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153. Find Minimum in Rotated Sorted Array

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.

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.

Solutions

  • class Solution { public int findMin(int[] nums) { int n = nums.length; if (nums[0] <= nums[n - 1]) { return nums[0]; } int left = 0, right = n - 1; while (left < right) { int mid = (left + right) >> 1; if (nums[0] <= nums[mid]) { left = mid + 1; } else { right = mid; } } return nums[left]; } } 
  • class Solution { public: int findMin(vector<int>& nums) { int n = nums.size(); if (nums[0] <= nums[n - 1]) return nums[0]; int left = 0, right = n - 1; while (left < right) { int mid = (left + right) >> 1; if (nums[0] <= nums[mid]) left = mid + 1; else right = mid; } return nums[left]; } }; 
  • class Solution: def findMin(self, nums: List[int]) -> int: if nums[0] <= nums[-1]: return nums[0] left, right = 0, len(nums) - 1 while left < right: mid = (left + right) >> 1 if nums[0] <= nums[mid]: left = mid + 1 else: right = mid return nums[left] 
  • func findMin(nums []int) int { n := len(nums) if nums[0] <= nums[n-1] { return nums[0] } left, right := 0, n-1 for left < right { mid := (left + right) >> 1 if nums[0] <= nums[mid] { left = mid + 1 } else { right = mid } } return nums[left] } 
  • function findMin(nums: number[]): number { let left = 0; let right = nums.length - 1; while (left < right) { const mid = (left + right) >>> 1; if (nums[mid] > nums[right]) { left = mid + 1; } else { right = mid; } } return nums[left]; } 
  • /** * @param {number[]} nums * @return {number} */ var findMin = function (nums) { const n = nums.length; if (nums[0] <= nums[n - 1]) { return nums[0]; } let left = 0, right = n - 1; while (left < right) { const mid = (left + right) >> 1; if (nums[0] <= nums[mid]) { left = mid + 1; } else { right = mid; } } return nums[left]; }; 
  • impl Solution { pub fn find_min(nums: Vec<i32>) -> i32 { let mut left = 0; let mut right = nums.len() - 1; while left < right { let mid = left + (right - left) / 2; if nums[mid] > nums[right] { left = mid + 1; } else { right = mid; } } nums[left] } }

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