Search in Rotated Sorted Array

LeetCode 33 | Difficulty: Medium

Medium

Problem 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`

- `-10^4 <= nums[i] <= 10^4`

- All values of `nums` are **unique**.

- `nums` is an ascending array that is possibly rotated.

- `-10^4 <= target <= 10^4`

Topics: Array, Binary Search

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Approach

Binary Search

Binary search reduces the search space by half at each step. The key insight is identifying the monotonic property — what condition lets you decide to go left or right?

When to use

Sorted array, or searching for a value in a monotonic function/space.

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Solutions

Solution 1: C# (Best: 142 ms)

Metric	Value
Runtime	142 ms
Memory	N/A
Date	2017-11-04

Solution

public class Solution {
    public int Search(int[] nums, int target) {
 
        if(nums.Length == 0) return -1;
        int lo = 0, high = nums.Length - 1;
    while (lo <= high)
    {
        int mid = (lo + high)/2;
        int num = (nums[mid] < nums[0]) == (target < nums[0])
                    ? nums[mid]
                    : target < nums[0] ? Int32.MinValue : Int32.MaxValue;
        if (num < target)
        {
            lo = mid+1;
        }
        else if (num > target)
        {
            high = mid-1;
        }
        else return mid;
    }
    return -1;
    }
}

📜 3 more C# submission(s)

Submission (2022-02-23) — 158 ms, 37.9 MB

public class Solution {
    public int Search(int[] nums, int target) {
 
        int l = 0, r = nums.Length-1;
    while(l<=r)
    {
        int mid = l + (r-l)/2;
        if(target == nums[mid])
        {
            return mid;
        }
        
        if(nums[l] <= nums[mid])
        {
            if(target >= nums[l] && target < nums[mid])
            {
                r = mid-1;
            }
            else
            {
                l = mid+1;
            }
        }
        else if(nums[mid] < nums[r])
        {
            if(target > nums[mid] && target <= nums[r])
            {
                l = mid+1;
            }
            else
            {
                r = mid-1;
            }
        }       
    }
    // Write your code here.
    return -1;
    }
}

Submission (2017-11-04) — 192 ms, N/A

public class Solution {
    public int Search(int[] nums, int target) {
 int lo = 0, high = nums.Length - 1;
    while (lo <= high)
    {
        int mid = (lo + high)/2;
        int num = (nums[mid] < nums[0]) == (target < nums[0])
                    ? nums[mid]
                    : target < nums[0] ? Int32.MinValue : Int32.MaxValue;
        if (num < target)
        {
            lo = mid+1;
        }
        else if (num > target)
        {
            high = mid-1;
        }
        else return mid;
    }
    return -1;
    }
}

Submission (2017-11-04) — 192 ms, N/A

public class Solution {
    public int Search(int[] nums, int target) {
 if(nums.Length<1) return -1;
    int lo = 0, high = nums.Length - 1;
    while (lo < high)
    {
        int mid = 1+(lo + high)/2;
        if(nums[mid] == target) return mid;

        if (nums[lo] < nums[mid])
        {
            if(target >= nums[lo] && target < nums[mid])
            high = mid -1;
            else
            lo = mid+1;
        }
        else
        {
            if (target > nums[mid] && target <= nums[high])
            {
                lo = mid+1;
            }
            else
            {
                high = mid-1;
            }
        }
    }
    
    return (lo>=0 && lo<nums.Length && nums[lo]==target) ? lo : -1;
    }
}

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Complexity Analysis

Approach	Time	Space
Binary Search	$O(log n)$	$O(1)$

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Interview Tips

Key Points

• Discuss the brute force approach first, then optimize. Explain your thought process.
• Precisely define what the left and right boundaries represent, and the loop invariant.