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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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