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MinimumInRotatedSortedArray.java
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MinimumInRotatedSortedArray.java
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package Leetcode;
/**
* @author kalpak
*
* 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, return the minimum element of this array.
*
* 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.
*/
public class MinimumInRotatedSortedArray {
public static int findMin(int[] nums) {
int result = Integer.MIN_VALUE;
int left = 0;
int right = nums.length - 1;
while(left <= right) {
int mid = left + (right - left)/2;
if(nums[left] <= nums[right])
return nums[left];
// Prevents overflow
int next = (mid + 1) % nums.length;
int prev = (mid - 1 + nums.length) % nums.length;
// Check for minimum element
if(nums[mid] <= nums[prev] && nums[mid] <= nums[next]) {
result = nums[mid];
break;
}
// Need to go in the direction of the unsorted array
if(nums[left] <= nums[mid])
left = mid + 1;
else if(nums[mid] <= nums[right])
right = mid - 1;
}
return result;
}
public static void main(String[] args) {
int[] arr = new int[]{3,4,5,1,2};
System.out.println(findMin(arr));
}
}