Top 30 Most Common Array Questions You Should Prepare For

Technical interviews often involve solving problems related to data structures, and arrays are fundamental. Mastering array questions is crucial for demonstrating your problem-solving abilities and understanding of basic algorithms. Whether you're a fresh graduate or an experienced professional, you'll likely encounter array-based problems across various difficulty levels. These questions test your logical thinking, efficiency in handling data, and ability to implement solutions using simple yet powerful concepts like pointers, sorting, hashing, and dynamic programming. Preparing for these common array questions will significantly boost your confidence and performance in coding interviews, laying a strong foundation for tackling more complex challenges.

What Are array questions?

array questions are coding challenges that involve manipulating or querying data stored in an array. Arrays are contiguous blocks of memory used to store collections of elements of the same data type, accessed via an index. Common array questions include tasks like finding elements, sorting, searching, manipulating elements based on certain conditions, finding subarrays, or performing operations on 2D arrays (matrices). These problems often require understanding time and space complexity trade-offs. Interviewers use array questions to evaluate candidates' ability to use loops, conditional statements, and data structures effectively to solve problems efficiently within constraints. Solving array questions is a core skill for any programmer.

Why Do Interviewers Ask array questions?

Interviewers ask array questions to assess a candidate's foundational programming skills. Arrays are simple yet powerful data structures, making them ideal for testing basic algorithmic thinking. Solving array questions requires understanding iteration, indexing, and basic data manipulation. They also serve as building blocks for more complex problems involving other data structures or algorithms. How a candidate approaches and optimizes solutions for array questions reveals their problem-solving process, ability to handle edge cases, and understanding of efficiency (time and space complexity). Proficiency in array questions indicates a solid grasp of programming fundamentals necessary for real-world software development tasks.

Preview List

1. Find the Maximum Element in an Array

2. Check if an Element Exists in an Array

3. Count the Frequency of an Element

4. Merge Two Sorted Arrays

5. Reverse an Array

6. Find the First Repeating Element

7. Find Common Elements in Three Sorted Arrays

8. Move All Zeros to the End of an Array

9. Find the Second Largest Element in an Array

10. Rotate a Matrix (2D Array) Clockwise

11. Find the Missing Number in an Array (1 to n)

12. Solve the Stock Buy and Sell Problem

13. Find the Longest Subarray with a Given Sum

14. Find the Maximum Subarray Sum

15. Find the Pair with a Given Sum (Two Sum Problem)

16. Minimum Window Substring

17. First and Last Position of an Element in a Sorted Array

18. Maximum Subarray Sum with at Least One Element

19. Maximum Average Subarray

20. Subarray with Given Sum

21. Minimum Size Subarray Sum

22. Jump Game

23. Climbing Stairs

24. Unique Paths

25. Minimum Deletions to Make Sorted

26. Maximum Length of Subarray with Equal Numbers of 0s and 1s

27. Find the Majority Element (appears more than n/2 times)

28. Find All Duplicate Elements

29. Rotate Array by K Steps

30. Find the Longest Increasing Subsequence

1. Find the Maximum Element in an Array

Why you might get asked this:

Tests basic iteration, comparison, and variable tracking skills. It's a fundamental array question.

How to answer:

Iterate through the array, keeping track of the largest element seen so far and updating it as needed.

Example answer:

function findMax(arr) {
  if (arr.length === 0) return undefined;
  let max = arr[0];
  for (let i = 1; i < arr.length; i++) {
    if (arr[i] > max) max = arr[i];
  }
  return max;
}

2. Check if an Element Exists in an Array

Why you might get asked this:

Assesses basic searching techniques and loop usage. A common array question for beginners.

How to answer:

Loop through the array and check if the current element matches the target value.

Example answer:

function elementExists(arr, target) {
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] === target) return true;
  }
  return false;
}

3. Count the Frequency of an Element

Why you might get asked this:

Tests iteration and using a counter variable. Simple but practical for analyzing data in arrays.

How to answer:

Initialize a counter to zero and increment it each time the target element is found during iteration.

Example answer:

function countFrequency(arr, target) {
  let count = 0;
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] === target) count++;
  }
  return count;
}

4. Merge Two Sorted Arrays

Why you might get asked this:

Evaluates pointer usage and logic for combining sorted data structures efficiently. A classic merge problem.

How to answer:

Use two pointers, one for each array, comparing elements and adding the smaller one to a new result array until one array is exhausted, then add remaining elements from the other.

Example answer:

function mergeSortedArrays(arr1, arr2) {
  let result = [];
  let i = 0, j = 0;
  while (i < arr1.length || j < arr2.length) {
    if (j >= arr2.length || (i < arr1.length && arr1[i] <= arr2[j])) {
      result.push(arr1[i++]);
    } else {
      result.push(arr2[j++]);
    }
  }
  return result;
}

5. Reverse an Array

Why you might get asked this:

Tests in-place modification and understanding of array boundaries and swapping. A fundamental array manipulation question.

How to answer:

Use two pointers, one starting at the beginning and one at the end. Swap the elements they point to, then move the pointers towards the center until they meet.

Example answer:

function reverseArray(arr) {
  let start = 0;
  let end = arr.length - 1;
  while (start < end) {
    [arr[start], arr[end]] = [arr[end], arr[start]]; // ES6 swap
    start++;
    end--;
  }
  return arr;
}

6. Find the First Repeating Element

Why you might get asked this:

Tests using auxiliary data structures (like hash maps) to track seen elements for efficient lookups.

How to answer:

Iterate through the array, storing each element in a hash map. If an element is already in the map, return it immediately.

Example answer:

function firstRepeatingElement(arr) {
  let map = new Map();
  for (let num of arr) {
    if (map.has(num)) return num;
    map.set(num, true);
  }
  return null; // Or throw error, depending on requirements
}

7. Find Common Elements in Three Sorted Arrays

Why you might get asked this:

Extends the two-pointer concept to multiple arrays, assessing complex conditional logic and pointer management.

How to answer:

Use three pointers, one for each array. Increment the pointer for the smallest element. If elements at all three pointers are equal, add to result and increment all pointers.

Example answer:

function commonElements(arr1, arr2, arr3) {
  let i = 0, j = 0, k = 0;
  let result = [];
  while (i < arr1.length && j < arr2.length && k < arr3.length) {
    if (arr1[i] === arr2[j] && arr2[j] === arr3[k]) {
      result.push(arr1[i]);
      i++; j++; k++;
    } else if (arr1[i] < arr2[j]) i++;
    else if (arr2[j] < arr3[k]) j++;
    else k++;
  }
  return result;
}

8. Move All Zeros to the End of an Array

Why you might get asked this:

Assesses in-place modification using two pointers to partition the array. A common array manipulation task.

How to answer:

Use a 'write' pointer starting at index 0. Iterate through the array with a 'read' pointer. If the element is non-zero, swap it with the element at the 'write' pointer and increment the 'write' pointer.

Example answer:

function moveZerosToEnd(arr) {
  let writeIdx = 0;
  for (let readIdx = 0; readIdx < arr.length; readIdx++) {
    if (arr[readIdx] !== 0) {
      [arr[writeIdx], arr[readIdx]] = [arr[readIdx], arr[writeIdx]];
      writeIdx++;
    }
  }
  return arr;
}

9. Find the Second Largest Element in an Array

Why you might get asked this:

Tests careful tracking of multiple values during iteration and handling edge cases (duplicates, array size).

How to answer:

Iterate through the array, maintaining variables for the largest and second-largest elements found so far, updating them based on comparisons. Handle initialization carefully.

Example answer:

function secondLargestElement(arr) {
  if (arr.length < 2) return -1; // Or appropriate indicator
  let max = -Infinity, secondMax = -Infinity;
  for (let num of arr) {
    if (num > max) {
      secondMax = max;
      max = num;
    } else if (num > secondMax && num !== max) {
      secondMax = num;
    }
  }
  return secondMax === -Infinity ? -1 : secondMax;
}

10. Rotate a Matrix (2D Array) Clockwise

Why you might get asked this:

Tests understanding of 2D array indexing and transformations. Often solved with transpose and reverse operations.

How to answer:

First, transpose the matrix (swap rows and columns). Then, reverse each row to complete the clockwise rotation.

Example answer:

function rotateMatrix(matrix) {
  let n = matrix.length;
  for (let i = 0; i < n; i++) {
    for (let j = i; j < n; j++) {
      [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]]; // Transpose
    }
  }
  for (let i = 0; i < n; i++) {
    matrix[i].reverse(); // Reverse each row
  }
  return matrix;
}

11. Find the Missing Number in an Array (1 to n)

Why you might get asked this:

Evaluates understanding of mathematical properties (summation) or bit manipulation for efficient solutions.

How to answer:

Calculate the expected sum of numbers from 1 to n using the formula n*(n+1)/2. Subtract the actual sum of the array elements from the expected sum.

Example answer:

function findMissingNumber(arr) {
  let n = arr.length + 1;
  let expectedSum = n * (n + 1) / 2;
  let actualSum = 0;
  for (let num of arr) {
    actualSum += num;
  }
  return expectedSum - actualSum;
}

12. Solve the Stock Buy and Sell Problem

Why you might get asked this:

A classic dynamic programming or greedy problem on arrays, testing logic for tracking maximum profit over time.

How to answer:

Iterate through the prices, keeping track of the minimum price encountered so far. At each day, calculate the potential profit if selling today (current price - minimum price) and update the maximum profit found.

Example answer:

function maxProfit(prices) {
  if (!prices || prices.length < 2) return 0;
  let minPrice = prices[0];
  let maxP = 0;
  for (let i = 1; i < prices.length; i++) {
    maxP = Math.max(maxP, prices[i] - minPrice);
    minPrice = Math.min(minPrice, prices[i]);
  }
  return maxP;
}

13. Find the Longest Subarray with a Given Sum

Why you might get asked this:

Tests prefix sums and hash maps for efficient lookup of previous sums to determine subarray lengths.

How to answer:

Use a hash map to store the cumulative sum encountered so far and the index where that sum occurred. Iterate through the array, calculate the current sum, and check if currentsum - targetsum exists in the map. Update max length.

Example answer:

function longestSubarrayWithSum(arr, targetSum) {
  let sumMap = new Map(); // sum -> index
  sumMap.set(0, -1); // Base case
  let currentSum = 0;
  let maxLen = 0;
  for (let i = 0; i < arr.length; i++) {
    currentSum += arr[i];
    if (sumMap.has(currentSum - targetSum)) {
      maxLen = Math.max(maxLen, i - sumMap.get(currentSum - targetSum));
    }
    if (!sumMap.has(currentSum)) {
      sumMap.set(currentSum, i);
    }
  }
  return maxLen;
}

14. Find the Maximum Subarray Sum

Why you might get asked this:

A standard dynamic programming problem (Kadane's algorithm), testing efficient computation of sums over changing windows.

How to answer:

Use Kadane's algorithm: iterate through the array, maintaining the maximum sum ending at the current position and the overall maximum sum found so far.

Example answer:

function maxSubarraySum(arr) {
  let maxSoFar = arr[0];
  let maxEndingHere = arr[0];
  for (let i = 1; i < arr.length; i++) {
    maxEndingHere = Math.max(arr[i], maxEndingHere + arr[i]);
    maxSoFar = Math.max(maxSoFar, maxEndingHere);
  }
  return maxSoFar;
}

15. Find the Pair with a Given Sum (Two Sum Problem)

Why you might get asked this:

A very common question testing the use of hash maps for quick lookups to find complements.

How to answer:

Use a hash map to store numbers encountered. Iterate through the array. For each number, check if targetsum - currentnumber exists in the map. If yes, return the pair. Otherwise, add the current number to the map.

Example answer:

function twoSum(arr, targetSum) {
  let numMap = new Map();
  for (let num of arr) {
    let complement = targetSum - num;
    if (numMap.has(complement)) {
      return [complement, num];
    }
    numMap.set(num, true); // Store number as seen
  }
  return null; // No such pair found
}

16. Minimum Window Substring

Why you might get asked this:

Involves sliding window and hash map techniques for string/array problems. It tests managing window states and counts.

How to answer:

Use a sliding window with two pointers (left, right). Use hash maps to track character counts needed from the target string and characters within the current window. Shrink the window from the left when it's valid, expanding from the right.

Example answer:

function minWindow(s, t) {
  let tCounts = {}; for(let c of t) tCounts[c] = (tCounts[c] || 0) + 1;
  let formed = 0, required = Object.keys(tCounts).length;
  let windowCounts = {};
  let minLen = Infinity, minStart = 0;
  let l = 0;
  for (let r = 0; r < s.length; r++) {
    let char = s[r]; windowCounts[char] = (windowCounts[char] || 0) + 1;
    if (char in tCounts && windowCounts[char] === tCounts[char]) formed++;
    while (l <= r && formed === required) {
      if (r - l + 1 < minLen) { minLen = r - l + 1; minStart = l; }
      char = s[l]; windowCounts[char]--;
      if (char in tCounts && windowCounts[char] < tCounts[char]) formed--;
      l++;
    }
  }
  return minLen === Infinity ? "" : s.substring(minStart, minStart + minLen);
}

17. First and Last Position of an Element in a Sorted Array

Why you might get asked this:

Tests binary search application. Requires understanding how to modify binary search to find bounds rather than just existence.

How to answer:

Perform two binary searches: one to find the first occurrence (adjusting search to the left if found) and another to find the last occurrence (adjusting search to the right if found).

Example answer:

function searchRange(nums, target) {
  const findBound = (lower) => {
    let left = 0, right = nums.length - 1;
    let idx = -1;
    while (left <= right) {
      let mid = Math.floor((left + right) / 2);
      if (nums[mid] === target) { idx = mid; if (lower) right = mid - 1; else left = mid + 1; }
      else if (nums[mid] < target) left = mid + 1;
      else right = mid - 1;
    }
    return idx;
  };
  return [findBound(true), findBound(false)];
}

18. Maximum Subarray Sum with at Least One Element

Why you might get asked this:

Often used interchangeably with the standard Maximum Subarray Sum problem, as Kadane's algorithm naturally handles this.

How to answer:

Same as the standard Maximum Subarray Sum (Kadane's algorithm). Initialize max sums with the first element to ensure at least one element is included.

Example answer:

function maxSubarraySumWithAtLeastOne(arr) {
  if (arr.length === 0) return -Infinity; // Or handle appropriately
  let maxSoFar = arr[0];
  let maxEndingHere = arr[0];
  for (let i = 1; i < arr.length; i++) {
    maxEndingHere = Math.max(arr[i], maxEndingHere + arr[i]);
    maxSoFar = Math.max(maxSoFar, maxEndingHere);
  }
  return maxSoFar;
}

19. Maximum Average Subarray

Why you might get asked this:

Applies the sliding window concept to find an optimal window based on a metric (average).

How to answer:

Calculate the sum of the first window of size k. Then, slide the window, updating the sum by subtracting the element leaving the window and adding the element entering. Track the maximum sum found.

Example answer:

function maxAverage(arr, k) {
  let currentSum = 0;
  for (let i = 0; i < k; i++) currentSum += arr[i];
  let maxSum = currentSum;
  for (let i = k; i < arr.length; i++) {
    currentSum += arr[i] - arr[i - k]; // Slide window sum
    maxSum = Math.max(maxSum, currentSum);
  }
  return maxSum / k;
}

20. Subarray with Given Sum

Why you might get asked this:

Tests either brute force, sliding window (for non-negative numbers), or prefix sum + hash map for general cases.

How to answer:

For positive numbers, use a sliding window. For general integers (positive/negative), use prefix sums and a hash map to track sum occurrences.

Example answer:

function subarrayWithSum(arr, targetSum) {
  let sumMap = new Map();
  sumMap.set(0, -1);
  let currentSum = 0;
  for (let i = 0; i < arr.length; i++) {
    currentSum += arr[i];
    if (sumMap.has(currentSum - targetSum)) {
      let start = sumMap.get(currentSum - targetSum) + 1;
      return arr.slice(start, i + 1); // Return the subarray
    }
    if (!sumMap.has(currentSum)) {
      sumMap.set(currentSum, i);
    }
  }
  return null; // No subarray found
}

21. Minimum Size Subarray Sum

Why you might get asked this:

A classic sliding window problem focusing on minimizing the window size while meeting a condition (sum >= target).

How to answer:

Use a sliding window. Expand the window from the right, adding elements to a current sum. When the sum reaches or exceeds the target, shrink the window from the left while maintaining the condition, tracking the minimum valid window length.

Example answer:

function minSizeSubarraySum(arr, targetSum) {
  let left = 0, currentSum = 0;
  let minLen = Infinity;
  for (let right = 0; right < arr.length; right++) {
    currentSum += arr[right];
    while (currentSum >= targetSum) {
      minLen = Math.min(minLen, right - left + 1);
      currentSum -= arr[left];
      left++;
    }
  }
  return minLen === Infinity ? 0 : minLen;
}

22. Jump Game

Why you might get asked this:

Tests greedy algorithms or dynamic programming on arrays. Evaluates logic for reachability.

How to answer:

Use a greedy approach. Iterate through the array, keeping track of the maximum reachable index. If at any point the current index is greater than the maximum reachable index, you cannot reach the end.

Example answer:

function canJump(nums) {
  let maxReach = 0;
  for (let i = 0; i < nums.length; i++) {
    if (i > maxReach) return false;
    maxReach = Math.max(maxReach, i + nums[i]);
    if (maxReach >= nums.length - 1) return true;
  }
  return false; // Should be true if loop finishes and maxReach >= last index
}

23. Climbing Stairs

Why you might get asked this:

A fundamental dynamic programming problem illustrating how to break down problems into overlapping subproblems.

How to answer:

This is a variation of the Fibonacci sequence. The number of ways to reach step n is the sum of ways to reach step n-1 (taking one step) and ways to reach step n-2 (taking two steps). Use DP or memoization.

Example answer:

function climbStairs(n) {
  if (n === 1) return 1;
  let dp = new Array(n + 1);
  dp[1] = 1;
  dp[2] = 2;
  for (let i = 3; i <= n; i++) {
    dp[i] = dp[i - 1] + dp[i - 2];
  }
  return dp[n];
}

24. Unique Paths

Why you might get asked this:

Another classic DP problem on a grid (2D array), testing combinations and path counting.

How to answer:

Use a 2D DP array where dp[i][j] is the number of unique paths to cell (i, j). dp[i][j] = dp[i-1][j] + dp[i][j-1]. Base cases are 1 for cells in the first row/column.

Example answer:

function uniquePaths(m, n) {
  let dp = Array(m).fill(0).map(() => Array(n).fill(1));
  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
    }
  }
  return dp[m - 1][n - 1];
}

25. Minimum Deletions to Make Sorted

Why you might get asked this:

Relates to the Longest Increasing Subsequence (LIS) problem. The minimum deletions is total length minus LIS length.

How to answer:

Find the length of the Longest Increasing Subsequence (LIS) of the given array. The minimum number of deletions required is array.length - LIS_length.

Example answer:

function minDeletions(arr) {
  if (arr.length <= 1) return 0;
  let tails = []; // Store smallest tail of all increasing subsequences of length i+1
  for (let x of arr) {
    let left = 0, right = tails.length;
    while (left < right) {
      let mid = Math.floor((left + right) / 2);
      if (tails[mid] < x) left = mid + 1;
      else right = mid;
    }
    if (left === tails.length) tails.push(x);
    else tails[left] = x;
  }
  return arr.length - tails.length;
}

26. Maximum Length of Subarray with Equal Numbers of 0s and 1s

Why you might get asked this:

Tests using prefix sums and hash maps to solve problems involving balanced counts (often by transforming 0s to -1s).

How to answer:

Transform 0s to -1s. The problem becomes finding the longest subarray with sum 0. Use prefix sums and a hash map to store the first index where each sum is encountered.

Example answer:

function maxSubarrayWithEqualZerosAndOnes(arr) {
  let sumMap = new Map(); // sum -> index
  sumMap.set(0, -1);
  let currentSum = 0;
  let maxLen = 0;
  for (let i = 0; i < arr.length; i++) {
    currentSum += arr[i] === 0 ? -1 : 1;
    if (sumMap.has(currentSum)) {
      maxLen = Math.max(maxLen, i - sumMap.get(currentSum));
    } else {
      sumMap.set(currentSum, i);
    }
  }
  return maxLen;
}

27. Find the Majority Element (appears more than n/2 times)

Why you might get asked this:

A classic problem with an efficient, space-saving solution (Boyer-Moore). Tests understanding of frequency counts.

How to answer:

Use Boyer-Moore Majority Vote algorithm. Initialize a candidate and counter. Iterate through the array: if counter is 0, set current element as candidate; if current element matches candidate, increment counter; otherwise, decrement counter.

Example answer:

function majorityElement(nums) {
  let count = 0;
  let candidate = null;
  for (let num of nums) {
    if (count === 0) {
      candidate = num;
    }
    count += (num === candidate) ? 1 : -1;
  }
  // A second pass is needed to confirm majority if problem didn't guarantee one exists
  return candidate;
}

28. Find All Duplicate Elements

Why you might get asked this:

Tests using hash maps or modifying the array in-place (if constraints allow) to find elements appearing more than once.

How to answer:

Use a hash map to count element frequencies. Iterate through the array and add elements to a result list if their count in the map exceeds 1 upon seeing them.

Example answer:

function findDuplicates(arr) {
  let map = new Map();
  let duplicates = [];
  for (let num of arr) {
    map.set(num, (map.get(num) || 0) + 1);
  }
  for (let [num, count] of map) {
    if (count > 1) {
      duplicates.push(num);
    }
  }
  return duplicates;
}

29. Rotate Array by K Steps

Why you might get asked this:

Tests array manipulation techniques, including handling wraps around the array boundaries.

How to answer:

Several methods: use an auxiliary array, reverse sections of the array, or use slicing and concatenation (less memory efficient but simpler). Handle k larger than array length using modulo.

Example answer:

function rotateArray(nums, k) {
  k = k % nums.length;
  if (k === 0) return nums;
  // Using reverse approach (in-place)
  nums.reverse(); // Reverse entire array
  // Reverse first k elements
  let left = 0, right = k - 1;
  while (left < right) { [nums[left], nums[right]] = [nums[right], nums[left]]; left++; right--; }
  // Reverse remaining elements
  left = k, right = nums.length - 1;
  while (left < right) { [nums[left], nums[right]] = [nums[right], nums[left]]; left++; right--; }
  return nums;
}

30. Find the Longest Increasing Subsequence

Why you might get asked this:

A fundamental dynamic programming problem, testing array state management and optimization.

How to answer:

Use DP where dp[i] is the length of the LIS ending at index i. Iterate from i=1 to n-1, then from j=0 to i-1. If arr[i] > arr[j], dp[i] = max(dp[i], dp[j] + 1). The answer is the maximum value in the dp array.

Example answer:

function longestIncreasingSubsequence(arr) {
  if (arr.length === 0) return 0;
  let dp = Array(arr.length).fill(1);
  for (let i = 1; i < arr.length; i++) {
    for (let j = 0; j < i; j++) {
      if (arr[i] > arr[j]) {
        dp[i] = Math.max(dp[i], dp[j] + 1);
      }
    }
  }
  return Math.max(...dp);
}

Other Tips to Prepare for a array questions

Practicing common array questions is key, but effective preparation involves more than just coding. Focus on understanding the underlying concepts. Ask clarifying questions about constraints (size, value range, time/space limits). "Understanding the problem is half the battle," as the saying goes. Before coding, discuss your approach, including edge cases and complexity analysis. Dry run your code with sample inputs. For array questions, drawing diagrams for pointers or array states can be very helpful. Consider variations of standard problems – slight changes can require different solutions. Using tools that simulate interview environments can refine your communication under pressure. A tool like the Verve AI Interview Copilot can provide real-time feedback on your communication and coding style, helping you articulate your thought process clearly. Remember, "Practice makes perfect." Leverage resources like Verve AI Interview Copilot (https://vervecopilot.com) to polish your technical communication, especially when explaining your logic for challenging array questions. Continuous practice and focused review of array questions will build confidence. Use Verve AI Interview Copilot to evaluate your explanations and get tips on improving your approach to various array questions.

Frequently Asked Questions

Q1: What are the typical complexities for array questions?
A1: Solutions for array questions often range from O(N) to O(N^2) time complexity, and O(1) to O(N) space complexity, depending on the algorithm.

Q2: Should I use a hash map or sort the array first?
A2: It depends on the problem. Hash maps offer O(1) average lookup, useful for frequency or pair problems. Sorting takes O(N log N) but enables O(N) two-pointer or O(log N) binary search techniques.

Q3: How do I handle edge cases in array questions?
A3: Always consider empty arrays, single-element arrays, arrays with duplicates, negative numbers, and extreme values.

Q4: What's the difference between an array and a linked list?
A4: Arrays store elements contiguously in memory (O(1) access by index), while linked lists use nodes with pointers (O(N) access).

Q5: Are 2D arrays (matrices) common in interviews?
A5: Yes, problems involving grids, paths, or transformations are common 2D array questions.

Q6: How important is space complexity for array questions?
A6: Very important. Interviewers often seek in-place (O(1) space) solutions for array questions when possible.