Search in Rotated Sorted Array

Solution uses a modified version of Binary Search.
Javascript (ES6)

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
const search = (nums, target) => {
  let low = 0;
  let high = nums.length - 1;
  while (low <= high) {
    const mid = low + Math.floor((high - low) / 2);
    if (nums[mid] === target) {
      return mid;
    }
    if (nums[low] <= nums[mid]) {
      // Left half is sorted.
      if (target >= nums[low] && target < nums[mid]) {
        high = mid - 1;
      } else {
        low = mid + 1;
      }
    } else {
      // Right half is sorted.
      if (target > nums[mid] && target <= nums[high]) {
        low = mid + 1;
      } else {
        high = mid - 1;
      }
    }
  }
  return -1;
};

console.log(search([4, 5, 1, 2, 3], 2)); // 3

Demo

Java

Time: O(nlogn)
Space: O(1)

class Solution {
 public int search(int[] nums, int target) {
  int low = 0, high = nums.length - 1;
  int mid;
  while (low <= high) {
   mid = low + (high - low) / 2;
   if (nums[mid] == target)
    return mid;
   else if (nums[mid] < nums[high]) {
    if (target > nums[mid] && target <= nums[high])
     low = mid + 1;
    else
     high = mid - 1;
   } else {
    if (target >= nums[low] && target < nums[mid])
     high = mid - 1;
    else
     low = mid + 1;
   }
  }
  return -1;
 }
}