Data Structures and Algorithms

DSA Interview Questions

HashMap & HashSet Problems: Common Interview Questions

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In the competitive world of software engineering interviews, especially at top tech companies, questions revolving around HashMap and HashSet frequently appear. These collections are fundamental in Java, offering efficient ways to handle key-value pairs and unique elements. This comprehensive guide dives deep into their workings, differences, and common interview problems. Whether you’re a beginner brushing up on basics or an experienced developer preparing for FAANG-level interviews, you’ll find actionable insights, code examples, and strategies to tackle these topics confidently. We’ll explore real-world applications, performance considerations, and over 30 actual interview questions drawn from sources like LeetCode, GeeksforGeeks, and reported FAANG experiences, complete with in-depth solutions.

Understanding HashMap in Java

HashMap is a cornerstone of Java’s Collections Framework, implementing the Map interface to store data in key-value pairs. It’s widely used for fast lookups, insertions, and deletions, making it a favorite in interview scenarios.

What is HashMap?

At its core, HashMap uses a hashing mechanism to map keys to values. Unlike arrays or lists, it doesn’t maintain order, but it excels in scenarios requiring quick access. According to Java documentation and expert analyses, HashMap’s average time complexity for basic operations is O(1), assuming a good hash function. This makes it ideal for caching, frequency counting, and more.

Internal Working of HashMap

Internally, HashMap employs an array of buckets (initially 16) where each bucket can hold a linked list or, in Java 8+, a balanced tree for collision resolution. When you call put(key, value), it computes the hash code of the key, determines the bucket index via (hash & (capacity – 1)), and stores the entry. If a collision occurs (same hash code), it chains entries in a list. Resizing happens when the load factor (default 0.75) is exceeded, doubling the capacity to maintain efficiency. For deeper dives into data structures, explore our DSA course.

Key Features and Best Practices

Null Handling: Allows one null key and multiple null values.
Thread Safety: Not thread-safe; use ConcurrentHashMap for multi-threaded environments.
Performance Tips: Choose initial capacity wisely to minimize resizing. For example, new HashMap<>(100) for expected 75 entries. Actionable advice: Always override equals() and hashCode() for custom keys to avoid unexpected behavior.

Understanding HashSet in Java

HashSet, implementing the Set interface, is perfect for storing unique elements without duplicates. It’s backed by a HashMap internally, where elements are keys and a dummy object is the value.

What is HashSet?

HashSet ensures no duplicates by leveraging hashing. It’s unordered and offers constant-time performance for add, remove, and contains operations. Statistics from developer surveys show it’s commonly used in 40% of Java projects for deduplication tasks.

Internal Working of HashSet

HashSet uses a HashMap under the hood: private transient HashMap<E,Object> map;. Additions call map.put(e, PRESENT), where PRESENT is a constant object. This inherits HashMap’s hashing and collision handling. Load factor and capacity mirror HashMap’s defaults.

Key Features and Best Practices

Null Handling: Allows one null element.
Thread Safety: Not synchronized; wrap with Collections.synchronizedSet() if needed.
Iteration: Use iterators or enhanced for-loops, but avoid modifications during iteration to prevent ConcurrentModificationException. For web development contexts where sets manage unique IDs, check our web development course.

Key Differences Between HashMap and HashSet

Understanding the nuances between HashMap and HashSet is crucial, as interviewers often probe this to test conceptual clarity.

Storage: HashMap stores key-value pairs; HashSet stores unique objects (internally as keys in a HashMap).
Duplicates: HashMap allows duplicate values but unique keys; HashSet allows no duplicates.
Nulls: HashMap: one null key, multiple null values; HashSet: one null element.
Interfaces: HashMap implements Map; HashSet implements Set.
Use Cases: Use HashMap for associations (e.g., employee ID to name); HashSet for uniqueness (e.g., unique visitors).
Performance: Both O(1) average case, but HashSet is slightly more memory-efficient for single elements.

Feature	HashMap	HashSet
Data Structure	Key-Value Pairs	Unique Elements
Implements	Map	Set
Duplicates Allowed	Values: Yes, Keys: No	No
Null Support	One null key, multiple null values	One null element
Internal Backing	Array + LinkedList/Tree	HashMap
Common Use	Caching, Mapping	Deduplication

For mastering these in system design, consider our master DSA, web dev, and system design course.

Common Theoretical Interview Questions

Before diving into coding problems, let’s address frequently asked conceptual questions from actual interviews at companies like Google and Amazon.

How does put() work in HashMap? It hashes the key, finds the bucket, and adds or updates the entry, handling collisions via chaining. In-depth: If the bucket has >8 entries (Java 8+), it converts to a tree for O(log n) access.
What happens if hashCode() is not overridden? Poor distribution leads to collisions, degrading to O(n) performance.
Difference between HashMap and Hashtable? Hashtable is synchronized and doesn’t allow nulls; HashMap is faster but not thread-safe.
Why is HashSet backed by HashMap? To reuse hashing logic and ensure uniqueness efficiently.
How to make HashSet thread-safe? Use Collections.synchronizedSet(new HashSet<>()) or ConcurrentHashMap-based sets.
What is load factor? Ratio triggering resize (default 0.75). Lower values reduce collisions but increase memory use.
Can HashMap have duplicate keys? No; it overrides existing ones.
Difference between HashSet and TreeSet? HashSet is unordered; TreeSet is sorted.
How does contains() work in HashSet? It checks if the element is a key in the backing HashMap.
What is rehashing? Resizing and redistributing entries when capacity doubles.

For data science applications using these structures, see our data science course.

30+ Coding Problems: Actual Interview Questions with Solutions

Drawing from LeetCode, GeeksforGeeks, and FAANG reports, here are 30+ problems commonly asked in interviews. Each includes a description, approach, time/space complexity, and Java code. These have been reported in interviews at Meta, Amazon, and Google. We’ve prioritized high-quality, in-depth ones for thorough preparation. Practice these on our crash course.

Easy Problems

Find if Array is Subset of Another (Asked at Amazon) Description: Check if all elements of arr2 are in arr1. Approach: Use HashSet for arr1 elements, check arr2 against it. Complexity: O(m + n) time, O(n) space.

				
					
boolean isSubset(int[] arr1, int[] arr2) {
    Set<Integer> set = new HashSet<>();
    for (int num : arr1) set.add(num);
    for (int num : arr2) if (!set.contains(num)) return false;
    return true;
}

In-depth: Handles duplicates implicitly; optimize for large arrays by choosing larger as set.

Frequency of Each Character in String (Google Interview) Description: Count occurrences of each char. Approach: Use HashMap<char, int> to tally. Complexity: O(n) time, O(k) space (k=unique chars).

				
					Map<Character, Integer> countFreq(String s) {
    Map<Character, Integer> map = new HashMap<>();
    for (char c : s.toCharArray()) map.put(c, map.getOrDefault(c, 0) + 1);
    return map;
}

In-depth: getOrDefault() is Java 8+; for older, use containsKey().

First Non-Repeating Character (Microsoft) Description: Find first unique char in string. Approach: HashMap for counts, then iterate string. Complexity: O(n) time, O(n) space.

				
					
char firstUnique(String s) {
    Map<Character, Integer> map = new LinkedHashMap<>();
    for (char c : s.toCharArray()) map.put(c, map.getOrDefault(c, 0) + 1);
    for (Map.Entry<Character, Integer> entry : map.entrySet()) if (entry.getValue() == 1) return entry.getKey();
    return ' ';
}

In-depth: LinkedHashMap preserves insertion order.

Intersection of Two Arrays (FAANG Common) Description: Find common elements. Approach: HashSet for one array, check other. Complexity: O(m + n) time, O(min(m,n)) space.

				
					int[] intersection(int[] nums1, int[] nums2) {
    Set<Integer> set = new HashSet<>(), res = new HashSet<>();
    for (int num : nums1) set.add(num);
    for (int num : nums2) if (set.contains(num)) res.add(num);
    return res.stream().mapToInt(Integer::intValue).toArray();

In-depth: Use sets for uniqueness.

Missing Elements in Range (Amazon) Description: Find missing nums in [low, high] from array. Approach: HashSet for array, iterate range. Complexity: O(n + r) time (r=range).

				
					int[] intersection(int[] nums1, int[] nums2) {
    Set<Integer> set = new HashSet<>(), res = new HashSet<>();
    for (int num : nums1) set.add(num);
    for (int num : nums2) if (set.contains(num)) res.add(num);
    return res.stream().mapToInt(Integer::intValue).toArray();

In-depth: Efficient for small ranges.

Pair with Given Sum (Google) Description: Check if two nums sum to target. Approach: HashSet, add and check complement. Complexity: O(n) time, O(n) space.

				
					boolean hasPair(int[] arr, int target) {
    Set<Integer> set = new HashSet<>();
    for (int num : arr) {
        if (set.contains(target - num)) return true;
        set.add(num);
    }
    return false;
}

In-depth: Efficient for small ranges.

Pair with Given Sum (Google) Description: Check if two nums sum to target. Approach: HashSet, add and check complement. Complexity: O(n) time, O(n) space.

				
					
int distinct(int[] arr) {
    return new HashSet<>(Arrays.stream(arr).boxed().toList()).size();
}

In-depth: Stream for conciseness.

Count Distinct Substrings (Advanced Variant, Asked at ByteDance) Description: Count unique substrings. Approach: Generate all substrings, add to HashSet. Complexity: O(n^2) time/space (naive).

				
					int countSubs(String s) {
    Set<String> set = new HashSet<>();
    for (int i = 0; i < s.length(); i++) for (int j = i + 1; j <= s.length(); j++) set.add(s.substring(i, j));
    return set.size();
}

In-depth: Use suffix tree for optimization.

Print Distinct Words in Paragraph (Amazon) Description: Unique words from text. Approach: Split, add to HashSet. Complexity: O(n) time.

				
					Set<String> uniqueWords(String para) {
    Set<String> set = new HashSet<>();
    Collections.addAll(set, para.split("\\s+"));
    return set;
}

In-depth: Handle punctuation with regex.

Pairs with Difference K (Google) Description: Count pairs |a-b| == k. Approach: HashSet, check num + k and num – k. Complexity: O(n) time.

				
					int countPairs(int[] arr, int k) {
    Set<Integer> set = new HashSet<>();
    int count = 0;
    for (int num : arr) {
        if (set.contains(num + k)) count++;
        if (set.contains(num - k)) count++;
        set.add(num);
    }
    return count;
}

In-depth: Avoid double-counting.

Medium Problems

Longest Consecutive Sequence (FAANG Staple) Description: Find longest consecutive nums chain. Approach: HashSet, start from non-preceding nums. Complexity: O(n) time.

				
					int longestConsecutive(int[] nums) {
    Set<Integer> set = new HashSet<>();
    for (int num : nums) set.add(num);
    int max = 0;
    for (int num : nums) {
        if (!set.contains(num - 1)) {
            int curr = num, len = 1;
            while (set.contains(++curr)) len++;
            max = Math.max(max, len);
        }
    }
    return max;
}

In-depth: O(n) by visiting each once.

All Pairs with Given Sum (Extension of Easy) Description: Print all pairs summing to target. Approach: HashMap for counts, handle duplicates. Complexity: O(n) time.

				
					List<List<String>> groupAnagrams(String[] strs) {
    Map<String, List<String>> map = new HashMap<>();
    for (String s : strs) {
        char[] chars = s.toCharArray();
        Arrays.sort(chars);
        String key = new String(chars);
        map.computeIfAbsent(key, k -> new ArrayList<>()).add(s);
    }
    return new ArrayList<>(map.values());
}

In-depth: Alt key: char count array.

Subarray Sum Equals K (Google) Description: Count subarrays summing to k. Approach: HashMap for prefix sums. Complexity: O(n) time.

				
					
int subarraySum(int[] nums, int k) {
    Map<Integer, Integer> map = new HashMap<>();
    map.put(0, 1);
    int sum = 0, count = 0;
    for (int num : nums) {
        sum += num;
        count += map.getOrDefault(sum - k, 0);
        map.put(sum, map.getOrDefault(sum, 0) + 1);
    }
    return count;
}

In-depth: Handles negatives.

Two Sum (LeetCode #1, FAANG) Description: Indices of two nums summing to target. Approach: HashMap for num to index. Complexity: O(n) time.

				
					int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int comp = target - nums[i];
        if (map.containsKey(comp)) return new int[]{map.get(comp), i};
        map.put(nums[i], i);
    }
    return null;
}

In-depth: One-pass efficient.

Contains Duplicate (Easy Variant, Meta) Description: Check for duplicates. Approach: Add to HashSet, check add return. Complexity: O(n) time.

				
					
boolean containsDuplicate(int[] nums) {
    Set<Integer> set = new HashSet<>();
    for (int num : nums) if (!set.add(num)) return true;
    return false;
}

In-depth: add() returns false on duplicate.

Longest Substring Without Repeating Chars (Google) Description: Max length substring no repeats. Approach: Sliding window with HashSet. Complexity: O(n) time.

				
					int lengthOfLongestSubstring(String s) {
    Set<Character> set = new HashSet<>();
    int max = 0, left = 0;
    for (int right = 0; right < s.length(); right++) {
        while (set.contains(s.charAt(right))) set.remove(s.charAt(left++));
        set.add(s.charAt(right));
        max = Math.max(max, right - left + 1);
    }
    return max;
}

In-depth: Optimize with map for last index.

Top K Frequent Elements (Amazon) Description: K most frequent nums. Approach: HashMap freq, then priority queue. Complexity: O(n log k) time.

				
					int[] topKFrequent(int[] nums, int k) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int num : nums) map.put(num, map.getOrDefault(num, 0) + 1);
    PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> map.get(a) - map.get(b));
    for (int key : map.keySet()) {
        pq.offer(key);
        if (pq.size() > k) pq.poll();
    }
    int[] res = new int[k];
    for (int i = 0; i < k; i++) res[i] = pq.poll();
    return res;
}

In-depth: Min-heap for efficiency.

Valid Sudoku (Google) Description: Check if 9×9 Sudoku valid. Approach: HashSets for rows, cols, boxes. Complexity: O(1) time (fixed size).

				
					boolean isValidSudoku(char[][] board) {
    Set<String> set = new HashSet<>();
    for (int i = 0; i < 9; i++) for (int j = 0; j < 9; j++) {
        char c = board[i][j];
        if (c != '.') {
            if (!set.add(c + "row" + i) || !set.add(c + "col" + j) || !set.add(c + "box" + (i/3)*3 + j/3)) return false;
        }
    }
    return true;
}

In-depth: String keys for uniqueness.

Single Number (XOR Alt, but HashMap Variant) Description: Find num appearing once (others twice). Approach: HashMap count, find odd. Complexity: O(n) time.

				
					int singleNumber(int[] nums) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int num : nums) map.put(num, map.getOrDefault(num, 0) + 1);
    for (Map.Entry<Integer, Integer> entry : map.entrySet()) if (entry.getValue() == 1) return entry.getKey();
    return -1;

In-depth: XOR optimal, but map versatile.

Hard Problems

LRU Cache (FAANG Design) Description: Implement LRU with get/put O(1). Approach: HashMap + Doubly Linked List. Complexity: O(1) per op.

				
					class LRUCache {
    class Node { int key, val; Node prev, next; }
    Map<Integer, Node> map = new HashMap<>();
    Node head = new Node(), tail = new Node();
    int cap;
    public LRUCache(int capacity) {
        cap = capacity;
        head.next = tail;
        tail.prev = head;
    }
    public int get(int key) {
        if (!map.containsKey(key)) return -1;
        Node node = map.get(key);
        remove(node);
        add(node);
        return node.val;
    }
    public void put(int key, int value) {
        if (map.containsKey(key)) {
            Node node = map.get(key);
            remove(node);
            node.val = value;
            add(node);
        } else {
            if (map.size() == cap) {
                Node del = tail.prev;
                remove(del);
                map.remove(del.key);
            }
            Node node = new Node();
            node.key = key; node.val = value;
            map.put(key, node);
            add(node);
        }
    }
    private void remove(Node node) {
        node.prev.next = node.next;
        node.next.prev = node.prev;
    }
    private void add(Node node) {
        node.next = head.next;
        node.next.prev = node;
        head.next = node;
        node.prev = head;
    }
}

In-depth: Head/tail sentinels simplify.

Design HashMap (LeetCode #706, Google) Description: Implement MyHashMap. Approach: Array of lists for chaining. Complexity: O(1) average.

				
					class MyHashMap {
    ListNode[] nodes = new ListNode[1000];
    public void put(int key, int value) {
        int i = key % nodes.length;
        if (nodes[i] == null) nodes[i] = new ListNode(-1, -1);
        ListNode prev = find(nodes[i], key);
        if (prev.next == null) prev.next = new ListNode(key, value);
        else prev.next.val = value;
    }
    public int get(int key) {
        int i = key % nodes.length;
        if (nodes[i] == null) return -1;
        ListNode node = find(nodes[i], key);
        return node.next == null ? -1 : node.next.val;
    }
    public void remove(int key) {
        int i = key % nodes.length;
        if (nodes[i] == null) return;
        ListNode prev = find(nodes[i], key);
        if (prev.next == null) return;
        prev.next = prev.next.next;
    }
    ListNode find(ListNode bucket, int key) {
        ListNode node = bucket, prev = null;
        while (node != null && node.key != key) {
            prev = node;
            node = node.next;
        }
        return prev;
    }
    class ListNode {
        int key, val;
        ListNode next;
        ListNode(int key, int val) { this.key = key; this.val = val; }
    }
}

In-depth: Modulo for bucketing.

Minimum Insertions for Palindrome (With Perms, Asked at Adobe) Description: Min inserts to make palindrome, perms allowed. Approach: HashMap char counts, count odds. Complexity: O(n) time.

				
					
int minInsertions(String s) {
    Map<Character, Integer> map = new HashMap<>();
    for (char c : s.toCharArray()) map.put(c, map.getOrDefault(c, 0) + 1);
    int odds = 0;
    for (int count : map.values()) if (count % 2 == 1) odds++;
    return odds == 0 ? 0 : odds - 1;
}

In-depth: Palindrome allows one odd.

Find Repetitive Element 1 to N-1 (Amazon) Description: Find duplicate in 1 to n-1. Approach: HashSet, check add. Complexity: O(n) time.

				
					int findDuplicate(int[] arr) {
    Set<Integer> set = new HashSet<>();
    for (int num : arr) if (!set.add(num)) return num;
    return -1;
}

In-depth: Assume one duplicate.

Remove Duplicates from Array (Meta) Description: Remove dups in-place. Approach: HashSet, then copy back. Complexity: O(n) time.

				
					int removeDuplicates(int[] nums) {
    Set<Integer> set = new LinkedHashSet<>();
    for (int num : nums) set.add(num);
    int i = 0;
    for (int num : set) nums[i++] = num;
    return set.size();
}

In-depth: -1 for full array.

Count Subarrays with XOR K (Advanced, ByteDance) Description: Subarrays XOR equals k. Approach: Prefix XOR map. Complexity: O(n) time.

				
					
int countXOR(int[] arr, int k) {
    Map<Integer, Integer> map = new HashMap<>();
    map.put(0, 1);
    int xor = 0, count = 0;
    for (int num : arr) {
        xor ^= num;
        count += map.getOrDefault(xor ^ k, 0);
        map.put(xor, map.getOrDefault(xor, 0) + 1);
    }
    return count;
}

In-depth: Similar to sum k.

Find Itinerary from Tickets (Google) Description: Reconstruct path from from-to pairs. Approach: HashMap adjacency, Euler path. Complexity: O(n) time.

				
					List<String> findItinerary(List<List<String>> tickets) {
    Map<String, PriorityQueue<String>> map = new HashMap<>();
    for (List<String> t : tickets) map.computeIfAbsent(t.get(0), k -> new PriorityQueue<>()).add(t.get(1));
    List<String> res = new ArrayList<>();
    dfs("JFK", map, res);
    Collections.reverse(res);
    return res;
}
private void dfs(String from, Map<String, PriorityQueue<String>> map, List<String> res) {
    while (map.containsKey(from) && !map.get(from).isEmpty()) dfs(map.get(from).poll(), map, res);
    res.add(from);
}

In-depth: PQ for lexical order.

Union and Intersection of Lists (Amazon) Description: Union/intersect two lists. Approach: HashSets for both. Complexity: O(m + n) time.

				
					Set<Integer> union(List<Integer> l1, List<Integer> l2) {
    Set<Integer> set = new HashSet<>(l1);
    set.addAll(l2);
    return set;
}
Set<Integer> intersect(List<Integer> l1, List<Integer> l2) {
    Set<Integer> set1 = new HashSet<>(l1), set2 = new HashSet<>(l2);
    set1.retainAll(set2);
    return set1;

In-depth: retainAll for intersect.

Minimum Elements to Remove for No Common (Variant, Asked at Flipkart) Description: Min removes so no common in two arrays. Approach: HashMaps for freq, calculate min. Complexity: O(m + n) time.

				
					int minRemove(int[] a1, int[] a2) {
    Map<Integer, Integer> m1 = new HashMap<>(), m2 = new HashMap<>();
    for (int n : a1) m1.put(n, m1.getOrDefault(n, 0) + 1);
    for (int n : a2) m2.put(n, m2.getOrDefault(n, 0) + 1);
    int min = 0;
    for (int key : m1.keySet()) if (m2.containsKey(key)) min += Math.min(m1.get(key), m2.get(key));
    return min;
}

In-depth: retainAll for intersect.

Minimum Elements to Remove for No Common (Variant, Asked at Flipkart) Description: Min removes so no common in two arrays. Approach: HashMaps for freq, calculate min. Complexity: O(m + n) time.

				
					int[] sortByFreq(int[] arr) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int n : arr) map.put(n, map.getOrDefault(n, 0) + 1);
    return Arrays.stream(arr).boxed().sorted((a, b) -> {
        int fa = map.get(a), fb = map.get(b);
        return fa != fb ? fb - fa : a - b;
    }).mapToInt(Integer::intValue).toArray();
}

In-depth: Stable with value tiebreaker.

K Closest Points to Origin (Amazon) Description: K points closest to (0,0). Approach: Max-heap with dist, but map variant possible. Complexity: O(n log k).

				
					int[][] kClosest(int[][] points, int k) {
    PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> b[0]*b[0] + b[1]*b[1] - a[0]*a[0] - a[1]*a[1]);
    for (int[] p : points) {
        pq.offer(p);
        if (pq.size() > k) pq.poll();
    }
    return pq.toArray(new int[0][]);
}

In-depth: HashMap not direct, but for coords.

These problems cover a broad spectrum, with many directly from interview experiences. Practice varying difficulties to build confidence.

Preparation Tips and Strategies

To ace these in interviews:

Understand Internals: Know hashing, collisions, and Java 8 changes.
Practice Coding: Solve on LeetCode; aim for 100+ hash problems.
Edge Cases: Test nulls, empties, large inputs.
Time Management: Explain trade-offs verbally.
Mock Interviews: Simulate with peers.

Expert quote from a FAANG engineer: “HashMap questions test not just code, but optimization thinking.”

Ready to level up? Dive into our courses for structured learning.

Conclusion

HashMap and HashSet are powerful tools that can make or break your interview performance. By understanding their mechanics and practicing these 30+ problems, you’ll be well-equipped. Remember, consistency is key—practice daily and analyze solutions deeply. What’s your next step? Share in the comments or sign up for updates to get more resources. Happy coding!

(Word count: ~2,200)

FAQs

What are common HashMap interview questions for Java developers?

Common questions include internal working, collision handling, and differences from HashSet, often with coding like Two Sum.

How does HashSet ensure uniqueness in Java collections?

HashSet uses a backing HashMap to store elements as keys, leveraging hashCode() and equals() for duplicate checks.

What are top hashing problems for coding interviews?

Top problems involve frequency counting, subarray sums, and anagram grouping using HashMap or HashSet for O(1) operations.

Differences between HashMap and HashSet in Java?

HashMap stores key-value pairs with unique keys; HashSet stores unique elements without values or order.

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HashMap & HashSet Problems: Common Interview Questions

Understanding HashMap in Java

What is HashMap?

Internal Working of HashMap

Key Features and Best Practices

Understanding HashSet in Java

What is HashSet?

Internal Working of HashSet

Key Features and Best Practices

Key Differences Between HashMap and HashSet

Common Theoretical Interview Questions

30+ Coding Problems: Actual Interview Questions with Solutions

Easy Problems

Find if Array is Subset of Another (Asked at Amazon) Description: Check if all elements of arr2 are in arr1. Approach: Use HashSet for arr1 elements, check arr2 against it. Complexity: O(m + n) time, O(n) space.

Medium Problems

Hard Problems

Preparation Tips and Strategies

Conclusion

FAQs

What are common HashMap interview questions for Java developers?

How does HashSet ensure uniqueness in Java collections?

What are top hashing problems for coding interviews?

Differences between HashMap and HashSet in Java?

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