Maximum and minimum of an array using minimum number of comparisons

Software Development
#1

Hello Everyone,

Write a C function to return minimum and maximum in an array. Your program should make the minimum number of comparisons.

First of all, how do we return multiple values from a C 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;

};

And the function declaration becomes: struct pair getMinMax(int arr[], int n) where arr[] is the array of size n whose minimum and maximum are needed.

METHOD (Simple 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)

// 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;

};

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 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;

}

Output:

Minimum element is 1 Maximum element is 3000

Time Complexity: O(n)

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.