mirror of
https://github.com/prdlk/leetcode.git
synced 2026-09-17 07:26:27 +00:00
docs(docs): delete obsolete Leetcode documentation files
This commit is contained in:
Prad Nukala
@@ -1,55 +0,0 @@
|
|||||||
---
|
|
||||||
title: LC-1365
|
|
||||||
description: How Many Numbers Are Smaller Than the Current Number
|
|
||||||
---
|
|
||||||
|
|
||||||
## Examples
|
|
||||||
|
|
||||||
| Input | Output |
|
|
||||||
| ---------------- | -------------------- |
|
|
||||||
| `[8,1,2,2,3]` | `[4,0,1,1,3]` |
|
|
||||||
| `[6,5,4,8]` | `[2,1,0,3]` |
|
|
||||||
| `[7,7,7,7]` | `[0,0,0,0]` |
|
|
||||||
|
|
||||||
## Hints
|
|
||||||
|
|
||||||
<Accordion type="single">
|
|
||||||
<AccordionItem title="Constraints">
|
|
||||||
Constraints:
|
|
||||||
• `2 <= nums.length <= 500`
|
|
||||||
• `0 <= nums[i] <= 100`
|
|
||||||
</AccordionItem>
|
|
||||||
<AccordionItem title="Problem Statement">
|
|
||||||
Given the array nums, for each `nums[i]` find out how many numbers in the array are smaller than it.
|
|
||||||
That is, for each `nums[i]` you have to count the number of valid `j`'s such that `j != i` and `nums[j] < nums[i]`.
|
|
||||||
</AccordionItem>
|
|
||||||
</Accordion>
|
|
||||||
|
|
||||||
## Solution
|
|
||||||
|
|
||||||
```js
|
|
||||||
/**
|
|
||||||
* @param {number[]} nums
|
|
||||||
* @return {number[]}
|
|
||||||
*/
|
|
||||||
|
|
||||||
var smallerNumbersThanCurrent = function (nums) {
|
|
||||||
// Default for Hammer
|
|
||||||
const freq = {};
|
|
||||||
for (let n of nums) freq[n] = (freq[n] || 0) + 1;
|
|
||||||
const sorted = Object.keys(freq).sort((a, b) => a - b);
|
|
||||||
|
|
||||||
// Problem Specific
|
|
||||||
let count = 0;
|
|
||||||
let smaller = {};
|
|
||||||
|
|
||||||
// Iterate over sorted.
|
|
||||||
for (let num of sorted) {
|
|
||||||
smaller[num] = count;
|
|
||||||
count += freq[num];
|
|
||||||
}
|
|
||||||
|
|
||||||
// Map output
|
|
||||||
return nums.map((n) => smaller[n]);
|
|
||||||
};
|
|
||||||
```
|
|
||||||
@@ -1,46 +0,0 @@
|
|||||||
---
|
|
||||||
title: LC-1636
|
|
||||||
description: Sort Array by Increasing Frequency
|
|
||||||
---
|
|
||||||
|
|
||||||
## Examples
|
|
||||||
|
|
||||||
| Input | Output |
|
|
||||||
| ---------------------- | ---------------------- |
|
|
||||||
| `[1,1,2,2,2,3]` | `[3,1,1,2,2,2]` |
|
|
||||||
| `[2,3,1,3,2]` | `[1,3,3,2,2]` |
|
|
||||||
| `[-1,1,-6,4,5,-6,1,4,1]` | `[5,-1,4,4,-6,-6,1,1,1]` |
|
|
||||||
|
|
||||||
## Hints
|
|
||||||
|
|
||||||
<Accordion type="single">
|
|
||||||
<AccordionItem title="Constraints">
|
|
||||||
Constraints:
|
|
||||||
• `1 <= nums.length <= 100`
|
|
||||||
• `-100 <= nums[i] <= 100`
|
|
||||||
</AccordionItem>
|
|
||||||
<AccordionItem title="Problem Statement">
|
|
||||||
Given an array of integers nums, sort the array in increasing order based on the frequency of the values.
|
|
||||||
If multiple values have the same frequency, sort them in decreasing order.
|
|
||||||
Return the sorted array.
|
|
||||||
</AccordionItem>
|
|
||||||
</Accordion>
|
|
||||||
|
|
||||||
## Solution
|
|
||||||
|
|
||||||
```js
|
|
||||||
/**
|
|
||||||
* @param {number[]} nums
|
|
||||||
* @return {number[]}
|
|
||||||
*/
|
|
||||||
|
|
||||||
var frequencySort = function (nums) {
|
|
||||||
// Default for Hammer
|
|
||||||
const freq = {};
|
|
||||||
for (let n of nums) freq[n] = (freq[n] || 0) + 1;
|
|
||||||
|
|
||||||
// Problem Specific
|
|
||||||
// Sort ascending by frequency; ties break by larger value first
|
|
||||||
return nums.sort((a, b) => freq[a] - freq[b] || b - a);
|
|
||||||
};
|
|
||||||
```
|
|
||||||
@@ -1,53 +0,0 @@
|
|||||||
---
|
|
||||||
title: LC-347
|
|
||||||
description: Top K Frequent Elements
|
|
||||||
---
|
|
||||||
|
|
||||||
## Examples
|
|
||||||
|
|
||||||
| Input | Output |
|
|
||||||
| ---------------------------------- | -------- |
|
|
||||||
| `nums = [1,1,1,2,2,3], k = 2` | `[1,2]` |
|
|
||||||
| `nums = [1], k = 1` | `[1]` |
|
|
||||||
| `nums = [1,2,1,2,1,2,3,1,3,2], k = 2` | `[1,2]` |
|
|
||||||
|
|
||||||
## Hints
|
|
||||||
|
|
||||||
<Accordion type="single">
|
|
||||||
<AccordionItem title="Constraints">
|
|
||||||
Constraints:
|
|
||||||
• `1 <= nums.length <= 10^5`
|
|
||||||
• `-10^4 <= nums[i] <= 10^4`
|
|
||||||
• `k` is in the range `[1, the number of unique elements in the array]`
|
|
||||||
• It is guaranteed that the answer is unique.
|
|
||||||
</AccordionItem>
|
|
||||||
<AccordionItem title="Problem Statement">
|
|
||||||
Given an integer array nums and an integer `k`, return the `k` most frequent elements.
|
|
||||||
You may return the answer in any order.
|
|
||||||
|
|
||||||
Follow up: Your algorithm's time complexity must be better than `O(n log n)`, where n is the array's size.
|
|
||||||
</AccordionItem>
|
|
||||||
</Accordion>
|
|
||||||
|
|
||||||
## Solution
|
|
||||||
|
|
||||||
```js
|
|
||||||
/**
|
|
||||||
* @param {number[]} nums
|
|
||||||
* @param {number} k
|
|
||||||
* @return {number[]}
|
|
||||||
*/
|
|
||||||
|
|
||||||
var topKFrequent = function (nums, k) {
|
|
||||||
// Default for Hammer
|
|
||||||
const freq = {};
|
|
||||||
for (let n of nums) freq[n] = (freq[n] || 0) + 1;
|
|
||||||
|
|
||||||
// Problem Specific
|
|
||||||
// Sort unique values by descending frequency, take the first k
|
|
||||||
return Object.keys(freq)
|
|
||||||
.map(Number)
|
|
||||||
.sort((a, b) => freq[b] - freq[a])
|
|
||||||
.slice(0, k);
|
|
||||||
};
|
|
||||||
```
|
|
||||||
@@ -1,50 +0,0 @@
|
|||||||
---
|
|
||||||
title: LC-387
|
|
||||||
description: First Unique Character in a String
|
|
||||||
---
|
|
||||||
|
|
||||||
## Examples
|
|
||||||
|
|
||||||
| Input | Output |
|
|
||||||
| ---------------- | ------ |
|
|
||||||
| `"leetcode"` | `0` |
|
|
||||||
| `"loveleetcode"` | `2` |
|
|
||||||
| `"aabb"` | `-1` |
|
|
||||||
|
|
||||||
## Hints
|
|
||||||
|
|
||||||
<Accordion type="single">
|
|
||||||
<AccordionItem title="Constraints">
|
|
||||||
Constraints:
|
|
||||||
• `1 <= s.length <= 10^5`
|
|
||||||
• `s` consists of only lowercase English letters.
|
|
||||||
</AccordionItem>
|
|
||||||
<AccordionItem title="Problem Statement">
|
|
||||||
Given a string `s`, find the first non-repeating character in it and return its index.
|
|
||||||
If it does not exist, return `-1`.
|
|
||||||
</AccordionItem>
|
|
||||||
</Accordion>
|
|
||||||
|
|
||||||
## Solution
|
|
||||||
|
|
||||||
```js
|
|
||||||
/**
|
|
||||||
* @param {string} s
|
|
||||||
* @return {number}
|
|
||||||
*/
|
|
||||||
|
|
||||||
var firstUniqChar = function (s) {
|
|
||||||
// Default for Hammer
|
|
||||||
const freq = {};
|
|
||||||
for (let c of s) freq[c] = (freq[c] || 0) + 1;
|
|
||||||
|
|
||||||
// Problem Specific
|
|
||||||
// Loop the ORIGINAL string — we want an index, so no sorting
|
|
||||||
for (let i = 0; i < s.length; i++) {
|
|
||||||
if (freq[s[i]] === 1) {
|
|
||||||
return i;
|
|
||||||
}
|
|
||||||
}
|
|
||||||
return -1;
|
|
||||||
};
|
|
||||||
```
|
|
||||||
@@ -1,49 +0,0 @@
|
|||||||
---
|
|
||||||
title: LC-451
|
|
||||||
description: Sort Characters By Frequency
|
|
||||||
---
|
|
||||||
|
|
||||||
## Examples
|
|
||||||
|
|
||||||
| Input | Output |
|
|
||||||
| ---------- | -------- |
|
|
||||||
| `"tree"` | `"eert"` |
|
|
||||||
| `"cccaaa"` | `"aaaccc"` |
|
|
||||||
| `"Aabb"` | `"bbAa"` |
|
|
||||||
|
|
||||||
## Hints
|
|
||||||
|
|
||||||
<Accordion type="single">
|
|
||||||
<AccordionItem title="Constraints">
|
|
||||||
Constraints:
|
|
||||||
• `1 <= s.length <= 5 * 10^5`
|
|
||||||
• `s` consists of uppercase and lowercase English letters and digits.
|
|
||||||
</AccordionItem >
|
|
||||||
<AccordionItem title="Problem Statement">
|
|
||||||
Given a string `s`, sort it in decreasing order based on the frequency of the characters.
|
|
||||||
The frequency of a character is the number of times it appears in the string.
|
|
||||||
Return the sorted string. If there are multiple answers, return any of them.
|
|
||||||
</AccordionItem >
|
|
||||||
</Accordion>
|
|
||||||
|
|
||||||
## Solution
|
|
||||||
|
|
||||||
```js
|
|
||||||
/**
|
|
||||||
* @param {string} s
|
|
||||||
* @return {string}
|
|
||||||
*/
|
|
||||||
|
|
||||||
var frequencySort = function (s) {
|
|
||||||
// Default for Hammer
|
|
||||||
const freq = {};
|
|
||||||
for (let c of s) freq[c] = (freq[c] || 0) + 1;
|
|
||||||
|
|
||||||
// Problem Specific
|
|
||||||
// Sort characters by descending frequency; same chars stay grouped
|
|
||||||
return s
|
|
||||||
.split("")
|
|
||||||
.sort((a, b) => freq[b] - freq[a] || a.localeCompare(b))
|
|
||||||
.join("");
|
|
||||||
};
|
|
||||||
```
|
|
||||||
@@ -1,68 +0,0 @@
|
|||||||
---
|
|
||||||
title: Introduction
|
|
||||||
description: Prad's Random Leetcode Notes
|
|
||||||
---
|
|
||||||
|
|
||||||
## Overview
|
|
||||||
|
|
||||||
This is a collection of random CS notes that I've made over the years. I've been learning CS for a while now and I thought it would be a good idea to make a public repository of my notes. This is the first iteration of the repository and I plan to add more content to it over time.
|
|
||||||
|
|
||||||
## The Golden Rule (Memorize This)
|
|
||||||
|
|
||||||
**Only sort if the problem asks for ORDER or RANKING.**
|
|
||||||
|
|
||||||
If the problem asks for:
|
|
||||||
- "first"
|
|
||||||
- "index"
|
|
||||||
- "position"
|
|
||||||
- "does it exist"
|
|
||||||
- "true/false"
|
|
||||||
- "count how many"
|
|
||||||
|
|
||||||
→ **DO NOT SORT**. Just use the frequency map and loop.
|
|
||||||
|
|
||||||
---
|
|
||||||
|
|
||||||
## Decision Cheat Sheet
|
|
||||||
|
|
||||||
| What the problem wants | Do you need to sort? | What to do instead | Example Problem |
|
|
||||||
|-----------------------------------------|----------------------|-------------------------------------|---------------------|
|
|
||||||
| Sort characters by how often they appear | YES | Freq + Sort | LC 451 |
|
|
||||||
| Top K most frequent | YES | Freq + Sort + slice | LC 347 |
|
|
||||||
| Sort array by frequency | YES | Freq + Sort | LC 1636 |
|
|
||||||
| First unique character (index) | NO | Freq + loop original string | LC 387 |
|
|
||||||
| Is it an anagram? | NO | Freq + check counts | LC 242 |
|
|
||||||
| How many numbers smaller than current | Kind of | Freq + prefix count | LC 1365 |
|
|
||||||
| Most frequent element | NO | Freq + find max | LC 169 |
|
|
||||||
|
|
||||||
---
|
|
||||||
|
|
||||||
## Mental Checklist (Say this out loud every time)
|
|
||||||
|
|
||||||
1. "Did I count the frequencies?" → Yes
|
|
||||||
2. "Does the problem care about the **original order**?"
|
|
||||||
- Yes → Loop the original string/array and check freq
|
|
||||||
- No → You can sort
|
|
||||||
3. "Does it want an **index** or a **boolean**?"
|
|
||||||
- Yes → Almost never sort
|
|
||||||
4. "Does it want the elements **re-ordered**?"
|
|
||||||
- Yes → Sort by frequency
|
|
||||||
|
|
||||||
---
|
|
||||||
|
|
||||||
## Super Short Version to Tattoo in Your Brain
|
|
||||||
|
|
||||||
**"If it wants the first / index / true-false → loop original.
|
|
||||||
If it wants sorted / top k / reordered → sort."**
|
|
||||||
|
|
||||||
---
|
|
||||||
|
|
||||||
## Practice Trigger Words
|
|
||||||
|
|
||||||
- "first non-repeating" → Loop original
|
|
||||||
- "return its index" → Loop original
|
|
||||||
- "sort by frequency" → Sort
|
|
||||||
- "top k frequent" → Sort + slice
|
|
||||||
- "valid anagram" → Just check counts
|
|
||||||
|
|
||||||
|
|
||||||
@@ -1,2 +0,0 @@
|
|||||||
.blume/
|
|
||||||
dist/
|
|
||||||
@@ -1,11 +0,0 @@
|
|||||||
---
|
|
||||||
title: Guides
|
|
||||||
description: News and writing from the team.
|
|
||||||
type: blog
|
|
||||||
---
|
|
||||||
|
|
||||||
<CardGroup cols={2}>
|
|
||||||
<Card title="Introducing Blume" href="/blog/introducing-blume">
|
|
||||||
Why we built a markdown-first docs framework.
|
|
||||||
</Card>
|
|
||||||
</CardGroup>
|
|
||||||
@@ -1,8 +0,0 @@
|
|||||||
---
|
|
||||||
title: Introducing Blume
|
|
||||||
type: blog
|
|
||||||
date: 2026-06-22
|
|
||||||
description: Why we built a markdown-first docs framework.
|
|
||||||
---
|
|
||||||
|
|
||||||
Documentation should be fast, AI-ready, and zero-config — down to not needing a starter template at all. Here's the thinking behind Blume.
|
|
||||||
Reference in New Issue
Block a user