Search in Rotated Sorted Array LeetCode Solutions | Easy Approach

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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.

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:

  • 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

Search in Rotated Sorted Array Solutions

Time: O(log n)
Space: O(1)

C++

 class Solution {
 public:
  int search(vector<int>& nums, int target) {
    int l = 0;
    int r = nums.size() - 1;

    while (l <= r) {
      const int m = l + (r - l) / 2;
      if (nums[m] == target)
        return m;
      if (nums[l] <= nums[m]) {  // nums[l..m] are sorted
        if (nums[l] <= target && target < nums[m])
          r = m - 1;
        else
          l = m + 1;
      } else {  // nums[m..n - 1] are sorted
        if (nums[m] < target && target <= nums[r])
          l = m + 1;
        else
          r = m - 1;
      }
    }

    return -1;
  }
};

Java

 
class Solution {
  public int search(int[] nums, int target) {
    int l = 0;
    int r = nums.length - 1;

    while (l <= r) {
      final int m = l + (r - l) / 2;
      if (nums[m] == target)
        return m;
      if (nums[l] <= nums[m]) { // nums[l..m] are sorted
        if (nums[l] <= target && target < nums[m])
          r = m - 1;
        else
          l = m + 1;
      } else { // nums[m..n - 1] are sorted
        if (nums[m] < target && target <= nums[r])
          l = m + 1;
        else
          r = m - 1;
      }
    }

    return -1;
  }
}

Python

 class Solution:
  def search(self, nums: List[int], target: int) -> int:
    l = 0
    r = len(nums) - 1

    while l <= r:
      m = (l + r) // 2
      if nums[m] == target:
        return m
      if nums[l] <= nums[m]:  # nums[l..m] are sorted
        if nums[l] <= target < nums[m]:
          r = m - 1
        else:
          l = m + 1
      else:  # nums[m..n - 1] are sorted
        if nums[m] < target <= nums[r]:
          l = m + 1
        else:
          r = m - 1

    return -1

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