LeetCode #1381 — MEDIUM

Design a Stack With Increment Operation

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Stack · Design

Example 1

["CustomStack","push","push","pop","push","push","push","increment","increment","pop","pop","pop","pop"]
[[3],[1],[2],[],[2],[3],[4],[5,100],[2,100],[],[],[],[]]

Step 02

Core Insight

What unlocks the optimal approach

Use an array to represent the stack. Push will add new integer to the array. Pop removes the last element in the array and increment will add val to the first k elements of the array.
This solution run in O(1) per push and pop and O(k) per increment.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1381: Design a Stack With Increment Operation
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);
 */

// Accepted solution for LeetCode #1381: Design a Stack With Increment Operation
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);
 */

# Accepted solution for LeetCode #1381: Design a Stack With Increment Operation
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)

// Accepted solution for LeetCode #1381: Design a Stack With Increment Operation
struct CustomStack {
    stack: Vec<i32>,
    inc: Vec<i32>,
    n: usize,
    max_size: usize,
}

impl CustomStack {
    fn new(max_size: i32) -> Self {
        let max_size = max_size as usize;
        let stack = vec![];
        let inc = vec![0; (1 + max_size) as usize];
        let n = 0;
        CustomStack {
            stack,
            inc,
            n,
            max_size,
        }
    }

    fn push(&mut self, x: i32) {
        if self.n != self.max_size {
            self.stack.push(x);
            self.n += 1;
        }
    }

    fn pop(&mut self) -> i32 {
        if let Some(mut top) = self.stack.pop() {
            self.inc[self.n - 1] += self.inc[self.n];
            top += self.inc[self.n];
            self.inc[self.n] = 0;
            self.n -= 1;
            top
        } else {
            -1
        }
    }

    fn increment(&mut self, k: i32, val: i32) {
        let k = k as usize;
        if k > self.n {
            self.inc[self.n] += val;
        } else {
            self.inc[k] += val;
        }
    }
}

#[test]
fn test() {
    let mut stack = CustomStack::new(3);
    stack.push(1);
    stack.push(2);
    assert_eq!(stack.pop(), 2);
    stack.push(2);
    stack.push(3);
    stack.push(4);
    stack.increment(5, 100);
    stack.increment(2, 100);
    assert_eq!(stack.pop(), 103);
    assert_eq!(stack.pop(), 202);
    assert_eq!(stack.pop(), 201);
    assert_eq!(stack.pop(), -1);
}

// Accepted solution for LeetCode #1381: Design a Stack With Increment Operation
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)
 */

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(1)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.