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Link for the Problem – Search in Rotated Sorted Array– LeetCode Problem
Search in Rotated Sorted Array– LeetCode Problem
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
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– LeetCode Solutions
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; } };
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; } }
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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