Next Greater Element II (Circular Array)

🎯

Next Greater Element II (Circular Array)

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Imagine you have a circular conveyor belt with numbers on packages, and you want to find the next bigger package for each one as the belt keeps rotating endlessly.

💡 This problem asks for the next greater element for each number in a circular array. Beginners often struggle because the circular nature means you can't just scan once linearly; you must consider wrap-around, which complicates naive solutions.

📋

Problem Statement

Given a circular array nums (the next element of the last element is the first element), return an array answer where answer[i] is the next greater number of nums[i]. If it doesn't exist, set answer[i] to -1.

1 ≤ nums.length ≤ 10^5-10^9 ≤ nums[i] ≤ 10^9

💡

Example

Input"[1, 2, 1]"

Output[2, -1, 2]

For nums[0] = 1, next greater is 2; for nums[1] = 2, no greater element; for nums[2] = 1, next greater is 2 (wraps around).

• Single element array → output [-1]
• All elements equal → output all -1
• Strictly decreasing array → all -1 except wrap-around
• Array with multiple duplicates → correctly find next greater ignoring duplicates

⚠️

Common Mistakes

Not simulating circular array correctly

Misses next greater elements that appear after wrapping around, leading to incorrect -1 results

✅ Iterate twice or use modulo indexing to simulate circularity

Pushing indices in second pass of iteration

Duplicates indices on stack causing incorrect updates or infinite loops

✅ Only push indices during the first pass (i < n)

Using values instead of indices in stack

Cannot update correct positions in answer array, leading to wrong outputs

✅ Always store indices in stack to map back to answer positions

Not initializing answer array with -1

Uninitialized values cause wrong outputs for elements with no next greater

✅ Initialize answer array with -1 to represent no next greater element

🧠

Brute Force (Nested Loops)

💡 This approach exists to build intuition by directly simulating the problem definition. It is simple but inefficient, helping beginners understand the problem deeply before optimizing.

Intuition

For each element, scan forward circularly until you find a greater element or return -1 if none exists.

Algorithm

1. Initialize an answer array with -1 for all elements.
2. For each index i in nums:
3. Scan forward circularly from i+1 to i (wrapping around) to find the next greater element.
4. If found, update answer[i] with that element; otherwise, leave it as -1.

💡 The nested loops and circular scanning make it hard to see efficiency, but this directly follows the problem statement.

</>

Code

from typing import List

def nextGreaterElements(nums: List[int]) -> List[int]:
    n = len(nums)
    answer = [-1] * n
    for i in range(n):
        for j in range(1, n):
            idx = (i + j) % n
            if nums[idx] > nums[i]:
                answer[i] = nums[idx]
                break
    return answer

# Driver code
if __name__ == '__main__':
    print(nextGreaterElements([1,2,1]))  # Output: [2, -1, 2]

Line Notes

n = len(nums)Store length once to avoid repeated calls

answer = [-1] * nInitialize answer with -1 assuming no greater element found

for i in range(n)Outer loop picks each element to find its next greater

for j in range(1, n)Inner loop scans forward circularly excluding current element

idx = (i + j) % nWrap index around to simulate circular array

if nums[idx] > nums[i]Check if current scanned element is greater

answer[i] = nums[idx]Update answer when next greater found

breakStop scanning once next greater is found

import java.util.*;

public class NextGreaterElementII {
    public static int[] nextGreaterElements(int[] nums) {
        int n = nums.length;
        int[] answer = new int[n];
        Arrays.fill(answer, -1);
        for (int i = 0; i < n; i++) {
            for (int j = 1; j < n; j++) {
                int idx = (i + j) % n;
                if (nums[idx] > nums[i]) {
                    answer[i] = nums[idx];
                    break;
                }
            }
        }
        return answer;
    }

    public static void main(String[] args) {
        int[] nums = {1, 2, 1};
        System.out.println(Arrays.toString(nextGreaterElements(nums)));
    }
}

Line Notes

int n = nums.length;Cache array length for efficiency

int[] answer = new int[n];Create answer array

Arrays.fill(answer, -1);Initialize with -1 assuming no greater element

for (int i = 0; i < n; i++)Outer loop over each element

for (int j = 1; j < n; j++)Inner loop scans forward circularly

int idx = (i + j) % n;Wrap index to simulate circular array

if (nums[idx] > nums[i])Check for next greater element

answer[i] = nums[idx];Update answer when found

break;Stop scanning after finding next greater

#include <iostream>
#include <vector>
using namespace std;

vector<int> nextGreaterElements(vector<int>& nums) {
    int n = nums.size();
    vector<int> answer(n, -1);
    for (int i = 0; i < n; i++) {
        for (int j = 1; j < n; j++) {
            int idx = (i + j) % n;
            if (nums[idx] > nums[i]) {
                answer[i] = nums[idx];
                break;
            }
        }
    }
    return answer;
}

int main() {
    vector<int> nums = {1, 2, 1};
    vector<int> res = nextGreaterElements(nums);
    for (int x : res) cout << x << ' ';
    cout << endl;
    return 0;
}

Line Notes

int n = nums.size();Store size once for efficiency

vector<int> answer(n, -1);Initialize answer with -1

for (int i = 0; i < n; i++)Outer loop over each element

for (int j = 1; j < n; j++)Inner loop scans circularly

int idx = (i + j) % n;Wrap index for circular array

if (nums[idx] > nums[i])Check for next greater element

answer[i] = nums[idx];Update answer when found

break;Stop scanning after finding next greater

function nextGreaterElements(nums) {
    const n = nums.length;
    const answer = new Array(n).fill(-1);
    for (let i = 0; i < n; i++) {
        for (let j = 1; j < n; j++) {
            const idx = (i + j) % n;
            if (nums[idx] > nums[i]) {
                answer[i] = nums[idx];
                break;
            }
        }
    }
    return answer;
}

// Test
console.log(nextGreaterElements([1, 2, 1])); // [2, -1, 2]

Line Notes

const n = nums.length;Cache length for efficiency

const answer = new Array(n).fill(-1);Initialize answer with -1

for (let i = 0; i < n; i++)Outer loop over each element

for (let j = 1; j < n; j++)Inner loop scans circularly

const idx = (i + j) % n;Wrap index for circular array

if (nums[idx] > nums[i])Check for next greater element

answer[i] = nums[idx];Update answer when found

break;Stop scanning after finding next greater

⏱

Complexity

TimeO(n^2)

SpaceO(n)

For each of the n elements, we scan up to n-1 elements circularly, resulting in O(n^2) time. Space is O(n) for the output array.

💡 For n=20, this means up to 400 operations, which is manageable, but for n=10^5, it becomes 10 billion operations, too slow for interviews.

Interview Verdict: TLE

This approach is too slow for large inputs but is useful to understand the problem and verify correctness on small inputs.

🧠

Monotonic Stack (Single Pass with Circular Simulation)

💡 This approach uses a stack to efficiently find the next greater element by maintaining a decreasing stack and simulating the circular array by iterating twice.

Intuition

Traverse the array twice, pushing indices onto a stack. For each element, pop smaller elements from the stack and assign the current element as their next greater.

Algorithm

1. Initialize an answer array with -1 and an empty stack.
2. Iterate i from 0 to 2*n - 1 (simulate circular array).
3. While stack not empty and nums[i % n] > nums[stack.top()], pop stack and update answer for popped index.
4. If i < n, push i onto stack to find its next greater later.

💡 The double iteration and stack pops can be tricky to visualize but ensure each element is processed efficiently.

</>

Code

from typing import List

def nextGreaterElements(nums: List[int]) -> List[int]:
    n = len(nums)
    answer = [-1] * n
    stack = []  # stores indices
    for i in range(2 * n):
        while stack and nums[i % n] > nums[stack[-1]]:
            idx = stack.pop()
            answer[idx] = nums[i % n]
        if i < n:
            stack.append(i)
    return answer

# Driver code
if __name__ == '__main__':
    print(nextGreaterElements([1,2,1]))  # Output: [2, -1, 2]

Line Notes

n = len(nums)Cache length for efficiency

answer = [-1] * nInitialize answer with -1

stack = []Stack to keep indices of elements waiting for next greater

for i in range(2 * n)Double iteration to simulate circular array

while stack and nums[i % n] > nums[stack[-1]]Pop smaller elements and update their answer

idx = stack.pop()Get index of element to update

answer[idx] = nums[i % n]Assign next greater element

if i < nOnly push indices from first pass to avoid duplicates

stack.append(i)Add current index to stack

import java.util.*;

public class NextGreaterElementII {
    public static int[] nextGreaterElements(int[] nums) {
        int n = nums.length;
        int[] answer = new int[n];
        Arrays.fill(answer, -1);
        Deque<Integer> stack = new ArrayDeque<>();
        for (int i = 0; i < 2 * n; i++) {
            while (!stack.isEmpty() && nums[i % n] > nums[stack.peek()]) {
                answer[stack.pop()] = nums[i % n];
            }
            if (i < n) {
                stack.push(i);
            }
        }
        return answer;
    }

    public static void main(String[] args) {
        int[] nums = {1, 2, 1};
        System.out.println(Arrays.toString(nextGreaterElements(nums)));
    }
}

Line Notes

int n = nums.length;Cache length for efficiency

int[] answer = new int[n];Create answer array

Arrays.fill(answer, -1);Initialize with -1

Deque<Integer> stack = new ArrayDeque<>();Stack to hold indices

for (int i = 0; i < 2 * n; i++)Double iteration for circular simulation

while (!stack.isEmpty() && nums[i % n] > nums[stack.peek()])Pop smaller elements and update answer

answer[stack.pop()] = nums[i % n];Assign next greater element

if (i < n)Push indices only in first pass

stack.push(i);Add current index to stack

#include <iostream>
#include <vector>
#include <stack>
using namespace std;

vector<int> nextGreaterElements(vector<int>& nums) {
    int n = nums.size();
    vector<int> answer(n, -1);
    stack<int> st;
    for (int i = 0; i < 2 * n; i++) {
        while (!st.empty() && nums[i % n] > nums[st.top()]) {
            answer[st.top()] = nums[i % n];
            st.pop();
        }
        if (i < n) st.push(i);
    }
    return answer;
}

int main() {
    vector<int> nums = {1, 2, 1};
    vector<int> res = nextGreaterElements(nums);
    for (int x : res) cout << x << ' ';
    cout << endl;
    return 0;
}

Line Notes

int n = nums.size();Cache size for efficiency

vector<int> answer(n, -1);Initialize answer with -1

stack<int> st;Stack to hold indices

for (int i = 0; i < 2 * n; i++)Double iteration simulates circular array

while (!st.empty() && nums[i % n] > nums[st.top()])Pop smaller elements and update answer

answer[st.top()] = nums[i % n];Assign next greater element

st.pop();Remove processed index

if (i < n) st.push(i);Push indices only in first pass

function nextGreaterElements(nums) {
    const n = nums.length;
    const answer = new Array(n).fill(-1);
    const stack = [];
    for (let i = 0; i < 2 * n; i++) {
        while (stack.length && nums[i % n] > nums[stack[stack.length - 1]]) {
            const idx = stack.pop();
            answer[idx] = nums[i % n];
        }
        if (i < n) stack.push(i);
    }
    return answer;
}

// Test
console.log(nextGreaterElements([1, 2, 1])); // [2, -1, 2]

Line Notes

const n = nums.length;Cache length for efficiency

const answer = new Array(n).fill(-1);Initialize answer with -1

const stack = [];Stack to hold indices

for (let i = 0; i < 2 * n; i++)Double iteration simulates circular array

while (stack.length && nums[i % n] > nums[stack[stack.length - 1]])Pop smaller elements and update answer

const idx = stack.pop();Get index to update

answer[idx] = nums[i % n];Assign next greater element

if (i < n) stack.push(i);Push indices only in first pass

⏱

Complexity

TimeO(n)

SpaceO(n)

Each element is pushed and popped at most once, and we iterate twice over the array, resulting in O(n) time. Space is O(n) for the stack and answer.

💡 For n=10^5, this means about 200,000 operations, which is efficient and acceptable for interviews.

Interview Verdict: Accepted

This is the optimal and standard approach for this problem, expected in interviews.

🧠

Monotonic Stack with Explicit Circular Array Copy

💡 This approach explicitly creates a doubled array to simulate circularity, then uses a monotonic stack to find next greater elements, trading space for clarity.

Intuition

Duplicate the array to handle circular wrap-around, then use a monotonic stack on the doubled array to find next greater elements for the original indices.

Algorithm

1. Create a new array by concatenating nums with itself.
2. Initialize an answer array with -1 and an empty stack.
3. Iterate over the doubled array indices.
4. While stack not empty and current element is greater than element at stack top, update answer for popped index if within original array.
5. Push current index if it is within original array length.

💡 Duplicating the array simplifies circular logic but doubles space usage, which may be acceptable depending on constraints.

</>

Code

from typing import List

def nextGreaterElements(nums: List[int]) -> List[int]:
    n = len(nums)
    doubled = nums + nums
    answer = [-1] * n
    stack = []
    for i, num in enumerate(doubled):
        while stack and doubled[stack[-1]] < num:
            idx = stack.pop()
            if idx < n:
                answer[idx] = num
        if i < n:
            stack.append(i)
    return answer

# Driver code
if __name__ == '__main__':
    print(nextGreaterElements([1,2,1]))  # Output: [2, -1, 2]

Line Notes

n = len(nums)Cache length for efficiency

doubled = nums + numsCreate doubled array to simulate circularity

answer = [-1] * nInitialize answer with -1

stack = []Stack to hold indices

for i, num in enumerate(doubled)Iterate over doubled array

while stack and doubled[stack[-1]] < numPop smaller elements and update answer

idx = stack.pop()Get index to update

if idx < nOnly update answer for original array indices

if i < nPush indices only from original array

stack.append(i)Add current index to stack

import java.util.*;

public class NextGreaterElementII {
    public static int[] nextGreaterElements(int[] nums) {
        int n = nums.length;
        int[] doubled = new int[2 * n];
        for (int i = 0; i < n; i++) {
            doubled[i] = nums[i];
            doubled[i + n] = nums[i];
        }
        int[] answer = new int[n];
        Arrays.fill(answer, -1);
        Deque<Integer> stack = new ArrayDeque<>();
        for (int i = 0; i < 2 * n; i++) {
            while (!stack.isEmpty() && doubled[stack.peek()] < doubled[i]) {
                int idx = stack.pop();
                if (idx < n) {
                    answer[idx] = doubled[i];
                }
            }
            if (i < n) {
                stack.push(i);
            }
        }
        return answer;
    }

    public static void main(String[] args) {
        int[] nums = {1, 2, 1};
        System.out.println(Arrays.toString(nextGreaterElements(nums)));
    }
}

Line Notes

int n = nums.length;Cache length for efficiency

int[] doubled = new int[2 * n];Create doubled array

for (int i = 0; i < n; i++)Copy original array twice

int[] answer = new int[n];Create answer array

Arrays.fill(answer, -1);Initialize with -1

Deque<Integer> stack = new ArrayDeque<>();Stack to hold indices

for (int i = 0; i < 2 * n; i++)Iterate over doubled array

while (!stack.isEmpty() && doubled[stack.peek()] < doubled[i])Pop smaller elements and update answer

int idx = stack.pop();Get index to update

if (idx < n)Update only original indices

if (i < n)Push indices only from original array

stack.push(i);Add current index to stack

#include <iostream>
#include <vector>
#include <stack>
using namespace std;

vector<int> nextGreaterElements(vector<int>& nums) {
    int n = nums.size();
    vector<int> doubled(2 * n);
    for (int i = 0; i < n; i++) {
        doubled[i] = nums[i];
        doubled[i + n] = nums[i];
    }
    vector<int> answer(n, -1);
    stack<int> st;
    for (int i = 0; i < 2 * n; i++) {
        while (!st.empty() && doubled[st.top()] < doubled[i]) {
            int idx = st.top();
            st.pop();
            if (idx < n) answer[idx] = doubled[i];
        }
        if (i < n) st.push(i);
    }
    return answer;
}

int main() {
    vector<int> nums = {1, 2, 1};
    vector<int> res = nextGreaterElements(nums);
    for (int x : res) cout << x << ' ';
    cout << endl;
    return 0;
}

Line Notes

int n = nums.size();Cache size for efficiency

vector<int> doubled(2 * n);Create doubled array

for (int i = 0; i < n; i++)Copy original array twice

vector<int> answer(n, -1);Initialize answer with -1

stack<int> st;Stack to hold indices

for (int i = 0; i < 2 * n; i++)Iterate over doubled array

while (!st.empty() && doubled[st.top()] < doubled[i])Pop smaller elements and update answer

int idx = st.top();Get index to update

st.pop();Remove processed index

if (idx < n) answer[idx] = doubled[i];Update only original indices

if (i < n) st.push(i);Push indices only from original array

function nextGreaterElements(nums) {
    const n = nums.length;
    const doubled = nums.concat(nums);
    const answer = new Array(n).fill(-1);
    const stack = [];
    for (let i = 0; i < 2 * n; i++) {
        while (stack.length && doubled[stack[stack.length - 1]] < doubled[i]) {
            const idx = stack.pop();
            if (idx < n) {
                answer[idx] = doubled[i];
            }
        }
        if (i < n) stack.push(i);
    }
    return answer;
}

// Test
console.log(nextGreaterElements([1, 2, 1])); // [2, -1, 2]

Line Notes

const n = nums.length;Cache length for efficiency

const doubled = nums.concat(nums);Create doubled array

const answer = new Array(n).fill(-1);Initialize answer with -1

const stack = [];Stack to hold indices

for (let i = 0; i < 2 * n; i++)Iterate over doubled array

while (stack.length && doubled[stack[stack.length - 1]] < doubled[i])Pop smaller elements and update answer

const idx = stack.pop();Get index to update

if (idx < n)Update only original indices

if (i < n) stack.push(i);Push indices only from original array

⏱

Complexity

TimeO(n)

SpaceO(n)

We iterate over a doubled array of size 2n, but each element is pushed and popped at most once, so time is O(n). Space is O(n) for the doubled array, stack, and answer.

💡 This approach uses more memory but is easier to understand, which can be helpful in interviews if time permits.

Interview Verdict: Accepted

This approach is acceptable but less space efficient than the previous monotonic stack approach.

📊

All Approaches - One-Glance Tradeoffs

💡 The monotonic stack with double iteration is the best approach to implement in interviews due to its optimal time and space and clarity.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n^2)	O(n)	No	N/A	Mention only - never code due to inefficiency
2. Monotonic Stack (Double Pass with Modulo)	O(n)	O(n)	No	N/A	Code this approach - optimal and standard
3. Monotonic Stack with Explicit Doubled Array	O(n)	O(n)	No	N/A	Mention as alternative if clarity preferred over space

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, practice coding the monotonic stack approach, and prepare to explain your reasoning clearly in interviews.

How to Present

Step 1: Clarify the problem and confirm circular array behavior.Step 2: Describe the brute force approach to show understanding.Step 3: Explain inefficiency and motivate the monotonic stack.Step 4: Present the monotonic stack solution with double iteration.Step 5: Code the solution and test with examples.Step 6: Discuss edge cases and possible follow-ups.

Time Allocation

Clarify: 3min → Approach: 5min → Code: 10min → Test & Discuss: 7min. Total ~25min

What the Interviewer Tests

Interviewer tests your ability to handle circular arrays, use stacks effectively, optimize from brute force to O(n), and write clean, bug-free code.

Common Follow-ups

• What if the array was not circular? → Single pass monotonic stack suffices.
• Can you find next smaller elements instead? → Similar approach with reversed comparison.

💡 These follow-ups test your flexibility with the monotonic stack pattern and understanding of circular vs linear arrays.

🔍

Pattern Recognition

When to Use

1) Asked for next greater/smaller element; 2) Circular array or wrap-around mentioned; 3) Need efficient O(n) solution; 4) Stack or monotonic stack hinted or applicable.

Signature Phrases

next greater numbercircular arrayif it doesn't exist, return -1

NOT This Pattern When

Problems asking for maximum subarray sums or sliding window maximums use different techniques despite some superficial similarity.

Similar Problems

Next Greater Element I - linear array versionDaily Temperatures - similar monotonic stack usageNext Smaller Element - inverse problem using similar stack logic

Practice

1. Given the following Python code for asteroid collision, what is the output when the input is [8, -8, 5, -10]?

def asteroidCollision(asteroids):
    stack = []
    for a in asteroids:
        alive = True
        while alive and stack and stack[-1] > 0 and a < 0:
            if stack[-1] < -a:
                stack.pop()
            elif stack[-1] == -a:
                stack.pop()
                alive = False
            else:
                alive = False
        if alive:
            stack.append(a)
    return stack

print(asteroidCollision([8, -8, 5, -10]))

easy

A. [5, -10]

B. [-10]

C. [8, 5, -10]

D. []

Solution

1. Step 1: Trace collision between 8 and -8

   8 and -8 are equal size but opposite direction, both explode -> stack empty.

2. Step 2: Process 5 and then -10

   5 is pushed; -10 collides with 5, since 10 > 5, 5 pops, -10 remains and is pushed.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Final stack contains only [-10] after resolving collisions [OK]

Hint: Equal size opposite asteroids both explode [OK]

Common Mistakes:

• Forgetting to pop both equal asteroids
• Appending asteroid after it exploded
• Not handling multiple collisions in while loop

2. Given the following code for sliding window maximum, what is the returned list when nums = [2, 1, 3] and k = 2?

easy

A. [3, 3]

B. [2, 1]

C. [1, 3]

D. [2, 3]

Solution

1. Step 1: Trace iteration i=0

   dq empty, append index 0 (value 2). No output yet since i < k-1.

2. Step 2: Trace iteration i=1

   nums[1]=1 < nums[0]=2, so append index 1. Window size reached, max is nums[dq[0]]=2.

3. Step 3: Trace iteration i=2

   nums[2]=3 > nums[1]=1 pop index 1, nums[2]=3 > nums[0]=2 pop index 0, append index 2. Max is nums[2]=3.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Output matches expected sliding window max [OK]

Hint: Pop smaller elements before appending new index [OK]

Common Mistakes:

• Forgetting to remove out-of-window indices
• Appending max too early or late

3. Consider the following buggy code for finding the first unique character in a stream. Which line contains the subtle bug that causes incorrect output when duplicates appear?

from collections import deque

def first_unique_stream(stream):
    freq = [0] * 26
    q = deque()
    result = []
    for ch in stream:
        idx = ord(ch) - ord('a')
        freq[idx] += 1
        if freq[idx] == 1:
            q.append(ch)
        # BUG: Missing removal of duplicates from queue front
        result.append(q[0] if q else '#')
    return result

medium

A. Line where freq[idx] is incremented

B. Line where character is appended to queue

C. Line where result is appended

D. Line where duplicates are removed from queue front (missing here)

Solution

1. Step 1: Identify missing logic

   The code does not remove characters from the queue front when their frequency exceeds 1.

2. Step 2: Understand impact

   Without popping duplicates, the queue front may be a repeated character, causing wrong first unique output.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Missing while loop to pop duplicates is the bug [OK]

Hint: Always remove duplicates from queue front to maintain correctness [OK]

Common Mistakes:

• Forgetting to pop duplicates from queue front
• Appending duplicates multiple times
• Not handling empty queue case

4. What is the time complexity of the optimal Remove K Digits algorithm using a monotonic stack on an input string of length n?

medium

A. O(n * k), because each digit can be popped up to k times

B. O(n), because each digit is pushed and popped at most once

C. O(n log n), due to sorting or priority queue operations inside the stack

D. O(n^2), because nested loops remove digits repeatedly

Solution

1. Step 1: Analyze push and pop operations

   Each digit is pushed exactly once onto the stack. Each digit can be popped at most once when a smaller digit arrives.

2. Step 2: Conclude overall complexity

   Since each digit is pushed and popped at most once, total operations are O(n), linear in input size.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Monotonic stack ensures linear time by amortized analysis [OK]

Hint: Each digit pushed/popped once -> O(n) time [OK]

Common Mistakes:

• Assuming worst case k pops per digit -> O(n*k)
• Thinking sorting or nested loops are involved

5. Suppose the calculator expression can contain multiplication '*' and division '/' operators in addition to plus, minus, and parentheses. Which modification to the stack-based approach correctly handles operator precedence and parentheses?

hard

A. Use two stacks: one for numbers and one for operators, applying operators with higher precedence immediately before pushing lower precedence operators

B. Continue using a single stack and apply all operators in the order they appear, ignoring precedence

C. Convert the expression to postfix notation first, then evaluate using a stack

D. Evaluate recursively without any stack, relying on Python's eval() function

Solution

1. Step 1: Understand operator precedence

   Multiplication and division have higher precedence than addition and subtraction, so evaluation order matters.

2. Step 2: Modify stack approach

   Using two stacks (numbers and operators) allows applying higher precedence operators immediately, preserving correct order.

3. Step 3: Why other options fail

   Single stack ignoring precedence leads to wrong results; postfix conversion is valid but more complex; relying on eval() is unsafe and not interview-appropriate.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Two-stack approach is standard for precedence and parentheses [OK]

Hint: Two stacks handle precedence and parentheses correctly [OK]

Common Mistakes:

• Ignoring precedence or relying on single stack only