Arrays are among the most frequently asked topics in coding interviews. They help interviewers evaluate logical thinking, problem-solving ability, and knowledge of Data Structures and Algorithms. Mastering array problems is essential for internships, placements, and software engineering interviews.
In this article, you will learn the most popular array interview questions along with Java solutions, outputs, explanations, and complexity analysis.

Table of Contents

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Find the Largest Element in an Array

Given an array of integers, find the largest element present in the array.
Example:

Input: [12, 45, 7, 89, 23]
Output: 89

// Java program to find largest element 
// in an array
public class LargestElement 
{
    public static void main(String[] args) 
    {
        int[] arr = {12, 45, 7, 89, 23};
        int largest = arr[0];

        for(int i = 1; i  arr.length; i++) 
        {
            if(arr[i]  largest) 
            {
                largest = arr[i];
            }
        }
        System.out.println("Largest Element: " + largest);
    }
}

Output:

Largest Element: 89

Explanation:
The program assumes the first element is the largest and then traverses the array. Whenever a larger value is found, the largest variable is updated.
Time Complexity: O(n)

Find the Second Largest Element in an Array

Find the second largest element without sorting the array.
Example:

Input: [10, 25, 40, 60, 80]
Output: 60

// Java program to find second largest element 
// in an array
public class SecondLargest 
{
    public static void main(String[] args) 
    {
        int[] arr = {10, 25, 40, 60, 80};
        int largest = Integer.MIN_VALUE;
        int secondLargest = Integer.MIN_VALUE;

        for(int num : arr) 
        {
            if(num  largest) 
            {
                secondLargest = largest;
                largest = num;
            }
            else if(num  secondLargest && num != largest) 
            {
                secondLargest = num;
            }
        }
        System.out.println("Second Largest Element: " + secondLargest);
    }
}

Output:

Second Largest Element: 60

Explanation:
The program maintains both the largest and second largest values while traversing the array only once.
Time Complexity: O(n)

Reverse an Array

Reverse all elements of an array.
Example:

Input: [1, 2, 3, 4, 5]
Output: [5, 4, 3, 2, 1]

// Java program to reverse an array
import java.util.Arrays;

public class ReverseArray 
{
    public static void main(String[] args) 
    {
        int[] arr = {1, 2, 3, 4, 5};
        int left = 0;
        int right = arr.length - 1;

        while(left  right) 
        {
            int temp = arr[left];
            arr[left] = arr[right];
            arr[right] = temp;
            left++;
            right--;
        }
        System.out.println(Arrays.toString(arr));
    }
}

Output:

[5, 4, 3, 2, 1]

Explanation:
Two pointers are used. One starts from the beginning and the other from the end. Elements are swapped until both pointers meet.
Time Complexity: O(n)

Find the Missing Number in an Array

Given an array containing numbers from 1 to n with one missing value, find the missing number.
Example:

Input: [1, 2, 3, 5]
Output: 4

// Java program to find missing 
// number in an array
public class MissingNumber 
{
    public static void main(String[] args) 
    {
        int[] arr = {1, 2, 3, 5};
        int n = 5;
        int expectedSum = n * (n + 1) / 2;
        int actualSum = 0;

        for(int num : arr) 
        {
            actualSum += num;
        }

        System.out.println("Missing Number: " + (expectedSum - actualSum));
    }
}

Output:

Missing Number: 4

Explanation:
The expected sum is calculated using the formula n(n+1)/2. Subtracting the actual sum from the expected sum gives the missing value.
Time Complexity: O(n)

Move All Zeros to the End

Move all zeros to the end while maintaining the order of non-zero elements.
Example:

Input: [0, 1, 0, 3, 12]
Output: [1, 3, 12, 0, 0]

// Java program to move all zeros 
// to the end
import java.util.Arrays;

public class MoveZeros 
{
    public static void main(String[] args) 
    {
        int[] arr = {0, 1, 0, 3, 12};
        int index = 0;

        for(int i = 0; i  arr.length; i++) 
        {
            if(arr[i] != 0) 
            {
                arr[index++] = arr[i];
            }
        }

        while(index  arr.length) 
        {
            arr[index++] = 0;
        }

        System.out.println(Arrays.toString(arr));
    }
}

Output:

[1, 3, 12, 0, 0]

Explanation:
All non-zero elements are moved to the front while maintaining their relative order. The remaining positions are filled with zeros.
Time Complexity: O(n)

Binary Search in a Sorted Array

Given a sorted array and a target element, find the index of the target element using Binary Search.
Example:

Input: [10, 20, 30, 40, 50]
Target: 30
Output: 2

// java program to implement binary 
// search in a sorted array
public class BinarySearchExample 
{
    public static void main(String[] args) 
    {
        int[] arr = {10, 20, 30, 40, 50};
        int target = 30;
        int left = 0;
        int right = arr.length - 1;
        int index = -1;

        while(left = right) 
        {
            int mid = (left + right) / 2;

            if(arr[mid] == target) 
            {
                index = mid;
                break;
            }
            else if(arr[mid]  target) 
            {
                left = mid + 1;
            }
            else 
            {
                right = mid - 1;
            }
        }
        System.out.println("Element found at index: " + index);
    }
}

Output:

Element found at index: 2

Explanation:
Binary Search repeatedly divides the search space into two halves. This makes it much faster than Linear Search for sorted arrays.
Time Complexity: O(log n)

Rotate an Array by K Positions

Rotate all elements of an array to the right by K positions.
Example:

Input: [1, 2, 3, 4, 5]
K = 2
Output: [4, 5, 1, 2, 3]

// Java program to rotate an array 
// by k-positions
import java.util.Arrays;

public class RotateArray 
{
    public static void main(String[] args) 
    {
        int[] arr = {1, 2, 3, 4, 5};
        int k = 2;
        int n = arr.length;
        int[] result = new int[n];
        
        for(int i = 0; i  n; i++) 
        {
            result[(i + k) % n] = arr[i];
        }
        System.out.println(Arrays.toString(result));
    }
}

Output:

[4, 5, 1, 2, 3]

Explanation:
Each element is moved to its new position using modular arithmetic. This ensures elements wrap around correctly.
Time Complexity: O(n)

Find Duplicate Elements in an Array

Find duplicate elements present in an array.
Example:

Input: [1, 2, 3, 2, 4, 1]
Output: [2, 1]

// Java program to find duplicate elements 
// in an array
import java.util.HashSet;

public class FindDuplicates 
{
    public static void main(String[] args) 
    {
        int[] arr = {1, 2, 3, 2, 4, 1};
        HashSetInteger seen = new HashSet();
        HashSetInteger duplicates = new HashSet();

        for(int num : arr) 
        {
            if(!seen.add(num)) 
            {
                duplicates.add(num);
            }
        }
        System.out.println("Duplicate Elements: " + duplicates);
    }
}

Output:

Duplicate Elements: [1, 2]

Explanation:
A HashSet stores elements that have already been visited. If an element appears again, it is identified as a duplicate.
Time Complexity: O(n)

Merge Two Sorted Arrays

Merge two sorted arrays into a single sorted array.
Example:

Input:
[1, 3, 5]
[2, 4, 6]
Output:
[1, 2, 3, 4, 5, 6]

// Java program to merge two sorted arrays
import java.util.Arrays;

public class MergeSortedArrays 
{
    public static void main(String[] args) 
    {
        int[] arr1 = {1, 3, 5};
        int[] arr2 = {2, 4, 6};
        int[] merged = new int[arr1.length + arr2.length];
        int i = 0, j = 0, k = 0;

        while(i  arr1.length && j  arr2.length) 
        {
            if(arr1[i]  arr2[j]) 
            {
                merged[k++] = arr1[i++];
            }
            else 
            {
                merged[k++] = arr2[j++];
            }
        }

        while(i  arr1.length) 
        {
            merged[k++] = arr1[i++];
        }

        while(j  arr2.length) 
        {
            merged[k++] = arr2[j++];
        }

        System.out.println(Arrays.toString(merged));
    }
}

Output:

[1, 2, 3, 4, 5, 6]

Explanation:
Two pointers compare elements from both arrays and place the smaller element into the merged array.
Time Complexity: O(n + m)

Maximum Subarray Sum (Kadane’s Algorithm)

Find the contiguous subarray with the maximum sum.
Example:

Input:
[-2, 1, -3, 4, -1, 2, 1, -5, 4]
Output:
6

// Java program to find maximum subarray sum
public class KadanesAlgorithm 
{
    public static void main(String[] args) 
    {
        int[] arr = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
        int maxSum = arr[0];
        int currentSum = arr[0];

        for(int i = 1; i  arr.length; i++) 
        {
            currentSum = Math.max(arr[i], currentSum + arr[i]);
            maxSum = Math.max(maxSum, currentSum);
        }
        System.out.println("Maximum Subarray Sum: " + maxSum);
    }
}

Output:

Maximum Subarray Sum: 6

Explanation:
Kadane’s Algorithm decides whether to start a new subarray or extend the current one. It keeps track of the maximum sum found so far.
Time Complexity: O(n)

Common Mistakes While Solving Array Questions

1. Ignoring Edge Cases: Many candidates forget to test empty arrays, arrays with one element, or arrays containing duplicate values.
2. Using Inefficient Solutions: A working solution may still be rejected if it has poor time complexity.
3. Incorrect Loop Boundaries: Loop-related errors can cause missed elements or Array Index Out of Bounds exceptions.
4. Not Understanding the Problem Clearly: Jumping directly into coding without understanding the problem often leads to incorrect solutions.
5. Forgetting Sorted Array Requirements: Some algorithms, such as Binary Search, only work correctly on sorted arrays.

Tips for Cracking Array Interviews

1. Learn Common Problem-Solving Patterns: Many array questions are based on techniques such as Two Pointers, Sliding Window, Prefix Sum, and Binary Search.
2. Focus on Optimization: Always think about how to improve the efficiency of your solution.
3. Practice Consistently: Regular practice helps you recognize patterns and solve problems faster.
4. Analyze Time Complexity: Understanding complexity analysis helps you choose better approaches.
5. Explain Your Approach: Interviewers often evaluate your thought process in addition to the final solution.

Conclusion

Array questions are among the most important topics in coding interviews because they test problem-solving ability, optimization skills, and programming fundamentals. By practicing these ten popular interview questions, you can strengthen your understanding of arrays and improve your confidence during technical interviews.
The key to success is not memorizing solutions but understanding the underlying concepts and techniques. With consistent practice and proper analysis, you will be able to solve a wide variety of array problems efficiently.

Frequently Asked Questions 

1. Which array interview questions are asked most frequently?

Finding the largest element, reversing an array, Binary Search, Move Zeros, and Maximum Subarray Sum are among the most frequently asked questions.

2. What techniques are useful for solving array problems?

Two Pointers, Sliding Window, Prefix Sum, Binary Search, Hashing, and Kadane’s Algorithm are commonly used techniques.

3. How can I improve my array problem-solving skills?

Practice array questions regularly, learn common patterns, and analyze the time complexity of every solution.

4. Are array questions important for placement interviews?

Yes. Array-based problems are among the most common questions asked in internships, placements, and software engineering interviews.

5. What should I learn after arrays?

After arrays, you can move on to strings, linked lists, stacks, queues, trees, graphs, and dynamic programming.

6. Is Kadane’s Algorithm important for interviews?

Yes. Kadane’s Algorithm is one of the most frequently asked array algorithms because it efficiently solves the Maximum Subarray Sum problem in O(n) time.

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