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Maximum Width Ramp

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ArraysTwo PointersStacks

A ramp in an integer array nums is a pair (i, j) for which i < j and nums[i] <= nums[j]. The width of such a ramp is j - i.

Given an integer array nums, return the maximum width of a ramp in nums. If there is no ramp in nums, return 0.

Example 1:

Input: nums = [6,0,8,2,1,5]
Output: 4
Explanation: The maximum width ramp is achieved at (i, j) = (1, 5): nums[1] = 0 and nums[5] = 5.

Example 2:

Input: nums = [9,8,1,0,1,9,4,0,4,1]
Output: 7
Explanation: The maximum width ramp is achieved at (i, j) = (2, 9): nums[2] = 1 and nums[9] = 1.

Constraints:

2 <= nums.length <= 5 * 10⁴
0 <= nums[i] <= 5 * 10⁴

Try coding it in LeetCode

Maximum Width Ramp

Medium

Asked by:

26 views

Topics:

ArraysTwo PointersStacks

A ramp in an integer array nums is a pair (i, j) for which i < j and nums[i] <= nums[j]. The width of such a ramp is j - i.

Given an integer array nums, return the maximum width of a ramp in nums. If there is no ramp in nums, return 0.

Example 1:

Input: nums = [6,0,8,2,1,5]
Output: 4
Explanation: The maximum width ramp is achieved at (i, j) = (1, 5): nums[1] = 0 and nums[5] = 5.

Example 2:

Input: nums = [9,8,1,0,1,9,4,0,4,1]
Output: 7
Explanation: The maximum width ramp is achieved at (i, j) = (2, 9): nums[2] = 1 and nums[9] = 1.

Constraints:

2 <= nums.length <= 5 * 10⁴
0 <= nums[i] <= 5 * 10⁴

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 is the range of values within the input array, and can the array contain negative numbers, zeros, or non-integer values?
What should I return if no ramp exists (i.e., there are no i < j such that A[i] <= A[j])?
Are there any constraints on the size of the input array `A`?
If multiple ramps with the same maximum width exist, is it acceptable to return any one of them, or should a specific one be returned based on some criteria (e.g., the one with the smallest starting index)?
Can the input array be empty or null?

Brute Force Solution

Approach

The goal is to find the widest 'ramp' in a collection of numbers. The brute force method checks every possible pair of numbers to see if they form a valid ramp, and then finds the widest one.

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

Take the first number in the collection.
Compare it to every other number that comes after it in the collection.
For each comparison, check if the second number is greater than or equal to the first number. If it is, you've found a possible ramp.
If it is a ramp, measure the distance between the first number and the second number.
Repeat this process, starting with the second number in the collection and comparing it to all the numbers after it.
Continue doing this for every number in the collection.
Keep track of the distance between each pair that forms a valid ramp.
Finally, find the largest distance among all the valid ramps you found. That's the width of the maximum width ramp.

Code Implementation

def maximum_width_ramp_brute_force(numbers):
    max_ramp_width = 0

    for left_index in range(len(numbers)):

        # Starting from the left, iterate
        # through the rest of the array
        for right_index in range(left_index + 1, len(numbers)):

            # Found valid ramp
            if numbers[left_index] <= numbers[right_index]:
                current_ramp_width = right_index - left_index
                
                # Update max ramp width
                # if current is larger
                if current_ramp_width > max_ramp_width:
                    max_ramp_width = current_ramp_width

    return max_ramp_width

Big(O) Analysis

Time Complexity

O(n²) – The brute force approach iterates through each of the n elements in the input array. For each element, it then iterates through the remaining elements to its right, comparing them to form a potential ramp. The number of comparisons decreases with each subsequent starting element. This nested iteration results in roughly n * (n-1)/2 comparisons, which simplifies to O(n²).

Space Complexity

O(1) – The brute force algorithm described only uses a few constant space variables to store the current indices being compared and the maximum ramp width found so far. These variables do not depend on the input size N (the number of elements in the collection). Thus, the auxiliary space complexity is constant.

Optimal Solution

Approach

The goal is to find the widest ramp in a sequence of numbers. We can efficiently do this by focusing on finding potential starting points and then quickly checking how far we can extend the ramp. We achieve this by preprocessing the data to identify key points of interest.

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

First, identify all the potential starting points for a ramp. These are the positions where the number is smaller than all the numbers before it.
Next, identify all the potential ending points for a ramp. These are the positions where the number is larger than all the numbers after it.
Now, examine the potential starting points and ending points together. For each starting point, find the furthest ending point that creates a valid ramp (meaning the number at the starting point is less than or equal to the number at the ending point).
Keep track of the widest ramp you find during this process. By only considering these key starting and ending positions, you avoid checking many unnecessary combinations and quickly identify the maximum width ramp.

Code Implementation

def maximum_width_ramp(numbers):
    potential_starting_indices = []
    potential_ending_indices = []
    array_length = len(numbers)

    minimum_so_far = float('inf')
    # Identify potential ramp starts. 
    for i in range(array_length):
        if numbers[i] <= minimum_so_far:
            potential_starting_indices.append(i)
            minimum_so_far = numbers[i]

    maximum_so_far = float('-inf')
    # Identify potential ramp ends.
    for i in range(array_length - 1, -1, -1):
        if numbers[i] >= maximum_so_far:
            potential_ending_indices.append(i)
            maximum_so_far = numbers[i]

    potential_ending_indices.reverse()
    maximum_width = 0
    start_index_pointer = 0
    end_index_pointer = 0

    # Find the widest valid ramp.
    while start_index_pointer < len(potential_starting_indices) and end_index_pointer < len(potential_ending_indices):
        if numbers[potential_starting_indices[start_index_pointer]] <= numbers[potential_ending_indices[end_index_pointer]]:
            current_width = potential_ending_indices[end_index_pointer] - potential_starting_indices[start_index_pointer]
            
            # Update max width if needed
            maximum_width = max(maximum_width, current_width)
            end_index_pointer += 1
        else:
            # Increment the start pointer to find a valid start.
            start_index_pointer += 1

    return maximum_width

Big(O) Analysis

Time Complexity

O(n) – The algorithm first iterates through the input array of size n to identify potential starting points, taking O(n) time. It then iterates again to identify potential ending points, also taking O(n) time. Finally, it iterates through the starting and ending points to find the widest ramp, which in the worst case involves comparing each starting point with each ending point, but crucially, due to the pre-processing steps of identifying local minimums and maximums, this final iteration is also effectively O(n) in the best possible implementation (as we only iterate through local minimums and maximums). Therefore, the overall time complexity is O(n) + O(n) + O(n) which simplifies to O(n).

Space Complexity

O(N) – The algorithm creates two lists, potential starting points and potential ending points. In the worst-case scenario, all elements in the input array could be potential starting or ending points. Therefore, the space used by these lists scales linearly with the input size, N, where N is the number of elements in the input array. Thus, the space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 0 if the input array is empty since no ramp can exist.
Array with only one element	Return 0 as a ramp needs at least two elements.
Array with all elements in descending order	Return 0 because no i < j exists where A[i] <= A[j].
Array with all identical elements	Return array length -1, which is j-i with the last element as j and first as i.
Large array size nearing memory limits	Ensure the algorithm uses memory efficiently to avoid memory exhaustion errors (e.g., avoids unnecessary data structures).
Input array containing negative numbers	The algorithm should correctly handle negative numbers as they can form valid ramps.
Integer overflow in calculating width (j - i)	While unlikely given typical array size limits, consider using a larger integer type or handling potential overflow if necessary (e.g., use long or check for overflow before subtraction).
Array is already sorted in ascending order	Return array length -1 which is the maximum possible width in this case.

Empty input array

How to Handle:

Return 0 if the input array is empty since no ramp can exist.

Array with only one element

How to Handle:

Return 0 as a ramp needs at least two elements.

Array with all elements in descending order

How to Handle:

Return 0 because no i < j exists where A[i] <= A[j].

Array with all identical elements

How to Handle:

Return array length -1, which is j-i with the last element as j and first as i.

Large array size nearing memory limits

How to Handle:

Ensure the algorithm uses memory efficiently to avoid memory exhaustion errors (e.g., avoids unnecessary data structures).

Input array containing negative numbers

How to Handle:

The algorithm should correctly handle negative numbers as they can form valid ramps.

Integer overflow in calculating width (j - i)

How to Handle:

While unlikely given typical array size limits, consider using a larger integer type or handling potential overflow if necessary (e.g., use long or check for overflow before subtraction).

Array is already sorted in ascending order

How to Handle:

Return array length -1 which is the maximum possible width in this case.

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 is the range of values within the input array, and can the array contain negative numbers, zeros, or non-integer values?
What should I return if no ramp exists (i.e., there are no i < j such that A[i] <= A[j])?
Are there any constraints on the size of the input array `A`?
If multiple ramps with the same maximum width exist, is it acceptable to return any one of them, or should a specific one be returned based on some criteria (e.g., the one with the smallest starting index)?
Can the input array be empty or null?

Brute Force Solution

Approach

The goal is to find the widest 'ramp' in a collection of numbers. The brute force method checks every possible pair of numbers to see if they form a valid ramp, and then finds the widest one.

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

Take the first number in the collection.
Compare it to every other number that comes after it in the collection.
For each comparison, check if the second number is greater than or equal to the first number. If it is, you've found a possible ramp.
If it is a ramp, measure the distance between the first number and the second number.
Repeat this process, starting with the second number in the collection and comparing it to all the numbers after it.
Continue doing this for every number in the collection.
Keep track of the distance between each pair that forms a valid ramp.
Finally, find the largest distance among all the valid ramps you found. That's the width of the maximum width ramp.

Code Implementation

def maximum_width_ramp_brute_force(numbers):
    max_ramp_width = 0

    for left_index in range(len(numbers)):

        # Starting from the left, iterate
        # through the rest of the array
        for right_index in range(left_index + 1, len(numbers)):

            # Found valid ramp
            if numbers[left_index] <= numbers[right_index]:
                current_ramp_width = right_index - left_index
                
                # Update max ramp width
                # if current is larger
                if current_ramp_width > max_ramp_width:
                    max_ramp_width = current_ramp_width

    return max_ramp_width

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, identify all the potential starting points for a ramp. These are the positions where the number is smaller than all the numbers before it.
Next, identify all the potential ending points for a ramp. These are the positions where the number is larger than all the numbers after it.
Now, examine the potential starting points and ending points together. For each starting point, find the furthest ending point that creates a valid ramp (meaning the number at the starting point is less than or equal to the number at the ending point).
Keep track of the widest ramp you find during this process. By only considering these key starting and ending positions, you avoid checking many unnecessary combinations and quickly identify the maximum width ramp.

Code Implementation

def maximum_width_ramp(numbers):
    potential_starting_indices = []
    potential_ending_indices = []
    array_length = len(numbers)

    minimum_so_far = float('inf')
    # Identify potential ramp starts. 
    for i in range(array_length):
        if numbers[i] <= minimum_so_far:
            potential_starting_indices.append(i)
            minimum_so_far = numbers[i]

    maximum_so_far = float('-inf')
    # Identify potential ramp ends.
    for i in range(array_length - 1, -1, -1):
        if numbers[i] >= maximum_so_far:
            potential_ending_indices.append(i)
            maximum_so_far = numbers[i]

    potential_ending_indices.reverse()
    maximum_width = 0
    start_index_pointer = 0
    end_index_pointer = 0

    # Find the widest valid ramp.
    while start_index_pointer < len(potential_starting_indices) and end_index_pointer < len(potential_ending_indices):
        if numbers[potential_starting_indices[start_index_pointer]] <= numbers[potential_ending_indices[end_index_pointer]]:
            current_width = potential_ending_indices[end_index_pointer] - potential_starting_indices[start_index_pointer]
            
            # Update max width if needed
            maximum_width = max(maximum_width, current_width)
            end_index_pointer += 1
        else:
            # Increment the start pointer to find a valid start.
            start_index_pointer += 1

    return maximum_width

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0 if the input array is empty since no ramp can exist.
Array with only one element	Return 0 as a ramp needs at least two elements.
Array with all elements in descending order	Return 0 because no i < j exists where A[i] <= A[j].
Array with all identical elements	Return array length -1, which is j-i with the last element as j and first as i.
Large array size nearing memory limits	Ensure the algorithm uses memory efficiently to avoid memory exhaustion errors (e.g., avoids unnecessary data structures).
Input array containing negative numbers	The algorithm should correctly handle negative numbers as they can form valid ramps.
Integer overflow in calculating width (j - i)	While unlikely given typical array size limits, consider using a larger integer type or handling potential overflow if necessary (e.g., use long or check for overflow before subtraction).
Array is already sorted in ascending order	Return array length -1 which is the maximum possible width in this case.

Empty input array

How to Handle:

Return 0 if the input array is empty since no ramp can exist.

Array with only one element

How to Handle:

Return 0 as a ramp needs at least two elements.

Array with all elements in descending order

How to Handle:

Return 0 because no i < j exists where A[i] <= A[j].

Array with all identical elements

How to Handle:

Return array length -1, which is j-i with the last element as j and first as i.

Large array size nearing memory limits

How to Handle:

Ensure the algorithm uses memory efficiently to avoid memory exhaustion errors (e.g., avoids unnecessary data structures).

Input array containing negative numbers

How to Handle:

The algorithm should correctly handle negative numbers as they can form valid ramps.

Integer overflow in calculating width (j - i)

How to Handle:

While unlikely given typical array size limits, consider using a larger integer type or handling potential overflow if necessary (e.g., use long or check for overflow before subtraction).

Array is already sorted in ascending order

How to Handle:

Return array length -1 which is the maximum possible width in this case.