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

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

<script>

class pair
{
  constructor(){
    this.min = null;
    this.max = null;
  }
};

// This code contributed by Saurabh Jaiswal

</script>

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 <algorithm>
#include <iostream>
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<iostream>
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<stdio.h>
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<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 */
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

<script>
// JavaScript program of above implementation

/* Class Pair is used to return two values from getMinMax() */
    function getMinMax(arr, n)
    {
        minmax = new  Array();
        var i;
        var min;
        var max;

        /*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 */

        var arr = [1000, 11, 445, 1, 330, 3000];
        var arr_size = 6;
        minmax = getMinMax(arr, arr_size);
        document.write("\nMinimum element is " ,minmax.min +"<br>");
        document.write("\nMaximum element is " , minmax.max);

// This code is contributed by shivanisinghss2110
</script>

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 – 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 <iostream>
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 <stdio.h>
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

<script>
// Javascript program of above implementation

    /* Class Pair is used to return two values from getMinMax() */
     class Pair {
         constructor(){
        this.min = -1;
        this.max = 10000000;
         }
    }

     function getMinMax(arr , low , high) {
        var minmax = new Pair();
        var mml = new Pair();
        var mmr = new Pair();
        var 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 = parseInt((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 */
        var arr = [ 1000, 11, 445, 1, 330, 3000 ];
        var arr_size = 6;
        var minmax = getMinMax(arr, 0, arr_size - 1);
        document.write("\nMinimum element is ", minmax.min);
        document.write("<br/>Maximum element is ", minmax.max);

// This code is contributed by Rajput-Ji
</script>

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: 
Algorithmic Paradigm: Divide and Conquer 

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<iostream>
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<stdio.h>

/* 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

<script>

// JavaScript program of above implementation
function getMinMax(arr){

    let n = arr.length
    let mx,mn,i

    // If array has even number of elements then
    // initialize the first two elements as minimum
    // and maximum
    if(n % 2 == 0){
        mx = Math.max(arr[0], arr[1])
        mn = Math.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 = Math.max(mx, arr[i + 1])
            mn = Math.min(mn, arr[i])
        }
        else{
            mx = Math.max(mx, arr[i])
            mn = Math.min(mn, arr[i + 1])
        }

        // Increment the index by 2 as two
        // elements are processed in loop
        i += 2
    }

    return [mx, mn]
}

// Driver Code

let arr = [1000, 11, 445, 1, 330, 3000]
let mx = getMinMax(arr)[0]
let mn = getMinMax(arr)[1]
document.write("Minimum element is", mn,"</br>")
document.write("Maximum element is", mx,"</br>")

// This code is contributed by shinjanpatra

</script>

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.