Formatted question description: https://leetcode.ca/all/1381.html

1381. Design a Stack With Increment Operation (Medium)

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.

Related Topics:
Stack, Design

Solution 1. Brute Force

// Time:
//      CustomStack, push, pop: O(1)
//      increment: O(k)
// Space: O(N)
class CustomStack {
    int N;
    vector<int> v;
public:
    CustomStack(int maxSize): N(maxSize) {}
    void push(int x) {
        if (v.size() >= N) return;
        v.push_back(x);
    }
    int pop() {
        if (v.empty()) return -1;
        int n = v.back();
        v.pop_back();
        return n;
    }
    void increment(int k, int val) {
        k = min(k, (int)v.size());
        for (int i = 0; i < k; ++i) v[i] += val;
    }
};

Solution 1.1. Queue

Use a double-ended queue deque. Also maintain a capacity and a size.

For the constructor, initialize deque, initialize capacity = maxSize and initialize size = 0.

For method push, if size < capacity, then add x to the end of deque and increase size by 1.

For method pop, if size > 0, then decrease size by 1, poll the first element of deque and return. Otherwise, return -1.

For method increment, set k to the minimum of k and size. For the k elements at the head of deque, poll them out, add val to each of them, and offer them back at the head of deque. Use a stack to temporally store the elements.

  • class CustomStack {
        Deque<Integer> deque;
        int capacity;
        int size;
    
        public CustomStack(int maxSize) {
            deque = new LinkedList<Integer>();
            capacity = maxSize;
            size = 0;
        }
        
        public void push(int x) {
            if (size < capacity) {
                deque.offerLast(x);
                size++;
            }
        }
        
        public int pop() {
            if (size > 0) {
                size--;
                return deque.pollLast();
            } else
                return -1;
        }
        
        public void increment(int k, int val) {
            k = Math.min(k, size);
            Stack<Integer> tempStack = new Stack<Integer>();
            for (int i = 0; i < k; i++) {
                int num = deque.pollFirst() + val;
                tempStack.push(num);
            }
            for (int i = 0; i < k; i++)
                deque.offerFirst(tempStack.pop());
        }
    }
    
  • // OJ: https://leetcode.com/problems/design-a-stack-with-increment-operation/
    // Time:
    //      CustomStack, push, pop: O(1)
    //      increment: O(k)
    // Space: O(N)
    class CustomStack {
        int N;
        vector<int> v;
    public:
        CustomStack(int maxSize): N(maxSize) {}
        void push(int x) {
            if (v.size() >= N) return;
            v.push_back(x);
        }
        int pop() {
            if (v.empty()) return -1;
            int n = v.back();
            v.pop_back();
            return n;
        }
        void increment(int k, int val) {
            k = min(k, (int)v.size());
            for (int i = 0; i < k; ++i) v[i] += val;
        }
    };
    
  • print("Todo!")
    

Solution 2. Lazy Propogation

  • // Time:
    //      CustomStack: O(N)
    //      push, pop, increment: O(1)
    // Space: O(N)
    // Ref: https://leetcode.com/problems/design-a-stack-with-increment-operation/discuss/539716/JavaC%2B%2BPython-Lazy-increment-O(1)
    class CustomStack {
        stack<int> s;
        vector<int> update;
    public:
        CustomStack(int maxSize) : update(maxSize) {}
        
        void push(int x) {
            if (s.size() == update.size()) return;
            s.push(x);
        }
        
        int pop() {
            if (s.empty()) return -1;
            int ans = s.top(), index = s.size() - 1;
            s.pop();
            ans += update[index];
            if (index - 1 >= 0) update[index - 1] += update[index];
            update[index] = 0;
            return ans;
        }
        
        void increment(int k, int val) {
            k = min(k, (int)s.size()) - 1;
            if (k >= 0) update[k] += val;
        }
    };
    
  • class CustomStack:
    
        def __init__(self, maxSize: int):
            self.stack = []
            self.max_size = maxSize
            self.inc = []
    
        def push(self, x: int) -> None:
            if len(self.stack) < self.max_size:
                self.stack.append(x)
                self.inc.append(0)
    
        def pop(self) -> int:
            if not self.stack: return -1
            if len(self.inc) > 1:
                self.inc[-2] += self.inc[-1]
            return self.inc.pop() + self.stack.pop()
    
        def increment(self, k: int, val: int) -> None:
            if self.inc:
                self.inc[min(k, len(self.inc))-1] += val
    
  • // todo
    

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