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"]
[[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 incrementpush 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).

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