Search in Rotated Sorted Array

Search in Rotated Sorted Array - LeetCodeLeetCode

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 -