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‎README.md

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> ["For coding interview preparation, LeetCode is one of the best online resource providing a rich library of more than 300 real coding interview questions for you to practice from using one of the 7 supported languages - C, C++, Java, Python, C#, JavaScript, Ruby."](https://www.quora.com/How-effective-is-Leetcode-for-preparing-for-technical-interviews)
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##
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* [SQL I](#sql-i)
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* [Level 1](#level-1)
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* [Level 2](#level-2)
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* [Udemy](#udemy)
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* [Programming Skills I](#programming-skills-i)
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* [Programming Skills II](#programming-skills-ii)
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* [Graph Theory I](#graph-theory-i)
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### SQL I
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#### Day 1 Select
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-|-
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#### Day 2 Select and Order
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|-|-|-|-|-|-|-
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#### Day 3 String Processing Functions
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|-|-|-|-|-|-|-
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#### Day 4 Union and Select
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|-|-|-|-|-|-|-
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#### Day 5 Union
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#### Day 6 Union
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-|-
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#### Day 7 Function
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#### Day 8 Function
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#### Day 9 Control of Flow
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#### Day 10 Where
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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* [SQL I](#sql-i)
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### Level 1
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-|-
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### SQL I
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#### Day 1 Select
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-|-
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#### Day 2 Select and Order
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-|-
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#### Day 3 String Processing Functions
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-|-
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#### Day 4 Union and Select
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-|-
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#### Day 5 Union
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-|-
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#### Day 6 Union
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-|-
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#### Day 7 Function
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-|-
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#### Day 8 Function
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-|-
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#### Day 9 Control of Flow
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-|-
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#### Day 10 Where
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-|-
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## Algorithms
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| # | Title | Language | Difficulty | Tag | Time, ms | Time, %

‎src/main/swift/g0001_0100/s0002_add_two_numbers/Solution.swift

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// #Data_Structure_II_Day_10_Linked_List #Programming_Skills_II_Day_15
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// #Big_O_Time_O(max(N,M))_Space_O(max(N,M)) #2024_06_17_Time_17_ms_(76.59%)_Space_15.7_MB_(14.57%)
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/*
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* public var val: Int
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* }
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*/
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class Solution {
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func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
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return addNumbers(l1, l2, 0)
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}

‎src/main/swift/g0001_0100/s0002_add_two_numbers/readme.md

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Here's the implementation:
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```swift
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/*
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* public var val: Int

‎src/main/swift/g0001_0100/s0004_median_of_two_sorted_arrays/Solution.swift

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return 0.0
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}
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}
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‎src/main/swift/g0001_0100/s0005_longest_palindromic_substring/Solution.swift

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return String(s[s.index(s.startIndex, offsetBy: start)..<s.index(s.startIndex, offsetBy: end)])
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}
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}
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‎src/main/swift/g0001_0100/s0019_remove_nth_node_from_end_of_list/Solution.swift

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// #Algorithm_I_Day_5_Two_Pointers #Level_2_Day_3_Linked_List #Big_O_Time_O(L)_Space_O(L)
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// #2024_06_18_Time_0_ms_(100.00%)_Space_15.5_MB_(48.03%)
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/*
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* public var val: Int

‎src/main/swift/g0001_0100/s0021_merge_two_sorted_lists/Solution.swift

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// #Level_1_Day_3_Linked_List #Udemy_Linked_List #Big_O_Time_O(m+n)_Space_O(m+n)
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// #2024_06_18_Time_3_ms_(96.41%)_Space_15.9_MB_(6.39%)
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/*
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* public var val: Int

‎src/main/swift/g0001_0100/s0023_merge_k_sorted_lists/Solution.swift

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// #Divide_and_Conquer #Merge_Sort #Big_O_Time_O(k*n*log(k))_Space_O(log(k))
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// #2024_06_19_Time_25_ms_(94.57%)_Space_17.3_MB_(6.09%)
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/*
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* public var val: Int

‎src/main/swift/g0001_0100/s0024_swap_nodes_in_pairs/Solution.swift

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// #Udemy_Linked_List #Big_O_Time_O(n)_Space_O(1)
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// #2024_06_19_Time_0_ms_(100.00%)_Space_15.8_MB_(10.29%)
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/*
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* public var val: Int

‎src/main/swift/g0001_0100/s0024_swap_nodes_in_pairs/readme.md

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```swift
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/*
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* public var val: Int

‎src/main/swift/g0001_0100/s0025_reverse_nodes_in_k_group/Solution.swift

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// #Udemy_Linked_List #Big_O_Time_O(n)_Space_O(k)
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// #2024_06_19_Time_12_ms_(87.05%)_Space_15.9_MB_(48.20%)
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/*
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* public var val: Int

‎src/main/swift/g0001_0100/s0025_reverse_nodes_in_k_group/readme.md

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```swift
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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‎src/main/swift/g0001_0100/s0042_trapping_rain_water/Solution.swift

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}
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}
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return ans
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}
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}
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}

‎src/main/swift/g0001_0100/s0062_unique_paths/Solution.swift

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// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Dynamic_Programming #Math
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// #Combinatorics #Algorithm_II_Day_13_Dynamic_Programming #Dynamic_Programming_I_Day_15
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// #Level_1_Day_11_Dynamic_Programming #Big_O_Time_O(m*n)_Space_O(m*n)
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// #2024_06_24_Time_0_ms_(100.00%)_Space_15.4_MB_(70.05%)
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public class Solution {
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public func uniquePaths(_ m: Int, _ n: Int) -> Int {
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var dp = Array(repeating: Array(repeating: 0, count: n), count: m)
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for i in 0..<m {
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dp[i][0] = 1
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}
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for j in 0..<n {
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dp[0][j] = 1
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}
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for i in 1..<m {
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for j in 1..<n {
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dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
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}
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}
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return dp[m - 1][n - 1]
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}
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}

‎src/main/swift/g0001_0100/s0062_unique_paths/readme.md

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62\. Unique Paths
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Medium
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A robot is located at the top-left corner of a `m x n` grid (marked 'Start' in the diagram below).
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The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
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How many possible unique paths are there?
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**Example 1:**
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![](https://assets.leetcode.com/uploads/2018/10/22/robot_maze.png)
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**Input:** m = 3, n = 7
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**Output:** 28
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**Example 2:**
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**Input:** m = 3, n = 2
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**Output:** 3
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**Explanation:**
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From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
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1. Right -> Down -> Down
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2. Down -> Down -> Right
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3. Down -> Right -> Down
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**Example 3:**
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**Input:** m = 7, n = 3
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**Output:** 28
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**Example 4:**
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**Input:** m = 3, n = 3
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**Output:** 6
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**Constraints:**
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* `1 <= m, n <= 100`
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* It's guaranteed that the answer will be less than or equal to <code>2 * 10<sup>9</sup></code>.
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To solve the "Unique Paths" problem in Swift with the Solution class, follow these steps:
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1. Define a method `uniquePaths` in the `Solution` class that takes two integers `m` and `n` as input and returns the number of unique paths from the top-left corner to the bottom-right corner of an `m x n` grid.
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2. Initialize a 2D array `dp` of size `m x n` to store the number of unique paths for each position in the grid.
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3. Initialize the first row and first column of `dp` to 1 since there is only one way to reach any position in the first row or column (by moving only right or down).
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4. Iterate over each position `(i, j)` in the grid, starting from the second row and second column:
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- Update `dp[i][j]` by adding the number of unique paths from the cell above `(i-1, j)` and the cell to the left `(i, j-1)`.
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5. Return the value of `dp[m-1][n-1]`, which represents the number of unique paths to reach the bottom-right corner of the grid.
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Here's the implementation of the `uniquePaths` method in Swift:
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```swift
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public class Solution {
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public func uniquePaths(_ m: Int, _ n: Int) -> Int {
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var dp = Array(repeating: Array(repeating: 0, count: n), count: m)
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for i in 0..<m {
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dp[i][0] = 1
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}
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for j in 0..<n {
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dp[0][j] = 1
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}
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for i in 1..<m {
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for j in 1..<n {
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dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
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}
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}
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return dp[m - 1][n - 1]
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}
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}
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```
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This implementation efficiently calculates the number of unique paths using dynamic programming, with a time complexity of O(m * n) and a space complexity of O(m * n).

‎src/main/swift/g0001_0100/s0064_minimum_path_sum/Solution.swift

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// #Medium #Top_100_Liked_Questions #Array #Dynamic_Programming #Matrix
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// #Dynamic_Programming_I_Day_16 #Udemy_Dynamic_Programming #Big_O_Time_O(m*n)_Space_O(m*n)
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// #2024_06_24_Time_17_ms_(92.63%)_Space_16_MB_(21.05%)
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class Solution {
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func minPathSum(_ grid: [[Int]]) -> Int {
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var matrix: [[Int]] = grid
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let n = grid.count - 1
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let m = grid[0].count - 1
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for i in 0...n {
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for j in 0...m {
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var step = matrix[i][j]
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if i > 0 && j > 0 {
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step += min(matrix[i - 1][j], matrix[i][j - 1])
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} else if i > 0 && j == 0 {
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step += matrix[i - 1][j]
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} else if j > 0 && i == 0 {
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step += matrix[i][j - 1]
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}
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matrix[i][j] = step
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}
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}
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return matrix[n][m]
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}
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}

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