Description
Design a stack that supports increment operations on its elements.
Implement the CustomStack class:
CustomStack(int maxSize)Initializes the object withmaxSizewhich is the maximum number of elements in the stack.void push(int x)Addsxto the top of the stack if the stack has not reached themaxSize.int pop()Pops and returns the top of the stack or-1if the stack is empty.void inc(int k, int val)Increments the bottomkelements of the stack byval. If there are less thankelements in the stack, 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 stk = new CustomStack(3); // Stack is Empty [] stk.push(1); // stack becomes [1] stk.push(2); // stack becomes [1, 2] stk.pop(); // return 2 --> Return top of the stack 2, stack becomes [1] stk.push(2); // stack becomes [1, 2] stk.push(3); // stack becomes [1, 2, 3] stk.push(4); // stack still [1, 2, 3], Do not add another elements as size is 4 stk.increment(5, 100); // stack becomes [101, 102, 103] stk.increment(2, 100); // stack becomes [201, 202, 103] stk.pop(); // return 103 --> Return top of the stack 103, stack becomes [201, 202] stk.pop(); // return 202 --> Return top of the stack 202, stack becomes [201] stk.pop(); // return 201 --> Return top of the stack 201, stack becomes [] stk.pop(); // return -1 --> Stack is empty return -1.
Constraints:
1 <= maxSize, x, k <= 10000 <= val <= 100- At most
1000calls will be made to each method ofincrement,pushandpopeach separately.
Solutions
Solution 1: Array Simulation
We can use an array stk to simulate the stack, and an integer i to represent the position of the next element to be pushed into the stack. In addition, we need another array add to record the cumulative increment value at each position.
When calling push(x), if i < maxSize, we put x into stk[i] and increment i by one.
When calling pop(), if i ≤ 0, it means the stack is empty, so we return -1. Otherwise, we decrement i by one, and the answer is stk[i] + add[i]. Then we add add[i] to add[i - 1], and set add[i] to zero. Finally, we return the answer.
When calling increment(k, val), if i > 0, we add val to add[min(i, k) - 1].
The time complexity is O(1), and the space complexity is O(n). Where n is the maximum capacity of the stack.
class CustomStack: def __init__(self, maxSize: int): self.stk = [0] * maxSize self.add = [0] * maxSize self.i = 0 def push(self, x: int) -> None: if self.i < len(self.stk): self.stk[self.i] = x self.i += 1 def pop(self) -> int: if self.i <= 0: return -1 self.i -= 1 ans = self.stk[self.i] + self.add[self.i] if self.i > 0: self.add[self.i - 1] += self.add[self.i] self.add[self.i] = 0 return ans def increment(self, k: int, val: int) -> None: i = min(k, self.i) - 1 if i >= 0: self.add[i] += val # Your CustomStack object will be instantiated and called as such: # obj = CustomStack(maxSize) # obj.push(x) # param_2 = obj.pop() # obj.increment(k,val)(code-box)
