Maximum Difference Between Increasing Elements

To solve this problem, we can use a simple one-pass algorithm. We maintain a variable min_val to store the minimum value found so far and max_diff to store the maximum difference found so far.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
class Solution:
    def maximumDifference(self, nums: List[int]) -> int:
        min_val = nums[0]
        max_diff = -1
        for i in range(1, len(nums)):
            if nums[i] > min_val:
                max_diff = max(max_diff, nums[i] - min_val)
            else:
                min_val = nums[i]
        return max_diff

Here’s what this code does:

We first initialize min_val with the first element of the array and max_diff as -1 because it’s the smallest possible output value.
We then iterate over the array starting from the second element (nums[1]).
If the current element is greater than min_val, we calculate the difference (nums[i] - min_val) and update max_diff if this difference is greater than the current max_diff.
If the current element is not greater than min_val, it becomes the new min_val.
We repeat this process for all elements in the array. At the end, max_diff will hold the maximum difference as per the problem statement.

This works because we are essentially looking for the greatest upward ‘jump’ in the array. By maintaining the minimum value seen so far (min_val), we are always ready to calculate the maximum difference when we see a larger value.

Building Blocks

The building blocks of this solution are:

Array traversal: This solution requires a single pass through the given array, from the second element to the last. This is a common technique used in many problems involving arrays.
Tracking minimum value: As we move through the array, we continually update the minimum value seen so far. This helps us to calculate the maximum difference when we encounter a larger value.
Tracking maximum difference: This involves constantly comparing and updating a stored value (max_diff) based on current computations (nums[i] - min_val), another common technique in a variety of problems.
Comparison operations: The solution involves comparing numbers to determine if one is greater than the other (nums[i] > min_val) and to find the maximum difference (max_diff = max(max_diff, nums[i] - min_val)).
Conditional logic: Depending on whether the current number is greater than the minimum value seen so far, we choose to either update max_diff or min_val.
Return the result: Finally, the maximum difference found during the traversal is returned as the result.

These building blocks are utilized together to build the solution for the problem.

I see Running Maximum Difference, Running Minimum. Is there others?

Yes, in addition to “Running Minimum” and “Running Maximum Difference”, there are a few more concepts at play here:

Sequential Access: We traverse the list in a sequential manner, one element after another. This is an essential concept for a majority of array-based problems.
Greedy Approach: The idea of always taking the best option available at the moment (updating max_diff whenever we see a larger difference, or updating min_val when we see a smaller value) is a hallmark of the greedy approach.
State Tracking: We maintain and update certain ‘state’ variables (min_val and max_diff) as we progress through the array.

While the problem’s solution may primarily hinge on the concepts of “Running Minimum” and “Running Maximum Difference”, these other conceptual building blocks also play crucial roles in forming a complete, working solution.

Abstract Representation of the Problem

Formulating an abstract representation can help in identifying the underlying structure and recognizing similar problems.

Here’s an abstract representation of this problem:

Given a sequence of integers, find the maximum positive difference between two elements such that the smaller element comes before the larger element in the sequence. If there is no such pair of elements, return a specific value to indicate this.

Key elements:

The sequence of integers, which we traverse in a linear manner.
The “maximum positive difference” we’re looking to identify, which is analogous to a running maximum difference.
The condition that the smaller element must come before the larger one, which introduces the concept of order.
The return value when there’s no valid pair, which provides a way to handle edge cases.

In this abstract representation, we’ve stripped away the specific real-world context and focused on the structural essence of the problem. We can see how this formulation could apply to various scenarios beyond stock prices or the specific number array given in this problem.

Drills

Exercise 1 - Working with lists in Python:
We use a loop to iterate over each element in the list and print it.
1 2 3
def print_list(nums): for num in nums: print(num)

Exercise 2 - Find the minimum value in a list:

We initialize a variable minimum with the first element in the list, then iterate over the list. If we encounter a number smaller than minimum, we update minimum.

1
2
3
4
5
6
def find_minimum(nums):
    minimum = nums[0]
    for num in nums:
        if num < minimum:
            minimum = num
    return minimum

Exercise 3 - Find the maximum difference in a list:

We initialize a variable max_difference as zero, then use a nested loop to find the maximum difference between any two numbers.

1
2
3
4
5
6
7
def find_max_difference(nums):
    max_difference = 0
    for i in range(len(nums)):
        for j in range(i+1, len(nums)):
            if nums[j] - nums[i] > max_difference:
                max_difference = nums[j] - nums[i]
    return max_difference

Exercise 4 - Combining the concepts:

Here we use the concepts from the previous exercises. We keep track of the minimum value found so far and the maximum difference found so far. By iterating only once through the list, we achieve a much more efficient solution than in Exercise 3.

1
2
3
4
5
6
7
8
9
def maximum_difference(nums):
    min_value = nums[0]
    max_difference = -1
    for i in range(1, len(nums)):
        if nums[i] > min_value:
            max_difference = max(max_difference, nums[i] - min_value)
        else:
            min_value = nums[i]
    return max_difference

Please note that in Exercise 4, we return -1 when no maximum difference is found according to the problem’s requirement. This is different from Exercise 3 where we returned 0 when no difference was found.

Q&A

Is this an example of sliding window?

No, this is not an example of the sliding window technique. The sliding window technique typically involves maintaining a ‘window’ of elements in an array or list that moves, or ‘slides’, over the course of the algorithm. The window’s size might remain fixed or vary depending on the problem’s requirements.

In this problem, we are not maintaining a window of elements. Instead, we are traversing the entire array once, tracking the minimum value found so far and the maximum difference found so far. These two variables (min_val and max_diff) are updated as needed while scanning through the array, but they do not represent a ‘window’ of elements in the array.

Exactly the same code as “best time to buy and sell stocks”. Do you agree?

Yes, I agree. The logic behind this problem is indeed very similar to the “Best Time to Buy and Sell Stocks” problem. Both problems require finding the maximum difference between two numbers in an array where the smaller number comes before the larger one.

In “Best Time to Buy and Sell Stocks”, the numbers are prices of a stock, and you’re trying to maximize your profit by choosing the best time to buy (the lowest price) and sell (the highest price after the buy).

Here, the problem is abstracted a bit more, but the core idea remains the same: find the largest difference between two numbers where the smaller number appears before the larger one in the array. The same technique of maintaining a running minimum and a running maximum difference can be used to solve both problems efficiently.

Maximum Difference Between Increasing Elements

Building Blocks

Abstract Representation of the Problem

Drills

Q&A

Similar Problems