Leetcode #1381: Design a Stack With Increment Operation
In this guide, we solve Leetcode #1381 Design a Stack With Increment Operation in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Design a stack that supports increment operations on its elements. Implement the CustomStack class: CustomStack(int maxSize) Initializes the object with maxSize which is the maximum number of elements in the stack.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Stack, Design, Array
Intuition
The problem has a natural nested or last-in-first-out structure.
A stack lets us resolve matches in the correct order as we scan.
Approach
Push items as they appear and pop when you can finalize a decision.
The stack captures the unresolved part of the input.
Steps:
- Push elements as you scan.
- Pop when a rule or match is satisfied.
- Use the stack to compute results.
Example
Input
["CustomStack","push","push","pop","push","push","push","increment","increment","pop","pop","pop","pop"]
[[3],[1],[2],[],[2],[3],[4],[5,100],[2,100],[],[],[],[]]
Output
[null,null,null,2,null,null,null,null,null,103,202,201,-1]
Explanation
CustomStack stk = new CustomStack(3); // Stack is Empty []
stk.push(1); // stack becomes [1]
stk.push(2); // stack becomes [1, 2]
stk.pop(); // return 2 --> Return top of the stack 2, stack becomes [1]
stk.push(2); // stack becomes [1, 2]
stk.push(3); // stack becomes [1, 2, 3]
stk.push(4); // stack still [1, 2, 3], Do not add another elements as size is 4
stk.increment(5, 100); // stack becomes [101, 102, 103]
stk.increment(2, 100); // stack becomes [201, 202, 103]
stk.pop(); // return 103 --> Return top of the stack 103, stack becomes [201, 202]
stk.pop(); // return 202 --> Return top of the stack 202, stack becomes [201]
stk.pop(); // return 201 --> Return top of the stack 201, stack becomes []
stk.pop(); // return -1 --> Stack is empty return -1.
Python Solution
class CustomStack:
def __init__(self, maxSize: int):
self.stk = [0] * maxSize
self.add = [0] * maxSize
self.i = 0
def push(self, x: int) -> None:
if self.i < len(self.stk):
self.stk[self.i] = x
self.i += 1
def pop(self) -> int:
if self.i <= 0:
return -1
self.i -= 1
ans = self.stk[self.i] + self.add[self.i]
if self.i > 0:
self.add[self.i - 1] += self.add[self.i]
self.add[self.i] = 0
return ans
def increment(self, k: int, val: int) -> None:
i = min(k, self.i) - 1
if i >= 0:
self.add[i] += val
# Your CustomStack object will be instantiated and called as such:
# obj = CustomStack(maxSize)
# obj.push(x)
# param_2 = obj.pop()
# obj.increment(k,val)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.