20. Design a Stack With Increment Operation

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.

Solution: (Using Two Stack: O(n))

class CustomStack
{
public:
    int ms;
    stack<int> s;

    CustomStack(int maxSize)
    {
        ms = maxSize;
    }

    void push(int x)
    {
        if (s.size() < ms)
        {
            s.push(x);
        }
        return;
    }

    int pop()
    {
        if (!s.empty())
        {
            int x = s.top();
            s.pop();
            return x;
        }

        return -1;
    }

    void increment(int k, int val)
    {

        stack<int> temp;
        while (!s.empty())
        {
            int n = s.top();
            s.pop();
            temp.push(n);
        }

        int count = 1;
        while (!temp.empty() && count <= k)
        {
            int n = temp.top();
            n += val;
            s.push(n);
            temp.pop();
            count++;
        }

        while (!temp.empty())
        {
            int n = temp.top();
            s.push(n);
            temp.pop();
        }
    }
};

Solution: (Using Vector: O(k))

class CustomStack
{
public:
    vector<int> v;
    int ms;
    CustomStack(int maxSize)
    {
        ms = maxSize;
    }

    void push(int x)
    {
        if (v.size() < ms)
        {
            v.push_back(x);
        }

        return;
    }

    int pop()
    {

        if (v.size() > 0)
        {
            int n = v[v.size() - 1];
            v.pop_back();
            return n;
        }
        return -1;
    }

    void increment(int k, int val)
    {
        int i = 0;
        while (i < v.size() && i < k)
        {
            v[i] += val;
            i++;
        }

        return;
    }
};

Solution: (Using 2 vector : O(1))

Approach: mainiting another vector to store increment

class CustomStack
{
public:
    
    vector<int> v;
    vector<int> inc;
    
    int ms;

    CustomStack(int maxSize)
    {
        ms = maxSize;
    }

    void push(int x)
    {
        if (v.size() < ms)
        {
            v.push_back(x);
            inc.push_back(0);
        }

        return;
    }

    int pop()
    {

        if (v.size() > 0)
        {
            int n = v[v.size() - 1];
            n += inc[inc.size() - 1];

            if (inc.size() > 1)
            {
                inc[inc.size() - 2] += inc[inc.size() - 1];
            }

            v.pop_back();
            inc.pop_back();
            return n;
        }
        return -1;
    }

    void increment(int k, int val)
    {
        int s = v.size();
        int idx = min(k, s) - 1;
        if (idx > -1)
        {
            inc[idx] += val;
        }
        return;
    }
};