C Program For Maximum and Minimum of an Array

Last Updated : 14 Jul, 2023

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.

struct pair {
    int min;
    int max;
};

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:

/* 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();
}

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:

/* 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();
}

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:

#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("nMinimum element is %d", minmax.min);
printf("nMaximum element is %d", minmax.max);
getchar();
}

Output:

Minimum element is 1
Maximum element is 3000

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

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

Second and third approaches make the 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.