Hashing Fundamentals
What is this?
This is where you learn the three everyday jobs a hash structure does. Use a set to check "have I seen this before?", a Counter to answer "how many of each do I have?", and a canonical key to lump together things that are secretly the same (like words that are anagrams). Each one replaces a slow, item-by-item comparison with one quick pass.
💡 Fun fact: The "group things that are secretly the same" trick has a famous twist — to detect anagrams you can multiply a prime number per letter, since every word maps to a unique product. This is essentially the Fundamental Theorem of Arithmetic, a 2000-year-old idea from Euclid, repurposed as a hash key.
🔓 The 5 problems in this chapter are free. Sign in with Google or Microsoft to start solving.
Core idea: Before the clever "what to store" tricks, master the three everyday uses of a hash structure: a set to test distinctness, a Counter to tally "how many of each", and a canonical key to group items that are secretly the same. Each turns an O(n²) scan into one O(n) pass.
The three reflexes
| Reflex | Trigger phrase | Tool |
|---|---|---|
| Set | "distinct / unique / seen" | set() |
| Frequency | "how many of each / most common / counts" | collections.Counter |
| Canonical key | "group the ones that are the same under …" | dict keyed by a normalized form |
The five problems
- Distribute Candies — a set gives the distinct-type count; answer
min(distinct, n//2). - Minimum Steps to Make Anagram — subtract two frequency tables; the deficit is the answer.
- Bulls and Cows — bulls by position, then count the leftover digits and sum
minof the two tallies for cows. - Longest Harmonious Subsequence — count values; for each
v, combinecount[v] + count[v+1]. - Group Shifted Strings — build a canonical key (mod-26 difference tuple) and group by it.
The pattern
Counting collapses "compare every element to every other" into "tally once, then read the tallies." A set answers membership in O(1). And when items are equivalent under some transformation (a shift, a sort, a rotation), compute a canonical form and let the map group them — the same trick behind Group Anagrams.
📓 Draw it yourself
- Two tally tables. For Bulls and Cows or the anagram problem, draw both frequency tables and read the answer off the per-letter differences.
- Canonical key. For Group Shifted Strings, write a few strings and their difference-tuple keys; watch equivalent ones land in the same bucket.
Snap photos and embed them with the /host-diagrams skill.
Key takeaways
- Three reflexes: set (distinct), Counter (how many), canonical key (group equivalents).
- Counting beats nested comparison — tally once in O(n), then answer questions off the tallies.
- A canonical key turns "are these equivalent?" into "do they hash the same?"
- Why it matters: these are the bread-and-butter hash moves every harder problem composes from.
Order: Distribute Candies → Min Steps to Make Anagram → Bulls and Cows → Longest Harmonious Subsequence → Group Shifted Strings.
Core idea: A harmonious subsequence is built from exactly two adjacent values
vandv+1, so you never reason about elements one at a time — you just count how often each value appears and, for everyv, addcount[v] + count[v+1].
Problem, rephrased
LeetCode 594. We call an array harmonious when the difference between its largest and smallest element is exactly 1 — not zero, not two, exactly one. Given an integer array, return the length of the longest harmonious subsequence.
A subsequence keeps the original order but lets you skip elements — they need not be contiguous. The practical consequence: if you decide your harmonious subsequence will be made of the values v and v+1, the best you can do is grab every occurrence of v and every occurrence of v+1. There's no reason to leave any behind, and you can't include any third value without breaking the "max − min = 1" rule.
So the whole problem collapses to: pick the pair of adjacent values whose combined occurrence count is largest.
| Input | Output | Why |
|---|---|---|
[1,3,2,2,5,2,3,7] |
5 |
Pick values 2 and 3: three 2s + two 3s = 5 → subsequence [2,2,2,3,3] |
[1,2,3,4] |
2 |
Every adjacent pair (1,2), (2,3), (3,4) gives 1+1 = 2 |
[1,1,1,1] |
0 |
Only one distinct value; no v has its neighbor v+1 present → no harmonious pair |
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