Median of Two Sorted Array of the Same Size

Understanding The Problem

Problem Description

There are two sorted arrays nums1 and nums2 of size n. Find the median of the two sorted arrays.

You may assume nums1 and nums2 cannot be both empty.

Example 1:

nums1 = [1, 3]
nums2 = [2, 4]
The median is 2.5 as (2 + 3) / 2 = 2.5

Example 2:

nums1 = [1, 2, 9]
nums2 = [3, 4, 7]
The median is (3 + 4)/2 = 3.5
What is Median?

The median is the value separating the higher half of a data sample, a population, or a probability distribution, from the lower half. For a data set, it may be thought of as the “middle” value.

For odd n, Median (M) = value of ((n + 1)/2)th item term.

For even n, Median (M) = value of {(n/2)th item term + (n/2 + 1)th item term}/2

Note → Since the size of the two input arrays for which we are looking for the median is even(2n), we need to take the average of the middle two numbers and return it.


We will be discussing linear and logarithmic solutions for the problem

  1. Counting while comparing — Find the middle elements after merging the sorted array using the merge procedure of merge sort.
  2. Comparing medians — Get the medians of two sorted arrays and compare the medians and move accordingly until the two medians become equal following divide and conquer approach.

1. Counting While Comparing

For this approach, we actually have to find those two elements that could be in the middle when the two arrays are merged in sorted order.

If we follow the merge procedure like the one in merge sort then we could just compare the values at the two pointers pointing two the indexes of the two arrays and then increment the pointers accordingly while keeping track of the count. 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.

Solution Steps
  1. Create two pointers i and j pointing two nums1 and nums2 respectively and initialize them with 0
  2. If the value at nums1[i] < nums2[j] then increment i otherwise, increment j
  3. Repeat step 2 until i or j becomes i+j equals to n
  4. Return median of the two elements at ith and jth indexes.
Pseudo Code
float getMedian(int num1[], int num2[], int size) { 
   int i = 0
   int j = 0
   int m1 = -1, m2 = -1 
   for (count = 0 to size) {
      if (i == size) {
         m1 = m2
         m2 = num2[0]
      else if (j == size) {
         m1 = m2
         m2 = num1[0]
      if (num1[i] < num2[j]) { 
         m1 = m2
         m2 = num1[i] 
         i = i + 1
      } else { 
         m1 = m2
         m2 = num2[j]
         j = j + 1
   return (m1 + m2)/2
Complexity Analysis

Time Complexity: O(n), where is the size of both the input arrays

Space Complexity: O(1)

Critical Ideas to Think
  • Did you recognize that the merge approach is similar to the one used in merge sort?
  • Why we are returning (m1 + m2)/ 2 in the end?
  • If we have created an auxiliary array and save both the arrays in a sorted manner and then return its median, will it give the correct result? If yes then what would be its time and space complexity?

2. Comparing medians

This approach works by comparing the two medians of both the array following the Divide & Conquer paradigm. If the first median is greater than the second median then we will look for the median of the two sorted arrays in num1[0 to median1] and num2[median2 to n]. (Why?)

We know that median1 is larger than or equal to the first half of num1. And the same for median2 and num2. At the same time, we also know that median1 is smaller than or equal to the second half of num1.

Let’s consider the case median1 > median2

num1:      [.....median1.....]
num2:      [.....median2.....]
num1+num2: [.....median2...median1........]

Let us call the median of the merged array m*. Since the first halves of num1 and num2 come before median1. So, median1 >= m*, this means that no values larger than median1 need to be considered in num1. So we only need to look in the first half or ar1.

Similarly, since the second halves of num1 and num2 come after median1, we have that median2 <= m*, this means that no values smaller than median2 need to be considered and we only need to look in the second half of num2.

Example →

num1 = {1, 12, 15, 26, 38}
num2 = {2, 13, 17, 30, 45}
med1 = 15 and med2 = 17 for num1[0 to 4] and num2[0 to 4]
med1 < med2. So median is present in {15, 26, 38} and {2, 13, 17}
now, med1 = 26 and med2 = 13
med1 > med2. So median is present in {15, 26} and {13, 17}
The size of both the subarrays are now 2. So,
median = (max(15,13) + min(26, 17))/2 = 16
Solution Steps
  1. Calculate the median of both the arrays, say m1 and m2 for num1[] and num2[].
  2. Repeat till the size of num1 and num2 becomes two.
  • If m1 == m2 then return m1.
  • If m1 > m2 then median will be present in either of the sub-arrays num1[0..m1] and num2[m2 to n].
  • If m2 > m1 then median will be present in either of the sub-arrays num1[m1 to n] and num2[0 to m2].

3. Return Median = (max(array1[0],array2[0]) + min(array1[1],array2[1]))/2

Pseudo Code
float getMedian(int num1[], int num2[], int size) { 
   // base cases
   if (size <= 0)
      return -1
   if (size == 1)
      return (num1[0] + num2[0]) / 2
   if (size == 2)
      return (max(num1[0], num2[0]) + min(num1[1], num2[1])) / 2
   int med1 = median(num1, size)
   int med2 = median(num2, size)
   // compare the medians
   if (med1 == med2)
      return med1 
   if (med1 < med2) {
      // recurse for the subarray num1[m1 to size] and num2[0 to m2]
      if (size % 2 == 0)
         return getMedian(num1 + size/2 - 1, num2, size - size/2 + 1)
         return getMedian(num1 + size / 2, num2, size - size / 2)
   } else {
      // recurse for the subarray num1[0 to m1] and num2[m2 to n]
      if (size % 2 == 0)
         return getMedian(num2 + size / 2 - 1, num1, size - size / 2 + 1) 
         return getMedian(num2 + size / 2, num1, size - size / 2) 
// Utility function to find median of array
int median(int arr[], int array_size) { 
   if (array_size % 2 == 0)
      return (arr[array_size / 2] + arr[array_size / 2 - 1]) / 2
      return arr[array_size / 2]
Complexity Analysis

Time Complexity: O(log n), where n is the size of both the arrays

Space Complexity: O(log n) (Why?)

Critical Ideas to Think
  • Can you code this approach in an iterative manner?
  • Why does the comparison of medians and moving accordingly will reach the correct result?
  • What would be the median when the two sorted arrays are of length two?

Comparison of Different Solutions

Suggested Problems to Solve

  • Median of two sorted arrays of different sizes
  • Median in a row-wise sorted matrix