Find minimum and maximum element in an array

Problem

Given an array A of size N of integers. Your task is to find the minimum and maximum elements in the array.

Note: Return an array that contains 2 elements the first one will be a minimum element and the second will be a maximum of an array.

Examples:

Example 1:

Input:
N = 6
A[] = {3, 2, 1, 56, 10000, 167}
Output: 1 10000
Explanation: minimum and maximum elements of array are 1 and 10000.

Example 2:

Input:
N = 5
A[]  = {1, 345, 234, 21, 56789}
Output: 1 56789
Explanation: minimum and maximum element of array are 1 and 56789.

Your task:

You don't need to read input or print anything. Your task is to complete the function getMinMax() which takes the array A[] and its size N as inputs and returns the minimum and maximum element of the array.

• Expected Time Complexity: $O(N)$
• Expected Auxiliary Space: $O(1)$

Constraints:

• $1 <= N <= 10^5$
• $1 <= A_i <=10^{12}$

Solution

Python

def getMinMax( a, n):
    if n == 0:
        return None 
    min_value = float('inf')  
    max_value = float('-inf')  
    for num in a:
        if num < min_value:
            min_value = num
        if num > max_value:
            max_value = num
    minmax = [min_value, max_value]
    return minmax

Java

static Pair getMinMax(long a[], long n) {
        if (n == 0) {
            return null; 
        }
        long min_value = Long.MAX_VALUE;
        long max_value = Long.MIN_VALUE; 
        for (long num : a) {
            if (num < min_value) {
                min_value = num;
            }
            if (num > max_value) {
                max_value = num;
            }
        }
        return new Pair(min_value, max_value);
}

C++

public:
    pair<long long, long long> getMinMax(long long a[], int n) {
        if (n == 0) {
            return {0, 0}; 
        }
        long long min_value = LLONG_MAX; 
        long long max_value = LLONG_MIN;
        for (int i = 0; i < n; ++i) {
            if (a[i] < min_value) {
                min_value = a[i];
            }
            if (a[i] > max_value) {
                max_value = a[i];
            }
        }
        return {min_value, max_value};
    }
};

C

struct pair getMinMax(long long int arr[], long long int n) {
    struct pair minmax;     
    int i;
    if (n == 1) {
         minmax.max = arr[0];
         minmax.min = arr[0];     
         return minmax;
    }
    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;
}

• Time Complexity: $O(N)$
• Auxiliary Space: $O(1)$