C / C++ Program for Median of two sorted arrays of same size - GeeksforGeeks

11 Dec, 2018
There are 2 sorted arrays A and B of size n each. Write an algorithm to find the median of the array obtained merging the above 2 arrays(i.e. array of length 2n). The complexity should be O(log(n)).
Note : Since size of the set for which we are looking for median is even (2n), we need take average of middle two numbers and return floor of the average.
Method 1 (Simply count while Merging)Use merge procedure of merge sort. Keep track of count while comparing elements of two arrays. If count becomes n(For 2n elements), we have reached the median. Take the average of the elements at indexes n-1 and n in the merged array. See the below implementation.
C++
C
// A Simple Merge based O(n)// solution to find median of// two sorted arrays#include <bits/stdc++.h>using namespace std;  /* This function returns median of ar1[] and ar2[].Assumptions in this function:Both ar1[] and ar2[] are sorted arraysBoth have n elements */int getMedian(int ar1[],              int ar2[], int n){    int i = 0; /* Current index of                   i/p array ar1[] */    int j = 0; /* Current index of                   i/p array ar2[] */    int count;    int m1 = -1, m2 = -1;      /* Since there are 2n elements,     median will be average of elements     at index n-1 and n in the array     obtained after merging ar1 and ar2 */    for (count = 0; count <= n; count++) {        /* Below is to handle case where            all elements of ar1[] are           smaller than smallest(or first)           element of ar2[]*/        if (i == n) {            m1 = m2;            m2 = ar2[0];            break;        }          /*Below is to handle case where           all elements of ar2[] are          smaller than smallest(or first)          element of ar1[]*/        else if (j == n) {            m1 = m2;            m2 = ar1[0];            break;        }          if (ar1[i] < ar2[j]) {            /* Store the prev median */            m1 = m2;            m2 = ar1[i];            i++;        }        else {            /* Store the prev median */            m1 = m2;            m2 = ar2[j];            j++;        }    }      return (m1 + m2) / 2;}  // Driver Codeint main(){    int ar1[] = { 1, 12, 15, 26, 38 };    int ar2[] = { 2, 13, 17, 30, 45 };      int n1 = sizeof(ar1) / sizeof(ar1[0]);    int n2 = sizeof(ar2) / sizeof(ar2[0]);    if (n1 == n2)        cout << "Median is "             << getMedian(ar1, ar2, n1);    else        cout << "Doesn't work for arrays"             << " of unequal size";    getchar();    return 0;}  // This code is contributed// by Shivi_Aggarwal
// A Simple Merge based O(n) solution to find median of// two sorted arrays#include <stdio.h>  /* This function returns median of ar1[] and ar2[].   Assumptions in this function:   Both ar1[] and ar2[] are sorted arrays   Both have n elements */int getMedian(int ar1[], int ar2[], int n){    int i = 0; /* Current index of i/p array ar1[] */    int j = 0; /* Current index of i/p array ar2[] */    int count;    int m1 = -1, m2 = -1;      /* Since there are 2n elements, median will be average     of elements at index n-1 and n in the array obtained after     merging ar1 and ar2 */    for (count = 0; count <= n; count++) {        /*Below is to handle case where all elements of ar1[] are          smaller than smallest(or first) element of ar2[]*/        if (i == n) {            m1 = m2;            m2 = ar2[0];            break;        }          /*Below is to handle case where all elements of ar2[] are          smaller than smallest(or first) element of ar1[]*/        else if (j == n) {            m1 = m2;            m2 = ar1[0];            break;        }          if (ar1[i] < ar2[j]) {            m1 = m2; /* Store the prev median */            m2 = ar1[i];            i++;        }        else {            m1 = m2; /* Store the prev median */            m2 = ar2[j];            j++;        }    }      return (m1 + m2) / 2;}  /* Driver program to test above function */int main(){    int ar1[] = { 1, 12, 15, 26, 38 };    int ar2[] = { 2, 13, 17, 30, 45 };      int n1 = sizeof(ar1) / sizeof(ar1[0]);    int n2 = sizeof(ar2) / sizeof(ar2[0]);    if (n1 == n2)        printf("Median is %d", getMedian(ar1, ar2, n1));    else        printf("Doesn't work for arrays of unequal size");    getchar();    return 0;}
Median is 16

Method 2 (By comparing the medians of two arrays)This method works by first getting medians of the two sorted arrays and then comparing them.
C++
C
// A divide and conquer based// efficient solution to find// median of two sorted arrays// of same size.#include <bits/stdc++.h>using namespace std;  /* to get median of a   sorted array */int median(int[], int);  /* This function returns median    of ar1[] and ar2[].Assumptions in this function:    Both ar1[] and ar2[] are     sorted arrays    Both have n elements */int getMedian(int ar1[],              int ar2[], int n){    /* return -1 for        invalid input */    if (n <= 0)        return -1;    if (n == 1)        return (ar1[0] + ar2[0]) / 2;    if (n == 2)        return (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1])) / 2;      /* get the median of        the first array */    int m1 = median(ar1, n);      /* get the median of        the second array */    int m2 = median(ar2, n);      /* If medians are equal then        return either m1 or m2 */    if (m1 == m2)        return m1;      /* if m1 < m2 then median must        exist in ar1[m1....] and                ar2[....m2] */    if (m1 < m2) {        if (n % 2 == 0)            return getMedian(ar1 + n / 2 - 1,                             ar2, n - n / 2 + 1);        return getMedian(ar1 + n / 2,                         ar2, n - n / 2);    }      /* if m1 > m2 then median must        exist in ar1[....m1] and                 ar2[m2...] */    if (n % 2 == 0)        return getMedian(ar2 + n / 2 - 1,                         ar1, n - n / 2 + 1);    return getMedian(ar2 + n / 2,                     ar1, n - n / 2);}  /* Function to get median    of a sorted array */int median(int arr[], int n){    if (n % 2 == 0)        return (arr[n / 2] + arr[n / 2 - 1]) / 2;    else        return arr[n / 2];}  // Driver codeint main(){    int ar1[] = { 1, 2, 3, 6 };    int ar2[] = { 4, 6, 8, 10 };    int n1 = sizeof(ar1) / sizeof(ar1[0]);    int n2 = sizeof(ar2) / sizeof(ar2[0]);    if (n1 == n2)        cout << "Median is "             << getMedian(ar1, ar2, n1);    else        cout << "Doesn't work for arrays "             << "of unequal size";    return 0;}  // This code is contributed// by Shivi_Aggarwal
// A divide and conquer based efficient solution to find median// of two sorted arrays of same size.#include <bits/stdc++.h>using namespace std;  int median(int[], int); /* to get median of a sorted array */  /* This function returns median of ar1[] and ar2[].   Assumptions in this function:   Both ar1[] and ar2[] are sorted arrays   Both have n elements */int getMedian(int ar1[], int ar2[], int n){    /* return -1  for invalid input */    if (n <= 0)        return -1;    if (n == 1)        return (ar1[0] + ar2[0]) / 2;    if (n == 2)        return (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1])) / 2;      int m1 = median(ar1, n); /* get the median of the first array */    int m2 = median(ar2, n); /* get the median of the second array */      /* If medians are equal then return either m1 or m2 */    if (m1 == m2)        return m1;      /* if m1 < m2 then median must exist in ar1[m1....] and        ar2[....m2] */    if (m1 < m2) {        if (n % 2 == 0)            return getMedian(ar1 + n / 2 - 1, ar2, n - n / 2 + 1);        return getMedian(ar1 + n / 2, ar2, n - n / 2);    }      /* if m1 > m2 then median must exist in ar1[....m1] and        ar2[m2...] */    if (n % 2 == 0)        return getMedian(ar2 + n / 2 - 1, ar1, n - n / 2 + 1);    return getMedian(ar2 + n / 2, ar1, n - n / 2);}  /* Function to get median of a sorted array */int median(int arr[], int n){    if (n % 2 == 0)        return (arr[n / 2] + arr[n / 2 - 1]) / 2;    else        return arr[n / 2];}  /* Driver program to test above function */int main(){    int ar1[] = { 1, 2, 3, 6 };    int ar2[] = { 4, 6, 8, 10 };    int n1 = sizeof(ar1) / sizeof(ar1[0]);    int n2 = sizeof(ar2) / sizeof(ar2[0]);    if (n1 == n2)        printf("Median is %d", getMedian(ar1, ar2, n1));    else        printf("Doesn't work for arrays of unequal size");    return 0;}
Median is 5

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