Time-complexity:O(logn)
Auxiliary-space: O(1)
Implementation
def get_median(a,a1,lo1,hi1,lo2,hi2)
#Length of the arrays
len=hi1-lo1+1
return -1 if len<=0 # return -1 if array is empty or invalid
return (a[lo1]+a1[lo2])/2 if len==1 # return median if each array is of size 1
return ([a[lo1], a1[lo2]].max + [a[hi1], a1[hi2]].min)/2 if len==2 # return median if each array is of size 2
#Median and index of median of array 1
m1,m1_i=median(a,len,lo1)
#Median and index of median of array 2
m2,m2_i=median(a1,len,lo2)
#Compare medians if equal return either m1 or m2,
#else if m1<m2 median lies in m1,hi1 and lo1,m2
#else m1>m2 median lies in m1,hi1 and lo1,m2
if m1==m2
return m1
elsif m1<m2
return get_median(a,a1,m1_i,hi1,lo2,m2_i)
else
return get_median(a,a1,lo1,m1_i,m2_i,hi2)
end
end
# Utility function to find median of array
def median(arr,n,lo)
if n%2==0
return (arr[lo+(n/2)]+arr[lo+((n/2)-1)])/2,lo+(n/2)
else
return arr[lo+(n/2)],lo+(n/2)
end
end
get_median([1,2,3,5,5],[1,2,5,9,11],0,4,0,4) # => 4