Maximum and minimum of an array

Software Development
#1

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 1 (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.
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.

METHOD 2 (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

Implementation

// 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;

}

Output:

Minimum element is 1 Maximum element is 3000

Time Complexity: O(n)