1381.Design a Stack With Increment Operation

Tags: Medium Stack Design

Links: https://leetcode.com/problems/design-a-stack-with-increment-operation/

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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.

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class CustomStack {
    stack<int> s;
    int capacity;
public:
    CustomStack(int maxSize) {
        capacity = maxSize;
    }

    void push(int x) {
        if (s.size() == capacity) return;
        s.push(x);
    }

    int pop() {
        if (s.empty()) return -1;
        int res = s.top();
        s.pop();
        return res;
    }

    void increment(int k, int val) {
        stack<int> tmp;
        if (s.size() <= k) {
            while (!s.empty()) {
                tmp.push(s.top() + val);
                s.pop();
            }
            while (!tmp.empty()) {
                s.push(tmp.top());
                tmp.pop();
            }
            return;
        }
        int cnt = 0;
        int n = s.size();
        while (cnt < n - k) {
            tmp.push(s.top());
            s.pop();
            ++cnt;
        }
        stack<int> store;
        while (!s.empty()) {
            store.push(s.top() + val);
            s.pop();
        }
        while(!store.empty()) {
            s.push(store.top());
            store.pop();
        }
        while (!tmp.empty()) {
            s.push(tmp.top());
            tmp.pop();
        }
    }
};

/**
 * 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);
 */