How do you solve Maximum Width Ramp on LeetCode?

Maximum Width Ramp (LeetCode #962) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a stack can help efficiently manage indices and find valid pairs.

What is the brute force solution for Maximum Width Ramp?

The brute force approach for Maximum Width Ramp is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible pair (i, j) to see if it forms a valid ramp. This is simple to understand but can be slow for larger arrays since it examines all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Width Ramp?

Maximum Width Ramp uses the following concepts: Array, Two Pointers, Stack, Monotonic Stack. The recommended patterns to study are: Monotonic Stack, Two Pointers.

What are the interview tips for Maximum Width Ramp?

Always clarify the problem constraints and edge cases. Think about how you can reduce the number of comparisons needed. Practice explaining your thought process clearly during the interview.

What are common mistakes when solving Maximum Width Ramp?

Not considering edge cases where no ramp exists. Forgetting to check the condition nums[i] <= nums[j] correctly.

#962

Maximum Width Ramp

Medium

ArrayTwo PointersStackMonotonic StackMonotonic StackTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can use a monotonic stack to keep track of indices of the array in increasing order. This allows us to efficiently find the maximum width by leveraging the properties of the ramp.

⚙️

Algorithm

3 steps

1Step 1: Create an empty stack to hold indices.
2Step 2: Iterate through the array and push indices onto the stack where the values are in increasing order.
3Step 3: Iterate through the array in reverse to find the maximum width by popping from the stack when we find a valid ramp condition.

solution.py11 lines

1def maxWidthRamp(nums):
2    stack = []
3    n = len(nums)
4    for i in range(n):
5        if not stack or nums[stack[-1]] > nums[i]:
6            stack.append(i)
7    max_width = 0
8    for j in range(n - 1, -1, -1):
9        while stack and nums[stack[-1]] <= nums[j]:
10            max_width = max(max_width, j - stack.pop())
11    return max_width

ℹ

Complexity note: The time complexity is O(n) because we traverse the array a constant number of times. The space complexity is O(n) due to the stack storing indices.

1Using a stack can help efficiently manage indices and find valid pairs.
2Understanding the properties of the ramp allows for optimized searching.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.