# LeetCode 1381. Design a Stack With Increment Operation

## Description

https://leetcode.com/problems/design-a-stack-with-increment-operation/

Design a stack which supports the following operations.

Implement the `CustomStack` class:

• `CustomStack(int maxSize)` Initializes the object with `maxSize` which is the maximum number of elements in the stack or do nothing if the stack reached the `maxSize`.
• `void push(int x)` Adds `x` to the top of the stack if the stack hasn’t reached the `maxSize`.
• `int pop()` Pops and returns the top of 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, just increment all the elements in the stack.

Example 1:

```Input
["CustomStack","push","push","pop","push","push","push","increment","increment","pop","pop","pop","pop"]
[,,,[],,,,[5,100],[2,100],[],[],[],[]]
Output
```

[null,null,null,2,null,null,null,null,null,103,202,201,-1]

Explanation CustomStack customStack = new CustomStack(3); // Stack is Empty [] customStack.push(1); // stack becomes  customStack.push(2); // stack becomes [1, 2] customStack.pop(); // return 2 –> Return top of the stack 2, stack becomes  customStack.push(2); // stack becomes [1, 2] customStack.push(3); // stack becomes [1, 2, 3] customStack.push(4); // stack still [1, 2, 3], Don’t add another elements as size is 4 customStack.increment(5, 100); // stack becomes [101, 102, 103] customStack.increment(2, 100); // stack becomes [201, 202, 103] customStack.pop(); // return 103 –> Return top of the stack 103, stack becomes [201, 202] customStack.pop(); // return 202 –> Return top of the stack 102, stack becomes  customStack.pop(); // return 201 –> Return top of the stack 101, stack becomes [] customStack.pop(); // return -1 –> Stack is empty return -1.

Constraints:

• `1 <= maxSize <= 1000`
• `1 <= x <= 1000`
• `1 <= k <= 1000`
• `0 <= val <= 100`
• At most `1000` calls will be made to each method of `increment``push` and `pop` each separately.

## Explanation

Set max size for the stack and use the size of stack or k value to decide which elements need to be incremented.

## Python Solution

``````class CustomStack:

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

return self.stack.pop()

def increment(self, k: int, val: int) -> None:

n = min(k, len(self.stack))

for i in range(n):
self.stack[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)``````
• Time Complexity: O(N).
• Space Complexity: O(N).