# Solving Leetcode Interviews in Seconds with AI: Design a Stack With Increment Operation


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1381" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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. void push(int x) Adds x to the top of the stack if the stack has not reached the maxSize. int pop() Pops and returns the top of the stack or -1 if the stack is empty. void inc(int k, int val) Increments the bottom k elements of the stack by val. If there are less than k elements in the stack, increment all the elements in the stack.    Example 1:  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.    Constraints:  1 <= maxSize, x, k <= 1000 0 <= val <= 100 At most 1000 calls will be made to each method of increment, push and pop each separately.  

	# Explanation
	Here's a breakdown of the approach, complexity, and Python code for the `CustomStack` problem:

*   **Key Approach:**
    *   Use a list to represent the stack.
    *   Maintain a separate list of increments. This allows for efficient `inc` operations by storing increments at each index instead of modifying stack elements directly.
    *   During `pop`, apply the accumulated increment before returning the value.

*   **Complexity:**
    *   Runtime Complexity: O(1) for all operations (push, pop, inc).
    *   Storage Complexity: O(maxSize)

	
	# Code
	```python
	class CustomStack:

    def __init__(self, maxSize: int):
        self.max_size = maxSize
        self.stack = []
        self.increments = [0] * maxSize

    def push(self, x: int) -> None:
        if len(self.stack) < self.max_size:
            self.stack.append(x)

    def pop(self) -> int:
        if not self.stack:
            return -1

        index = len(self.stack) - 1
        val = self.stack.pop()
        increment = self.increments[index]
        self.increments[index] = 0

        return val + increment


    def increment(self, k: int, val: int) -> None:
        n = len(self.stack)
        end_index = min(k, n)
        for i in range(end_index):
          self.increments[i] += val
	```