Search in Rotated Sorted Array

Search in Rotated Sorted Array

Question:

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

Prior to being passed to your function, nums is rotated at an unknown pivot index k (0 <= 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 rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

How to Solve:

The solution consists of two phases. In phase one, we find the index of the minimum element in the array (i.e. the breakpoint of the rotated array), which is reusing the algorithm from the problem: "Find Minimum in Rotated Sorted Array".

After we have located the minimum index (i.e. the number of rotates this array has undergone), we compare and see if the target belongs to the right half of the breakpoint or the left half. Afterwards, we may perform a normal binary search for target in the correct half.

My C++ Solution:

class Solution {
 public:
  int search(vector<int> &nums, int target) {
    int n = nums.size();
    if (n == 0) return -1;
    int low = 0, high = n - 1, mid = low + (high - low) / 2, min_idx;
    if (nums[low] < nums[high]) {  // already sorted
      min_idx = 0;
    } else {
      while (low < high) {
        mid = low + (high - low) / 2;
        if (nums[mid] > nums[high]) {
          low = mid + 1;  // look at right half
        } else {
          high = mid;  // look at left half
        }
      }
      min_idx = low;
      // Reset low & high
      low = 0;
      high = n - 1;
    }
    // target happens to be the minimum num in the array
    if (target == nums[min_idx]) {
      return min_idx;
    }
    // target in right half
    else if (target > nums[min_idx] && target <= nums[high]) {
      low = min_idx + 1;
    }
    // target in left half
    else {
      high = min_idx - 1;
    }
    // Normal binary search after getting correct low & high
    while (low <= high) {
      mid = low + (high - low) / 2;
      if (nums[mid] == target) {
        return mid;
      } else if (nums[mid] < target) {
        low = mid + 1;
      } else {
        high = mid - 1;
      }
    }
    return -1;  // not found
  }
};