Stack & Queue Questions for Placement 2026 (with Solutions)

Last Updated: March 2026

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Overview

Stacks and Queues are fundamental linear data structures that form the backbone of many algorithms and system designs. A Stack follows LIFO (Last In, First Out) principle, while a Queue follows FIFO (First In, First Out). These structures are extensively tested in technical interviews at top tech companies.

Companies Focused on Stack & Queue

• Amazon: LRU Cache, Stock Span Problems
• Microsoft: Expression Evaluation, Parentheses Matching
• Google: Min Stack, Sliding Window Maximum
• Adobe: Implement Queue using Stacks
• Flipkart: Next Greater Element, LRU Implementation
• Uber: Ride-matching queue systems

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Core Concepts

Stack (LIFO - Last In First Out)

Operations:
- push(x): Add element to top    O(1)
- pop(): Remove top element      O(1)
- peek()/top(): View top element O(1)
- isEmpty(): Check if empty      O(1)

Visual Representation:

    [5]  <- Top (Last pushed)
    [3]
    [8]
    [2]  <- Bottom (First pushed)

Applications:

• Function call stack (recursion)
• Expression evaluation and conversion
• Backtracking algorithms
• Undo/Redo operations
• Browser history

Queue (FIFO - First In First Out)

Operations:
- enqueue(x): Add to rear         O(1)
- dequeue(): Remove from front    O(1)
- front(): View front element     O(1)
- isEmpty(): Check if empty       O(1)

Visual Representation:

Front -> [2] [8] [3] [5] <- Rear
         (First)     (Last)

Applications:

• CPU scheduling
• Printer spooling
• Breadth-First Search (BFS)
• Cache implementations (LRU)
• Message queues

Monotonic Stack/Queue

A monotonic stack/queue maintains elements in increasing or decreasing order. This is crucial for problems like:

• Next Greater Element
• Sliding Window Maximum
• Largest Rectangle in Histogram

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20 Frequently Asked Coding Questions

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Question 1: Implement Stack using Arrays

Problem: Implement a stack with push, pop, peek, and isEmpty operations.

class Stack:
    def __init__(self, capacity=1000):
        self.capacity = capacity
        self.arr = [None] * capacity
        self.top_idx = -1
    
    def push(self, x):
        if self.top_idx >= self.capacity - 1:
            raise Exception("Stack Overflow")
        self.top_idx += 1
        self.arr[self.top_idx] = x
    
    def pop(self):
        if self.isEmpty():
            raise Exception("Stack Underflow")
        val = self.arr[self.top_idx]
        self.top_idx -= 1
        return val
    
    def peek(self):
        if self.isEmpty():
            raise Exception("Stack is empty")
        return self.arr[self.top_idx]
    
    def isEmpty(self):
        return self.top_idx == -1
    
    def size(self):
        return self.top_idx + 1

# Driver Code
stack = Stack()
stack.push(10)
stack.push(20)
stack.push(30)
print(stack.peek())    # Output: 30
print(stack.pop())     # Output: 30
print(stack.size())    # Output: 2
print(stack.isEmpty()) # Output: False

Time Complexity: All operations O(1)
Space Complexity: O(n) where n is capacity

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Question 2: Implement Queue using Arrays (Circular Queue)

Problem: Implement a circular queue to efficiently use space.

class CircularQueue:
    def __init__(self, capacity=1000):
        self.capacity = capacity
        self.arr = [None] * capacity
        self.front = 0
        self.rear = -1
        self.size = 0
    
    def enqueue(self, x):
        if self.isFull():
            raise Exception("Queue is full")
        self.rear = (self.rear + 1) % self.capacity
        self.arr[self.rear] = x
        self.size += 1
    
    def dequeue(self):
        if self.isEmpty():
            raise Exception("Queue is empty")
        val = self.arr[self.front]
        self.front = (self.front + 1) % self.capacity
        self.size -= 1
        return val
    
    def front_elem(self):
        if self.isEmpty():
            raise Exception("Queue is empty")
        return self.arr[self.front]
    
    def isEmpty(self):
        return self.size == 0
    
    def isFull(self):
        return self.size == self.capacity

# Driver Code
queue = CircularQueue(5)
queue.enqueue(10)
queue.enqueue(20)
queue.enqueue(30)
print(queue.front_elem())  # Output: 10
print(queue.dequeue())     # Output: 10
queue.enqueue(40)
queue.enqueue(50)
print(queue.size)          # Output: 4

Time Complexity: All operations O(1)
Space Complexity: O(n)

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Question 3: Implement Queue using Two Stacks

Problem: Implement a queue using only two stacks.

class QueueUsingStacks:
    def __init__(self):
        self.stack1 = []  # For enqueue
        self.stack2 = []  # For dequeue
    
    def enqueue(self, x):
        self.stack1.append(x)
    
    def dequeue(self):
        if self.isEmpty():
            raise Exception("Queue is empty")
        
        # Transfer elements if stack2 is empty
        if not self.stack2:
            while self.stack1:
                self.stack2.append(self.stack1.pop())
        
        return self.stack2.pop()
    
    def front(self):
        if self.isEmpty():
            raise Exception("Queue is empty")
        
        if not self.stack2:
            while self.stack1:
                self.stack2.append(self.stack1.pop())
        
        return self.stack2[-1]
    
    def isEmpty(self):
        return not self.stack1 and not self.stack2

# Driver Code
queue = QueueUsingStacks()
queue.enqueue(1)
queue.enqueue(2)
queue.enqueue(3)
print(queue.front())    # Output: 1
print(queue.dequeue())  # Output: 1
print(queue.dequeue())  # Output: 2
queue.enqueue(4)
print(queue.dequeue())  # Output: 3

Time Complexity: Amortized O(1) for all operations
Space Complexity: O(n)

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Question 4: Implement Stack using Two Queues

Problem: Implement a stack using only two queues.

from collections import deque

class StackUsingQueues:
    def __init__(self):
        self.queue1 = deque()
        self.queue2 = deque()
    
    def push(self, x):
        # Add to queue2
        self.queue2.append(x)
        
        # Move all elements from queue1 to queue2
        while self.queue1:
            self.queue2.append(self.queue1.popleft())
        
        # Swap queues
        self.queue1, self.queue2 = self.queue2, self.queue1
    
    def pop(self):
        if self.isEmpty():
            raise Exception("Stack is empty")
        return self.queue1.popleft()
    
    def top(self):
        if self.isEmpty():
            raise Exception("Stack is empty")
        return self.queue1[0]
    
    def isEmpty(self):
        return not self.queue1

# Driver Code
stack = StackUsingQueues()
stack.push(1)
stack.push(2)
stack.push(3)
print(stack.top())     # Output: 3
print(stack.pop())     # Output: 3
print(stack.pop())     # Output: 2

Time Complexity: Push O(n), Pop O(1)
Space Complexity: O(n)

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Question 5: Valid Parentheses

Problem: Given a string containing brackets, determine if it's valid.

class Solution:
    def isValid(self, s: str) -> bool:
        stack = []
        mapping = {')': '(', '}': '{', ']': '['}
        
        for char in s:
            if char in mapping:  # Closing bracket
                if not stack or stack.pop() != mapping[char]:
                    return False
            else:  # Opening bracket
                stack.append(char)
        
        return not stack

# Driver Code
sol = Solution()
print(sol.isValid("()"))       # Output: True
print(sol.isValid("()[]{}"))   # Output: True
print(sol.isValid("(]"))       # Output: False
print(sol.isValid("([)]"))     # Output: False
print(sol.isValid("{[]}"))     # Output: True

Time Complexity: O(n)
Space Complexity: O(n)

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Question 6: Next Greater Element I

Problem: For each element in nums1, find the next greater element in nums2.

class Solution:
    def nextGreaterElement(self, nums1, nums2):
        # Monotonic decreasing stack
        stack = []
        next_greater = {}
        
        for num in nums2:
            while stack and stack[-1] < num:
                next_greater[stack.pop()] = num
            stack.append(num)
        
        # Remaining elements have no greater element
        while stack:
            next_greater[stack.pop()] = -1
        
        return [next_greater[num] for num in nums1]

# Driver Code
nums1 = [4, 1, 2]
nums2 = [1, 3, 4, 2]
sol = Solution()
print(sol.nextGreaterElement(nums1, nums2))  # Output: [-1, 3, -1]

Time Complexity: O(n)
Space Complexity: O(n)

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Question 7: Next Greater Element II (Circular Array)

Problem: Find next greater element for each element in circular array.

class Solution:
    def nextGreaterElements(self, nums):
        n = len(nums)
        result = [-1] * n
        stack = []  # Store indices
        
        # Traverse twice for circular effect
        for i in range(2 * n - 1, -1, -1):
            actual_idx = i % n
            
            while stack and nums[stack[-1]] <= nums[actual_idx]:
                stack.pop()
            
            if stack:
                result[actual_idx] = nums[stack[-1]]
            
            stack.append(actual_idx)
        
        return result

# Driver Code
nums = [1, 2, 1]
sol = Solution()
print(sol.nextGreaterElements(nums))  # Output: [2, -1, 2]

Time Complexity: O(n)
Space Complexity: O(n)

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Question 8: Min Stack

Problem: Design a stack that supports push, pop, top, and getMin in constant time.

class MinStack:
    def __init__(self):
        self.stack = []
        self.min_stack = []
    
    def push(self, val: int) -> None:
        self.stack.append(val)
        # Push to min_stack if empty or val <= current min
        if not self.min_stack or val <= self.min_stack[-1]:
            self.min_stack.append(val)
    
    def pop(self) -> None:
        val = self.stack.pop()
        if val == self.min_stack[-1]:
            self.min_stack.pop()
    
    def top(self) -> int:
        return self.stack[-1]
    
    def getMin(self) -> int:
        return self.min_stack[-1]

# Driver Code
minStack = MinStack()
minStack.push(-2)
minStack.push(0)
minStack.push(-3)
print(minStack.getMin())  # Output: -3
minStack.pop()
print(minStack.top())     # Output: 0
print(minStack.getMin())  # Output: -2

Time Complexity: All operations O(1)
Space Complexity: O(n)

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Question 9: Sliding Window Maximum

Problem: Find maximum in every window of size k.

from collections import deque

class Solution:
    def maxSlidingWindow(self, nums, k):
        result = []
        dq = deque()  # Store indices, maintain decreasing order
        
        for i, num in enumerate(nums):
            # Remove elements outside window
            if dq and dq[0] < i - k + 1:
                dq.popleft()
            
            # Remove smaller elements from back
            while dq and nums[dq[-1]] < num:
                dq.pop()
            
            dq.append(i)
            
            # Add to result when window is full
            if i >= k - 1:
                result.append(nums[dq[0]])
        
        return result

# Driver Code
nums = [1, 3, -1, -3, 5, 3, 6, 7]
k = 3
sol = Solution()
print(sol.maxSlidingWindow(nums, k))  # Output: [3, 3, 5, 5, 6, 7]

Time Complexity: O(n)
Space Complexity: O(k)

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Question 10: Largest Rectangle in Histogram

Problem: Find the largest rectangle area in a histogram.

class Solution:
    def largestRectangleArea(self, heights):
        stack = []  # Store indices
        max_area = 0
        n = len(heights)
        
        for i in range(n):
            while stack and heights[stack[-1]] > heights[i]:
                height = heights[stack.pop()]
                width = i if not stack else i - stack[-1] - 1
                max_area = max(max_area, height * width)
            stack.append(i)
        
        # Process remaining elements
        while stack:
            height = heights[stack.pop()]
            width = n if not stack else n - stack[-1] - 1
            max_area = max(max_area, height * width)
        
        return max_area

# Driver Code
heights = [2, 1, 5, 6, 2, 3]
sol = Solution()
print(sol.largestRectangleArea(heights))  # Output: 10

Time Complexity: O(n)
Space Complexity: O(n)

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Question 11: Stock Span Problem

Problem: Calculate stock span for each day (consecutive days with price <= current).

class Solution:
    def calculateSpan(self, price, n):
        span = [0] * n
        stack = []  # Store indices
        
        for i in range(n):
            # Pop elements with price <= current
            while stack and price[stack[-1]] <= price[i]:
                stack.pop()
            
            if not stack:
                span[i] = i + 1
            else:
                span[i] = i - stack[-1]
            
            stack.append(i)
        
        return span

# Driver Code
price = [100, 80, 60, 70, 60, 75, 85]
n = len(price)
sol = Solution()
print(sol.calculateSpan(price, n))  # Output: [1, 1, 1, 2, 1, 4, 6]

Time Complexity: O(n)
Space Complexity: O(n)

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Question 12: LRU Cache Implementation

Problem: Design and implement an LRU (Least Recently Used) cache.

from collections import OrderedDict

class LRUCache:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.cache = OrderedDict()
    
    def get(self, key: int) -> int:
        if key not in self.cache:
            return -1
        # Move to end (most recently used)
        self.cache.move_to_end(key)
        return self.cache[key]
    
    def put(self, key: int, value: int) -> None:
        if key in self.cache:
            self.cache.move_to_end(key)
        self.cache[key] = value
        
        if len(self.cache) > self.capacity:
            # Remove first item (least recently used)
            self.cache.popitem(last=False)

# Without OrderedDict (using Doubly Linked List + HashMap)
class Node:
    def __init__(self, key=0, val=0):
        self.key = key
        self.val = val
        self.prev = None
        self.next = None

class LRUCacheOptimized:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.cache = {}
        self.head = Node()
        self.tail = Node()
        self.head.next = self.tail
        self.tail.prev = self.head
    
    def _remove(self, node):
        prev, nxt = node.prev, node.next
        prev.next, nxt.prev = nxt, prev
    
    def _add_to_front(self, node):
        node.next = self.head.next
        node.prev = self.head
        self.head.next.prev = node
        self.head.next = node
    
    def get(self, key: int) -> int:
        if key not in self.cache:
            return -1
        node = self.cache[key]
        self._remove(node)
        self._add_to_front(node)
        return node.val
    
    def put(self, key: int, value: int) -> None:
        if key in self.cache:
            self._remove(self.cache[key])
        
        node = Node(key, value)
        self.cache[key] = node
        self._add_to_front(node)
        
        if len(self.cache) > self.capacity:
            lru = self.tail.prev
            self._remove(lru)
            del self.cache[lru.key]

# Driver Code
cache = LRUCacheOptimized(2)
cache.put(1, 1)
cache.put(2, 2)
print(cache.get(1))    # Output: 1
cache.put(3, 3)        # Evicts key 2
print(cache.get(2))    # Output: -1
cache.put(4, 4)        # Evicts key 1
print(cache.get(1))    # Output: -1
print(cache.get(3))    # Output: 3
print(cache.get(4))    # Output: 4

Time Complexity: Get and Put O(1)
Space Complexity: O(capacity)

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Question 13: Evaluate Reverse Polish Notation

Problem: Evaluate the value of an arithmetic expression in Reverse Polish Notation.

class Solution:
    def evalRPN(self, tokens):
        stack = []
        
        for token in tokens:
            if token in "+-*/":
                b = stack.pop()
                a = stack.pop()
                
                if token == '+':
                    stack.append(a + b)
                elif token == '-':
                    stack.append(a - b)
                elif token == '*':
                    stack.append(a * b)
                else:  # Division
                    stack.append(int(a / b))
            else:
                stack.append(int(token))
        
        return stack[0]

# Driver Code
tokens = ["2", "1", "+", "3", "*"]
sol = Solution()
print(sol.evalRPN(tokens))  # Output: 9

tokens2 = ["4", "13", "5", "/", "+"]
print(sol.evalRPN(tokens2))  # Output: 6

Time Complexity: O(n)
Space Complexity: O(n)

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Question 14: Daily Temperatures

Problem: Find days until a warmer temperature for each day.

class Solution:
    def dailyTemperatures(self, temperatures):
        n = len(temperatures)
        result = [0] * n
        stack = []  # Store indices
        
        for i in range(n):
            while stack and temperatures[stack[-1]] < temperatures[i]:
                prev_idx = stack.pop()
                result[prev_idx] = i - prev_idx
            stack.append(i)
        
        return result

# Driver Code
temperatures = [73, 74, 75, 71, 69, 72, 76, 73]
sol = Solution()
print(sol.dailyTemperatures(temperatures))  # Output: [1, 1, 4, 2, 1, 1, 0, 0]

Time Complexity: O(n)
Space Complexity: O(n)

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Question 15: Online Stock Span

Problem: Design an algorithm that collects daily price quotes and returns the span.

class StockSpanner:
    def __init__(self):
        self.stack = []  # (price, span)
    
    def next(self, price: int) -> int:
        span = 1
        
        while self.stack and self.stack[-1][0] <= price:
            span += self.stack.pop()[1]
        
        self.stack.append((price, span))
        return span

# Driver Code
stockSpanner = StockSpanner()
print(stockSpanner.next(100))  # Output: 1
print(stockSpanner.next(80))   # Output: 1
print(stockSpanner.next(60))   # Output: 1
print(stockSpanner.next(70))   # Output: 2
print(stockSpanner.next(60))   # Output: 1
print(stockSpanner.next(75))   # Output: 4
print(stockSpanner.next(85))   # Output: 6

Time Complexity: Amortized O(1) per query
Space Complexity: O(n)

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Question 16: Simplify Path

Problem: Convert Unix-style absolute path to simplified canonical path.

class Solution:
    def simplifyPath(self, path: str) -> str:
        stack = []
        components = path.split('/')
        
        for comp in components:
            if comp == '' or comp == '.':
                continue
            elif comp == '..':
                if stack:
                    stack.pop()
            else:
                stack.append(comp)
        
        return '/' + '/'.join(stack)

# Driver Code
sol = Solution()
print(sol.simplifyPath("/home/"))                    # Output: "/home"
print(sol.simplifyPath("/../"))                      # Output: "/"
print(sol.simplifyPath("/home//foo/"))               # Output: "/home/foo"
print(sol.simplifyPath("/a/./b/../../c/"))          # Output: "/c"

Time Complexity: O(n)
Space Complexity: O(n)

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Question 17: Implement Stack with Increment Operation

Problem: Design a stack with push, pop, and increment operations.

class CustomStack:
    def __init__(self, maxSize: int):
        self.stack = []
        self.maxSize = maxSize
        self.inc = []  # Increment tracking
    
    def push(self, x: int) -> None:
        if len(self.stack) < self.maxSize:
            self.stack.append(x)
            self.inc.append(0)
    
    def pop(self) -> int:
        if not self.stack:
            return -1
        
        # Propagate increment to below element
        if len(self.inc) > 1:
            self.inc[-2] += self.inc[-1]
        
        return self.stack.pop() + self.inc.pop()
    
    def increment(self, k: int, val: int) -> None:
        if self.inc:
            idx = min(k, len(self.inc)) - 1
            self.inc[idx] += val

# Driver Code
customStack = CustomStack(3)
customStack.push(1)
customStack.push(2)
print(customStack.pop())         # Output: 2
customStack.push(2)
customStack.push(3)
customStack.push(4)              # Stack is full
customStack.increment(5, 100)
customStack.increment(2, 100)
print(customStack.pop())         # Output: 103
print(customStack.pop())         # Output: 202
print(customStack.pop())         # Output: 201

Time Complexity: Push O(1), Pop O(1), Increment O(1)
Space Complexity: O(n)

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Question 18: Number of Recent Calls

Problem: Count recent calls within 3000 milliseconds window.

from collections import deque

class RecentCounter:
    def __init__(self):
        self.queue = deque()
    
    def ping(self, t: int) -> int:
        self.queue.append(t)
        
        # Remove pings older than 3000ms
        while self.queue[0] < t - 3000:
            self.queue.popleft()
        
        return len(self.queue)

# Driver Code
recentCounter = RecentCounter()
print(recentCounter.ping(1))     # Output: 1
print(recentCounter.ping(100))   # Output: 2
print(recentCounter.ping(3001))  # Output: 3
print(recentCounter.ping(3002))  # Output: 3

Time Complexity: Amortized O(1) per ping
Space Complexity: O(3000) = O(1)

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Question 19: Maximum Frequency Stack

Problem: Implement a stack that pops the most frequent element.

from collections import defaultdict

class FreqStack:
    def __init__(self):
        self.freq = defaultdict(int)      # element -> frequency
        self.group = defaultdict(list)    # frequency -> list of elements
        self.max_freq = 0
    
    def push(self, val: int) -> None:
        self.freq[val] += 1
        f = self.freq[val]
        self.max_freq = max(self.max_freq, f)
        self.group[f].append(val)
    
    def pop(self) -> int:
        val = self.group[self.max_freq].pop()
        self.freq[val] -= 1
        
        if not self.group[self.max_freq]:
            self.max_freq -= 1
        
        return val

# Driver Code
freqStack = FreqStack()
freqStack.push(5)
freqStack.push(7)
freqStack.push(5)
freqStack.push(7)
freqStack.push(4)
freqStack.push(5)
print(freqStack.pop())  # Output: 5
print(freqStack.pop())  # Output: 7
print(freqStack.pop())  # Output: 5
print(freqStack.pop())  # Output: 4

Time Complexity: Push O(1), Pop O(1)
Space Complexity: O(n)

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Question 20: Trapping Rain Water

Problem: Calculate how much water can be trapped after raining.

class Solution:
    def trap(self, height):
        if not height:
            return 0
        
        n = len(height)
        left, right = 0, n - 1
        left_max, right_max = 0, 0
        water = 0
        
        while left < right:
            if height[left] < height[right]:
                if height[left] >= left_max:
                    left_max = height[left]
                else:
                    water += left_max - height[left]
                left += 1
            else:
                if height[right] >= right_max:
                    right_max = height[right]
                else:
                    water += right_max - height[right]
                right -= 1
        
        return water

# Alternative: Stack-based approach
def trap_stack(height):
    stack = []
    water = 0
    
    for i, h in enumerate(height):
        while stack and height[stack[-1]] < h:
            top = stack.pop()
            if not stack:
                break
            distance = i - stack[-1] - 1
            bounded_height = min(height[stack[-1]], h) - height[top]
            water += distance * bounded_height
        stack.append(i)
    
    return water

# Driver Code
height = [0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1]
sol = Solution()
print(sol.trap(height))  # Output: 6

Time Complexity: O(n)
Space Complexity: O(1) for two-pointer, O(n) for stack

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MCQ Questions

Question 1

What is the time complexity of push operation in a stack?

• A) O(n)
• B) O(log n)
• C) O(1)
• D) O(n²)

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Question 2

Which data structure is used to implement recursion?

• A) Queue
• B) Stack
• C) Array
• D) Linked List

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Question 3

What principle does a Queue follow?

• A) LIFO
• B) FIFO
• C) LILO
• D) FILO

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Question 4

In a monotonic decreasing stack, elements are stored in:

• A) Increasing order from bottom to top
• B) Decreasing order from bottom to top
• C) Random order
• D) Sorted order

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Question 5

Which problem can be efficiently solved using a monotonic queue?

• A) Sorting
• B) Sliding Window Maximum
• C) Binary Search
• D) Matrix Multiplication

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Interview Tips for Stack & Queue

1. Understand the fundamentals: Know LIFO vs FIFO thoroughly

2. Practice these patterns:

   • Monotonic stack for next greater/smaller element
   • Two stacks for implementing queues
   • Stack for expression evaluation
   • Queue for sliding window problems

3. Common interview scenarios:

   • Implementing data structures from scratch
   • Solving problems with space constraints
   • Time complexity analysis

4. Edge cases to handle:

   • Empty stack/queue operations
   • Overflow conditions
   • Single element
   • Maximum capacity

5. System Design applications:

   • Browser history (stack)
   • Print spooling (queue)
   • Undo/Redo operations (stack)
   • Rate limiting (queue)

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Frequently Asked Questions

Q1: When to use Stack vs Queue?

A: Use Stack for LIFO operations (backtracking, expression evaluation, undo). Use Queue for FIFO operations (BFS, scheduling, caching, level-order traversal).

Q2: How do I implement a Queue using Stacks efficiently?

A: Use two stacks. Push to stack1. When popping, if stack2 is empty, transfer all elements from stack1 to stack2. This gives amortized O(1) time complexity.

Q3: What is a Monotonic Stack?

A: A stack that maintains elements in either strictly increasing or decreasing order. Used for Next Greater Element, Largest Rectangle in Histogram, etc.

Q4: How does LRU Cache work?

A: LRU Cache uses a combination of HashMap (for O(1) access) and Doubly Linked List (for O(1) removal/addition). Most recently used items move to front, least recently used at back.

Q5: What is the advantage of Circular Queue?

A: Circular Queue efficiently utilizes space by reusing empty locations created after dequeue operations, preventing "false overflow" in regular queues.

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Master these stack and queue problems to excel in technical interviews. Focus on understanding patterns rather than memorizing solutions.