length of nums will always be between 2 and 1000 (inclusive), so we don't have to handle arrays of length 0 or 1.
3. Non-Increasing Array: If the array is sorted in non-increasing order (e.g., [9, 4, 3, 2]), there will be no i and j such that nums[i] < nums[j]. The algorithm should return -1.
4. All Elements are Equal: If all elements in the array are the same (e.g. [5, 5, 5, 5]), the algorithm will also return -1 because no i and j will satisfy nums[i] < nums[j].

The provided optimal solution correctly handles all these cases.