NeetCode 150 in Java: Collections, Patterns and Practice

The NeetCode 150 in Java is not just a list of algorithms to translate into another language. Java’s collections, primitive types, comparator rules and call-stack behaviour all affect whether a solution is easy to write, test and explain. A correct pattern can still fail when a sum overflows, a heap orders values incorrectly or a recursive traversal goes too deep.

This guide shows how to work through the NeetCode 150 list, practised on thita. It focuses on Java coding interview habits: selecting the right standard-library type, making numeric assumptions explicit and testing the edge cases that expose implementation bugs. The examples are original, compact solutions to problems available on thita. They do not reproduce NeetCode or LeetCode problem statements, solutions or video content.

The NeetCode 150 contains 150 problems: 28 Easy, 101 Medium and 21 Hard. Thita is not affiliated with NeetCode. Start with the NeetCode 150 by pattern, where each problem is listed under a thita pattern and tagged with its NeetCode topic.

How should you approach the NeetCode 150 in Java?

Treat the list as a pattern curriculum rather than a queue of answers to memorise. For each problem, identify the invariant first, then choose the Java representation that makes that invariant visible. Explain the straightforward approach, state why its time or space use is unsuitable, and only then write the improved version.

A practical loop for each problem is:

Read the input contract and write down empty, single-item, duplicate and extreme-value cases where they apply.
Name the pattern: frequency map, sliding window, monotonic stack, binary search on an answer, traversal or dynamic programming.
Choose Java types before coding. Decide whether a value needs long, whether a queue may contain null, and whether ordering is natural or comparator-defined.
Implement one small invariant and say what it means while you code.
Run the solution against the test cases, then add a mental test for the boundary that is easiest to overlook.
Re-solve it without looking at the previous implementation and give the time and space complexity aloud.

A useful progression is to begin with arrays, hashing, two pointers and sliding windows. Move next to stacks, binary search and linked lists, then trees and heaps. Backtracking, tries and graph traversal build on those state-management habits. Dynamic programming, greedy methods, intervals, mathematics, geometry and bit manipulation then give you a broader set of ways to represent state and prove an invariant. This is not a demand to solve topics in a rigid order; it is a way to make each Java decision easier to place.

Thita groups problems into 99 DSA patterns in 16 families through its DSA patterns sheet. That view is useful when a NeetCode topic is broad: a frequency-map problem and a top-k problem may both involve arrays, but their Java data structures and invariants are different.

Which Java collections should you know cold for the NeetCode 150 in Java?

The standard-library choices are small enough to rehearse deliberately:

HashMap<K, V> maps a key to a value and is the usual choice for frequencies, indices and adjacency data.
HashSet<T> answers membership questions and helps remove repeated work.
ArrayList<T> is a flexible indexed result container.
ArrayDeque<T> handles queue and stack behaviour. Use offer, poll, peek, push and pop instead of reaching for the legacy Stack class.
PriorityQueue<T> supplies heap operations. Its natural ordering is a min-heap, so provide a comparator when the problem needs another order.
Arrays.sort sorts primitive arrays and can sort int[][] rows with a comparator.
StringBuilder is the natural tool for repeated string construction.

Collections store objects, so an int becomes an Integer when it is placed in a map, set, deque or heap. Java handles most boxing and unboxing automatically, but remember that an unboxing operation cannot turn null into a number. ArrayDeque also rejects null, so a tree traversal should track levels by queue size or store an explicit state object rather than using a null sentinel.

For map updates, getOrDefault is clear when you need the current value, merge is concise for accumulation, and computeIfAbsent is useful when the value is a mutable collection. The following two examples put those choices into real list problems.

How do HashMap idioms simplify Java frequency problems?

Group Anagrams

Group each word by a sorted-character signature. Words with the same signature belong to the same result group.

Practise Group Anagrams on thita

JAVA
import java.util.*;

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        Map<String, List<String>> groups = new HashMap<>();
        for (String word : strs) {
            char[] letters = word.toCharArray();
            Arrays.sort(letters);
            String key = new String(letters);
            groups.computeIfAbsent(key, ignored -> new ArrayList<>()).add(word);
        }
        return new ArrayList<>(groups.values());
    }
}

The method handles an empty array by returning an empty result and keeps duplicate words in their appropriate group. If there are n words and m is the relevant word length, the time complexity is O(n × m log m). The space complexity is O(n × m) for the keys and returned groups.

Top K Frequent Elements

Count repeated integers, then keep the k values with the greatest frequencies in a bounded min-heap.

Practise Top K Frequent Elements on thita

JAVA
import java.util.*;

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> frequency = new HashMap<>();
        for (int value : nums) {
            frequency.merge(value, 1, Integer::sum);
        }

        PriorityQueue<Integer> heap = new PriorityQueue<>(
            (a, b) -> Integer.compare(
                frequency.getOrDefault(a, 0),
                frequency.getOrDefault(b, 0)
            )
        );
        for (int value : frequency.keySet()) {
            heap.offer(value);
            if (heap.size() > k) {
                heap.poll();
            }
        }

        int[] answer = new int[k];
        for (int i = k - 1; i >= 0; i--) {
            answer[i] = heap.poll();
        }
        return answer;
    }
}

Here merge adds one to an existing count or creates the first count, while getOrDefault supplies a safe fallback when reading a key. The comparator uses Integer.compare, not subtraction. If n is the input length and u is the number of distinct values, the total time complexity is O(n + u log k), and the space complexity is O(u + k). The usual problem contract supplies a valid k; duplicates are counted rather than treated as separate heap entries.

Why is ArrayDeque preferable to Stack in Java?

Stack is a legacy synchronised type. ArrayDeque communicates the intended operation more directly: use push and pop for last-in, first-out work, or offer and poll for first-in, first-out work. It also avoids the temptation to mix stack and queue operations without naming the invariant.

Binary Tree Level Order Traversal

Visit a binary tree one level at a time, placing non-null children into an ArrayDeque for the next level.

Practise Binary Tree Level Order Traversal on thita

JAVA
import java.util.*;

class Solution {
    static class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode() {}
        TreeNode(int val) { this.val = val; }
        TreeNode(int val, TreeNode left, TreeNode right) {
            this.val = val;
            this.left = left;
            this.right = right;
        }
    }

    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> answer = new ArrayList<>();
        if (root == null) return answer;

        ArrayDeque<TreeNode> queue = new ArrayDeque<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int levelSize = queue.size();
            List<Integer> level = new ArrayList<>(levelSize);
            for (int i = 0; i < levelSize; i++) {
                TreeNode node = queue.poll();
                level.add(node.val);
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
            answer.add(level);
        }
        return answer;
    }
}

The null-root case returns an empty list, and the queue never receives a null value. The time complexity is O(n), and the space complexity is O(n), including the returned levels and the queue. The helper TreeNode is defined inside the file, so the block compiles independently.

This iterative form also illustrates the recursion-depth rule. Recursive tree code uses call-stack space proportional to the tree height. A deeply unbalanced tree can therefore exhaust the Java call stack even when the algorithmic idea is correct. When the input depth is not safely bounded, replace recursive DFS with an explicit ArrayDeque containing the node and any state the recursive call would have carried.

What should you watch for with PriorityQueue comparators?

Never write (a, b) -> a - b as a general integer comparator. Subtraction can overflow, producing the wrong sign and therefore the wrong ordering. Use Integer.compare(a, b) for int values and Long.compare(a, b) for long values. The same rule applies when ordering array rows, nodes or custom records by a numeric field.

In a min-heap solution, ask what the root means before coding. In Top K Frequent Elements, the root is the least frequent retained value, so it is the first candidate to remove when the heap grows beyond k. That one sentence is more useful than memorising a particular constructor.

How do overflow and character arithmetic affect Java solutions?

Java evaluates int arithmetic as int unless an operand is promoted first. A midpoint written as (left + right) / 2 can overflow before the division. Prefer left + (right - left) / 2; if the bounds themselves or the computation around them need a wider type, use long deliberately.

The same issue appears in totals. Assigning an already-overflowed int expression to a long is too late. Start with long total = 0 and promote an operand before addition or multiplication. Binary search on an answer is a good place to practise both rules.

Koko Eating Bananas

Find the smallest eating speed whose total required hours fits the allowed limit; the feasibility test is monotonic.

Practise Koko Eating Bananas on thita

JAVA
import java.util.*;

class Solution {
    public int minEatingSpeed(int[] piles, int h) {
        if (piles.length == 0) return 0;

        int low = 1;
        int high = 0;
        for (int pile : piles) high = Math.max(high, pile);
        if (high == 0) return 0;

        while (low < high) {
            int mid = low + (high - low) / 2;
            long hours = 0;
            for (int pile : piles) {
                hours += (pile + (long) mid - 1) / mid;
            }
            if (hours <= h) high = mid;
            else low = mid + 1;
        }
        return low;
    }
}

The cast before the addition prevents the ceiling-division expression from overflowing. The empty-input guard also avoids returning an unexplained speed for an input with no piles. For n piles and maximum pile size M, the time complexity is O(n log M), and the space complexity is O(1).

For fixed lowercase input, c - 'a' converts a character into an integer offset suitable for an int[26] frequency table. It is a Java-specific detail worth saying aloud: the subtraction is character arithmetic, and the result is an integer index. If a problem permits a wider character set, use a map or another representation instead of assuming this alphabet.

Valid Anagram

Compare two strings by adding the characters from one to a fixed frequency table and removing the characters from the other.

Practise Valid Anagram on thita

JAVA
import java.util.*;

class Solution {
    public boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) return false;

        int[] count = new int[26];
        for (int i = 0; i < s.length(); i++) {
            count[s.charAt(i) - 'a']++;
            count[t.charAt(i) - 'a']--;
        }
        for (int value : count) {
            if (value != 0) return false;
        }
        return true;
    }
}

This handles empty strings and repeated characters. Its time complexity is O(n), and its auxiliary space is O(1) because the table has a fixed alphabet size. The implementation intentionally relies on the input being restricted to a through z; a different input contract requires a different counting structure.

Why does StringBuilder matter in Java string problems?

A Java String is immutable. Repeated concatenation in a loop can create temporary strings, while StringBuilder provides a mutable buffer with explicit append operations. It is particularly useful when the encoding includes a delimiter or length prefix, because the construction and parsing rules remain visible.

Encode and Decode Strings

Write each string as its length, a separator and its contents, then read those length-prefixed records from left to right.

Practise Encode and Decode Strings on thita

JAVA
import java.util.*;

class Solution {
    public String encode(List<String> strs) {
        StringBuilder result = new StringBuilder();
        for (String value : strs) {
            result.append(value.length()).append(':').append(value);
        }
        return result.toString();
    }

    public List<String> decode(String encoded) {
        List<String> result = new ArrayList<>();
        int index = 0;
        while (index < encoded.length()) {
            int separator = encoded.indexOf(':', index);
            int length = Integer.parseInt(encoded.substring(index, separator));
            int start = separator + 1;
            result.add(encoded.substring(start, start + length));
            index = start + length;
        }
        return result;
    }
}

An empty list encodes to an empty string and decodes back to an empty list. Because the decoder uses the recorded length, a string can contain the separator without breaking the format. If T is the total number of input characters, encoding and decoding take O(T) time; the representation and returned values use O(T) space.

How should you sort int[][] intervals in Java?

A primitive int[] cannot receive a comparator, but an int[][] is an array of int[] references. That means Arrays.sort(intervals, comparator) can order its rows. Compare the first endpoint with Integer.compare; use the second endpoint as a tie-breaker when it makes the ordering explicit.

Merge Intervals

Sort intervals by their start, then extend the current interval while the next interval overlaps it.

Practise Merge Intervals on thita

JAVA
import java.util.*;

class Solution {
    public int[][] merge(int[][] intervals) {
        if (intervals.length < 2) return intervals;

        Arrays.sort(intervals, (a, b) -> {
            int byStart = Integer.compare(a[0], b[0]);
            return byStart != 0 ? byStart : Integer.compare(a[1], b[1]);
        });

        List<int[]> merged = new ArrayList<>();
        int start = intervals[0][0];
        int end = intervals[0][1];

        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] <= end) {
                end = Math.max(end, intervals[i][1]);
            } else {
                merged.add(new int[] {start, end});
                start = intervals[i][0];
                end = intervals[i][1];
            }
        }
        merged.add(new int[] {start, end});
        return merged.toArray(new int[merged.size()][]);
    }
}

The method returns an empty two-dimensional array unchanged and merges overlapping or touching intervals according to the usual interval condition. Duplicate intervals do not create duplicate output ranges. The time complexity is O(n log n) for sorting, and the space complexity is O(n) for the returned result and sort-related storage.

Which NeetCode topics are easier or harder in Java?

Arrays, hashing, two pointers and sliding windows become comfortable once HashMap, HashSet, Arrays and primitive arrays are familiar. Java’s static types make the state explicit, and int[] is a direct choice for fixed alphabets or numeric tables. The main risk is not the pattern; it is accidentally choosing a boxed or nullable representation when a primitive array would make the invariant clearer.

Linked-list problems are conceptually small but pointer updates must be ordered carefully. Save the next node before changing a link, and use a dummy node when it removes a special case at the head. Heaps and intervals are similarly approachable, but Java makes the ordering decision visible through a comparator. State what the heap root represents before calling offer or poll.

Trees and graphs can be harder because Java gives you several valid representations. An adjacency list might be a List<List<Integer>> when vertices are dense and indexed, or a Map<Integer, List<Integer>> when keys are naturally sparse. computeIfAbsent makes the latter concise, but you still need to explain whether a missing key means an empty neighbour list or an invalid vertex.

Backtracking is less about a special Java library and more about object lifetimes. Add a choice, recurse, then undo exactly that choice. When a completed path is added to the answer, copy it rather than storing a list that will continue to change. For deep trees or graphs, recursion can be less suitable than an explicit ArrayDeque because Java’s call stack has finite depth.

Dynamic programming is also mostly a modelling challenge, but numeric range matters. A table of int values is not automatically safe because every input fits in an int; sums, products and counts can grow beyond that range. Decide the cell type from the result bound, not only from the input type.

What Java-specific bugs should you rehearse?

Writing a - b in a comparator and allowing overflow to corrupt ordering.
Calculating a binary-search midpoint with left + right.
Letting an int sum overflow before assigning it to a long; promote an operand before the operation.
Calling queue.add(null) or queue.offer(null) on an ArrayDeque.
Forgetting that PriorityQueue is a min-heap by default.
Comparing strings with == instead of .equals.
Mutating a list while iterating over it with a for-each loop.
Using string concatenation in a large construction loop instead of StringBuilder.
Saving a mutable backtracking path without copying it into the result.
Recursing through an input whose depth may be large without considering an iterative traversal.
Relying on Integer object identity instead of value comparison.

These checks make a useful final pass before you describe a solution. They connect the abstract algorithm to Java’s actual types and library contracts, which is exactly where a language-specific implementation can diverge from language-neutral pseudocode.

How can you practise the NeetCode 150 in Java on thita?

Open the NeetCode 150 by pattern, choose a problem that is on thita and select Java in the browser editor. Run the code against the problem’s test cases. When a case fails, treat the failure as evidence about an invariant, boundary or type choice rather than immediately replacing the implementation. Re-run after each targeted fix.

After the solution works, rehearse the interview version: state the approach, name the invariant, justify the collection, give the time and space complexity, and discuss an edge case. Then practise the same problem in an AI mock interview, where the session is spoken, runs your code and asks follow-up questions. This is a useful test of whether you can explain why a comparator is safe, why a total needs long or why an iterative traversal is preferable to recursion.

The aim of NeetCode 150 in Java practice is not to memorise a file for every problem. It is to make the Java decisions automatic enough that you can focus on the pattern, test the edge cases and communicate your reasoning clearly.

The NeetCode 150 in Java: How to Work Through the List on thita