Design a Stack With Increment Operation solution and explanation
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The problem
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.
Approach overview
Problem Overview: Design a custom stack that supports push, pop, and increment(k, val). The increment operation adds val to the bottom k elements of the stack without exceeding the stack's maximum capacity.
Approach 1: Direct Array-Based Implementation (Push/Pop O(1), Increment O(k))
The simplest solution stores elements in a fixed-size array representing the stack. push appends to the end if capacity allows, and pop removes the last element. For increment(k, val), iterate from index 0 to min(k, size) - 1 and add val to each element. This approach uses basic array operations and mirrors how a standard stack behaves internally. Time complexity is O(1) for push/pop and O(k) for increment, with O(n) space for the stack.
Approach 2: Optimized Increment Using Lazy Propagation (All Operations O(1))
The bottleneck is repeatedly updating up to k elements. Instead, maintain a second array inc[] where inc[i] stores pending increments for the stack prefix ending at index i. When increment(k, val) is called, add val to inc[min(k, size) - 1] instead of updating each element. During pop, propagate the pending increment down to the next index before removing the element. This technique behaves like lazy propagation used in advanced design problems. Every operation—push, pop, and increment—runs in O(1) time with O(n) space.
Recommended for interviews: Start with the direct array approach to show the straightforward stack simulation. Then optimize by avoiding repeated updates with the lazy increment array. Interviewers usually expect the O(1) lazy propagation solution because it demonstrates strong data structure design and the ability to optimize repeated operations.
Approach 1: Direct Array-Based Implementation
This approach uses a direct array implementation to handle all operations. A list or array is used as the underlying data structure for the stack, and operations are directly performed on the stack as specified. This results in a direct and straightforward implementation, where increment operations iterate over the bottom k elements to increment them by the given value.
class CustomStack:
def __init__(self, maxSize: int):
self.stack = []
self.maxSize = maxSize
def push(self, x: int) -> None:
if len(self.stack) < self.maxSize:
self.stack.append(x)
def pop(self) -> int:
if len(self.stack) == 0:
return -1
return self.stack.pop()
def increment(self, k: int, val: int) -> None:
for i in range(min(k, len(self.stack))):
self.stack[i] += valclass CustomStack {
constructor(maxSize) {
this.stack = [];
this.maxSize = maxSize;
}
push(x) {
if (this.stack.length < this.maxSize) {
this.stack.push(x);
}
}
pop() {
if (this.stack.length === 0) {
return -1;
}
return this.stack.pop();
}
increment(k, val) {
for (let i = 0; i < Math.min(k, this.stack.length); i++) {
this.stack[i] += val;
}
}
}The CustomStack class keeps a list stack which is the container for stack elements. maxSize is the maximum capacity for the stack. In the push method, the element is added if the stack has not exceeded the set maxSize. The pop method removes and returns the top element or -1 if the stack is empty. In the increment method, the bottom k elements are incremented by iterating through each and adding the given val.
Time Complexity: O(k) for the increment operation, O(1) for the push and pop operations.
Space Complexity: O(n) where n is the max size of the stack.
Approach 2: Optimized Increment Using Lazy Propagation
This approach uses an auxiliary array to optimize the increment operation by storing lazy increments and applying them during the pop operation. This lazy approach ensures that increments do not have to iterate over the stack directly, thus reducing potential redundant operations.
class CustomStack {
private int[] stack;
private int[] increment;
private int top;
public CustomStack(int maxSize) {
stack = new int[maxSize];
increment = new int[maxSize];
top = 0;
}
public void push(int x) {
if (top < stack.length) {
stack[top++] = x;
}
}
public int pop() {
if (top == 0) return -1;
int index = --top;
int ret = stack[index] + increment[index];
if (index > 0) increment[index - 1] += increment[index];
increment[index] = 0;
return ret;
}
public void increment(int k, int val) {
int i = Math.min(k, top) - 1;
if (i >= 0) increment[i] += val;
}
}#include <vector>
class CustomStack {
std::vector<int> stack;
std::vector<int> increment;
int maxSize;
public:
CustomStack(int maxSize) : maxSize(maxSize) {
increment.resize(maxSize, 0);
}
void push(int x) {
if (stack.size() < maxSize) {
stack.push_back(x);
}
}
int pop() {
int idx = stack.size() - 1;
if (idx < 0) return -1;
int result = stack[idx] + increment[idx];
if (idx > 0) increment[idx - 1] += increment[idx];
increment[idx] = 0;
stack.pop_back();
return result;
}
void increment(int k, int val) {
int i = std::min(k, (int)stack.size()) - 1;
if (i >= 0) increment[i] += val;
}
};The Java CustomStack class uses arrays for both stack and increment. Top is an index marker for the top of the stack. Lazy increments are applied when pop is called using the same logic as in the C++ solution, where each element’s increment value is propagated down as necessary.
Time Complexity: O(1) for push and pop; O(1) amortized for increment.
Space Complexity: O(n) where n is the max size of the stack.
Approach 3: 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.
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)
*/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);
*/Complexity comparison
| Approach | Complexity |
|---|---|
| Direct Array-Based Implementation | Time Complexity: |
| Optimized Increment Using Lazy Propagation | Time Complexity: |
| Array Simulation | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Direct Array-Based Implementation | Push/Pop: O(1), Increment: O(k) | O(n) | Good for understanding the stack behavior and simple implementations |
| Lazy Propagation Increment Optimization | O(1) per operation | O(n) | Best for interviews and large operation counts where repeated increments must be efficient |
Frequently asked questions
Is Design a Stack With Increment Operation easy or hard?
The problem is rated Medium because the straightforward solution is simple, but achieving optimal O(1) increments requires recognizing the lazy propagation idea. Understanding how to defer updates makes the design more challenging.
Design a Stack With Increment Operation Python/Java solution
Python solutions typically implement the direct array approach for simplicity, while Java or C++ versions often demonstrate the optimized lazy propagation technique using an additional increment array. Both versions maintain O(n) space and efficient stack operations.
How to solve Design a Stack With Increment Operation in O(n)?
The stack stores up to n elements, so total space is O(n). Achieve constant-time operations by maintaining a second array that tracks pending increments for stack prefixes. When popping, propagate the stored increment to the next index and apply it to the popped value.
What is the best approach for Design a Stack With Increment Operation?
The optimal approach uses a lazy propagation technique with an additional increment array. Instead of updating the bottom k elements directly, store pending increments and apply them when elements are popped. This keeps push, pop, and increment operations all at O(1) time with O(n) space.
Is Design a Stack With Increment Operation asked at Google/Amazon/Meta?
Stack design problems frequently appear in interviews at companies like Amazon, Google, and Meta. Variants that require optimizing stack operations or designing custom data structures are common in system design and data structure interview rounds.
What data structure is used in Design a Stack With Increment Operation?
The problem primarily uses a stack implemented with an array or dynamic list. The optimized solution adds another array to store deferred increments, enabling lazy updates and constant-time operations.
What is the time complexity of Design a Stack With Increment Operation?
A basic array-based implementation runs push and pop in O(1) time but increment in O(k). The optimized lazy propagation design improves this so every operation—push, pop, and increment—runs in O(1) time with O(n) space.
