Find Minimum in Rotated Sorted Array

https://leetcode.com/problems/find-minimum-in-rotated-sorted-array

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. 

Key Observations:

1. A rotated sorted array is a sorted array that has been rotated at some pivot point. For example:
   • Original array: [1, 2, 3, 4, 5]
   • Rotated arrays: [4, 5, 1, 2, 3], [3, 4, 5, 1, 2]
2. Minimum Element:
   • The minimum element is the only element in the rotated array that is smaller than its predecessor.
   • Alternatively, in the case of no rotation (e.g., [1, 2, 3, 4, 5]), the first element is the smallest.
3. Binary Search Approach:
   • Since the array is sorted and rotated, we can use binary search to identify the minimum element in O(log n) time.

Steps to Solve:

1. Binary Search:
   • Set two pointers: low = 0 (start of array) and high = n-1 (end of array).
   • While low < high, compute the middle index: mid = low + (high - low) / 2.
   • Depending on the relationship between arr[mid] and arr[high]:
     • If arr[mid] > arr[high]: The minimum must lie in the right half of the array (set low = mid + 1).
     • Otherwise, the minimum must be in the left half or at mid (set high = mid).
   • When low == high, the minimum element is at index low.
2. Return the Result:
   • Output the element at arr[low].

Algorithm Complexity:

• Time Complexity:
  Binary search operates in O(log n) time due to halving the search space in every iteration.
• Space Complexity:
  The algorithm uses O(1) space as we only manipulate indices and access the array directly.

Implementation in C++:

Here is the efficient implementation:

#include <iostream>
#include <vector>
using namespace std;

int findMinimum(const vector<int>& arr) {
    int low = 0, high = arr.size() - 1;

    // Binary search to find the minimum element
    while (low < high) {
        int mid = low + (high - low) / 2;

        // If middle element is greater than the last element,
        // the minimum must be in the right half
        if (arr[mid] > arr[high]) {
            low = mid + 1;
        } 
        // Otherwise, the minimum is in the left half (or could be at mid)
        else {
            high = mid;
        }
    }

    // At this point, low == high and points to the minimum element
    return arr[low];
}

int main() {
    vector<int> arr = {4, 5, 6, 7, 0, 1, 2}; // Example rotated array
    cout << "The minimum element is: " << findMinimum(arr) << endl;

    return 0;
}