1381. Design a Stack With Increment Operation

Description

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.

 

Solutions

Solution: Lazy Increment

• Time complexity: O(1)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number} maxSize
 */
const CustomStack = function (maxSize) {
  this.stack = [];
  this.increments = new Array(maxSize).fill(0);
  this.maxSize = maxSize;
};

/**
 * @param {number} x
 * @return {void}
 */
CustomStack.prototype.push = function (x) {
  if (this.stack.length >= this.maxSize) return;

  this.stack.push(x);
};

/**
 * @return {number}
 */
CustomStack.prototype.pop = function () {
  const n = this.stack.length;

  if (!n) return -1;
  const value = this.stack.pop();
  const increment = this.increments[n - 1];

  this.increments[n - 2] += increment;
  this.increments[n - 1] = 0;

  return value + increment;
};

/**
 * @param {number} k
 * @param {number} val
 * @return {void}
 */
CustomStack.prototype.increment = function (k, val) {
  const n = Math.min(k, this.stack.length);

  if (!n) return;
  this.increments[n - 1] += val;
};

/**
 * Your CustomStack object will be instantiated and called as such:
 * var obj = new CustomStack(maxSize)
 * obj.push(x)
 * var param_2 = obj.pop()
 * obj.increment(k,val)
 */