## C# || How To Find Minimum In Rotated Sorted Array With Duplicates Using C#

The following is a module with functions which demonstrates how to find the minimum value in a rotated sorted array with duplicates using C#.

1. Find Min Duplicates – Problem Statement

Suppose an array of length **n** sorted in ascending order is **rotated** between **1** and **n** times. For example, the array **nums = [0,1,4,4,5,6,7]** might become:

**[4,5,6,7,0,1,4]**if it was rotated**4**times.**[0,1,4,4,5,6,7]**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** that may contain **duplicates**, return *the minimum element of this array*.

You must decrease the overall operation steps as much as possible.

**Example 1:**

Input: nums = [1,3,5]

Output: 1

**Example 2:**

Input: nums = [2,2,2,0,1]

Output: 0

2. Find Min Duplicates – Solution

The following is a solution which demonstrates how to find the minimum value in a rotated sorted array with duplicates.

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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; // In cases where the front and back of the array are the // same (ex: [1,19,28,87,91,0,1,1,1]), decrease the end of the array while (nums[lo] == nums[hi] && lo < hi) { --hi; } // Perform binary search 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; // Duplicate element found (nums[mid] == nums[hi]) } else { --hi; } } 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

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