Design a Stack With Increment Operation

Problem

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 (Java)

class CustomStack {
    private int top;
    private int maxSize;
    private int[] stack;

    public CustomStack(int maxSize) {
        this.maxSize = maxSize;
        this.stack = new int[maxSize];
    }

    public void push(int x) {
        if (top == maxSize) {
            return;
        }
        stack[top] = x;
        top++;
    }

    public int pop() {
        if (top == 0) {
            return -1;
        }
        int popValue = stack[top - 1];
        stack[top - 1] = 0;
        top--;
        return popValue;
    }

    public void increment(int k, int val) {
        if (top == 0 || k == 0) {
            return;
        }
        for (int i = 0; i < k; i++) {
            if (i == top) {
                break;
            }
            stack[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);
 */

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).