leetcode_1381

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

1. vector of pair

2. reference: https://leetcode-cn.com/problems/design-a-stack-with-increment-operation/solution/python-jie-fa-pop-shi-lei-jia-by-amir-6/
3. Use pair<int, int> to store the increment in the last element will be covered.

4. Whenever an element is popped, accumulate the increment at the inc of the top element.

class CustomStack {
public:
    vector<pair<int, int>> s;
    int maxsize = 0;
    CustomStack(int maxSize) {
        maxsize = maxSize;
    }

    void push(int x) {
        if (s.size() != maxsize)
            s.push_back({x, 0});
    }

    int pop() {
        if (!s.size())
            return -1;
        auto [val, inc] = s.back();
        s.pop_back();
        if (s.size())
            s.back().second += inc;

        return val + inc;
    }

    void increment(int k, int val) {
        k = min(k - 1, (int)s.size() - 1);
        if (k < 0) return;
        s[k].second += 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);
 */