feat(problems): add solutions for Search in Rotated Sorted Array and Koko Eating Bananas

work/Medium/Array/33.search-in-rotated-sorted-array.js

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+/*
+ * 33. Search in Rotated Sorted Array
+ * Difficulty: Medium
+ * https://leetcode.com/problems/search-in-rotated-sorted-array/
+ * 
+ * ──────────────────────────────────────────────────
+ * 
+ * There is an integer array nums sorted in ascending order (with
+ * distinct values).
+ * 
+ * Prior to being passed to your function, nums is possibly left rotated
+ * at an unknown index k (1 <= k < nums.length) such that the resulting
+ * array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ...,
+ * nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be left
+ * rotated by 3 indices and become [4,5,6,7,0,1,2].
+ * 
+ * Given the array nums after the possible rotation and an integer
+ * target, return the index of target if it is in nums, or -1 if it is
+ * not in nums.
+ * 
+ * You must write an algorithm with O(log n) runtime complexity.
+ * 
+ *  
+ * 
+ * Example 1:
+ * 
+ * Input: nums = [4,5,6,7,0,1,2], target = 0
+ * Output: 4
+ * 
+ * Example 2:
+ * 
+ * Input: nums = [4,5,6,7,0,1,2], target = 3
+ * Output: -1
+ * 
+ * Example 3:
+ * 
+ * Input: nums = [1], target = 0
+ * Output: -1
+ * 
+ *  
+ * 
+ * Constraints:
+ * 
+ * 	• 1 <= nums.length <= 5000
+ * 
+ * 	• -10^4 <= nums[i] <= 10^4
+ * 
+ * 	• All values of nums are unique.
+ * 
+ * 	• nums is an ascending array that is possibly rotated.
+ * 
+ * 	• -10^4 <= target <= 10^4
+*/
+
+/**
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number}
+ */
+var search = function(nums, target) {
+  let l = 0, r = nums.length - 1;   
+
+  while (l <= r){
+    let mid = Math.floor((l + r)/2);
+    if (target === nums[mid]) {
+      return mid
+    }
+
+    // Left sorted portion
+    if (nums[l] <= nums[mid]){
+      if (target > nums[mid] || target < nums[l]){
+        l = mid + 1;
+      }else{
+        r = mid - 1;
+      }
+    }
+    // Right sorted portion
+    else{
+      if(target < nums[mid] || target > nums[r]){
+        r = mid - 1; 
+      }else{
+        l = mid + 1;
+      }
+    }
+  }
+  return -1;
+};