153. Find Minimum in Rotated Sorted Array
1. Description
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.
2. Example
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.
3. 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.
4. Solutions
My Accepted Solution
n = i_nums.size()
Time complexity: O($log_2n$)
Space complexity: O(1)
class Solution {
public:
// int findMin(vector<int>& nums)
int findMin(const vector<int>& i_nums) {
if(i_nums.front() < i_nums.back()) {
// the first element is the min value
return i_nums.front();
} else if(i_nums.size() == 1 || i_nums[i_nums.size() - 1] < i_nums[i_nums.size() - 2]) {
// the last element is the min value
return i_nums.back();
}
int minValue = INT_MAX;
for(int left = 0, right = i_nums.size() - 1; left <= right;) {
int middle = left + (right - left) / 2;
if(i_nums[middle] < i_nums[middle - 1] && i_nums[middle] < i_nums[middle + 1]) {
minValue = i_nums[middle];
break;
}
i_nums[middle] > i_nums.front() ? left = middle + 1 : right = middle;
}
return minValue;
}
};
4.1 Binary Search
n = i_nums.size()
Time complexity: O($log_2n$)
Space complexity: O(1)
class Solution {
public:
// int findMin(vector<int>& nums)
int findMin(const vector<int>& i_nums) {
int left = 0, right = i_nums.size() - 1;
while(left < right) {
int middle = left + (right - left) / 2;
if (i_nums[middle] < i_nums[right]) {
right = middle;
} else {
left = middle + 1;
}
}
return i_nums[left];
}
};