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Maximum Difference Between Increasing Elements

Easy

Cisco

1 view

Topics:

Arrays

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.

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

Try coding it in LeetCode

Maximum Difference Between Increasing Elements

Easy

Cisco

1 view

Topics:

Arrays

Return the maximum difference. If no such i and j exists, return -1.

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the input array `nums`?
Can the input array `nums` contain negative numbers, zero, or only positive integers?
If there are multiple pairs that satisfy the condition with the same maximum difference, can I return any one of them?
Is the input array guaranteed to be non-empty?
Can you provide some examples of edge cases or particularly tricky inputs I should be aware of?

Brute Force Solution

Approach

The brute force approach for finding the maximum difference between increasing numbers is straightforward. It involves checking every possible pair of numbers to find the largest difference where the second number is bigger than the first.

Here's how the algorithm would work step-by-step:

Start by picking the very first number in the list.
Compare this number to every number that comes after it in the list.
If a later number is bigger than the first number you picked, calculate the difference between them.
Keep track of the biggest difference you've found so far.
Now, repeat the process, but this time start with the second number in the list.
Again, compare this number to every number that comes after it.
If a later number is bigger, calculate the difference and see if it's the biggest one yet.
Keep repeating this process, moving through each number in the list and comparing it to all the numbers that come after it.
Once you've checked every possible pair, the largest difference you've kept track of is your answer.

Code Implementation

def maximum_difference_brute_force(numbers):
    maximum_difference = -1

    # Iterate through each number as a potential smaller number
    for i in range(len(numbers)):
        for j in range(i + 1, len(numbers)):

            # Check if the number at the later index is greater.
            if numbers[j] > numbers[i]:

                #We found a valid increasing pair, calculate the difference
                current_difference = numbers[j] - numbers[i]

                #Update maximum difference if needed
                if current_difference > maximum_difference:
                    maximum_difference = current_difference
    return maximum_difference

Big(O) Analysis

Time Complexity

O(n²) – The provided brute force approach involves iterating through the array and, for each element, comparing it with all subsequent elements. This creates a nested loop structure where the outer loop iterates n times, and the inner loop iterates approximately n/2 times on average. Consequently, we perform roughly n * (n/2) comparisons. Discarding the constant factor, the time complexity is O(n²).

Space Complexity

O(1) – The brute force approach described iterates through the input array, keeping track of only the maximum difference found so far. No auxiliary data structures like arrays, hash maps, or trees are created. Therefore, the space required remains constant regardless of the input size N, making the space complexity O(1).

Optimal Solution

Approach

The goal is to find the biggest difference between two numbers where the larger number comes after the smaller number. The best approach is to keep track of the smallest number we've seen so far and compare it to each number as we go along to potentially update the largest difference found.

Here's how the algorithm would work step-by-step:

Start by assuming the biggest difference we've found is zero.
Also, start by assuming the smallest number we've seen is the very first number we're given.
Go through the rest of the numbers, one at a time.
For each number, check if it's smaller than the smallest number we've seen so far. If it is, update the smallest number.
If the current number is bigger than the smallest number we've seen, calculate the difference between the current number and the smallest number.
If this difference is bigger than the biggest difference we've found so far, update the biggest difference.
After checking all the numbers, the biggest difference we've tracked is the answer.

Code Implementation

def maximum_difference_between_increasing_elements(numbers):
    maximum_difference = 0
    smallest_number_seen_so_far = numbers[0]

    # Initialize smallest number to first element

    for current_number in numbers:
        # Update the smallest number if a smaller number is encountered
        if current_number < smallest_number_seen_so_far:

            smallest_number_seen_so_far = current_number

        # If the current number is greater, calculate the difference
        if current_number > smallest_number_seen_so_far:

            difference = current_number - smallest_number_seen_so_far
            # Update the maximum difference if a larger difference is found
            if difference > maximum_difference:

                maximum_difference = difference

    return maximum_difference

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array nums of size n once. In each iteration, it performs a constant amount of work: comparing the current element with the current minimum and potentially updating the maximum difference or the minimum. Therefore, the time complexity is directly proportional to the number of elements in the array, resulting in O(n) time complexity.

Space Complexity

O(1) – The algorithm uses a constant amount of extra space. It only stores two variables: one for the current smallest number seen so far and another for the maximum difference found. The space used by these variables does not depend on the size of the input array, N. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty or null input array	Return -1 immediately as no valid difference can exist.
Input array with only one element	Return -1 as there's no pair to calculate the difference.
Array with all elements in descending order	The algorithm should return -1 since no increasing pair exists.
Array with all identical values	The algorithm should return -1 as no increasing pair exists.
Array with negative numbers	The solution should correctly handle negative numbers in the array.
Integer overflow when calculating the difference	Use a data type with a larger range like long to prevent overflow if the difference might exceed the range of int.
Array with extreme large positive and negative numbers	Ensure the comparison and difference calculation handle large numbers correctly without causing overflow or unexpected behavior.
Large input array to test for performance	The algorithm should have a time complexity of O(n) for optimal performance.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the input array `nums`?
Can the input array `nums` contain negative numbers, zero, or only positive integers?
If there are multiple pairs that satisfy the condition with the same maximum difference, can I return any one of them?
Is the input array guaranteed to be non-empty?
Can you provide some examples of edge cases or particularly tricky inputs I should be aware of?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start by picking the very first number in the list.
Compare this number to every number that comes after it in the list.
If a later number is bigger than the first number you picked, calculate the difference between them.
Keep track of the biggest difference you've found so far.
Now, repeat the process, but this time start with the second number in the list.
Again, compare this number to every number that comes after it.
If a later number is bigger, calculate the difference and see if it's the biggest one yet.
Keep repeating this process, moving through each number in the list and comparing it to all the numbers that come after it.
Once you've checked every possible pair, the largest difference you've kept track of is your answer.

Code Implementation

def maximum_difference_brute_force(numbers):
    maximum_difference = -1

    # Iterate through each number as a potential smaller number
    for i in range(len(numbers)):
        for j in range(i + 1, len(numbers)):

            # Check if the number at the later index is greater.
            if numbers[j] > numbers[i]:

                #We found a valid increasing pair, calculate the difference
                current_difference = numbers[j] - numbers[i]

                #Update maximum difference if needed
                if current_difference > maximum_difference:
                    maximum_difference = current_difference
    return maximum_difference

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start by assuming the biggest difference we've found is zero.
Also, start by assuming the smallest number we've seen is the very first number we're given.
Go through the rest of the numbers, one at a time.
For each number, check if it's smaller than the smallest number we've seen so far. If it is, update the smallest number.
If the current number is bigger than the smallest number we've seen, calculate the difference between the current number and the smallest number.
If this difference is bigger than the biggest difference we've found so far, update the biggest difference.
After checking all the numbers, the biggest difference we've tracked is the answer.

Code Implementation

def maximum_difference_between_increasing_elements(numbers):
    maximum_difference = 0
    smallest_number_seen_so_far = numbers[0]

    # Initialize smallest number to first element

    for current_number in numbers:
        # Update the smallest number if a smaller number is encountered
        if current_number < smallest_number_seen_so_far:

            smallest_number_seen_so_far = current_number

        # If the current number is greater, calculate the difference
        if current_number > smallest_number_seen_so_far:

            difference = current_number - smallest_number_seen_so_far
            # Update the maximum difference if a larger difference is found
            if difference > maximum_difference:

                maximum_difference = difference

    return maximum_difference

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty or null input array	Return -1 immediately as no valid difference can exist.
Input array with only one element	Return -1 as there's no pair to calculate the difference.
Array with all elements in descending order	The algorithm should return -1 since no increasing pair exists.
Array with all identical values	The algorithm should return -1 as no increasing pair exists.
Array with negative numbers	The solution should correctly handle negative numbers in the array.
Integer overflow when calculating the difference	Use a data type with a larger range like long to prevent overflow if the difference might exceed the range of int.
Array with extreme large positive and negative numbers	Ensure the comparison and difference calculation handle large numbers correctly without causing overflow or unexpected behavior.
Large input array to test for performance	The algorithm should have a time complexity of O(n) for optimal performance.