153. Find Minimum in Rotated Sorted Array (Solution || Leetcode medium || Java)

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.

SOLUTION:

class Solution {
public int findMin(int[] nums) {
int start = 0;//range for binary searchc
int end = nums.length — 1;

if(nums[0] < nums[end] || nums.length < 2){//if the array is not rotated
return nums[0];
}

while(start <= end){//applying binary search if array is rotated
int mid = start + (end — start)/2;

if(nums[mid] > nums[mid+1]){//The number just after pivot is the minimum
return nums[mid +1];
}else if(nums[mid] < nums[mid — 1]){
return nums[mid];
}else if(nums[mid] > nums[start]){
start = mid + 1;
}else{
end = mid — 1;
}
}

return -1;
}
}

TIME COMPLEXITY :- O(log(n))

SPACE COMMPLEXITY :- O(1)

Thank you for reading. If you have any queries, please let me know in the comment section, I will surely response toward that.