Description#
Design a stack that supports increment operations on its elements.
Implement the CustomStack
class:
CustomStack(int maxSize)
Initializes the object with maxSize
which is the maximum number of elements in the stack.void push(int x)
Adds x
to the top of the stack if the stack has not reached the maxSize
.int pop()
Pops and returns the top of the 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, 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 <= 1000
0 <= val <= 100
- At most
1000
calls will be made to each method of increment
, push
and pop
each 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 $i \leq 0$, 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.
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| 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)
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| class CustomStack {
private int[] stk;
private int[] add;
private int i;
public CustomStack(int maxSize) {
stk = new int[maxSize];
add = new int[maxSize];
}
public void push(int x) {
if (i < stk.length) {
stk[i++] = x;
}
}
public int pop() {
if (i <= 0) {
return -1;
}
int ans = stk[--i] + add[i];
if (i > 0) {
add[i - 1] += add[i];
}
add[i] = 0;
return ans;
}
public void increment(int k, int val) {
if (i > 0) {
add[Math.min(i, k) - 1] += 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);
*/
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| class CustomStack {
public:
CustomStack(int maxSize) {
stk.resize(maxSize);
add.resize(maxSize);
i = 0;
}
void push(int x) {
if (i < stk.size()) {
stk[i++] = x;
}
}
int pop() {
if (i <= 0) {
return -1;
}
int ans = stk[--i] + add[i];
if (i > 0) {
add[i - 1] += add[i];
}
add[i] = 0;
return ans;
}
void increment(int k, int val) {
if (i > 0) {
add[min(k, i) - 1] += val;
}
}
private:
vector<int> stk;
vector<int> add;
int i;
};
/**
* 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);
*/
|
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| type CustomStack struct {
stk []int
add []int
i int
}
func Constructor(maxSize int) CustomStack {
return CustomStack{make([]int, maxSize), make([]int, maxSize), 0}
}
func (this *CustomStack) Push(x int) {
if this.i < len(this.stk) {
this.stk[this.i] = x
this.i++
}
}
func (this *CustomStack) Pop() int {
if this.i <= 0 {
return -1
}
this.i--
ans := this.stk[this.i] + this.add[this.i]
if this.i > 0 {
this.add[this.i-1] += this.add[this.i]
}
this.add[this.i] = 0
return ans
}
func (this *CustomStack) Increment(k int, val int) {
if this.i > 0 {
this.add[min(k, this.i)-1] += val
}
}
/**
* Your CustomStack object will be instantiated and called as such:
* obj := Constructor(maxSize);
* obj.Push(x);
* param_2 := obj.Pop();
* obj.Increment(k,val);
*/
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| class CustomStack {
private stk: number[];
private add: number[];
private i: number;
constructor(maxSize: number) {
this.stk = Array(maxSize).fill(0);
this.add = Array(maxSize).fill(0);
this.i = 0;
}
push(x: number): void {
if (this.i < this.stk.length) {
this.stk[this.i++] = x;
}
}
pop(): number {
if (this.i <= 0) {
return -1;
}
const ans = this.stk[--this.i] + this.add[this.i];
if (this.i > 0) {
this.add[this.i - 1] += this.add[this.i];
}
this.add[this.i] = 0;
return ans;
}
increment(k: number, val: number): void {
if (this.i > 0) {
this.add[Math.min(this.i, k) - 1] += val;
}
}
}
/**
* Your CustomStack object will be instantiated and called as such:
* var obj = new CustomStack(maxSize)
* obj.push(x)
* var param_2 = obj.pop()
* obj.increment(k,val)
*/
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