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‎README.md‎

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@@ -277,6 +277,7 @@ implementation 'com.github.javadev:leetcode-in-kotlin:1.8'
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-
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| 0542 |[01 Matrix](src/main/kotlin/g0501_0600/s0542_01_matrix/Solution.kt)| Medium | Array, Dynamic_Programming, Breadth_First_Search, Matrix | 441 | 94.06
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#### Day 6 Matrix Related Problems
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-
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| 0542 |[01 Matrix](src/main/kotlin/g0501_0600/s0542_01_matrix/Solution.kt)| Medium | Array, Dynamic_Programming, Breadth_First_Search, Matrix | 441 | 94.06
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| 0994 |[Rotting Oranges](src/main/kotlin/g0901_1000/s0994_rotting_oranges/Solution.kt)| Medium | Array, Breadth_First_Search, Matrix | 308 | 57.93
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#### Day 10 Recursion Backtracking
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| <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- --> | <!-- -->
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|-|-|-|-|-|-
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| 0540 |[Single Element in a Sorted Array](src/main/kotlin/g0501_0600/s0540_single_element_in_a_sorted_array/Solution.kt)| Medium | Array, Binary_Search | 274 | 86.67
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#### Day 10
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| 0647 |[Palindromic Substrings](src/main/kotlin/g0601_0700/s0647_palindromic_substrings/Solution.kt)| Medium | Top_100_Liked_Questions, String, Dynamic_Programming | 266 | 67.83
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| 0560 |[Subarray Sum Equals K](src/main/kotlin/g0501_0600/s0560_subarray_sum_equals_k/Solution.kt)| Medium | Top_100_Liked_Questions, Array, Hash_Table, Prefix_Sum, Data_Structure_II_Day_5_Array | 692 | 53.27
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| 0543 |[Diameter of Binary Tree](src/main/kotlin/g0501_0600/s0543_diameter_of_binary_tree/Solution.kt)| Easy | Top_100_Liked_Questions, Depth_First_Search, Tree, Binary_Tree, Level_2_Day_7_Tree, Udemy_Tree_Stack_Queue | 307 | 43.93
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| 0542 |[01 Matrix](src/main/kotlin/g0501_0600/s0542_01_matrix/Solution.kt)| Medium | Array, Dynamic_Programming, Breadth_First_Search, Matrix, Algorithm_I_Day_9_Breadth_First_Search_Depth_First_Search, Graph_Theory_I_Day_5_Matrix_Related_Problems | 441 | 94.06
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| 0541 |[Reverse String II](src/main/kotlin/g0501_0600/s0541_reverse_string_ii/Solution.kt)| Easy | String, Two_Pointers | 200 | 83.33
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| 0540 |[Single Element in a Sorted Array](src/main/kotlin/g0501_0600/s0540_single_element_in_a_sorted_array/Solution.kt)| Medium | Array, Binary_Search, Binary_Search_II_Day_9 | 274 | 86.67
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| 0539 |[Minimum Time Difference](src/main/kotlin/g0501_0600/s0539_minimum_time_difference/Solution.kt)| Medium | Array, String, Math, Sorting | 183 | 100.00
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| 0538 |[Convert BST to Greater Tree](src/main/kotlin/g0501_0600/s0538_convert_bst_to_greater_tree/Solution.kt)| Medium | Depth_First_Search, Tree, Binary_Tree, Binary_Search_Tree | 252 | 77.78
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| 0537 |[Complex Number Multiplication](src/main/kotlin/g0501_0600/s0537_complex_number_multiplication/Solution.kt)| Medium | String, Math, Simulation | 171 | 75.00
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| 0535 |[Encode and Decode TinyURL](src/main/kotlin/g0501_0600/s0535_encode_and_decode_tinyurl/Codec.kt)| Medium | String, Hash_Table, Design, Hash_Function | 183 | 81.25

‎src/main/kotlin/g0501_0600/s0539_minimum_time_difference/Solution.kt‎

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package g0501_0600.s0539_minimum_time_difference
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// #Medium #Array #String #Math #Sorting #2023_01_16_Time_183_ms_(100.00%)_Space_36_MB_(96.23%)
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class Solution {
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fun findMinDifference(timePoints: List<String>): Int {
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return if (timePoints.size < 300) {
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smallInputSize(timePoints)
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} else largeInputSize(timePoints)
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}
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private fun largeInputSize(timePoints: List<String>): Int {
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val times = BooleanArray(60 * 24)
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for (time in timePoints) {
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val hours = time.substring(0, 2).toInt()
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val minutes = time.substring(3, 5).toInt()
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if (times[hours * 60 + minutes]) {
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return 0
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}
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times[hours * 60 + minutes] = true
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}
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var prev = -1
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var min = 60 * 24
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for (i in 0 until times.size + times.size / 2) {
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if (i < times.size) {
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if (times[i] && prev == -1) {
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prev = i
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} else if (times[i]) {
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min = Math.min(min, i - prev)
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prev = i
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}
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} else {
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if (times[i - times.size] && prev == -1) {
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prev = i
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} else if (times[i - times.size]) {
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min = Math.min(min, i - prev)
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prev = i
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}
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}
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}
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return min
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}
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private fun smallInputSize(timePoints: List<String>): Int {
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val times = IntArray(timePoints.size)
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var j = 0
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for (time in timePoints) {
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val hours = time.substring(0, 2).toInt()
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val minutes = time.substring(3, 5).toInt()
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times[j++] = hours * 60 + minutes
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}
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times.sort()
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var min = 60 * 24
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for (i in 1..times.size) {
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min = if (i == times.size) {
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Math.min(min, times[0] + 60 * 24 - times[times.size - 1])
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} else {
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Math.min(min, times[i] - times[i - 1])
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}
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if (min == 0) {
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return 0
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}
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}
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return min
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}
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}

‎src/main/kotlin/g0501_0600/s0539_minimum_time_difference/readme.md‎

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539\. Minimum Time Difference
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Medium
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Given a list of 24-hour clock time points in **"HH:MM"** format, return _the minimum **minutes** difference between any two time-points in the list_.
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**Example 1:**
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**Input:** timePoints = ["23:59","00:00"]
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**Output:** 1
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**Example 2:**
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**Input:** timePoints = ["00:00","23:59","00:00"]
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**Output:** 0
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**Constraints:**
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* <code>2 <= timePoints.length <= 2 * 10<sup>4</sup></code>
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* `timePoints[i]` is in the format **"HH:MM"**.

‎src/main/kotlin/g0501_0600/s0540_single_element_in_a_sorted_array/Solution.kt‎

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package g0501_0600.s0540_single_element_in_a_sorted_array
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// #Medium #Array #Binary_Search #Binary_Search_II_Day_9
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// #2023_01_16_Time_274_ms_(86.67%)_Space_44.4_MB_(83.33%)
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class Solution {
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fun singleNonDuplicate(nums: IntArray): Int {
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var start = 0
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var end = nums.size - 1
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while (start < end) {
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val mid = start + (end - start) / 2
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if (mid + 1 < nums.size && nums[mid] != nums[mid + 1] && mid - 1 >= 0 && nums[mid] != nums[mid - 1]) {
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return nums[mid]
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} else if (mid + 1 < nums.size && nums[mid] == nums[mid + 1] && mid % 2 == 0) {
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start = mid + 1
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} else if (mid - 1 >= 0 && nums[mid] == nums[mid - 1] && mid % 2 == 1) {
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start = mid + 1
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} else {
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end = mid - 1
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}
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}
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return nums[start]
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}
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}

‎src/main/kotlin/g0501_0600/s0540_single_element_in_a_sorted_array/readme.md‎

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540\. Single Element in a Sorted Array
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Medium
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You are given a sorted array consisting of only integers where every element appears exactly twice, except for one element which appears exactly once.
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Return _the single element that appears only once_.
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Your solution must run in `O(log n)` time and `O(1)` space.
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**Example 1:**
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**Input:** nums = [1,1,2,3,3,4,4,8,8]
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**Output:** 2
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**Example 2:**
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**Input:** nums = [3,3,7,7,10,11,11]
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**Output:** 10
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**Constraints:**
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* <code>1 <= nums.length <= 10<sup>5</sup></code>
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* <code>0 <= nums[i] <= 10<sup>5</sup></code>

‎src/main/kotlin/g0501_0600/s0541_reverse_string_ii/Solution.kt‎

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package g0501_0600.s0541_reverse_string_ii
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// #Easy #String #Two_Pointers #2023_01_17_Time_200_ms_(83.33%)_Space_35.7_MB_(86.67%)
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class Solution {
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fun reverseStr(s: String, k: Int): String {
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val res = StringBuilder()
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var p1 = 0
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var p2 = 2 * k - 1
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if (s.length < k) {
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res.append(reverse(s))
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return res.toString()
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}
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while (p1 < s.length) {
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if (s.length < p1 + k) {
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res.append(reverse(s.substring(p1)))
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} else {
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res.append(reverse(s.substring(p1, p1 + k)))
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if (s.length < p2 + 1) {
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res.append(s.substring(p1 + k))
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} else {
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res.append(s, p1 + k, p2 + 1)
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}
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}
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p1 = p1 + 2 * k
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p2 = p2 + 2 * k
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}
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return res.toString()
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}
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private fun reverse(s: String): String {
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val sb = StringBuilder()
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sb.append(s)
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sb.reverse()
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return sb.toString()
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}
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}

‎src/main/kotlin/g0501_0600/s0541_reverse_string_ii/readme.md‎

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541\. Reverse String II
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Easy
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Given a string `s` and an integer `k`, reverse the first `k` characters for every `2k` characters counting from the start of the string.
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If there are fewer than `k` characters left, reverse all of them. If there are less than `2k` but greater than or equal to `k` characters, then reverse the first `k` characters and leave the other as original.
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**Example 1:**
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**Input:** s = "abcdefg", k = 2
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**Output:** "bacdfeg"
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**Example 2:**
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**Input:** s = "abcd", k = 2
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**Output:** "bacd"
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**Constraints:**
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* <code>1 <= s.length <= 10<sup>4</sup></code>
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* `s` consists of only lowercase English letters.
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* <code>1 <= k <= 10<sup>4</sup></code>

‎src/main/kotlin/g0501_0600/s0542_01_matrix/Solution.kt‎

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package g0501_0600.s0542_01_matrix
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// #Medium #Array #Dynamic_Programming #Breadth_First_Search #Matrix
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// #Algorithm_I_Day_9_Breadth_First_Search_Depth_First_Search
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// #Graph_Theory_I_Day_5_Matrix_Related_Problems
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// #2023_01_17_Time_441_ms_(94.06%)_Space_54_MB_(78.22%)
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import java.util.Arrays
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class Solution {
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fun updateMatrix(mat: Array<IntArray>): Array<IntArray> {
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val dist = Array(mat.size) { IntArray(mat[0].size) }
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for (i in mat.indices) {
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Arrays.fill(dist[i], Int.MAX_VALUE - 100000)
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}
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for (i in mat.indices) {
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for (j in mat[0].indices) {
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if (mat[i][j] == 0) {
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dist[i][j] = 0
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} else {
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if (i > 0) {
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dist[i][j] = Math.min(dist[i][j], dist[i - 1][j] + 1)
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}
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if (j > 0) {
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dist[i][j] = Math.min(dist[i][j], dist[i][j - 1] + 1)
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}
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}
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}
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}
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for (i in mat.indices.reversed()) {
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for (j in mat[0].indices.reversed()) {
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if (i < mat.size - 1) {
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dist[i][j] = Math.min(dist[i][j], dist[i + 1][j] + 1)
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}
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if (j < mat[0].size - 1) {
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dist[i][j] = Math.min(dist[i][j], dist[i][j + 1] + 1)
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}
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}
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}
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return dist
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}
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}

‎src/main/kotlin/g0501_0600/s0542_01_matrix/readme.md‎

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542\. 01 Matrix
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Medium
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Given an `m x n` binary matrix `mat`, return _the distance of the nearest_ `0` _for each cell_.
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The distance between two adjacent cells is `1`.
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**Example 1:**
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![](https://assets.leetcode.com/uploads/2021/04/24/01-1-grid.jpg)
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**Input:** mat = [[0,0,0],[0,1,0],[0,0,0]]
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**Output:** [[0,0,0],[0,1,0],[0,0,0]]
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**Example 2:**
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![](https://assets.leetcode.com/uploads/2021/04/24/01-2-grid.jpg)
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**Input:** mat = [[0,0,0],[0,1,0],[1,1,1]]
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**Output:** [[0,0,0],[0,1,0],[1,2,1]]
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**Constraints:**
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* `m == mat.length`
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* `n == mat[i].length`
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* <code>1 <= m, n <= 10<sup>4</sup></code>
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* <code>1 <= m * n <= 10<sup>4</sup></code>
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* `mat[i][j]` is either `0` or `1`.
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* There is at least one `0` in `mat`.

‎src/test/kotlin/g0501_0600/s0539_minimum_time_difference/SolutionTest.kt‎

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package g0501_0600.s0539_minimum_time_difference
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import org.hamcrest.CoreMatchers.equalTo
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import org.hamcrest.MatcherAssert.assertThat
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import org.junit.jupiter.api.Test
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import java.util.stream.IntStream
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internal class SolutionTest {
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@Test
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fun findMinDifference() {
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assertThat(Solution().findMinDifference(listOf("23:59", "00:00")), equalTo(1))
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}
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@Test
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fun findMinDifference2() {
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assertThat(
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Solution().findMinDifference(listOf("00:00", "23:59", "00:00")),
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equalTo(0)
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)
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}
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@Test
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fun findMinDifference3() {
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val timePoints: MutableList<String> = ArrayList()
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IntStream.rangeClosed(0, 310 / 2)
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.forEach { index: Int ->
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timePoints.add("23:59")
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timePoints.add("00:00")
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}
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assertThat(Solution().findMinDifference(timePoints), equalTo(0))
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}
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@Test
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fun findMinDifference4() {
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val timePoints: MutableList<String> = ArrayList()
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var index = 0
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while (index < 301) {
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val hour = index / 60 % 24
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val minute = index % 60
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timePoints.add(
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String.format(
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"%s:%s",
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if (hour < 10) "0$hour" else hour, if (minute < 10) "0$minute" else minute
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)
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)
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index++
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}
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assertThat(Solution().findMinDifference(timePoints), equalTo(1))
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}
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}

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