Find Minimum in Rotated Sorted Array

LeetCode 153 | Difficulty: Medium

Medium

Problem Description

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

- `[4,5,6,7,0,1,2]` if it was rotated `4` times.

- `[0,1,2,4,5,6,7]` if it was rotated `7` times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

Example 1:

Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:

Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:

Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times.

Constraints:

- `n == nums.length`

- `1 <= n <= 5000`

- `-5000 <= nums[i] <= 5000`

- All the integers of `nums` are **unique**.

- `nums` is sorted and rotated between `1` and `n` times.

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: 96 ms)

Metric	Value
Runtime	96 ms
Memory	N/A
Date	2018-07-04

Solution

public class Solution {
    public int FindMin(int[] nums) {
       int lo = 0, hi = nums.Length - 1;
    while (lo < hi)
    {
        if(nums[lo]<nums[hi])
        {
            return nums[lo];
        }
        int mid = lo + (hi - lo) / 2;
        if (nums[mid] > nums[hi])
        {
            lo = mid+1;
        }
        else if (nums[mid] < nums[lo]){
            hi = mid;
        }
        else
        {
            hi--;
        }       
    }
    return nums[lo]; 
    }
}

📜 1 more C# submission(s)

Submission (2021-11-24) — 100 ms, 36.5 MB

public class Solution {
    public int FindMin(int[] nums) {
        int l=0, r = nums.Length-1;
    while(l<r)
    {
        if(nums[l] < nums[r])
        return nums[l];
        
        int mid = l + (r-l)/2;
        if(nums[l]>nums[mid])
        {
            r = mid;
        }
        else
        {
            l = mid+1;
        }
    }
    return nums[l];
    }
}

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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.
• LeetCode provides 3 hint(s) for this problem — try solving without them first.

💡 Hints

Hint 1: Array was originally in ascending order. Now that the array is rotated, there would be a point in the array where there is a small deflection from the increasing sequence. eg. The array would be something like [4, 5, 6, 7, 0, 1, 2].

Hint 2: You can divide the search space into two and see which direction to go. Can you think of an algorithm which has O(logN) search complexity?

Hint 3: - All the elements to the left of inflection point > first element of the array.

• All the elements to the right of inflection point < first element of the array.