Find Minimum in Rotated Sorted Array
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# LeetCode #153: Find Minimum in Rotated Sorted Array (C/C++).

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source: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array
C/C++ Solution to LeetCode problem 153. Find Minimum in Rotated Sorted Array.

## Problem

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, a, a, ..., a[n-1]] 1 time results in the array [a[n-1], a, a, a, ..., 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.

## Examples

### 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.

## Solution

The logic is pretty similar to the problem #33:

• We identify the sorted part of the array:
• If the value at the lower index < current min value:
• Update minimum value.
• Next search will be in this part of the array.
• If not, then the minimum value is on the unsorted part.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 class Solution { public: int findMin(vector<int>& nums) { int min = nums; int l = 0; int h = nums.size() - 1; int m; while (l <= h) { m = (l + h) / 2; if (nums[m] < min) min = nums[m]; if (nums[l] <= nums[m]) { if (nums[l] < min) h = m - 1; else l = m + 1; } else { if (nums[l] < min) l = m + 1; else h = m - 1; } } return min; } };