Python Program for Median of two sorted arrays of same size

Write a Python program for a given 2 sorted arrays A and B of size n each. the task is to find the median of the array obtained by merging the above 2 arrays(i.e. array of length 2n). The complexity should be O(log(n)).

Examples:

Input: ar1[] = {1, 12, 15, 26, 38}
ar2[] = {2, 13, 17, 30, 45}
Output: 16
Explanation:
After merging two arrays, we get {1, 2, 12, 13, 15, 17, 26, 30, 38, 45}
The middle two elements are 15 and 17
The average of middle elements is (15 + 17)/2 which is equal to 16

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.

Python Program for Median of two sorted arrays of same size using 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.

Below is the implementation of the above approach:

Python3

# A Simple Merge based O(n) Python 3 solution
# to find median of two sorted lists

# This function returns median of ar1[] and ar2[].
# Assumptions in this function:
# Both ar1[] and ar2[] are sorted arrays
# Both have n elements
def getMedian( ar1, ar2 , n):
    i = 0 # Current index of i/p list ar1[]

    j = 0 # Current index of i/p list ar2[]

    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
    count = 0
    while count < n + 1:
        count += 1

        # 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[]
        elif j == n:
            m1 = m2
            m2 = ar1[0]
            break
        # equals sign because if two
        # arrays have some common elements
        if ar1[i] <= ar2[j]:
            m1 = m2 # Store the prev median
            m2 = ar1[i]
            i += 1
        else:
            m1 = m2 # Store the prev median
            m2 = ar2[j]
            j += 1
    return (m1 + m2)/2

# Driver code to test above function
ar1 = [1, 12, 15, 26, 38]
ar2 = [2, 13, 17, 30, 45]
n1 = len(ar1)
n2 = len(ar2)
if n1 == n2:
    print("Median is ", getMedian(ar1, ar2, n1))
else:
    print("Doesn't work for arrays of unequal size")

# This code is contributed by "Sharad_Bhardwaj".

Output

Median is 16

Time Complexity: O(n)
Auxiliary Space: O(1)

Python Program for Median of two sorted arrays of same size (By comparing the medians of two arrays):

Step-by-step approach:

• Merge the two input arrays ar1[] and ar2[].
• Sort ar1[] and ar2[] respectively.
• The median will be the last element of ar1[] + the first
  • element of ar2[] divided by 2. [(ar1[n-1] + ar2[0])/2].

Below is the implementation of the above approach:

Python3

# Python program for above approach

# function to return median of the arrays
# both are sorted & of same size
def getMedian(ar1, ar2, n):
    i, j = n - 1, 0

    # while loop to swap all smaller numbers to arr1
    while(ar1[i] > ar2[j] and i > -1 and j < n):
        ar1[i], ar2[j] = ar2[j], ar1[i]
        i -= 1
        j += 1

    ar1.sort()
    ar2.sort()

    return (ar1[-1] + ar2[0]) >> 1


# Driver program
if __name__ == '__main__':
    ar1 = [1, 12, 15, 26, 38]
    ar2 = [2, 13, 17, 30, 45]

    n1, n2 = len(ar1), len(ar2)

    if(n1 == n2):
        print('Median is', getMedian(ar1, ar2, n1))
    else:
        print("Doesn't work for arrays of unequal size")

# This code is contributed by saitejagampala

Output

Median is 16

Time Complexity: O(nlogn)
Auxiliary Space: O(1)

Python Program for Median of two sorted arrays of same size using Binary Search:

Step-by-step approach:

We can find the kth element by using binary search on whole range of constraints of elements.

• Initialize ans = 0.0
• Initialize low = -10^9, high = 10^9 and pos = n
• Run a loop while(low <= high):
  • Calculate mid = (low + (high – low)>>1)
  • Find total elements less or equal to mid in the given arrays
  • If the count is less or equal to pos
    • Update low = mid + 1
    • Else high = mid – 1
• Store low in ans, i.e., ans = low.
• Again follow step3 with pos as n – 1
• Return (sum + low * 1.0)/2
• Median of two sorted arrays of same size

Below is the implementation of the above approach:

Python3

# Calculate the number of elements less than or equal to mid in the given arrays
def count_less_than_or_equal_to_mid(mid, arrays):
    count = 0
    for array in arrays:
        count += len([x for x in array if x <= mid])
    return count

def find_kth_element(arrays, n):
    ans = 0.0
    low = -1e9
    high = 1e9
    pos = n

    # Binary search to find the kth element
    while low <= high:
        mid = low + (high - low) // 2
        count = count_less_than_or_equal_to_mid(mid, arrays)
        if count <= pos:
            low = mid + 1
        else:
            high = mid - 1

    ans = low

    # Update pos and repeat the binary search to find the (n-1)th element
    pos = n - 1
    low = -1e9
    high = 1e9
    while low <= high:
        mid = low + (high - low) // 2
        count = count_less_than_or_equal_to_mid(mid, arrays)
        if count <= pos:
            low = mid + 1
        else:
            high = mid - 1

    ans += low

    # Return the average of the two elements
    return (ans / 2.0)

# Test with some arrays
arrays = [[1, 4, 5, 6, 10], [2, 3, 4, 5, 7]]
n = 5
print("Median in", find_kth_element(arrays, n))

#code is contributed by khushboogoyal499

Output

Median is 4.5

Time Complexity: O(log n)
Auxiliary Space: O(1)

Please refer complete article on Median of two sorted arrays of same size for more details!