1381. Design a Stack With Increment Operation

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

// OJ: https://leetcode.com/problems/design-a-stack-with-increment-operation/
// Author: github.com/lzl124631x
// 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 2. Lazy Propogation

// OJ: https://leetcode.com/problems/design-a-stack-with-increment-operation/
// Author: github.com/lzl124631x
// 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;
    }
};