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# Maximum and minimum of an array using minimum number of comparisons

Given an array of size N. The task is to find the maximum and the minimum element of the array using the minimum number of comparisons.

Examples:

Input: arr[] = {3, 5, 4, 1, 9}
Output: Minimum element is: 1
Maximum element is: 9

Input: arr[] = {22, 14, 8, 17, 35, 3}
Output:  Minimum element is: 3
Maximum element is: 35

Recommended Practice

First of all, how do we return multiple values from a function? We can do it either using structures or pointers.
We have created a structure named pair (which contains min and max) to return multiple values.

## C

 `struct` `pair {` `    ``int` `min;` `    ``int` `max;` `};`

## C++

 `struct` `pair {` `    ``int` `min;` `    ``int` `max;` `};`   `// This code contributed by Aarti_Rathi`

## Java

 `static` `class` `pair {` `    ``int` `min;` `    ``int` `max;` `};`   `// This code contributed by Rajput-Ji`

## Python3

 `# Python3 implementation`     `class` `pair:`   `    ``def` `__init__(``self``):` `        ``self``.``min` `=` `None` `        ``self``.``max` `=` `None`     `# This code contributed by phasing17`

## C#

 `public` `static` `class` `pair {` `    ``public` `int` `min;` `    ``public` `int` `max;` `};` `// This code contributed by Rajput-Ji`

## Javascript

 ``

## Maximum and minimum of an array using Sorting:

Approach:
One approach to find the maximum and minimum element in an array is to first sort the array in ascending order. Once the array is sorted, the first element of the array will be the minimum element and the last element of the array will be the maximum element.

Algorithm:

1. Initialize an array.
2. Sort the array in ascending order.
3. The first element of the array will be the minimum element.
4. The last element of the array will be the maximum element.
5. Print the minimum and maximum element.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach` `#include ` `#include ` `using` `namespace` `std;`   `struct` `Pair {` `    ``int` `min;` `    ``int` `max;` `};`   `Pair getMinMax(``int` `arr[], ``int` `n)` `{` `    ``Pair minmax;`   `    ``sort(arr, arr + n);`   `    ``minmax.min = arr[0];` `    ``minmax.max = arr[n - 1];`   `    ``return` `minmax;` `}`   `int` `main()` `{` `    ``int` `arr[] = { 1000, 11, 445, 1, 330, 3000 };` `    ``int` `arr_size = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``Pair minmax = getMinMax(arr, arr_size);`   `    ``cout << ``"Minimum element is "` `<< minmax.min << endl;` `    ``cout << ``"Maximum element is "` `<< minmax.max << endl;`   `    ``return` `0;` `}`   `// This code is contributed by Tapesh(tapeshdua420)`

## Java

 `import` `java.util.Arrays;`   `class` `Pair {` `    ``public` `int` `min;` `    ``public` `int` `max;` `}`   `class` `Main {` `    ``static` `Pair getMinMax(``int` `arr[], ``int` `n) {` `        ``Pair minmax = ``new` `Pair();` `        ``Arrays.sort(arr);` `        ``minmax.min = arr[``0``];` `        ``minmax.max = arr[n - ``1``];` `        ``return` `minmax;` `    ``}`   `    ``public` `static` `void` `main(String[] args) {` `        ``int` `arr[] = { ``1000``, ``11``, ``445``, ``1``, ``330``, ``3000` `};` `        ``int` `arr_size = arr.length;` `        ``Pair minmax = getMinMax(arr, arr_size);` `        ``System.out.println(``"Minimum element is "` `+ minmax.min);` `        ``System.out.println(``"Maximum element is "` `+ minmax.max);` `    ``}` `}`

## Python

 `def` `getMinMax(arr):` `    ``arr.sort()` `    ``minmax ``=` `{``"min"``: arr[``0``], ``"max"``: arr[``-``1``]}` `    ``return` `minmax`   `arr ``=` `[``1000``, ``11``, ``445``, ``1``, ``330``, ``3000``]` `minmax ``=` `getMinMax(arr)`   `print``(``"Minimum element is"``, minmax[``"min"``])` `print``(``"Maximum element is"``, minmax[``"max"``])`

Output

```Minimum element is 1
Maximum element is 3000```

### Complexity Analysis:

The time complexity of this approach is O(n log n), where n is the number of elements in the array, as we are using a sorting algorithm. The space complexity is O(1), as we are not using any extra space.

Number of Comparisons:
The number of comparisons made to find the minimum and maximum elements is equal to the number of comparisons made during the sorting process. For any comparison-based sorting algorithm, the minimum number of comparisons required to sort an array of n elements is O(n log n). Hence, the number of comparisons made in this approach is also O(n log n).

## Maximum and minimum of an array using Linear search:

Initialize values of min and max as minimum and maximum of the first two elements respectively. Starting from 3rd, compare each element with max and min, and change max and min accordingly (i.e., if the element is smaller than min then change min, else if the element is greater than max then change max, else ignore the element)

Below is the implementation of the above approach:

## C++

 `// C++ program of above implementation ` `#include` `using` `namespace` `std;`   `// Pair struct is used to return ` `// two values from getMinMax()` `struct` `Pair ` `{` `    ``int` `min;` `    ``int` `max;` `}; `   `Pair getMinMax(``int` `arr[], ``int` `n)` `{` `    ``struct` `Pair minmax;     ` `    ``int` `i;` `    `  `    ``// If there is only one element ` `    ``// then return it as min and max both` `    ``if` `(n == 1)` `    ``{` `        ``minmax.max = arr[0];` `        ``minmax.min = arr[0];     ` `        ``return` `minmax;` `    ``} ` `    `  `    ``// If there are more than one elements,` `    ``// then initialize min and max` `    ``if` `(arr[0] > arr[1]) ` `    ``{` `        ``minmax.max = arr[0];` `        ``minmax.min = arr[1];` `    ``} ` `    ``else` `    ``{` `        ``minmax.max = arr[1];` `        ``minmax.min = arr[0];` `    ``} ` `    `  `    ``for``(i = 2; i < n; i++)` `    ``{` `        ``if` `(arr[i] > minmax.max)     ` `            ``minmax.max = arr[i];` `            `  `        ``else` `if` `(arr[i] < minmax.min)     ` `            ``minmax.min = arr[i];` `    ``}` `    ``return` `minmax;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 1000, 11, 445, ` `                  ``1, 330, 3000 };` `    ``int` `arr_size = 6;` `    `  `    ``struct` `Pair minmax = getMinMax(arr, arr_size);` `    `  `    ``cout << ``"Minimum element is "` `         ``<< minmax.min << endl;` `    ``cout << ``"Maximum element is "` `         ``<< minmax.max;` `         `  `    ``return` `0;` `} `   `// This code is contributed by nik_3112`

## C

 `/* structure is used to return two values from minMax() */` `#include` `struct` `pair ` `{` `  ``int` `min;` `  ``int` `max;` `};  `   `struct` `pair getMinMax(``int` `arr[], ``int` `n)` `{` `  ``struct` `pair minmax;     ` `  ``int` `i;` `  `  `  ``/*If there is only one element then return it as min and max both*/` `  ``if` `(n == 1)` `  ``{` `     ``minmax.max = arr[0];` `     ``minmax.min = arr[0];     ` `     ``return` `minmax;` `  ``}    `   `  ``/* If there are more than one elements, then initialize min ` `      ``and max*/` `  ``if` `(arr[0] > arr[1])  ` `  ``{` `      ``minmax.max = arr[0];` `      ``minmax.min = arr[1];` `  ``}  ` `  ``else` `  ``{` `      ``minmax.max = arr[1];` `      ``minmax.min = arr[0];` `  ``}    `   `  ``for` `(i = 2; i  minmax.max)      ` `      ``minmax.max = arr[i];` `  `  `    ``else` `if` `(arr[i] <  minmax.min)      ` `      ``minmax.min = arr[i];` `  ``}` `  `  `  ``return` `minmax;` `}`   `/* Driver program to test above function */` `int` `main()` `{` `  ``int` `arr[] = {1000, 11, 445, 1, 330, 3000};` `  ``int` `arr_size = 6;` `  ``struct` `pair minmax = getMinMax (arr, arr_size);` `  ``printf``(``"nMinimum element is %d"``, minmax.min);` `  ``printf``(``"nMaximum element is %d"``, minmax.max);` `  ``getchar``();` `}  `

## Java

 `// Java program of above implementation` `public` `class` `GFG {` `/* Class Pair is used to return two values from getMinMax() */` `    ``static` `class` `Pair {`   `        ``int` `min;` `        ``int` `max;` `    ``}`   `    ``static` `Pair getMinMax(``int` `arr[], ``int` `n) {` `        ``Pair minmax = ``new`  `Pair();` `        ``int` `i;`   `        ``/*If there is only one element then return it as min and max both*/` `        ``if` `(n == ``1``) {` `            ``minmax.max = arr[``0``];` `            ``minmax.min = arr[``0``];` `            ``return` `minmax;` `        ``}`   `        ``/* If there are more than one elements, then initialize min ` `    ``and max*/` `        ``if` `(arr[``0``] > arr[``1``]) {` `            ``minmax.max = arr[``0``];` `            ``minmax.min = arr[``1``];` `        ``} ``else` `{` `            ``minmax.max = arr[``1``];` `            ``minmax.min = arr[``0``];` `        ``}`   `        ``for` `(i = ``2``; i < n; i++) {` `            ``if` `(arr[i] > minmax.max) {` `                ``minmax.max = arr[i];` `            ``} ``else` `if` `(arr[i] < minmax.min) {` `                ``minmax.min = arr[i];` `            ``}` `        ``}`   `        ``return` `minmax;` `    ``}`   `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main(String args[]) {` `        ``int` `arr[] = {``1000``, ``11``, ``445``, ``1``, ``330``, ``3000``};` `        ``int` `arr_size = ``6``;` `        ``Pair minmax = getMinMax(arr, arr_size);` `        ``System.out.printf(``"\nMinimum element is %d"``, minmax.min);` `        ``System.out.printf(``"\nMaximum element is %d"``, minmax.max);`   `    ``}`   `}`

## Python3

 `# Python program of above implementation`   `# structure is used to return two values from minMax()`   `class` `pair:` `    ``def` `__init__(``self``):` `        ``self``.``min` `=` `0` `        ``self``.``max` `=` `0`   `def` `getMinMax(arr: ``list``, n: ``int``) ``-``> pair:` `    ``minmax ``=` `pair()`   `    ``# If there is only one element then return it as min and max both` `    ``if` `n ``=``=` `1``:` `        ``minmax.``max` `=` `arr[``0``]` `        ``minmax.``min` `=` `arr[``0``]` `        ``return` `minmax`   `    ``# If there are more than one elements, then initialize min` `    ``# and max` `    ``if` `arr[``0``] > arr[``1``]:` `        ``minmax.``max` `=` `arr[``0``]` `        ``minmax.``min` `=` `arr[``1``]` `    ``else``:` `        ``minmax.``max` `=` `arr[``1``]` `        ``minmax.``min` `=` `arr[``0``]`   `    ``for` `i ``in` `range``(``2``, n):` `        ``if` `arr[i] > minmax.``max``:` `            ``minmax.``max` `=` `arr[i]` `        ``elif` `arr[i] < minmax.``min``:` `            ``minmax.``min` `=` `arr[i]`   `    ``return` `minmax`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[``1000``, ``11``, ``445``, ``1``, ``330``, ``3000``]` `    ``arr_size ``=` `6` `    ``minmax ``=` `getMinMax(arr, arr_size)` `    ``print``(``"Minimum element is"``, minmax.``min``)` `    ``print``(``"Maximum element is"``, minmax.``max``)`   `# This code is contributed by` `# sanjeev2552`

## C#

 `// C# program of above implementation` `using` `System;`   `class` `GFG ` `{` `    ``/* Class Pair is used to return ` `    ``two values from getMinMax() */` `    ``class` `Pair ` `    ``{` `        ``public` `int` `min;` `        ``public` `int` `max;` `    ``}`   `    ``static` `Pair getMinMax(``int` `[]arr, ``int` `n)` `    ``{` `        ``Pair minmax = ``new` `Pair();` `        ``int` `i;`   `        ``/* If there is only one element ` `        ``then return it as min and max both*/` `        ``if` `(n == 1)` `        ``{` `            ``minmax.max = arr[0];` `            ``minmax.min = arr[0];` `            ``return` `minmax;` `        ``}`   `        ``/* If there are more than one elements,` `        ``then initialize min and max*/` `        ``if` `(arr[0] > arr[1])` `        ``{` `            ``minmax.max = arr[0];` `            ``minmax.min = arr[1];` `        ``} ` `        ``else` `        ``{` `            ``minmax.max = arr[1];` `            ``minmax.min = arr[0];` `        ``}`   `        ``for` `(i = 2; i < n; i++)` `        ``{` `            ``if` `(arr[i] > minmax.max) ` `            ``{` `                ``minmax.max = arr[i];` `            ``} ` `            ``else` `if` `(arr[i] < minmax.min)` `            ``{` `                ``minmax.min = arr[i];` `            ``}` `        ``}` `        ``return` `minmax;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main(String []args) ` `    ``{` `        ``int` `[]arr = {1000, 11, 445, 1, 330, 3000};` `        ``int` `arr_size = 6;` `        ``Pair minmax = getMinMax(arr, arr_size);` `        ``Console.Write(``"Minimum element is {0}"``,` `                                   ``minmax.min);` `        ``Console.Write(``"\nMaximum element is {0}"``, ` `                                     ``minmax.max);` `    ``}` `}`   `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output

```Minimum element is 1
Maximum element is 3000```

Time Complexity: O(n)
Auxiliary Space: O(1) as no extra space was needed.

In this method, the total number of comparisons is 1 + 2(n-2) in the worst case and 1 + n â€“ 2 in the best case.
In the above implementation, the worst case occurs when elements are sorted in descending order and the best case occurs when elements are sorted in ascending order.

## Maximum and minimum of an array using the tournament method:

Divide the array into two parts and compare the maximums and minimums of the two parts to get the maximum and the minimum of the whole array.

Pair MaxMin(array, array_size)
if array_size = 1
return element as both max and min
else if arry_size = 2
one comparison to determine max and min
return that pair
else    /* array_size  > 2 */
recur for max and min of left half
recur for max and min of right half
one comparison determines true max of the two candidates
one comparison determines true min of the two candidates
return the pair of max and min

Below is the implementation of the above approach:

## C++

 `// C++ program of above implementation` `#include ` `using` `namespace` `std;`   `// structure is used to return` `// two values from minMax()` `struct` `Pair {` `    ``int` `min;` `    ``int` `max;` `};`   `struct` `Pair getMinMax(``int` `arr[], ``int` `low, ``int` `high)` `{` `    ``struct` `Pair minmax, mml, mmr;` `    ``int` `mid;`   `    ``// If there is only one element` `    ``if` `(low == high) {` `        ``minmax.max = arr[low];` `        ``minmax.min = arr[low];` `        ``return` `minmax;` `    ``}`   `    ``// If there are two elements` `    ``if` `(high == low + 1) {` `        ``if` `(arr[low] > arr[high]) {` `            ``minmax.max = arr[low];` `            ``minmax.min = arr[high];` `        ``}` `        ``else` `{` `            ``minmax.max = arr[high];` `            ``minmax.min = arr[low];` `        ``}` `        ``return` `minmax;` `    ``}`   `    ``// If there are more than 2 elements` `    ``mid = (low + high) / 2;` `    ``mml = getMinMax(arr, low, mid);` `    ``mmr = getMinMax(arr, mid + 1, high);`   `    ``// Compare minimums of two parts` `    ``if` `(mml.min < mmr.min)` `        ``minmax.min = mml.min;` `    ``else` `        ``minmax.min = mmr.min;`   `    ``// Compare maximums of two parts` `    ``if` `(mml.max > mmr.max)` `        ``minmax.max = mml.max;` `    ``else` `        ``minmax.max = mmr.max;`   `    ``return` `minmax;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 1000, 11, 445, 1, 330, 3000 };` `    ``int` `arr_size = 6;`   `    ``struct` `Pair minmax = getMinMax(arr, 0, arr_size - 1);`   `    ``cout << ``"Minimum element is "` `<< minmax.min << endl;` `    ``cout << ``"Maximum element is "` `<< minmax.max;`   `    ``return` `0;` `}`   `// This code is contributed by nik_3112`

## C

 `/* structure is used to return two values from minMax() */` `#include ` `struct` `pair {` `    ``int` `min;` `    ``int` `max;` `};`   `struct` `pair getMinMax(``int` `arr[], ``int` `low, ``int` `high)` `{` `    ``struct` `pair minmax, mml, mmr;` `    ``int` `mid;`   `    ``// If there is only one element` `    ``if` `(low == high) {` `        ``minmax.max = arr[low];` `        ``minmax.min = arr[low];` `        ``return` `minmax;` `    ``}`   `    ``/* If there are two elements */` `    ``if` `(high == low + 1) {` `        ``if` `(arr[low] > arr[high]) {` `            ``minmax.max = arr[low];` `            ``minmax.min = arr[high];` `        ``}` `        ``else` `{` `            ``minmax.max = arr[high];` `            ``minmax.min = arr[low];` `        ``}` `        ``return` `minmax;` `    ``}`   `    ``/* If there are more than 2 elements */` `    ``mid = (low + high) / 2;` `    ``mml = getMinMax(arr, low, mid);` `    ``mmr = getMinMax(arr, mid + 1, high);`   `    ``/* compare minimums of two parts*/` `    ``if` `(mml.min < mmr.min)` `        ``minmax.min = mml.min;` `    ``else` `        ``minmax.min = mmr.min;`   `    ``/* compare maximums of two parts*/` `    ``if` `(mml.max > mmr.max)` `        ``minmax.max = mml.max;` `    ``else` `        ``minmax.max = mmr.max;`   `    ``return` `minmax;` `}`   `/* Driver program to test above function */` `int` `main()` `{` `    ``int` `arr[] = { 1000, 11, 445, 1, 330, 3000 };` `    ``int` `arr_size = 6;` `    ``struct` `pair minmax = getMinMax(arr, 0, arr_size - 1);` `    ``printf``(``"nMinimum element is %d"``, minmax.min);` `    ``printf``(``"nMaximum element is %d"``, minmax.max);` `    ``getchar``();` `}`

## Java

 `// Java program of above implementation` `public` `class` `GFG {` `    ``/* Class Pair is used to return two values from` `     ``* getMinMax() */` `    ``static` `class` `Pair {`   `        ``int` `min;` `        ``int` `max;` `    ``}`   `    ``static` `Pair getMinMax(``int` `arr[], ``int` `low, ``int` `high)` `    ``{` `        ``Pair minmax = ``new` `Pair();` `        ``Pair mml = ``new` `Pair();` `        ``Pair mmr = ``new` `Pair();` `        ``int` `mid;`   `        ``// If there is only one element` `        ``if` `(low == high) {` `            ``minmax.max = arr[low];` `            ``minmax.min = arr[low];` `            ``return` `minmax;` `        ``}`   `        ``/* If there are two elements */` `        ``if` `(high == low + ``1``) {` `            ``if` `(arr[low] > arr[high]) {` `                ``minmax.max = arr[low];` `                ``minmax.min = arr[high];` `            ``}` `            ``else` `{` `                ``minmax.max = arr[high];` `                ``minmax.min = arr[low];` `            ``}` `            ``return` `minmax;` `        ``}`   `        ``/* If there are more than 2 elements */` `        ``mid = (low + high) / ``2``;` `        ``mml = getMinMax(arr, low, mid);` `        ``mmr = getMinMax(arr, mid + ``1``, high);`   `        ``/* compare minimums of two parts*/` `        ``if` `(mml.min < mmr.min) {` `            ``minmax.min = mml.min;` `        ``}` `        ``else` `{` `            ``minmax.min = mmr.min;` `        ``}`   `        ``/* compare maximums of two parts*/` `        ``if` `(mml.max > mmr.max) {` `            ``minmax.max = mml.max;` `        ``}` `        ``else` `{` `            ``minmax.max = mmr.max;` `        ``}`   `        ``return` `minmax;` `    ``}`   `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``int` `arr[] = { ``1000``, ``11``, ``445``, ``1``, ``330``, ``3000` `};` `        ``int` `arr_size = ``6``;` `        ``Pair minmax = getMinMax(arr, ``0``, arr_size - ``1``);` `        ``System.out.printf(``"\nMinimum element is %d"``,` `                          ``minmax.min);` `        ``System.out.printf(``"\nMaximum element is %d"``,` `                          ``minmax.max);` `    ``}` `}`

## Python3

 `# Python program of above implementation` `def` `getMinMax(low, high, arr):` `    ``arr_max ``=` `arr[low]` `    ``arr_min ``=` `arr[low]`   `    ``# If there is only one element` `    ``if` `low ``=``=` `high:` `        ``arr_max ``=` `arr[low]` `        ``arr_min ``=` `arr[low]` `        ``return` `(arr_max, arr_min)`   `    ``# If there is only two element` `    ``elif` `high ``=``=` `low ``+` `1``:` `        ``if` `arr[low] > arr[high]:` `            ``arr_max ``=` `arr[low]` `            ``arr_min ``=` `arr[high]` `        ``else``:` `            ``arr_max ``=` `arr[high]` `            ``arr_min ``=` `arr[low]` `        ``return` `(arr_max, arr_min)` `    ``else``:`   `        ``# If there are more than 2 elements` `        ``mid ``=` `int``((low ``+` `high) ``/` `2``)` `        ``arr_max1, arr_min1 ``=` `getMinMax(low, mid, arr)` `        ``arr_max2, arr_min2 ``=` `getMinMax(mid ``+` `1``, high, arr)`   `    ``return` `(``max``(arr_max1, arr_max2), ``min``(arr_min1, arr_min2))`     `# Driver code` `arr ``=` `[``1000``, ``11``, ``445``, ``1``, ``330``, ``3000``]` `high ``=` `len``(arr) ``-` `1` `low ``=` `0` `arr_max, arr_min ``=` `getMinMax(low, high, arr)` `print``(``'Minimum element is '``, arr_min)` `print``(``'nMaximum element is '``, arr_max)`   `# This code is contributed by DeepakChhitarka`

## C#

 `// C# implementation of the approach` `using` `System;`   `public` `class` `GFG {` `    ``/* Class Pair is used to return two values from` `     ``* getMinMax() */` `    ``public` `class` `Pair {`   `        ``public` `int` `min;` `        ``public` `int` `max;` `    ``}`   `    ``static` `Pair getMinMax(``int``[] arr, ``int` `low, ``int` `high)` `    ``{` `        ``Pair minmax = ``new` `Pair();` `        ``Pair mml = ``new` `Pair();` `        ``Pair mmr = ``new` `Pair();` `        ``int` `mid;`   `        ``// If there is only one element` `        ``if` `(low == high) {` `            ``minmax.max = arr[low];` `            ``minmax.min = arr[low];` `            ``return` `minmax;` `        ``}`   `        ``/* If there are two elements */` `        ``if` `(high == low + 1) {` `            ``if` `(arr[low] > arr[high]) {` `                ``minmax.max = arr[low];` `                ``minmax.min = arr[high];` `            ``}` `            ``else` `{` `                ``minmax.max = arr[high];` `                ``minmax.min = arr[low];` `            ``}` `            ``return` `minmax;` `        ``}`   `        ``/* If there are more than 2 elements */` `        ``mid = (low + high) / 2;` `        ``mml = getMinMax(arr, low, mid);` `        ``mmr = getMinMax(arr, mid + 1, high);`   `        ``/* compare minimums of two parts*/` `        ``if` `(mml.min < mmr.min) {` `            ``minmax.min = mml.min;` `        ``}` `        ``else` `{` `            ``minmax.min = mmr.min;` `        ``}`   `        ``/* compare maximums of two parts*/` `        ``if` `(mml.max > mmr.max) {` `            ``minmax.max = mml.max;` `        ``}` `        ``else` `{` `            ``minmax.max = mmr.max;` `        ``}`   `        ``return` `minmax;` `    ``}`   `    ``/* Driver program to test above function */` `    ``public` `static` `void` `Main(String[] args)` `    ``{` `        ``int``[] arr = { 1000, 11, 445, 1, 330, 3000 };` `        ``int` `arr_size = 6;` `        ``Pair minmax = getMinMax(arr, 0, arr_size - 1);` `        ``Console.Write(``"\nMinimum element is {0}"``,` `                      ``minmax.min);` `        ``Console.Write(``"\nMaximum element is {0}"``,` `                      ``minmax.max);` `    ``}` `}`   `// This code contributed by Rajput-Ji`

## Javascript

 ``

Output

```Minimum element is 1
Maximum element is 3000```

Time Complexity: O(n)
Auxiliary Space: O(log n) as the stack space will be filled for the maximum height of the tree formed during recursive calls same as a binary tree.

Total number of comparisons: let the number of comparisons be T(n). T(n) can be written as follows:

```T(n) = T(floor(n/2)) + T(ceil(n/2)) + 2
T(2) = 1
T(1) = 0```

If n is a power of 2, then we can write T(n) as:

`T(n) = 2T(n/2) + 2`

After solving the above recursion, we get

`T(n)  = 3n/2 -2`

Thus, the approach does 3n/2 -2 comparisons if n is a power of 2. And it does more than 3n/2 -2 comparisons if n is not a power of 2.

## Maximum and minimum of an array by comparing in pairs:

If n is odd then initialize min and max as the first element.
If n is even then initialize min and max as minimum and maximum of the first two elements respectively.
For the rest of the elements, pick them in pairs and compare their
maximum and minimum with max and min respectively.

Below is the implementation of the above approach:

## C++

 `// C++ program of above implementation ` `#include ` `using` `namespace` `std; `   `// Structure is used to return ` `// two values from minMax() ` `struct` `Pair ` `{ ` `    ``int` `min; ` `    ``int` `max; ` `}; `   `struct` `Pair getMinMax(``int` `arr[], ``int` `n) ` `{ ` `    ``struct` `Pair minmax;     ` `    ``int` `i; ` `    `  `    ``// If array has even number of elements ` `    ``// then initialize the first two elements ` `    ``// as minimum and maximum ` `    ``if` `(n % 2 == 0) ` `    ``{ ` `        ``if` `(arr[0] > arr[1])     ` `        ``{ ` `            ``minmax.max = arr[0]; ` `            ``minmax.min = arr[1]; ` `        ``} ` `        ``else` `        ``{ ` `            ``minmax.min = arr[0]; ` `            ``minmax.max = arr[1]; ` `        ``} ` `        `  `        ``// Set the starting index for loop ` `        ``i = 2; ` `    ``} ` `    `  `    ``// If array has odd number of elements ` `    ``// then initialize the first element as ` `    ``// minimum and maximum ` `    ``else` `    ``{ ` `        ``minmax.min = arr[0]; ` `        ``minmax.max = arr[0]; ` `        `  `        ``// Set the starting index for loop ` `        ``i = 1; ` `    ``} ` `    `  `    ``// In the while loop, pick elements in ` `    ``// pair and compare the pair with max ` `    ``// and min so far ` `    ``while` `(i < n - 1) ` `    ``{         ` `        ``if` `(arr[i] > arr[i + 1])         ` `        ``{ ` `            ``if``(arr[i] > minmax.max)     ` `                ``minmax.max = arr[i]; ` `                `  `            ``if``(arr[i + 1] < minmax.min)         ` `                ``minmax.min = arr[i + 1];     ` `        ``} ` `        ``else`        `        ``{ ` `            ``if` `(arr[i + 1] > minmax.max)     ` `                ``minmax.max = arr[i + 1]; ` `                `  `            ``if` `(arr[i] < minmax.min)         ` `                ``minmax.min = arr[i];     ` `        ``} ` `        `  `        ``// Increment the index by 2 as ` `        ``// two elements are processed in loop ` `        ``i += 2; ` `    ``}         ` `    ``return` `minmax; ` `} `   `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1000, 11, 445, ` `                ``1, 330, 3000 }; ` `    ``int` `arr_size = 6; ` `    `  `    ``Pair minmax = getMinMax(arr, arr_size); ` `    `  `    ``cout << ``"Minimum element is "` `        ``<< minmax.min << endl; ` `    ``cout << ``"Maximum element is "` `        ``<< minmax.max; ` `        `  `    ``return` `0; ` `} `   `// This code is contributed by nik_3112 `

## C

 `#include`   `/* structure is used to return two values from minMax() */` `struct` `pair ` `{` `  ``int` `min;` `  ``int` `max;` `};  `   `struct` `pair getMinMax(``int` `arr[], ``int` `n)` `{` `  ``struct` `pair minmax;     ` `  ``int` `i;  `   `  ``/* If array has even number of elements then ` `    ``initialize the first two elements as minimum and ` `    ``maximum */` `  ``if` `(n%2 == 0)` `  ``{         ` `    ``if` `(arr[0] > arr[1])     ` `    ``{` `      ``minmax.max = arr[0];` `      ``minmax.min = arr[1];` `    ``}  ` `    ``else` `    ``{` `      ``minmax.min = arr[0];` `      ``minmax.max = arr[1];` `    ``}` `    ``i = 2;  ``/* set the starting index for loop */` `  ``}  `   `   ``/* If array has odd number of elements then ` `    ``initialize the first element as minimum and ` `    ``maximum */` `  ``else` `  ``{` `    ``minmax.min = arr[0];` `    ``minmax.max = arr[0];` `    ``i = 1;  ``/* set the starting index for loop */` `  ``}` `  `  `  ``/* In the while loop, pick elements in pair and ` `     ``compare the pair with max and min so far */`    `  ``while` `(i < n-1)  ` `  ``{          ` `    ``if` `(arr[i] > arr[i+1])          ` `    ``{` `      ``if``(arr[i] > minmax.max)        ` `        ``minmax.max = arr[i];` `      ``if``(arr[i+1] < minmax.min)          ` `        ``minmax.min = arr[i+1];        ` `    ``} ` `    ``else`         `    ``{` `      ``if` `(arr[i+1] > minmax.max)        ` `        ``minmax.max = arr[i+1];` `      ``if` `(arr[i] < minmax.min)          ` `        ``minmax.min = arr[i];        ` `    ``}        ` `    ``i += 2; ``/* Increment the index by 2 as two ` `               ``elements are processed in loop */` `  ``}            `   `  ``return` `minmax;` `}    `   `/* Driver program to test above function */` `int` `main()` `{` `  ``int` `arr[] = {1000, 11, 445, 1, 330, 3000};` `  ``int` `arr_size = 6;` `  ``struct` `pair minmax = getMinMax (arr, arr_size);` `  ``printf``(``"Minimum element is %d"``, minmax.min);` `  ``printf``(``"\nMaximum element is %d"``, minmax.max);` `  ``getchar``();` `}`

## Java

 `// Java program of above implementation` `public` `class` `GFG {`   `/* Class Pair is used to return two values from getMinMax() */` `    ``static` `class` `Pair {`   `        ``int` `min;` `        ``int` `max;` `    ``}`   `    ``static` `Pair getMinMax(``int` `arr[], ``int` `n) {` `        ``Pair minmax = ``new` `Pair();` `        ``int` `i;` `        ``/* If array has even number of elements then  ` `    ``initialize the first two elements as minimum and  ` `    ``maximum */` `        ``if` `(n % ``2` `== ``0``) {` `            ``if` `(arr[``0``] > arr[``1``]) {` `                ``minmax.max = arr[``0``];` `                ``minmax.min = arr[``1``];` `            ``} ``else` `{` `                ``minmax.min = arr[``0``];` `                ``minmax.max = arr[``1``];` `            ``}` `            ``i = ``2``;` `            ``/* set the starting index for loop */` `        ``} ``/* If array has odd number of elements then  ` `    ``initialize the first element as minimum and  ` `    ``maximum */` `else` `{` `            ``minmax.min = arr[``0``];` `            ``minmax.max = arr[``0``];` `            ``i = ``1``;` `            ``/* set the starting index for loop */` `        ``}`   `        ``/* In the while loop, pick elements in pair and  ` `     ``compare the pair with max and min so far */` `        ``while` `(i < n - ``1``) {` `            ``if` `(arr[i] > arr[i + ``1``]) {` `                ``if` `(arr[i] > minmax.max) {` `                    ``minmax.max = arr[i];` `                ``}` `                ``if` `(arr[i + ``1``] < minmax.min) {` `                    ``minmax.min = arr[i + ``1``];` `                ``}` `            ``} ``else` `{` `                ``if` `(arr[i + ``1``] > minmax.max) {` `                    ``minmax.max = arr[i + ``1``];` `                ``}` `                ``if` `(arr[i] < minmax.min) {` `                    ``minmax.min = arr[i];` `                ``}` `            ``}` `            ``i += ``2``;` `            ``/* Increment the index by 2 as two  ` `               ``elements are processed in loop */` `        ``}`   `        ``return` `minmax;` `    ``}`   `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main(String args[]) {` `        ``int` `arr[] = {``1000``, ``11``, ``445``, ``1``, ``330``, ``3000``};` `        ``int` `arr_size = ``6``;` `        ``Pair minmax = getMinMax(arr, arr_size);` `        ``System.out.printf(``"Minimum element is %d"``, minmax.min);` `        ``System.out.printf(``"\nMaximum element is %d"``, minmax.max);`   `    ``}` `}`

## Python3

 `# Python3 program of above implementation ` `def` `getMinMax(arr):` `    `  `    ``n ``=` `len``(arr)` `    `  `    ``# If array has even number of elements then ` `    ``# initialize the first two elements as minimum ` `    ``# and maximum ` `    ``if``(n ``%` `2` `=``=` `0``):` `        ``mx ``=` `max``(arr[``0``], arr[``1``])` `        ``mn ``=` `min``(arr[``0``], arr[``1``])` `        `  `        ``# set the starting index for loop ` `        ``i ``=` `2` `        `  `    ``# If array has odd number of elements then ` `    ``# initialize the first element as minimum ` `    ``# and maximum ` `    ``else``:` `        ``mx ``=` `mn ``=` `arr[``0``]` `        `  `        ``# set the starting index for loop ` `        ``i ``=` `1` `        `  `    ``# In the while loop, pick elements in pair and ` `    ``# compare the pair with max and min so far ` `    ``while``(i < n ``-` `1``):` `        ``if` `arr[i] < arr[i ``+` `1``]:` `            ``mx ``=` `max``(mx, arr[i ``+` `1``])` `            ``mn ``=` `min``(mn, arr[i])` `        ``else``:` `            ``mx ``=` `max``(mx, arr[i])` `            ``mn ``=` `min``(mn, arr[i ``+` `1``])` `            `  `        ``# Increment the index by 2 as two ` `        ``# elements are processed in loop ` `        ``i ``+``=` `2` `    `  `    ``return` `(mx, mn)` `    `  `# Driver Code` `if` `__name__ ``=``=``'__main__'``:` `    `  `    ``arr ``=` `[``1000``, ``11``, ``445``, ``1``, ``330``, ``3000``]` `    ``mx, mn ``=` `getMinMax(arr)` `    ``print``(``"Minimum element is"``, mn)` `    ``print``(``"Maximum element is"``, mx)` `    `  `# This code is contributed by Kaustav`

## C#

 `// C# program of above implementation` `using` `System;` `    `  `class` `GFG ` `{`   `    ``/* Class Pair is used to return ` `       ``two values from getMinMax() */` `    ``public` `class` `Pair ` `    ``{` `        ``public` `int` `min;` `        ``public` `int` `max;` `    ``}`   `    ``static` `Pair getMinMax(``int` `[]arr, ``int` `n)` `    ``{` `        ``Pair minmax = ``new` `Pair();` `        ``int` `i;` `        `  `        ``/* If array has even number of elements ` `        ``then initialize the first two elements ` `        ``as minimum and maximum */` `        ``if` `(n % 2 == 0) ` `        ``{` `            ``if` `(arr[0] > arr[1])` `            ``{` `                ``minmax.max = arr[0];` `                ``minmax.min = arr[1];` `            ``} ` `            ``else` `            ``{` `                ``minmax.min = arr[0];` `                ``minmax.max = arr[1];` `            ``}` `            ``i = 2;` `        ``}` `        `  `        ``/* set the starting index for loop */` `        ``/* If array has odd number of elements then ` `        ``initialize the first element as minimum and ` `        ``maximum */` `        ``else` `        ``{` `            ``minmax.min = arr[0];` `            ``minmax.max = arr[0];` `            ``i = 1;` `            ``/* set the starting index for loop */` `        ``}`   `        ``/* In the while loop, pick elements in pair and ` `        ``compare the pair with max and min so far */` `        ``while` `(i < n - 1) ` `        ``{` `            ``if` `(arr[i] > arr[i + 1]) ` `            ``{` `                ``if` `(arr[i] > minmax.max) ` `                ``{` `                    ``minmax.max = arr[i];` `                ``}` `                ``if` `(arr[i + 1] < minmax.min)` `                ``{` `                    ``minmax.min = arr[i + 1];` `                ``}` `            ``} ` `            ``else` `            ``{` `                ``if` `(arr[i + 1] > minmax.max)` `                ``{` `                    ``minmax.max = arr[i + 1];` `                ``}` `                ``if` `(arr[i] < minmax.min)` `                ``{` `                    ``minmax.min = arr[i];` `                ``}` `            ``}` `            ``i += 2;` `            `  `            ``/* Increment the index by 2 as two ` `            ``elements are processed in loop */` `        ``}` `        ``return` `minmax;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main(String []args) ` `    ``{` `        ``int` `[]arr = {1000, 11, 445, 1, 330, 3000};` `        ``int` `arr_size = 6;` `        ``Pair minmax = getMinMax(arr, arr_size);` `        ``Console.Write(``"Minimum element is {0}"``,` `                                   ``minmax.min);` `        ``Console.Write(``"\nMaximum element is {0}"``, ` `                                     ``minmax.max);` `    ``}` `}`   `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output

```Minimum element is 1
Maximum element is 3000```

Time Complexity: O(n)
Auxiliary Space: O(1) as no extra space was needed.

The total number of comparisons: Different for even and odd n, see below:

```       If n is odd:    3*(n-1)/2
If n is even:   1 Initial comparison for initializing min and max,
and 3(n-2)/2 comparisons for rest of the elements
=  1 + 3*(n-2)/2 = 3n/2 -2```

The second and third approaches make an equal number of comparisons when n is a power of 2.
In general, method 3 seems to be the best.
Please write comments if you find any bug in the above programs/algorithms or a better way to solve the same problem.

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