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chore(notes): remove 1365 notes and 451.sort-characters-by-frequency.js file
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Prad Nukala
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/*
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* 451. Sort Characters By Frequency
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* Difficulty: Medium
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* https://leetcode.com/problems/sort-characters-by-frequency/
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*
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* ──────────────────────────────────────────────────
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*
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* Given a string s, sort it in decreasing order based on the frequency
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* of the characters. The frequency of a character is the number of times
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* it appears in the string.
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*
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* Return the sorted string. If there are multiple answers, return any
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* of them.
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*
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*
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*
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* Example 1:
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*
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* Input: s = "tree"
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* Output: "eert"
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* Explanation: 'e' appears twice while 'r' and 't' both appear once.
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* So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also
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* a valid answer.
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*
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* Example 2:
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*
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* Input: s = "cccaaa"
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* Output: "aaaccc"
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* Explanation: Both 'c' and 'a' appear three times, so both "cccaaa"
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* and "aaaccc" are valid answers.
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* Note that "cacaca" is incorrect, as the same characters must be
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* together.
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*
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* Example 3:
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*
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* Input: s = "Aabb"
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* Output: "bbAa"
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* Explanation: "bbaA" is also a valid answer, but "Aabb" is incorrect.
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* Note that 'A' and 'a' are treated as two different characters.
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*
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*
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*
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* Constraints:
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*
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* • 1 <= s.length <= 5 * 10^5
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*
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* • s consists of uppercase and lowercase English letters and digits.
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*/
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/**
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* @param {string} s
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* @return {string}
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*/
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var frequencySort = function (s) {
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const freq = {};
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for (let c of s) freq[c] = (freq[c] || 0) + 1;
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return s
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.split("")
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.sort((a, b) => freq[b] - freq[a] || a.localeCompare(b))
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.join("");
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};
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@@ -1,150 +0,0 @@
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# 1365. How Many Numbers Are Smaller Than the Current Number
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**Difficulty:** Easy
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**URL:** https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/
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**Topics:** Array, Hash Table, Sorting, Counting Sort
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---
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## The One Insight That Makes This Work
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> If I know how many times each value appears, and I add those counts up from left to right, then `prefix[v]` tells me **how many elements are ≤ v** — instantly, for any v.
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That's it. Everything below is just executing this idea.
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---
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## Step 1: Spot the Signal
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Read the constraints:
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```
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0 <= nums[i] <= 100
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```
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Values are bounded to a tiny range (0–100). This is the **flashing neon sign** that says: don't sort, don't nest loops — build a frequency array indexed by value.
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**Rule of thumb:** value range ≤ ~10⁶ and you need counting/ranking? Frequency array.
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---
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## Step 2: Count Every Value (the "bucket" pass)
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Make an array with one slot per possible value. Walk the input once. Each number votes for its own slot.
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```javascript
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const freq = new Array(101).fill(0); // slots for values 0..100
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for (const x of nums) freq[x]++;
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```
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For `nums = [8, 1, 2, 2, 3]`:
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```
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value: 0 1 2 3 4 5 6 7 8 ...
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freq: 0 1 2 1 0 0 0 0 1 ...
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```
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Read it as: "one 1, two 2s, one 3, one 8."
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---
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## Step 3: Prefix Sum (the magic pass)
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Now transform `freq` in place: each slot becomes itself **plus everything before it**.
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```javascript
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for (let i = 1; i < 101; i++) freq[i] += freq[i - 1];
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```
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Same example after the pass:
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```
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value: 0 1 2 3 4 5 6 7 8 ...
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freq: 0 1 3 4 4 4 4 4 5 ...
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```
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New meaning: `freq[v]` = **count of elements ≤ v**.
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- `freq[3] = 4` → four numbers are ≤ 3 (they are 1, 2, 2, 3) ✓
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- `freq[7] = 4` → still four numbers ≤ 7 ✓
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---
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## Step 4: Answer Queries in O(1)
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"How many numbers are **strictly smaller** than x?" is the same question as "how many numbers are **≤ x − 1**?"
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```javascript
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return nums.map((x) => (x === 0 ? 0 : freq[x - 1]));
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```
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The `x === 0` guard exists because nothing can be smaller than the minimum possible value — and `freq[-1]` would be `undefined`.
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Trace on `[8, 1, 2, 2, 3]`:
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| x | lookup | answer |
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|---|--------|--------|
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| 8 | freq[7] | 4 |
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| 1 | freq[0] | 0 |
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| 2 | freq[1] | 1 |
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| 2 | freq[1] | 1 |
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| 3 | freq[2] | 3 |
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→ `[4, 0, 1, 1, 3]` ✓
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---
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## Full Solution
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```javascript
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var smallerNumbersThanCurrent = function (nums) {
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// 1. Bucket counts
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const freq = new Array(101).fill(0);
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for (const x of nums) freq[x]++;
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// 2. Prefix sum: freq[v] = count of elements <= v
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for (let i = 1; i < 101; i++) freq[i] += freq[i - 1];
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// 3. Strictly smaller than x == count of elements <= x-1
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return nums.map((x) => (x === 0 ? 0 : freq[x - 1]));
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};
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```
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**Complexity:** O(n + k) time, O(k) space, where k = value range (101 here). No sort, no log factor.
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---
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## The Reusable Pattern (memorize this shape)
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```
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1. BUCKET — freq[value]++ for every element
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2. PREFIX — freq[i] += freq[i-1] left to right
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3. QUERY — freq[v] answers "how many ≤ v" in O(1)
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freq[v-1] answers "how many < v"
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n - freq[v] answers "how many > v"
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```
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### Where else this exact shape shows up
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| Problem | Same pattern, different query |
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|---|---|
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| **Counting Sort** | Prefix sums become final sorted positions |
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| **LC 315 / rank queries** | "How many smaller" is literally a rank |
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| **LC 1122 Relative Sort Array** | Bucket + walk buckets in order |
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| **Radix sort digit pass** | Bucket by digit, prefix for placement |
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| **Histogram percentiles** | freq[v] / n = percentile of v |
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| **"How many in range [a, b]?"** | freq[b] − freq[a−1] — the prefix subtraction trick |
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### The generalization ladder
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- Values bounded and small → **frequency array** (this pattern)
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- Values huge but few distinct → **coordinate compression** first, then this pattern
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- Need updates between queries → upgrade prefix array to a **Fenwick tree (BIT)** — same idea, log-time updates
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---
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## Common Mistakes
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1. **Returning `freq[x]` instead of `freq[x-1]`** — that counts elements ≤ x (including x itself and its duplicates). Off-by-one between "≤" and "<" is where this pattern bites.
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2. **Forgetting the `x === 0` edge** — smallest possible value has nothing below it.
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3. **Sizing the array to `nums.length` instead of the value range** — the buckets are indexed by *value*, not position.
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