How do you solve Build an Array With Stack Operations on LeetCode?

Build an Array With Stack Operations (LeetCode #1441) 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: Recognizing unnecessary operations can lead to a more efficient solution.

What is the time complexity of Build an Array With Stack Operations?

The optimal solution for Build an Array With Stack Operations runs in O(n) time and uses O(n) space. This complexity is optimal because we only iterate through the target array and perform a constant amount of work for each element.

What is the brute force solution for Build an Array With Stack Operations?

The brute force approach for Build an Array With Stack Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach simulates every possible push and pop operation while iterating through the stream of integers. It checks if the current stack matches the target after each operation, leading to many unnecessary operations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Build an Array With Stack Operations?

Build an Array With Stack Operations uses the following concepts: Array, Stack, Simulation. The recommended patterns to study are: Stack, Simulation.

What are the interview tips for Build an Array With Stack Operations?

Clarify the problem requirements before jumping into coding. Think about edge cases, such as when target is the first few numbers or the last number in the range. Practice simulating stack operations with pen and paper to build intuition.

What are common mistakes when solving Build an Array With Stack Operations?

Not considering how to handle numbers not in the target. Overcomplicating the logic by simulating every push/pop without optimization.

#1441

Build an Array With Stack Operations

Medium

ArrayStackSimulationStackSimulation

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)

The optimal approach directly constructs the operations needed by recognizing that we only need to push the numbers in the target array and pop the numbers in between them. This avoids unnecessary operations.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty list for operations.
2Step 2: Iterate through each number in the target array.
3Step 3: For each number, push it onto the stack and record the operation.
4Step 4: If the current number is not the last in the target, add a 'Pop' operation to discard the next number.

solution.py7 lines

1def buildArray(target, n):
2    operations = []
3    for i in range(1, target[-1] + 1):
4        operations.append('Push')
5        if i != target[len(operations) - 1]:
6            operations.append('Pop')
7    return operations

ℹ

Complexity note: This complexity is optimal because we only iterate through the target array and perform a constant amount of work for each element.

1Recognizing unnecessary operations can lead to a more efficient solution.
2Understanding stack behavior is crucial for simulating operations.

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