Design a Stack With Increment Operation
Tags: leetcode, dsa, theabbie
Published: 2022-06-06 13:00:49

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**
\["CustomStack","push","push","pop","push","push","push","increment","increment","pop","pop","pop","pop"\]
\[\[3\],\[1\],\[2\],\[\],\[2\],\[3\],\[4\],\[5,100\],\[2,100\],\[\],\[\],\[\],\[\]\]
**Output**
\[null,null,null,2,null,null,null,null,null,103,202,201,-1\]
**Explanation**
CustomStack customStack = new CustomStack(3); // Stack is Empty \[\]
customStack.push(1);                          // stack becomes \[1\]
customStack.push(2);                          // stack becomes \[1, 2\]
customStack.pop();                            // return 2 --> Return top of the stack 2, stack becomes \[1\]
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 \[201\]
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.

**SOLUTION:**

    class CustomStack:
    
        def __init__(self, maxSize: int):
            self.stack = []
            self.maxSize = maxSize
    
        def push(self, x: int) -> None:
            if len(self.stack) < self.maxSize:
                self.stack.append(x)
    
        def pop(self) -> int:
            if len(self.stack) > 0:
                return self.stack.pop()
            return -1
            
    
        def increment(self, k: int, val: int) -> None:
            i = 0
            while i < k and i < len(self.stack):
                self.stack[i] += val
                i += 1
    
    
    # Your CustomStack object will be instantiated and called as such:
    # obj = CustomStack(maxSize)
    # obj.push(x)
    # param_2 = obj.pop()
    # obj.increment(k,val)