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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.
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.length1 <= n <= 5000-5000 <= nums[i] <= 5000nums are unique.nums is sorted and rotated between 1 and n times.class Solution:
def findMin(self, nums: List[int]) -> int:
l, r = 0, len(nums) - 1
if nums[l] < nums[r]:
return nums[0]
while l < r:
m = (l + r) >> 1
if nums[m] > nums[r]:
l = m + 1
else:
r = m
return nums[l]
class Solution {
public int findMin(int[] nums) {
int l = 0, r = nums.length - 1;
if (nums[l] < nums[r]) return nums[0];
while (l < r) {
int m = (l + r) >>> 1;
if (nums[m] > nums[r]) l = m + 1;
else r = m;
}
return nums[l];
}
}
class Solution {
public:
int findMin(vector<int>& nums) {
int l = 0, r = nums.size() - 1;
if (nums[l] < nums[r]) return nums[0];
while (l < r) {
int m = (l + r) >> 1;
if (nums[m] > nums[r]) l = m + 1;
else r = m;
}
return nums[l];
}
};
func findMin(nums []int) int {
l, r := 0, len(nums) - 1
if nums[l] < nums[r] {
return nums[0]
}
for l < r {
m := (l + r) >> 1
if nums[m] > nums[r] {
l = m + 1
} else {
r = m
}
}
return nums[l]
}
/**
* @param {number[]} nums
* @return {number}
*/
var findMin = function (nums) {
let l = 0,
r = nums.length - 1;
if (nums[l] < nums[r]) return nums[0];
while (l < r) {
const m = (l + r) >> 1;
if (nums[m] > nums[r]) l = m + 1;
else r = m;
}
return nums[l];
};
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