Master Hashing for Coding Interviews: Top LeetCode Problems Explained

 Master Hashing for Coding Interviews: Top LeetCode Problems Explained

⏱️ Estimated reading time: 12 minutes

Data is transformed with a hash function to produce a substantial fixed-size value (or representation of the data). This makes access, insert, and deleting data using hashing very efficient compared with traditional data access methods. Hashing is one of the most widely used data techniques around; and gives applications the ability to handle increasing amounts of data (thus scalability), and supports speed by reducing the time complexity of access of data from linear (or O(n)) to constant time (or O(1)). By providing an index of digitally based storage, hash tables are used by virtually all modern technologies, from sharing passwords to creating databases, to providing a caching, search engine, and many other applications including blockchain systems (whereby every block's content can be found with a simple hash).

🟢 EASY (Foundational + Pattern Building)

I.               Two Sum






🧠 Explanation

To use a hash map to keep track of an array’s numbers and their respective indexes as you traverse through the array. At every element, a hash map lookup is performed to check if there is already an entry in the table; namely, a key whose value equals (target - current number). If a match is found, the pair will be found immediately without entering additional loops.

⚙️ Logic in Steps

1.     Create an empty hashmap `num_map`.

2.     Iterate through array and for each item do the following:

i.               Calculate complement = target – num

ii.              Check if complement is already in the map; if it is then return indices

iii.            If not stored the current number along with its index

3.     Return result when pair is found.

⏱ Time and Space Complexity

Time Complexity O(n): Single pass through array

Space Complexity O(n): Hashmap stores elements

📝 Key Takeaways

1.     Hashing reduces brute force O(n²) to O(n) 

2.     Complement concept is key insight 

3.     One of the most asked interview problems

II.               Contains Duplicate






🧠 Explanation

To find duplicate values we first sort the list so that all duplicates will be positioned together and then check the sorted list for correspond to duplicates. Duplicates will occur in pairs because there would be two equal values that occur next to each other when the list is scanned.

⚙️ Logic in Steps

1.     You must sort array nums, and then compare some of the values:

2.     Compare nums[i] to nums[i+1] for i = 0 to n – 2

3.     If they are equal to each other, then you have found duplicates.

4.     If you found duplicates, then return True; otherwise, return False.

⏱ Time and Space Complexity

Time Complexity O(n log n): Due to sorting

Space Complexity O(1): No extra space used

📝 Key Takeaways

1.     Sorting helps bring duplicates together 

2.     Simple and intuitive approach 

3.     Not optimal, hashing can solve it in O(n) time

III.               Valid Anagram






🧠 Explanation

A program that determines if two string inputs are anagrams uses a hash table that counts each of the strings' character occurrences to compare the overall count for each string. If both strings contain the same overall number of occurrences for each character, then they are considered to be anagrams.

⚙️ Logic in Steps

1.     Create frequency maps using `Counter` for both strings 

2.     Compare the two frequency maps 

3.     If equal, return `True`, else `False`

⏱ Time and Space Complexity

Time Complexity O(n): Traverse both strings once

Space Complexity O(k): Store character frequencies

📝 Key Takeaways

1.     Anagrams have identical character frequencies 

2.     Hashing simplifies comparison 

3.     Clean and efficient approach using built-in tools

IV.               Intersection of Two Arrays






🧠 Explanation

Using a set to get the intersection between two arrays can be accomplished by first converting both arrays into sets. This will eliminate any duplicates in the original data set, and the intersection of the two sets will provide all unique and common elements between both arrays.

⚙️ Logic in Steps

1.     Make two sets called `set_1` and `set_2` from each of the two arrays. 

2.     Get the intersection between them by using `set_1 & set_2`. 

3.     Convert your intersection into a list. 

4.     Return that final list.

⏱ Time and Space Complexity

Time Complexity O(n + m): Traversing both arrays

Space Complexity O(n + m): Storing elements in sets

📝 Key Takeaways

1.     Sets automatically remove duplicates 

2.     Built-in set operations make solution concise 

3.     Efficient way to find unique common elements

🟡 Medium (Interview-Level Skills)

V.               Longest Substring Without Repeating Characters






🧠 Explanation

We use a sliding shaft and a hash table to keep track of the unique substring. Each time a duplicate character is found by enlarging the window of right-side pointer, we will contract the need as far as required until all characters in window are unique. The largest window size recorded in the above process would be the answer to this problem.

⚙️ Logic in Steps

1.     Start out with `char_set`, `left = 0` and `max_len = 0` 

2.     Loop through `right` pointer: 

i.               If a duplicate is found then remove character at s[left] from character set and increment left pointer

ii.              Add the current character to the set

iii.            Set max_len = the larger of max_len or (right – left + 1)

3.     Return the max_len value.

⏱ Time and Space Complexity

Time Complexity O(n): Each character processed once

Space Complexity O(k): Set stores unique characters

📝 Key Takeaways

1.     Sliding window efficiently avoids brute force 

2.     Hash set enables O(1) duplicate checks 

3.     Core pattern for many substring problems

VI.               Group Anagrams






🧠 Explanation

The way to use this solution "create an anagram group" is through a hash map-based approach. A hash map allows for storing words as keys through their alphanumeric characters in an unsorted manner. Because all anagrams are composed of exactly the same characters or letters, anagrams will have the same sorted key and therefore will fall into the same bucket or grouping in an algorithm-designed hash map.

⚙️ Logic in Steps

1.     Using `defaultdict(list)`, create a hashmap called `groups`.  For all words in the list: 

i.               Create an ordered result for that word and make it the key.

ii.              Add that word (original form) to grouped values with that key (`groups[key]`).

2.     Return all grouped values (from groups) as a list.

⏱ Time and Space Complexity

Time Complexity O(n · k log k): Sorting each word of length k

Space Complexity O(n · k): Storing grouped anagrams

📝 Key Takeaways

1.     Anagrams have identical sorted forms 

2.     Hashing + sorting is a powerful grouping technique 

3.     Common interview pattern for string grouping

VII.               Subarray Sum Equals K






🧠 Explanation

Using both a hash map and prefix sum allows this solution to count the number of subarrays that total to k without having to check the value of every single subarray. Rather, it keeps a running total of all cumulative sums and uses previously encountered cumulative sums to determine quickly if a valid subarray exists. If prefix_sum - k is found in the hash map then that means there is a subarray ending at the current index with a value of k.

⚙️ Logic in Steps

1.     Set initial values `prefix_sum` to 0, `count` to 0, and hashmap `freq` equal to `{0:1}`.

2.     Loop through array by doing the following:

i.               Add current number to prefix_sum

ii.              Determine if prefix_sum − k exists in map

iii.            If it does, add its frequency to count

iv.            Update prefix_sum frequency in the map for the current prefix_sum

3.     At end, return count.

⏱ Time and Space Complexity

Time Complexity O(n): Single traversal

Space Complexity O(n): Storing prefix sums

📝 Key Takeaways

1.     Prefix sum converts subarray problem into lookup problem 

2.     Hashing enables constant-time checks 

3.     Very high-frequency interview pattern

VIII.               Top K Frequent Elements




🧠 Explanation

The way that this solution operates utilizes both a hashmap and a min-heap (heap size = k). The first of these operations is the frequency count; the second is maintaining a heap data structure of size k to store the top k most frequent elements. When calculating an element’s frequency would result in a heap containing more than k elements (heap size exceeds k), the least frequent element will be removed from the heap.

⚙️ Logic in Steps

1.     Using `Counter`, you can count frequencies 

2.     Next we need to initialize a new min-heap 

3.     We will go over the frequency map: 

i.               For each frequency pair, add (frequency -> number) to the heap

ii.              If the number of entries in the heap is greater than k, pop the smallest entry from the heap

4.     Finally, read from the heap and return results 

⏱ Time and Space Complexity

Time Complexity O(n log k): Heap operations for each element

Space Complexity O(n): Storing frequencies and heap

📝 Key Takeaways

1.     Heap helps maintain top `k` elements efficiently 

2.     Combining hashing + heap is a powerful pattern 

3.     Better than sorting entire array for large inputs

Conclusion

By converting difficult challenges into rapid searches, hashing constitutes one of the strongest resources for coding assessments both during the interview process and when developing real life applications. The problem types we've reviewed in this book share similar patterns including: counting frequencies, the sliding window approach, prefix sums, and optimising via a heap – all supporting the argument that improving your understanding of these ideas will help you solve problems faster and help you develop an instinct that will help you tackle new problems. Practicing on a regular basis will assist in making hashing less a 'topic' and more a 'reaction', which should assist you when answering questions during a technical assessment.

If you Missed the Previous Part

If you missed the previous part on linked lists, don’t worry, that foundation is already waiting for you. Go check it out to strengthen your core concepts before diving deeper into hashing

Coming Up

Up next, we dive into Prefix Sum, a technique that turns subarray problems into smooth, almost one-pass magic. Once you get it, many “hard-looking” problems suddenly feel… suspiciously easy

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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