🚀Day 3. Search in Rotated Sorted Array 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

Given a sorted (in ascending order) and rotated array arr of distinct elements which may be rotated at some point and given an element key, the task is to find the index of the given element key in the array arr. The whole array arr is given as the range to search. Rotation shifts elements of the array by a certain number of positions, with elements that fall off one end reappearing at the other end.

Note: 0-based indexing is followed & returns -1 if the key is not present.

🔍 Example Walkthrough:

Input:

arr[] = [5, 6, 7, 8, 9, 10, 1, 2, 3], key = 10

Output:

5

Explanation: 10 is found at index 5.

🎯 My Approach:

1. Initialization:

• Use a linear search approach to iterate through the array arr.

1. Search for Key:

• Iterate through each element in the array from i = 0 to n-1 (where n is the size of the array).

• Compare each element with the key.

• If a match is found, return the index i.

• If the end of the array is reached without finding the key, return -1.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n), as we iterate through the array once to find the key.

• Expected Auxiliary Space Complexity: O(1), as we only use a constant amount of additional space.

📝 Solution Code

Code (C++)

class Solution {
public:
    int search(vector<int>& arr, int key) {
        int n = arr.size();
        for(int i = 0; i < n; i++) {
            if(arr[i] == key) {
                return i;
            }
        }
        return -1;
    }
};

Code (Java)

class Solution {
    int search(int[] arr, int key) {
        int n = arr.length;
        for (int i = 0; i < n; i++) {
            if (arr[i] == key) {
                return i;
            }
        }
        return -1;
    }
}

Code (Python)

class Solution:
    def search(self, arr, key):
        n = len(arr)
        for i in range(n):
            if arr[i] == key:
                return i
        return -1

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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