# Find the minimum element in a sorted and rotated array

A sorted array is rotated at some unknown point, find the minimum element in it.

Following solution assumes that all elements are distinct.

Examples

```Input: {5, 6, 1, 2, 3, 4}
Output: 1

Input: {1, 2, 3, 4}
Output: 1

Input: {2, 1}
Output: 1```

A simple solution is to traverse the complete array and find minimum. This solution requires Θ(n) time.
We can do it in O(Logn) using Binary Search. If we take a closer look at above examples, we can easily figure out following pattern:

• The minimum element is the only element whose previous is greater than it. If there is no previous element element, then there is no rotation (first element is minimum). We check this condition for middle element by comparing it with (mid-1)’th and (mid+1)’th elements.
• If minimum element is not at middle (neither mid nor mid + 1), then minimum element lies in either left half or right half.
1. If middle element is smaller than last element, then the minimum element lies in left half
2. Else minimum element lies in right half.

We strongly recommend you to try it yourself before seeing the following implementation.

`// C program to find minimum element in a sorted and rotated array`
`#include <stdio.h>`
`int` `findMin(``int` `arr[], ``int` `low, ``int` `high)`
`{`
`    ``// This condition is needed to handle the case when array is not`
`    ``// rotated at all`
`    ``if` `(high < low)  ``return` `arr[0];`
`    ``// If there is only one element left`
`    ``if` `(high == low) ``return` `arr[low];`
`    ``// Find mid`
`    ``int` `mid = low + (high - low)/2; ``/*(low + high)/2;*/`
`    ``// Check if element (mid+1) is minimum element. Consider`
`    ``// the cases like {3, 4, 5, 1, 2}`
`    ``if` `(mid < high && arr[mid+1] < arr[mid])`
`       ``return` `arr[mid+1];`
`    ``// Check if mid itself is minimum element`
`    ``if` `(mid > low && arr[mid] < arr[mid - 1])`
`       ``return` `arr[mid];`
`    ``// Decide whether we need to go to left half or right half`
`    ``if` `(arr[high] > arr[mid])`
`       ``return` `findMin(arr, low, mid-1);`
`    ``return` `findMin(arr, mid+1, high);`
`}`
`// Driver program to test above functions`
`int` `main()`
`{`
`    ``int` `arr1[] =  {5, 6, 1, 2, 3, 4};`
`    ``int` `n1 = ``sizeof``(arr1)/``sizeof``(arr1[0]);`
`    ``printf``(``"The minimum element is %dn"``, findMin(arr1, 0, n1-1));`
`    ``int` `arr2[] =  {1, 2, 3, 4};`
`    ``int` `n2 = ``sizeof``(arr2)/``sizeof``(arr2[0]);`
`    ``printf``(``"The minimum element is %dn"``, findMin(arr2, 0, n2-1));`
`    ``int` `arr3[] =  {1};`
`    ``int` `n3 = ``sizeof``(arr3)/``sizeof``(arr3[0]);`
`    ``printf``(``"The minimum element is %dn"``, findMin(arr3, 0, n3-1));`
`    ``int` `arr4[] =  {1, 2};`
`    ``int` `n4 = ``sizeof``(arr4)/``sizeof``(arr4[0]);`
`    ``printf``(``"The minimum element is %dn"``, findMin(arr4, 0, n4-1));`
`    ``int` `arr5[] =  {2, 1};`
`    ``int` `n5 = ``sizeof``(arr5)/``sizeof``(arr5[0]);`
`    ``printf``(``"The minimum element is %dn"``, findMin(arr5, 0, n5-1));`
`    ``int` `arr6[] =  {5, 6, 7, 1, 2, 3, 4};`
`    ``int` `n6 = ``sizeof``(arr6)/``sizeof``(arr6[0]);`
`    ``printf``(``"The minimum element is %dn"``, findMin(arr6, 0, n6-1));`
`    ``int` `arr7[] =  {1, 2, 3, 4, 5, 6, 7};`
`    ``int` `n7 = ``sizeof``(arr7)/``sizeof``(arr7[0]);`
`    ``printf``(``"The minimum element is %dn"``, findMin(arr7, 0, n7-1));`
`    ``int` `arr8[] =  {2, 3, 4, 5, 6, 7, 8, 1};`
`    ``int` `n8 = ``sizeof``(arr8)/``sizeof``(arr8[0]);`
`    ``printf``(``"The minimum element is %dn"``, findMin(arr8, 0, n8-1));`
`    ``int` `arr9[] =  {3, 4, 5, 1, 2};`
`    ``int` `n9 = ``sizeof``(arr9)/``sizeof``(arr9[0]);`
`    ``printf``(``"The minimum element is %dn"``, findMin(arr9, 0, n9-1));`
`    ``return` `0;`
`}`

Output:

```The minimum element is 1
The minimum element is 1
The minimum element is 1
The minimum element is 1
The minimum element is 1
The minimum element is 1
The minimum element is 1
The minimum element is 1
The minimum element is 1```

How to handle duplicates?
It turned out that duplicates can’t be handled in O(Logn) time in all cases.  The special cases that cause problems are like {2, 2, 2, 2, 2, 2, 2, 2, 0, 1, 1, 2} and {2, 2, 2, 0, 2, 2, 2, 2, 2, 2, 2, 2}. It doesn’t look possible to go to left half or right half by doing constant number of comparisons at the middle. So the problem with repetition can be solved in O(n) worst case.

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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rakesh