CLRS 9.3.8 Find Median from Two Sorted Array

Exercise 9.3.8

Let $X[1..n]$ and $Y[1..n]$ be two arrays, each containing $n$ numbers already in sorted order. Give an $\O(\lg{n})$-time algorithm to find the median of all $2n$ elements in arrays $X$ and $Y$.

This was fun!

1. If the two arrays are of length $1$, we pick the lower of the two elements
2. We the two medians of the array
3. We take the lower part of the array with the greater median and the upper part of the array with the lesser median. If each array has $n$ elements, we take the first/last $\lfloor n / 2 \rfloor$ elements
4. We solve the problem for the new arrays

Let's reason about why this works. Since we have $2n$ elements, we know that the length is an even number and we're looking for a lower median. We need to observe that the median we're looking for is between the medians of the two arrays. Let's elaborate on that.
Let's assume that the median is at position $k$ in array $A$. This means that there are $k - 1$ elements less than the median in $A$ and $n - k$ elements greater than the median in $B$. If $k < n / 2$ then the median of $A$ will be greater than the final median, but the median of $B$ will be lesser than it. It's the other way around for $k \ge n / 2$. Thus the median of the two arrays is always between the medians of each.
Step 3 removes the same number of elements from each array, half of which are greater than the median and half of which are less than it. This reduces the subproblem to two smaller arrays that are sorted and their elements have the same median.

def two_array_median(a, b):
    if len(a) == 1:
        return min(a[0], b[0])

    m = median_index(len(a))
    i = m + 1
    if a[m] < b[m]:
        return two_array_median(a[-i:], b[:i])
    else:
        return two_array_median(a[:i], b[-i:])

def median_index(n):
    if n % 2:
        return n // 2
    else:
        return n // 2 - 1

https://code.google.com/p/clrs/source/browse/ch09/9.3-8.py

def find_median(A, B, lo, hi):

n = len(A)

if hi - lo <= 0: return None

k = (lo + hi) // 2

left = B[n-k-2] if n-k-2 >=0 else -1

rigt = B[n-k-1]

if left <= A[k] <= rigt:

return A[k]

elif A[k] > rigt:

return find_median(A, B, lo, k)

else:

return find_median(A, B, k + 1, hi)

def two_array_median(A, B):

n = len(A)

med = find_median(A, B, 0, n)

if med is None:

med = find_median(B, A, 0, n)

return med

https://code.google.com/p/rxwen-blog-stuff/source/browse/trunk/algorithm/i2a_ex_9.3-8/ex9_3_8.cpp

int getMedian(const int_vector& x, int lx, int hx, const int_vector& y, int ly, int hy)

{

int cx, cy, mx, my, z = 0;

cx = (lx+hx)/2;

cy = hy-(cx-lx);

mx = x[cx];

my = y[cy];

if(mx <= my)

{

if(cy < 1 || mx >= y[cy-1])

return mx;

else

return getMedian(x, cx+1, hx, y, ly, cy-1);

}

else

{

if(cx < 1 || my >= x[cx-1])

return my;

else

return getMedian(x, lx, cx-1, y, cy+1, hy);

}

}

http://rxwen.blogspot.com/2009/12/ex-93-8-of-introduction-to-algorithms.html
Read full article from Exercise 9.3.8