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1381. Design a Stack With Increment Operation

中文文档

Description

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.

Solutions

Python3

class CustomStack:

    def __init__(self, maxSize: int):
        self.s = [0] * maxSize
        self.tail = 0

    def push(self, x: int) -> None:
        if self.tail < len(self.s):
            self.s[self.tail] = x
            self.tail += 1

    def pop(self) -> int:
        if self.tail == 0:
            return -1
        self.tail -= 1
        return self.s[self.tail]

    def increment(self, k: int, val: int) -> None:
        for i in range(min(k, self.tail)):
            self.s[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)

Java

class CustomStack {
    private int[] s;
    private int tail;

    public CustomStack(int maxSize) {
        s = new int[maxSize];
    }
    
    public void push(int x) {
        if (tail < s.length) {
            s[tail++] = x;
        }
    }
    
    public int pop() {
        return tail == 0 ? -1 : s[--tail];
    }
    
    public void increment(int k, int val) {
        for (int i = 0; i < Math.min(k, tail); ++i) {
            s[i] += val;
        }
    }
}

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

...