How do you solve Steps to Make Array Non-decreasing on LeetCode?

Steps to Make Array Non-decreasing (LeetCode #2289) 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: An element is removed if there is a greater element to its left.

What is the brute force solution for Steps to Make Array Non-decreasing?

The brute force approach for Steps to Make Array Non-decreasing is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we repeatedly scan the array to remove elements that violate the non-decreasing property. This is simple but inefficient, as we may need to traverse the array multiple times. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Steps to Make Array Non-decreasing?

Always start with a brute force solution to understand the problem better. Think about how to optimize your solution by identifying patterns. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Steps to Make Array Non-decreasing?

Not considering edge cases where the array is already non-decreasing. Failing to correctly update the dp array and stack during the optimal solution.

#2289

Steps to Make Array Non-decreasing

Medium

ArrayLinked ListDynamic ProgrammingStackMonotonic StackSimulationStackDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

In the optimal solution, we recognize that each element can be removed based on how many steps it takes for it to be removed. We can use a stack to track the maximum number of steps required for each element efficiently.

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Algorithm

4 steps

1Step 1: Initialize a dp array to track the number of steps for each element.
2Step 2: Use a stack to keep track of indices of elements in a non-decreasing manner.
3Step 3: For each element, check the stack to determine how many steps it will take to remove it.
4Step 4: Update the dp array and the maximum steps accordingly.

solution.py14 lines

1# Full working Python code
2
3def steps_to_make_non_decreasing(nums):
4    n = len(nums)
5    dp = [0] * n
6    stack = []
7    for i in range(n):
8        while stack and nums[stack[-1]] > nums[i]:
9            dp[i] = max(dp[i], dp[stack.pop()] + 1)
10        stack.append(i)
11    return max(dp)
12
13# Example usage
14print(steps_to_make_non_decreasing([5,3,4,4,7,3,6,11,8,5,11]))

ℹ

Complexity note: The time complexity is O(n) because we traverse the array once, and each element is pushed and popped from the stack at most once. The space complexity is O(n) due to the dp array and stack.

1An element is removed if there is a greater element to its left.
2The number of steps for an element depends on the maximum steps of elements that precede it.

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