Product of maximum in first array and minimum in second in C

Given two arrays, we need to find the product of the maximum element from the first array and the minimum element from the second array. This is a common programming problem that demonstrates array traversal and basic mathematical operations.

Syntax

int findMaxElement(int arr[], int size);
int findMinElement(int arr[], int size);
int calculateProduct(int arr1[], int arr2[], int n1, int n2);

Method 1: Using Sorting Approach

In this approach, we sort both arrays and then multiply the last element of the first array with the first element of the second array −

#include <stdio.h>

void sortArray(int arr[], int n) {
    int temp;
    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {
            if (arr[i] > arr[j]) {
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
    }
}

int minMaxProduct(int arr1[], int arr2[], int n1, int n2) {
    sortArray(arr1, n1);
    sortArray(arr2, n2);
    return arr1[n1 - 1] * arr2[0];
}

int main() {
    int arr1[] = {2, 3, 9, 11, 1};
    int arr2[] = {5, 4, 2, 6, 9};
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
    
    printf("Array 1: ");
    for (int i = 0; i < n1; i++) {
        printf("%d ", arr1[i]);
    }
    printf("\nArray 2: ");
    for (int i = 0; i < n2; i++) {
        printf("%d ", arr2[i]);
    }
    
    int result = minMaxProduct(arr1, arr2, n1, n2);
    printf("\nProduct of maximum and minimum: %d<br>", result);
    
    return 0;
}

Array 1: 2 3 9 11 1 
Array 2: 5 4 2 6 9 
Product of maximum and minimum: 22

Method 2: Using Linear Search (More Efficient)

This approach finds the maximum and minimum elements without sorting, which is more efficient −

#include <stdio.h>

int findMax(int arr[], int n) {
    int max = arr[0];
    for (int i = 1; i < n; i++) {
        if (arr[i] > max) {
            max = arr[i];
        }
    }
    return max;
}

int findMin(int arr[], int n) {
    int min = arr[0];
    for (int i = 1; i < n; i++) {
        if (arr[i] < min) {
            min = arr[i];
        }
    }
    return min;
}

int main() {
    int arr1[] = {6, 2, 5, 4, 1};
    int arr2[] = {3, 7, 5, 9, 6};
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
    
    int max1 = findMax(arr1, n1);
    int min2 = findMin(arr2, n2);
    int product = max1 * min2;
    
    printf("Maximum in first array: %d<br>", max1);
    printf("Minimum in second array: %d<br>", min2);
    printf("Product: %d * %d = %d<br>", max1, min2, product);
    
    return 0;
}

Maximum in first array: 6
Minimum in second array: 3
Product: 6 * 3 = 18

Comparison

Conclusion

The linear search approach is more efficient with O(n) time complexity compared to the sorting method's O(n²). It also preserves the original arrays, making it the preferred solution for this problem.