0153-find-minimum-in-rotated-sorted-array.js

/**
 * 153. Find Minimum in Rotated Sorted Array
 * https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/
 * Difficulty: Medium
 *
 * Suppose an array of length n sorted in ascending order is rotated between 1 and n times.
 * For example, the array nums = [0,1,2,4,5,6,7] might become:
 * - [4,5,6,7,0,1,2] if it was rotated 4 times.
 * - [0,1,2,4,5,6,7] if it was rotated 7 times.
 *
 * Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array
 * [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
 *
 * Given the sorted rotated array nums of unique elements, return the minimum element of this
 * array.
 *
 * You must write an algorithm that runs in O(log n) time.
 */

/**
 * @param {number[]} nums
 * @return {number}
 */
var findMin = function(nums) {
  let left = 0;
  let right = nums.length - 1;

  while (left < right) {
    const mid = Math.floor((left + right) / 2);

    if (nums[mid] <= nums[right]) {
      right = mid;
    } else {
      left = mid + 1;
    }
  }

  return nums[left];
};

// var findMin = function(nums) {
//   let l = 0;
//   let r = nums.length - 1;
//   let res = nums[0];

//   while (l <= r) {
//       if (nums[l] <= nums[r]) {
//           res = Math.min(res, nums[l]);
//           break;
//       }

//       // let m = l + Math.floor((r - l) / 2);
//       let m = Math.floor((l + r) / 2);
//       res = Math.min(res, nums[m]);
//       if (nums[m] >= nums[l]) {
//           l = m + 1;
//       } else {
//           r = m - 1;
//       }
//   }
//   return res;
// };