How do you solve Design a Stack With Increment Operation on LeetCode?

Design a Stack With Increment Operation (LeetCode #1381) 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 an auxiliary array allows for efficient increment operations.

What is the brute force solution for Design a Stack With Increment Operation?

The brute force approach for Design a Stack With Increment Operation is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will use a simple list to represent the stack. For the increment operation, we will directly iterate through the bottom k elements and increment them, which can be inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Design a Stack With Increment Operation?

Design a Stack With Increment Operation uses the following concepts: Array, Stack, Design. The recommended patterns to study are: Array, Stack.

What are the interview tips for Design a Stack With Increment Operation?

Clarify the requirements and constraints of the problem before coding. Think about space and time complexity as you design your solution. Discuss your thought process and reasoning with the interviewer.

What are common mistakes when solving Design a Stack With Increment Operation?

Forgetting to handle edge cases like popping from an empty stack. Not updating the increment array correctly after a pop.

#1381

Design a Stack With Increment Operation

Medium

ArrayStackDesignArrayStack

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)

In the optimal approach, we maintain an array for the stack and an auxiliary array to track increments. This allows us to perform the increment operation in constant time, while still keeping track of the actual values in the stack.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty list for the stack and an auxiliary list for increments.
2Step 2: For push, append the element to the stack and 0 to increments if it hasn't reached maxSize.
3Step 3: For pop, return the top element and apply any increment from the auxiliary list.
4Step 4: For increment, update the increment array at the index k-1.

solution.py24 lines

1class CustomStack:
2    def __init__(self, maxSize: int):
3        self.stack = []
4        self.increments = [0] * maxSize
5        self.maxSize = maxSize
6
7    def push(self, x: int) -> None:
8        if len(self.stack) < self.maxSize:
9            self.stack.append(x)
10
11    def pop(self) -> int:
12        if not self.stack:
13            return -1
14        index = len(self.stack) - 1
15        increment = self.increments[index]
16        self.increments[index] = 0
17        value = self.stack.pop() + increment
18        if index > 0:
19            self.increments[index - 1] += increment
20        return value
21
22    def inc(self, k: int, val: int) -> None:
23        if k > 0:
24            self.increments[min(k - 1, len(self.stack) - 1)] += val

ℹ

Complexity note: Push and pop operations are O(1), while increment is O(1) due to the auxiliary array. Overall, space complexity is O(n) due to the storage of the stack and increments.

1Using an auxiliary array allows for efficient increment operations.
2Understanding the difference between stack size and the number of elements is crucial.

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