LeetCode Challenge #153. Find Minimum in Rotated Sorted Array

Day 24 of 100 Days LeetCode Challenge

LeetCode Challenge #153. 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.

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.

Video Solution

Java Solution

				
					class Solution {
    public int findMin(int[] nums) {
       if(nums.length==1){
           return nums[0];
       }else if ( nums[0]<nums[nums.length-1]){
           return nums[0];
       }

       int start = 0 ;
       int end = nums.length-1 ;

       while(start<=end){
           int mid = (start + end )/2;

           if(mid!=0 && nums[mid-1]>nums[mid]){
               return nums[mid];
           }else if( mid!=nums.length-1 && nums[mid]>nums[mid+1] ){
               return nums[mid+1];
           }else if(nums[start]<=nums[mid]){ // left part is sorted 
             start = mid+1 ;
           }else{ // right part is sorted ;
                end = mid-1 ;
           }
       }
       return -1 ;
    }
}

Day 24 of 100 Days LeetCode Challenge