Crack the Top 50 Java Data Structure Interview Questions

Java remains one of the most popular languages around the world, especially in financial fields. Companies like Goldman Sachs, eBay, Google, and Microsoft all hire Java developers.

Today, we’ll help you prepare for your next job interview at these and other popular companies by reviewing 50 of the most common Java data structure interview questions and answers.

By the end, you’ll have all the experience you need to answer the data structure questions that make up the bulk of most interviews.

Answer any Java interview problem by learning the patterns behind common questions.

Grokking the Coding Interview Patterns

With thousands of potential questions to account for, preparing for the coding interview can feel like an impossible challenge. Yet with a strategic approach, coding interview prep doesn’t have to take more than a few weeks. Stop drilling endless sets of practice problems, and prepare more efficiently by learning coding interview patterns. This course teaches you the underlying patterns behind common coding interview questions. By learning these essential patterns, you will be able to unpack and answer any problem the right way — just by assessing the problem statement. This approach was created by FAANG hiring managers to help you prepare for the typical rounds of interviews at major tech companies like Apple, Google, Meta, Microsoft, and Amazon. Before long, you will have the skills you need to unlock even the most challenging questions, grok the coding interview, and level up your career with confidence. This course is also available in JavaScript, Python, Go, and C++ — with more coming soon!

85hrs

Intermediate

460 Challenges

461 Quizzes

Array Data Structure Questions#

1. Array initialization#

Which of the following statements will initialize an array?

// Option 1
int a[] = new int[5];

// 2
int a[5] = {1,2,3,4,5};

// 3
int a[] = {1,2,3,4,5};

Java

class checkMergeArray { 
  // Merge arr1 and arr2 into resultantArray
  public static int[] mergeArrays(int[] arr1, int[] arr2) { 
    int s1 = arr1.length;
    int s2 = arr2.length;
    int[] resultantArray = new int[s1+s2];
    int i = 0, j = 0, k = 0;
    // Traverse both array 
    while (i < s1 && j < s2) { 
      // Check if current element of first 
      // array is smaller than current element 
      // of second array. If yes, store first 
      // array element and increment first array 
      // index. Otherwise do same with second array 
      if (arr1[i] < arr2[j]) 
        resultantArray[k++] = arr1[i++]; 
      else
        resultantArray[k++] = arr2[j++]; 
    } 
    // Store remaining elements of first array 
    while (i < s1) 
      resultantArray[k++] = arr1[i++]; 
    // Store remaining elements of second array 
    while (j < s2) 
      resultantArray[k++] = arr2[j++]; 
    return resultantArray;
  } 
  public static void main(String args[]) {
    int[] arr1 = {1,12,14,17,23}; // creating a sorted array called arr1
    int[] arr2 = {11,19,27};  // creating a sorted array called arr2
    int[] resultantArray = mergeArrays(arr1, arr2); // calling mergeArrays
    System.out.print("Arrays after merging: ");
    for(int i = 0; i < arr1.length + arr2.length; i++) {
      System.out.print(resultantArray[i] + " ");
    }
  }
}

Time Complexity: $O(n + m)$ where n and m are the sizes of arr1 and arr2.

In the solution above, we start by creating a new empty array of the size equal to the combined size of both input arrays.

Starting from the index 0 individually compare the elements at corresponding indexes of both arrays.

Place the element with a lower value in the resultant array, and increment the index of the array where you find the smallest element.

Keep repeating this until you hit the end of one array. Move the elements of the other array into the resultantArray as it is.

Java

class CheckSum{ 
  //Helper Function to sort given Array (Quick Sort)          
  public static int partition(int[] arr, int low, int high) { 
    int pivot = arr[high];  
    int i = (low - 1); // index of smaller element 
    for (int j = low; j < high; j++) { 
      // If current element is <= to pivot 
      if (arr[j] <= pivot) { 
        i++; 
        // swap arr[i] and arr[j] 
        int temp = arr[i]; 
        arr[i] = arr[j]; 
        arr[j] = temp; 
      } 
    } 
    // swap arr[i+1] and arr[high] (or pivot) 
    int temp = arr[i + 1]; 
    arr[i + 1] = arr[high]; 
    arr[high] = temp; 
    return i + 1; 
  } 
  public static void sort(int[] arr, int low, int high) { 
    if (low < high) { 
      int pi = partition(arr, low, high); 
      sort(arr, low, pi - 1); 
      sort(arr, pi + 1, high); 
    } 
  } 
  //Return 2 elements of arr that sum to the given value
  public static int[] findSum(int[] arr, int n) {
    //Helper sort function that uses the Quicksort Algorithm
    sort(arr, 0, arr.length - 1);   //Sort the array in Ascending Order
    int Pointer1 = 0;    //Pointer 1 -> At Start
    int Pointer2 = arr.length - 1;   //Pointer 2 -> At End
    int[] result = new int[2];
    int sum = 0;
    while (Pointer1 != Pointer2) {
      sum = arr[Pointer1] + arr[Pointer2];  //Calulate Sum of Pointer 1 and 2
      if (sum < n) 
        Pointer1++;  //if sum is less than given value => Move Pointer 1 to Right
      else if (sum > n) 
        Pointer2--; 
      else {
        result[0] = arr[Pointer1];
        result[1] = arr[Pointer2];
        return result; // containing 2 number 
      }
    }
    return arr; 
  } 
  public static void main(String args[]) {
    int n = 14;
    int[] arr1 = {1,2,3,4,5};
    if(arr1.length > 0){
      int[] arr2 = findSum(arr1, n);
      int num1 = arr2[0];
      int num2 = arr2[1];
      if((num1 + num2) != n)
        System.out.println("Not Found");
      else {
        System.out.println("Number 1 = " + num1);
        System.out.println("Number 2 = " + num2);
        System.out.println("Sum = " +  (n) );
      }
    } else {
      System.out.println("Input Array is Empty!");
    }
  }
}

Time Complexity: $O(nlogn)$

The best way to solve this is by first sorting the array.

Here, we use QuickSort to sort the array first. Then using two variables, one starting from the first index of the array and the second from size−1 index of the array.

If the sum of the elements on these indexes of the array is smaller than the given value n, then increment index from the start else decrement index from the end until the given value n is equal to the sum.

Store the elements on these indexes in the result array and return it.

Java

class CheckMinimum
{
  //Returns minimum value from given Array 
  public static int findMinimum(int[] arr) {
 
    int minimum = arr[0];
 
    //At every Index compare its value with minimum and if its less 
    //then make that index value new minimum value
    for (int i = 1; i < arr.length; i++) {
 
      if (arr[i] < minimum) {
        minimum = arr[i];
      }
    }
    return minimum;
  } //end of findMinimum
 
  public static void main(String args[]) {
 
    int[] arr = { 9, 2, 3, 6};
 
    System.out.print("Array : ");
    for(int i = 0; i < arr.length; i++)
      System.out.print(arr[i] + " ");
    System.out.println();
 
    int min = findMinimum(arr);
    System.out.println("Minimum in the Array: " + min);
 
  }
 
}

Java

class CheckReArrange
{
  //Rearrange Positive and Negative Values of Given Array 
  public static void reArrange(int[] arr) 
  {
    int j = 0; 
    for (int i = 0; i < arr.length; i++) { 
      if (arr[i] < 0) {   // if negative number found
        if (i != j) { 
          int temp = arr[i];
          arr[i] = arr[j]; // swapping with leftmost positive
          arr[j] = temp;
        }
        j++; 
      } 
    } 
  } //end of reArrange()
  public static void main(String args[]) {
    int[] arr = {2, 4, -6, 8, -5, -10};
    System.out.print("Array before re-arranging: ");
    for(int i = 0; i < arr.length; i++)
      System.out.print(arr[i] + " ");
    System.out.println();
    reArrange(arr);
    System.out.print("Array after rearranging: ");
    for(int i = 0; i < arr.length; i++)
      System.out.print(arr[i] + " ");
  }
}

Java

class CheckRotateArray
{
  //Rotates given Array by 1 position
  public static void rotateArray(int[] arr) {
 
    //Store last element of Array.
    //Start from the Second last and Right Shift the Array by one.
    //Store the last element saved on the first index of the Array.
    int lastElement = arr[arr.length - 1];
 
    for (int i = arr.length - 1; i > 0; i--) {
 
      arr[i] = arr[i - 1];
    }
 
    arr[0] = lastElement;
  } //end of rotateArray
 
  public static void main(String args[]) {
 
    int[] arr = {3, 6, 1, 8, 4, 2};
 
    System.out.print("Array before rotation: ");
    for(int i = 0; i < arr.length; i++) {
      System.out.print(arr[i] + " ");
    }
    System.out.println();
 
    rotateArray(arr);
 
    System.out.print("Array after rotation: ");
    for(int i = 0; i < arr.length; i++) {
      System.out.print(arr[i] + " ");
    }
  }
 
}

Time Complexity: $O(n)$

To rotate the array towards the right, we have to move the array elements towards the right by one index.

This means every element stored at index i will be moved to i + 1 position.

However, if we start shifting elements from the first element of the array, then the element at last index arr[arr.length - 1] is overwritten.

We fix this by saving the last element of the array in the variable lastElement.

Then we start shifting elements from index i - 1 to i, where the initial value of i will be arr.length - 1, and we will keep shifting the elements until i is greater than 0.

When the loop ends, we store the value of lastElement in arr[0].

Answer any Java interview problem by learning the patterns behind common questions

Grokking the Coding Interview Patterns

With thousands of potential questions to account for, preparing for the coding interview can feel like an impossible challenge. Yet with a strategic approach, coding interview prep doesn’t have to take more than a few weeks. Stop drilling endless sets of practice problems, and prepare more efficiently by learning coding interview patterns. This course teaches you the underlying patterns behind common coding interview questions. By learning these essential patterns, you will be able to unpack and answer any problem the right way — just by assessing the problem statement. This approach was created by FAANG hiring managers to help you prepare for the typical rounds of interviews at major tech companies like Apple, Google, Meta, Microsoft, and Amazon. Before long, you will have the skills you need to unlock even the most challenging questions, grok the coding interview, and level up your career with confidence. This course is also available in JavaScript, Python, Go, and C++ — with more coming soon!

85hrs

Intermediate

460 Challenges

461 Quizzes

Java

public class SinglyLinkedList<T> {
//Node inner class for SLL
    public class Node {
        public T data;
        public Node nextNode;
    }
    public Node headNode; //head node of the linked list
    public int size;      //size of the linked list
    //Constructor - initializes headNode and size
    public SinglyLinkedList() {
        headNode = null;
        size = 0;
    }
    //Helper Function that checks if List is empty or not 
    public boolean isEmpty() {
        if (headNode == null) return true;
        return false;
    }
    //Inserts new data at the start of the linked list
    public void insertAtHead(T data) {
        //Creating a new node and assigning it the new data value
        Node newNode = new Node();
        newNode.data = data;
        //Linking head to the newNode's nextNode
        newNode.nextNode = headNode;
        headNode = newNode;
        size++;
    }
    // Helper Function to printList
    public void printList() {
        if (isEmpty()) {
            System.out.println("List is Empty!");
            return;
        }
        Node temp = headNode;
        System.out.print("List : ");
        while (temp.nextNode != null) {
            System.out.print(temp.data.toString() + " -> ");
            temp = temp.nextNode;
        }
        System.out.println(temp.data.toString() + " -> null");
    }
    
    //Inserts new data at the end of the linked list
    public void insertAtEnd(T data) {
        //if the list is empty then call insertATHead()
        if (isEmpty()) {
            insertAtHead(data);
            return;
        }
        //Creating a new Node with value data 
        Node newNode = new Node();
        newNode.data = data;
        newNode.nextNode = null;
        Node last = headNode;
        //iterate to the last element 
        while (last.nextNode != null) {
            last = last.nextNode;
        }
        //make newNode the next element of the last node
        last.nextNode = newNode;
        size++;
    }
}

Java

class CheckSearch {  
    public static void main(String args[]) {
        SinglyLinkedList<Integer> sll = new SinglyLinkedList<Integer>(); 
        for (int i = 1; i <= 10; i++) {
            sll.insertAtEnd(i);
        }

        if(sll.searchNode(3)) {   // Calling search function
            System.out.println("Value: 3 is Found");
        }
        else {
            System.out.println("Value: 3 is not Found");
        }

        if(sll.searchNode(15)) {   // Calling search function
            System.out.println("Value: 15 is Found");
        }
        else {
            System.out.println("Value: 15 is not Found");
        }
    }
}

Java

 public static <T> Object nthElementFromEnd(SinglyLinkedList<T> list, int n) {
        int size = list.getSize();
        n = size - n + 1; //we can use the size variable to calculate distance from start
        if (size == 0 || n > size) {
            return null; //returns null if list is empty or n is greater than size
        }
        SinglyLinkedList.Node current = list.getHeadNode();
        int count = 1;
        //traverse until count is not equal to n
        while (current != null) {
            if (count == n)
                return current.data;
            count++;
            current = current.nextNode;
        }
        return null;
    }
    public static void main( String args[] ) {
        SinglyLinkedList<Integer> sll = new SinglyLinkedList<Integer>();
        sll.printList(); //list is empty
        System.out.println("3rd element from end : " + nthElementFromEnd(sll, 3)); //will return null
        for(int i=0; i<15; i+=2){
            sll.insertAtHead(i);
        }
        sll.printList(); // List is 14 -> 12 -> 10 -> 8 -> 6 -> 4 -> 2 -> 0 -> null
        System.out.println("3rd element from end : " + nthElementFromEnd(sll, 3)); //will return 4
        System.out.println("10th element from end : " + nthElementFromEnd(sll, 10));//will return null
    }
}

Java

class CheckReverse {
    public static void main( String args[] ) {
        SinglyLinkedList<Integer> list = new SinglyLinkedList<Integer>();
        for(int i = 0; i < 15; i+=2)
            list.insertAtEnd(i);
        System.out.print("Before ");
        list.printList();
        reverse(list);
        System.out.print("After ");
        list.printList();
    }
    public static <T> void reverse(SinglyLinkedList<T> list){
    SinglyLinkedList<T>.Node previous = null; //To keep track of the previous element, will be used in swapping links
            SinglyLinkedList<T>.Node current = list.headNode; //firstElement
            SinglyLinkedList<T>.Node next = null;
 
            //While Traversing the list, swap links
            while (current != null) {
                    next = current.nextNode;
                    current.nextNode = previous;
                    previous = current;
                    current = next;
            }
            //Linking Head Node with the new First Element
            list.headNode = previous;
    }
}

Time Complexity: $O(n)$

The loop that iterates through the list is the key to this solution. For any current node, its link with the previous node is reversed, and the variable next stores the next node in the list:

Store the current node’s nextNode in next
Set current node’s nextNode to previous (reversal)
Make the current node the new previous so that it can be used for the next iteration
Use next to move on to the next node

In the end, we simply point the head to the last node in our loop.

Java

public class DoublyLinkedList<T> {
    //Node inner class for DLL
    public class Node {
        public T data;
        public Node nextNode;
        public Node prevNode;
    }
    public Node headNode;
    public Node tailNode;
    public int size;
    public DoublyLinkedList() {
        this.headNode = null;
        this.tailNode = null;
    }
    public boolean isEmpty() {
        if (headNode == null && tailNode == null)
            return true;
        return false;
    }
    public Node getHeadNode() {
        return headNode;
    }
    public Node getTailNode() {
        return tailNode;
    }
    public int getSize() {
        return size;
    }
    public void insertAtHead(T data) {
        Node newNode = new Node();
        newNode.data = data;
        newNode.nextNode = this.headNode; //Linking newNode to head's nextNode
        newNode.prevNode = null;
        if (headNode != null)
            headNode.prevNode = newNode;
        else
            tailNode = newNode;
        this.headNode = newNode;
        size++;
    }
    public void insertAtEnd(T data) {
        if (isEmpty()) {
            insertAtHead(data);
            return;
        }
        Node newNode = new Node();
        newNode.data = data;
        newNode.nextNode = null;
        newNode.prevNode = tailNode;
        tailNode.nextNode = newNode;
        tailNode = newNode;
        size++;
    }
    public void printList() {
        if (isEmpty()) {
            System.out.println("List is Empty!");
            return;
        }
        Node temp = headNode;
        System.out.print("List : null <- ");
        while (temp.nextNode != null) {
            System.out.print(temp.data.toString() + " <-> ");
            temp = temp.nextNode;
        }
        System.out.println(temp.data.toString() + " -> null");
    }
    public void printListReverse() {
        if (isEmpty()) {
            System.out.println("List is Empty!");
            return;
        }
        Node temp = tailNode;
        System.out.print("List : null <- ");
        while (temp.prevNode != null) {
            System.out.print(temp.data.toString() + " <-> ");
            temp = temp.prevNode;
        }
        System.out.println(temp.data.toString() + " -> null");
    }
    public void deleteByValue(T data) {
        //if empty then simply return
        if (isEmpty())
            return;
        //Start from head node
        Node currentNode = this.headNode;
        if (currentNode.data.equals(data)) {
            //data is at head so delete from head
            deleteAtHead();
            return;
        }
        //traverse the list searching for the data to delete
        while (currentNode != null) {
            //node to delete is found
            if (data.equals(currentNode.data)) {
                currentNode.prevNode.nextNode = currentNode.nextNode;
                if (currentNode.nextNode != null)
                    currentNode.nextNode.prevNode = currentNode.prevNode;
                size--;
            }
            currentNode = currentNode.nextNode;
        }
    }
    public void deleteAtHead() {
        if (isEmpty())
            return;
        headNode = headNode.nextNode;
        if (headNode == null)
            tailNode = null;
        else
            headNode.prevNode = null;
        size--;
    }
    public void deleteAtTail() {
        if (isEmpty())
            return;
        tailNode = tailNode.prevNode;
        if (tailNode == null)
            headNode = null;
        else
            tailNode.nextNode = null;
        size--;
    }
}

Java

class ReverseWords {
  // Null terminating strings are not used in java
  public static void strRev(char[] str, int start, int end) {
    if (str == null || str.length < 2) {
      return;
    }
    while (start < end) {
      char temp = str[start];
      str[start] = str[end];
      str[end] = temp;
      start++;
      end--;
    }
  }
  public static void reverseWords(char[] sentence) {
    // Here sentence is a null-terminated string ending with char '\0'.
    if (sentence == null || sentence.length == 0) {
      return;
    }
    // To reverse all words in the string, we will first reverse
    // the string. Now all the words are in the desired location, but
    // in reverse order: "Hello World" -> "dlroW olleH".
    int len = sentence.length;
    strRev(sentence, 0, len - 2);
    // Now, let's iterate the sentence and reverse each word in place.
    // "dlroW olleH" -> "World Hello"
    int start = 0;
    int end;
    while (true) {
      // find the  start index of a word while skipping spaces.
      while (sentence[start] == ' ') {
        ++start;
      }
      if (start >= sentence.length - 1) {
        break;
      }
      // find the end index of the word.
      end = start + 1;
      while (end < sentence.length - 1 && sentence[end] != ' ') {
        ++end;
      }
      // let's reverse the word in-place.
      strRev(sentence, start, end - 1);
      start = end;
    }
  }
  static char[] getArray(String t) {
    char[] s = new char[t.length() + 1];
    int i = 0;
    for (; i < t.length(); ++i) {
      s[i] = t.charAt(i);
    }
    return s;
  }
  public static void main(String[] args) {
    char[] s = getArray("Hello World!");
    System.out.println(s);
    reverseWords(s);
    System.out.println(s);
  }
}

Java

class PalindromeSubStrings{
  public static boolean isPalindrome(String input, int i, int j) {
    while(j > i){
      if(input.charAt(i) != input.charAt(j))
        return false;
      i++;
      j--;
    }
    return true;
  }
  public static int findAllPalindromeSubstrings(String input) {
    int count = 0;
    for(int i = 0 ; i < input.length() ; i++) {
      for(int j = i + 1 ; j < input.length() ; j++) {
        if(isPalindrome(input,i,j)){
          System.out.println(input.substring(i,j+1));
          count++;
        }
      }
    }
  
    return count;
  }
  public static void main(String[] args) {
    String str = "aabbbaa";
    int count = findAllPalindromeSubstrings(str);
    System.out.println("Total palindrome substrings: " + count);
  }
}

Java

import java.util.*;
class LongestSubstringKDistinct {
  public static int findLength(String str, int k) {
    if (str == null || str.length() == 0 || str.length() < k)
      throw new IllegalArgumentException();
    int windowStart = 0, maxLength = 0;
    Map<Character, Integer> charFrequencyMap = new HashMap<>();
    // in the following loop we'll try to extend the range [windowStart, windowEnd]
    for (int windowEnd = 0; windowEnd < str.length(); windowEnd++) {
      char rightChar = str.charAt(windowEnd);
      charFrequencyMap.put(rightChar, charFrequencyMap.getOrDefault(rightChar, 0) + 1);
      // shrink the sliding window, until we are left with 'k' distinct characters in the frequency map
      while (charFrequencyMap.size() > k) {
        char leftChar = str.charAt(windowStart);
        charFrequencyMap.put(leftChar, charFrequencyMap.get(leftChar) - 1);
        if (charFrequencyMap.get(leftChar) == 0) {
          charFrequencyMap.remove(leftChar);
        }
        windowStart++; // shrink the window
      }
      maxLength = Math.max(maxLength, windowEnd - windowStart + 1); // remember the maximum length so far
    }
    return maxLength;
  }
  public static void main(String[] args) {
    System.out.println("Length of the longest substring: " + LongestSubstringKDistinct.findLength("araaci", 2));
    System.out.println("Length of the longest substring: " + LongestSubstringKDistinct.findLength("araaci", 1));
    System.out.println("Length of the longest substring: " + LongestSubstringKDistinct.findLength("cbbebi", 3));
  }
}

Time Complexity: $O(n)$

This problem follows the Sliding Window pattern.

We can use a HashMap to remember the frequency of each character we have processed.

First, we will insert characters from the beginning of the string until we have K distinct characters in the HashMap.
These characters will be our sliding window. We are asked to find the longest such window having no more than K distinct characters. We will remember the length of this window as the longest window so far.
After this, we will keep adding one character in the sliding window (i.e., slide the window ahead).
In each step, we will try to shrink the window from the beginning if the count of distinct characters in the HashMap is larger than K. We will shrink the window until we have no more than K distinct characters in the HashMap.
While shrinking, we’ll decrement the character’s frequency going out of the window and remove it from the HashMap if its frequency becomes zero.
At the end of each step, we’ll check if the current window length is the longest so far, and if so, remember its length.

29. Fruit Basket Problem#

Problem Statement

With a given array of characters where each character represents a fruit tree, place the maximum number of fruits in each of 2 baskets. The only restriction is that each basket can have only one type of fruit.

You can start with any tree, but you can’t skip a tree once you have started. You will pick one fruit from each tree until you cannot, i.e., you will stop when you have to pick from a third fruit type.

Write a function to return the maximum number of fruits in both the baskets.

Solution and Explanation:

Java

import java.util.*;
class MaxFruitCountOf2Types {
  public static int findLength(char[] arr) {
    int windowStart = 0, maxLength = 0;
    Map<Character, Integer> fruitFrequencyMap = new HashMap<>();
    // try to extend the range [windowStart, windowEnd]
    for (int windowEnd = 0; windowEnd < arr.length; windowEnd++) {
      fruitFrequencyMap.put(arr[windowEnd], fruitFrequencyMap.getOrDefault(arr[windowEnd], 0) + 1);
      // shrink the sliding window, until we are left with '2' fruits in the frequency map
      while (fruitFrequencyMap.size() > 2) {
        fruitFrequencyMap.put(arr[windowStart], fruitFrequencyMap.get(arr[windowStart]) - 1);
        if (fruitFrequencyMap.get(arr[windowStart]) == 0) {
          fruitFrequencyMap.remove(arr[windowStart]);
        }
        windowStart++; // shrink the window
      }
      maxLength = Math.max(maxLength, windowEnd - windowStart + 1);
    }
    return maxLength;
  }
  public static void main(String[] args) {
    System.out.println("Maximum number of fruits: " + 
                          MaxFruitCountOf2Types.findLength(new char[] { 'A', 'B', 'C', 'A', 'C' }));
    System.out.println("Maximum number of fruits: " + 
                          MaxFruitCountOf2Types.findLength(new char[] { 'A', 'B', 'C', 'B', 'B', 'C' }));
  }
}

Java

class AllBraces{
  static void print(ArrayList<ArrayList<Character>> arr) {
    for (int i = 0; i < arr.size(); i++) {
      System.out.println(arr.get(i).toString());
    }
  }
  static void printAllBacesRec(
    int n,
    int leftCount,
    int rightCount, ArrayList<Character> output,
    ArrayList<ArrayList<Character>> result) {
    if (leftCount >= n && rightCount >=n) {
      result.add((ArrayList)output.clone());
    }
    if (leftCount < n) {
      output.add('{');
      printAllBacesRec(n, leftCount + 1, rightCount, output, result);
      output.remove(output.size() - 1);
    }
    if (rightCount < leftCount) {
      output.add('}');
      printAllBacesRec(n, leftCount, rightCount + 1, output, result);
      output.remove(output.size() - 1);
    }
  }
  static ArrayList<ArrayList<Character>> printAllBraces(int n) {
    ArrayList<ArrayList<Character>> result = new ArrayList<ArrayList<Character>>();
    ArrayList<Character> output = new ArrayList<Character>();
    printAllBacesRec(n, 0, 0, output, result);
    return result;
  }
  
  public static void main(String[] args) {
    ArrayList<ArrayList<Character>> result = new ArrayList<ArrayList<Character>>();
    result =  printAllBraces(3);
    print (result);
  }
}

Stack and Queue Data Structure Questions

Implement Queue using Stacks
How does dequeue work for Queue elements?
Is a Queue Last in First Out (LIFO) or First in First Out (FIFO)?
What is a postfix expression?
Evaluate Stack prefix expressions
Generate Binary Numbers from 1 to n using Queue
Reverse the First K Elements of a Queue
Sort the Values in a Stack
Next Greater Element using Stack
How does a Priority Queue differ from a regular Queue?

Tree Data Structure Questions

Check if Two Binary Trees are Identical
What is the difference between a serialized and deserialized Binary Tree?
What types of solutions are suited for breadth-first search (BFS)?
How does post-order traversal compare with preorder traversal?
Nth Highest Number in Binary Search Tree (BST)
Print all Leaf Nodes of a Binary Tree
Find the Greatest Sum of a Path Beginning at the Root Node
Check if Left and Right Subtrees are Identical
Write an In-Order Iterator for a Binary Tree
Reverse Level Order Traversal

What to learn next#

Congratulations on finishing those 50 questions!

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Continue reading about Data Structures and Interview Prep#

Frequently Asked Questions

What are the Types of Data Structures in Java?

Some common types of data structures in Java: -Array -Linked List -Stack -Queue -Binary Tree -Binary Search Tree -Heap -Hashing -Graph

How to prepare for a DSA interview?

Data structures and algorithms (DSA) rely heavily on core programming concepts. It’s essential to have a strong base in basic coding to grasp them effectively. Start by practicing your code in a programming language that you’re comfortable with, and then steadily enhance your coding abilities.

How do I prepare for a data structure interview?

Review key concepts: Focus on fundamental data structures (arrays, linked lists, stacks, queues, trees, graphs, hash tables).
Practice coding: Solve problems on platforms like LeetCode, HackerRank, or CodeSignal.
Understand algorithms: Study sorting, searching, and traversal algorithms and practice writing them from scratch.
Mock interviews: Simulate real interview scenarios to improve problem-solving speed and communication.

Crack the Top 50 Java Data Structure Interview Questions

Practice 100+ Java data structure questions in one place Get hands-on experience with all the best Java questions from across our course library. Ace the Java Coding Interview

Array Data Structure Questions#

1. Array initialization#

2. Loops and Array size#

3. Linear or non-linear Data Structure#

4. Default Values#

5. Common Error Message#

6. Merge Two Sorted Arrays#

7. Find Two Numbers that Add up to n#

8. Find Minimum Value in Array#

9. Rearrange Positive & Negative Values#

10. Right Rotate the Array by One Index#

Linked List Data Structure Questions#

11. Difference between Arrays and Linked Lists#

12. Singly Linked List#

13. Mystery Code#

14. Circular Linked List#

15. Linked List Storage#

16. Insertion in a Singly Linked List (insert at End)#

17. Search in a Singly Linked List#

18. Return the Nth node from End#

19. Reverse a Linked List#

20. Find if Doubly Linked-list is a Palindrome#

String Data Structure Questions#

21. Creating a String Object#

22. Storage of Strings#

23. Advantage of Immutability#

24. Mutable Strings in Java#

25. Equals Behavior#

26. Reverse Words in a Sentence#

27. Find all Palindrome Substrings#

28. Longest Substring with K Distinct Characters#

29. Fruit Basket Problem#

30. Print All Combinations of Balanced Braces#

What to learn next#

Continue reading about Data Structures and Interview Prep#

Frequently Asked Questions

What are the Types of Data Structures in Java?

How to prepare for a DSA interview?

How do I prepare for a data structure interview?

Practice 100+ Java data structure questions in one place
Get hands-on experience with all the best Java questions from across our course library.

Ace the Java Coding Interview