LeetCode #33: Search in Rotated Sorted Array (C/C++).

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source: https://leetcode.com/problems/search-in-rotated-sorted-array
C/C++ Solution to LeetCode problem 33. Search in Rotated Sorted Array.

Problem

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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.

Examples

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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

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• 1 <= nums.length <= 5000
• -104 <= nums[i] <= 104
• All values of nums are unique.
• nums is an ascending array that is possibly rotated.
• -104 <= target <= 104

Solution

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The logic is as follows:

• Binary search splits the array in two.
• If the array is rotated, one part will be sorted, and the other not.
• First we identify the sorted half (the sorted part has the value at the lower index <= that the value at the middle index).
  • If the target is in the sorted part, then we update our indexes to perform the next binary search in that part.
  • If not, then it means the target is in the unsorted part, so we update indexes to perform next search there.

class Solution {
public:
  int search(vector<int>& nums, int target) {
    int l = 0;
    int h = nums.size() - 1;
    int m;
        
    while (l <= h) {
      m = (l + h) / 2;
      if (nums[m] == target)
        return m;

      if (nums[l] <= nums[m]) {
        if (target <= nums[m] && target >= nums[l])
          h = m - 1;
        else
          l = m + 1;
      } else {
        if (target >= nums[m] && target <= nums[h])
          l = m + 1;
        else
          h = m - 1;
      }
    }
    return -1;
  }
};