C# || How To Find Minimum In Rotated Sorted Array Using C#
The following is a module with functions which demonstrates how to find the minimum value in a rotated sorted array using C#.
1. Find Min – Problem Statement
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.
2. Find Min – Solution
The following is a solution which demonstrates how to find the minimum value in a rotated sorted array.
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// ============================================================================ // Author: Kenneth Perkins // Date: Oct 22, 2021 // Taken From: http://programmingnotes.org/ // File: Solution.cs // Description: Demonstrates how to find the minimum value in a sorted array // ============================================================================ public class Solution { public int FindMin(int[] nums) { var lo = 0; var hi = nums.Length - 1; while (lo < hi) { var mid = lo + (hi - lo) / 2; // Midpoint element is greater than right side of array, so // the minumum element is somewhere on the right side of array. // Advance lo value to equal midpoint + 1 if (nums[mid] > nums[hi]) { lo = mid + 1; // Midpoint element is less than right side of array, so // the minumum element is somewhere on the left side of array. // Decrease hi value to equal midpoint } else if (nums[mid] < nums[hi]) { hi = mid; // Minimum element found } else { lo = mid; break; } } return nums[lo]; } }// http://programmingnotes.org/ |
QUICK NOTES:
The highlighted lines are sections of interest to look out for.
The code is heavily commented, so no further insight is necessary. If you have any questions, feel free to leave a comment below.
Once compiled, you should get this as your output for the example cases:
1
0
11
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