Search in Rotated Sorted Array

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Problem

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is possibly rotated at an unknown pivot 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 rotated at pivot index 3 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

Pseudocode

- unique property is that array is sorted in ascending order
    - use binary search to look for values that isn't sorted

Solution

// from solutions
//https://leetcode.com/problems/search-in-rotated-sorted-array/solutions/273622/javascript-simple-o-log-n-binary-search-solution/
var search = function (nums, target) {
  let left = 0;
  let right = nums.length - 1;

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

    if (nums[mid] === target) {
      return mid;
    }

    // When dividing the roated array into two halves, one must be sorted.

    // Check if the left side is sorted
    if (nums[left] <= nums[mid]) {
      if (nums[left] <= target && target <= nums[mid]) {
        // target is in the left
        right = mid - 1;
      } else {
        // target is in the right
        left = mid + 1;
      }
    }

    // Otherwise, the right side is sorted
    else {
      if (nums[mid] <= target && target <= nums[right]) {
        // target is in the right
        left = mid + 1;
      } else {
        // target is in the left
        right = mid - 1;
      }
    }
  }

  return -1;
};

Time and Space Complexity

Time

• What did the code do

• Total -

Space

• What did the code do

• Total -