text: Median of two sorted arrays of same size | A divide and conquer based efficient solution to find median of two sorted arrays of same size . ; 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 ; get the median of the first array ; get the median of the second array ; If medians are equal then return either m1 or m2 ; if m1 < m2 then median must exist in ar1 [ m1 ... . ] and ar2 [ ... . m2 ] ; if m1 > m2 then median must exist in ar1 [ ... . m1 ] and ar2 [ m2 ... ] ; Function to get median of a sorted array ; Driver program to test above function
code: #include <bits/stdc++.h> NEW_LINE using namespace std ; int median ( int [ ] , int ) ; int getMedian ( int ar1 [ ] , int ar2 [ ] , int n ) { 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 ) ; int m2 = median ( ar2 , n ) ; if ( m1 == m2 ) return m1 ; 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 ( n % 2 == 0 ) return getMedian ( ar2 + n / 2 - 1 , ar1 , n - n / 2 + 1 ) ; return getMedian ( ar2 + n / 2 , ar1 , n - n / 2 ) ; } int median ( int arr [ ] , int n ) { if ( n % 2 == 0 ) return ( arr [ n / 2 ] + arr [ n / 2 - 1 ] ) / 2 ; else return arr [ n / 2 ] ; } 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 ; }