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. 

Solution

class Solution {

public int findMin(int[] nums) {

// If the list has just one element then return that element.

if (nums.length == 1) {

return nums[0];

}

// initializing left and right pointers.

int left = 0, right = nums.length - 1;

// if the last element is greater than the first element then there is no

// rotation.

// e.g. 1 < 2 < 3 < 4 < 5 < 7. Already sorted array.

// Hence the smallest element is first element. A[0]

if (nums[right] > nums[0]) {

return nums[0];

}

// Binary search way

while (right >= left) {

// Find the mid element

int mid = left + (right - left) / 2;

// if the mid element is greater than its next element then mid+1 element is the

// smallest

// This point would be the point of change. From higher to lower value.

if (nums[mid] > nums[mid + 1]) {

return nums[mid + 1];

}

// if the mid element is lesser than its previous element then mid element is

// the smallest

if (nums[mid - 1] > nums[mid]) {

return nums[mid];

}

// if the mid elements value is greater than the 0th element this means

// the least value is still somewhere to the right as we are still dealing with

// elements

// greater than nums[0]

if (nums[mid] > nums[0]) {

left = mid + 1;

} else {

// if nums[0] is greater than the mid value then this means the smallest value

// is somewhere to

// the left

right = mid - 1;

}

}

return Integer.MAX_VALUE;

}

}