C Exercises: Find the median of two sorted arrays of different size

C Array: Exercise-71 with Solution

Write a program in C to find the median of two sorted arrays of different sizes.

Expected Output:
The given first array is : 90 240 300
The given second array is : 10 13 14 20 25
The median of two different size arrays are : 22.500000

The task is to write a C program to find the median of two sorted arrays of different sizes. The program should merge the two arrays while maintaining the sorted order and then calculate the median value of the combined array. The median is the middle element if the total number of elements is odd, or the average of the two middle elements if the total number is even.

Sample Solution:

C Code:

#include <stdio.h>

// Function to find the maximum of two numbers
int findMax(int arr1, int arr2) {
    return arr1 > arr2 ? arr1 : arr2;
}

// Function to find the minimum of two numbers
int findMin(int arr1, int arr2) {
    return arr1 < arr2 ? arr1 : arr2; 
}

// Function to find the median of two sorted arrays of different sizes
double medianOfDiffSortArrays(int *arr1, int n, int *arr2, int m) {
    int indexMin = 0, indexMax = n, i, j, median;

    // Perform binary search to find the median
    while (indexMin <= indexMax) {
        i = (indexMin + indexMax) / 2;
        j = ((n + m + 1) / 2) - i;

        if (i < n && j > 0 && arr2[j - 1] > arr1[i])
            indexMin = i + 1;
        else if (i > 0 && j < m && arr2[j] < arr1[i - 1])
            indexMax = i - 1;
        else {
            if (i == 0)
                median = arr2[j - 1];
            else if (j == 0)
                median = arr1[i - 1];
            else
                median = findMax(arr1[i - 1], arr2[j - 1]);

            break;
        }
    }

    // Calculate median based on odd or even total elements
    if ((n + m) % 2 == 1)
        return (double)median;
    if (i == n)
        return (median + arr2[j]) / 2.0;
    if (j == m)
        return (median + arr1[i]) / 2.0;

    return (median + findMin(arr1[i], arr2[j])) / 2.0;
}

int main() {
    int arr1[] = {90, 240, 300};
    int arr2[] = {10, 13, 14, 20, 25};
    int n = sizeof(arr1) / sizeof(int);
    int m = sizeof(arr2) / sizeof(int);
    int i;

    // Print the given arrays
    printf("The given first array is :  ");
    for(i = 0; i < n; i++) {
        printf("%d  ", arr1[i]);
    } 
    printf("\n");

    printf("The given second array is :  ");
    for(i = 0; i < m; i++) {
        printf("%d  ", arr2[i]);
    } 
    printf("\n");	

    // Determine the median of two different-sized arrays and print the result
    if (n < m)
        printf("The median of two different size arrays is: %f", medianOfDiffSortArrays(arr1, n, arr2, m));
    else
        printf("The median of two different size arrays is: %f", medianOfDiffSortArrays(arr2, m, arr1, n));

    return 0;
}

Output:

The given first array is :  90  240  300  
The given second array is :  10  13  14  20  25  
The median of two different size arrays are : 22.500000

Flowchart:

C Programming Code Editor:

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