1+ package medium
2+ 3+ import ArraysTopic
4+ import BinarySearchTopic
5+ 6+ /* *
7+ * 81. Search in Rotated Sorted Array II
8+ * https://leetcode.com/problems/search-in-rotated-sorted-array/
9+ *
10+ There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values).
11+ Before being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length)
12+ such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed).
13+ For example, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4].
14+ Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums.
15+ You must decrease the overall operation steps as much as possible.
16+ BULLSHIT
17+ */
18+ 19+ class Medium81 : BinarySearchTopic , ArraysTopic {
20+ 21+ fun search (nums : IntArray , target : Int ): Boolean {
22+ val n = nums.size
23+ if (n == 0 ) return false
24+ var end = n - 1
25+ var start = 0
26+ while (start <= end) {
27+ val mid = start + (end - start) / 2
28+ if (nums[mid] == target) {
29+ return true
30+ }
31+ if (! isBinarySearchHelpful(nums, start, nums[mid])) {
32+ start++
33+ continue
34+ }
35+ // which array does pivot belong to.
36+ val pivotArray = existsInFirst(nums, start, nums[mid])
37+ 38+ // which array does target belong to.
39+ val targetArray = existsInFirst(nums, start, target)
40+ if (pivotArray xor targetArray) { // If pivot and target exist in different sorted arrays, recall that xor is true when both operands are distinct
41+ if (pivotArray) {
42+ start = mid + 1 // pivot in the first, target in the second
43+ } else {
44+ end = mid - 1 // target in the first, pivot in the second
45+ }
46+ } else { // If pivot and target exist in same sorted array
47+ if (nums[mid] < target) {
48+ start = mid + 1
49+ } else {
50+ end = mid - 1
51+ }
52+ }
53+ }
54+ return false
55+ }
56+ 57+ // returns true if we can reduce the search space in current binary search space
58+ private fun isBinarySearchHelpful (arr : IntArray , start : Int , element : Int ): Boolean {
59+ return arr[start] != element
60+ }
61+ 62+ // returns true if element exists in first array, false if it exists in second
63+ private fun existsInFirst (arr : IntArray , start : Int , element : Int ): Boolean {
64+ return arr[start] <= element
65+ }
66+ }
67+ 68+ fun main () {
69+ println (Medium81 ().search(intArrayOf(2 , 5 , 6 , 0 , 0 , 1 , 2 ), 0 ))
70+ println (Medium81 ().search(intArrayOf(2 , 5 , 6 , 0 , 0 , 1 , 2 ), 3 ))
71+ println (Medium81 ().search(intArrayOf(1 , 0 , 1 , 1 , 1 ), 0 ))
72+ }
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