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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

Solution 1

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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]

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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];
    }
}

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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];
    }
};

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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]
}

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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];
}

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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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/**
 * @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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