33. Search in Rotated Sorted Array

Description

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

Prior to being passed to your function, nums is possibly left rotated at an unknown 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 left rotated by 3 indices 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

Submitted Code

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        n = len(nums) - 1
        l, r = 0, n
        
        while l <= r:
            m = (l + r) // 2          # 중간값
            if nums[m] == target:
                return m
            elif nums[l] <= nums[m]:            # 왼쪽이 정렬되고
                if nums[l] <= target < nums[m]:   # 정렬된 왼쪽 범위에 target 존재
                    r = m - 1
                else:
                    l = m + 1                     # 오른쪽 범위에 target 존재
            elif nums[m] <= nums[r]:            # 오른쪽이 정렬되고
                if nums[m] < target <= nums[r]:   # 정렬된 오른쪽 범위에 target 존재
                    l = m + 1
                else:                             # 왼쪽 범위에 target 존재
                    r = m - 1
        
        return -1

Runtime: 0 ms | Beats 100.00%
Memory: 19.50 MB | Beats 30.45%

이진 탐색으로 두 포인터의 가운데를 찾으면, 최소한 한 쪽은 정렬되어 있다는 규칙을 이용할 수 있다. target이 속한 정렬된 범위를 찾을 때까지 포인터 m의 왼쪽 또는 오른쪽으로 이동하며 탐색하면 된다.

Other Solutions

1st

class Solution {
public:
    int search(int A[], int n, int target) {
        int lo=0,hi=n-1;
        // find the index of the smallest value using binary search.
        // Loop will terminate since mid < hi, and lo or hi will shrink by at least 1.
        // Proof by contradiction that mid < hi: if mid==hi, then lo==hi and loop would have been terminated.
        while(lo<hi){
            int mid=(lo+hi)/2;
            if(A[mid]>A[hi]) lo=mid+1;
            else hi=mid;
        }
        // lo==hi is the index of the smallest value and also the number of places rotated.
        int rot=lo;
        lo=0;hi=n-1;

        // The usual binary search and accounting for rotation.
        while(lo<=hi){
            int mid=(lo+hi)/2;
            int realmid=(mid+rot)%n;
            if(A[realmid]==target)return realmid;
            if(A[realmid]<target)lo=mid+1;
            else hi=mid-1;
        }
        return -1;
    }
};

time complexity: 𝑂(log𝑛)
space complexity: 𝑂(1)

최소값의 위치(회전된 시작점)를 먼저 찾은 후, 정렬된 배열처럼 인덱스를 보정하는 방법이다.