Find Minimum in Rotated Sorted Array - Array - Medium - LeetCode
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Find Minimum in Rotated Sorted Array - Array - Medium - LeetCode

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  • 1A rotated sorted array can be restored to its original order by identifying the minimum element after rotation.
  • 2The algorithm to find the minimum element operates in O(log n) time complexity using a binary search approach.
  • 3The solution requires only O(1) space complexity, making it efficient for large arrays.

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"A rotated sorted array can be restored to its original order by identifying the minimum element after rotation."

Find Minimum in Rotated Sorted Array - Array - Medium - LeetCode

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, 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.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.

public class Solution {
 public int FindMin(int[] nums) {
 if(nums.Length ==1){
 return nums[0];
 }
 
 int start = 0;
 int end = nums.Length-1;
 while(start < end){
 int mid = start + (end-start)/2;
 if(nums[mid] > nums[end]){
 start = mid+1;
 }
 else{
 end = mid;
 }
 }
 
 return nums[start];
 }
}

Time Complexity: O(logn)

Space Complexity: O(1)

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sumitc91

Published on 18 November 2020 · 1 min read · 217 words

Part of AskGif Blog · coding

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