This is an extension of the median of two sorted arrays of equal size problem. Here we handle arrays of unequal size also.
Examples:
Input: a[] = {1, 12, 15, 26, 38}
b[] = {2, 13, 24}
Output: 14
Explanation:
After merging arrays the result is
1, 2, 12, 13, 15, 24, 26, 38
median = (13 + 15)/2 = 14.
Input: a[] = {1, 3, 5, 7}
b[] = {2, 4, 6, 8}
Output: 5
Explanation :
After merging arrays the result is
1, 2, 3, 4, 5, 6, 7, 8, 9
median = 5
Approach:
- The approach discussed in this post is similar to method 1 of equal size median. Use merge procedure of merge sort.
- Keep a variable to track of count while comparing elements of two arrays.
- Perform merge of two sorted arrays. Increase the value of count as a new element is inserted.
- The most efficient way of merging two arrays is to compare the top of both arrays and pop the smaller value and increase the count.
- Continue comparing the top/first elements of both arrays until there are no elements left in any arrays
- Now to find the median there are two cases to be handled.
- If the sum of length of array sizes is even then store the elements when the count is of value ((n1 + n2)/2)-1 and (n1 + n2)/2 in the merged array add the elements and divide by 2 return the value as median.
- If the value(sum of sizes) is odd then return the (n1 + n2)/2 th element(when value of count is (n1 + n2)/2) as median.
C++
// A C++ program to find median of two sorted arrays of// unequal sizes#include <bits/stdc++.h>using namespace std;// A utility function to find median of two integers/* This function returns median of a[] and b[].Assumptions in this function: Both a[] and b[]are sorted arrays */float findmedian(int a[], int n1, int b[], int n2){ int i = 0; /* Current index of i/p array a[] */ int j = 0; /* Current index of i/p array b[] */ int k; int m1 = -1, m2 = -1; for (k = 0; k <= (n1 + n2) / 2; k++) { if (i < n1 && j < n2) { if (a[i] < b[j]) { m2 = m1; m1 = a[i]; i++; } else { m2 = m1; m1 = b[j]; j++; } } /* Below is to handle the case where all elements of a[] are smaller than smallest(or first) element of b[] or a[] is empty*/ else if (i == n1) { m2 = m1; m1 = b[j]; j++; } /* Below is to handle case where all elements of b[] are smaller than smallest(or first) element of a[] or b[] is empty*/ else if (j == n2) { m2 = m1; m1 = a[i]; i++; } } /*Below is to handle the case where sum of number of elements of the arrays is even */ if ((n1 + n2) % 2 == 0) return (m1 + m2) * 1.0 / 2; /* Below is to handle the case where sum of number of elements of the arrays is odd */ return m1;}// Driver program to test above functionsint main(){ int a[] = { 1, 12, 15, 26, 38 }; int b[] = { 2, 13, 24 }; int n1 = sizeof(a) / sizeof(a[0]); int n2 = sizeof(b) / sizeof(b[0]); printf("%f", findmedian(a, n1, b, n2)); return 0;} |
Java
// A Java program to find median of two sorted arrays of// unequal sizesimport java.io.*;class GFG { // A utility function to find median of two integers /* This function returns median of a[] and b[].Assumptions in this function: Both a[] and b[]are sorted arrays */ static float findmedian(int a[], int n1, int b[], int n2) { int i = 0; /* Current index of i/p array a[] */ int j = 0; /* Current index of i/p array b[] */ int k; int m1 = -1, m2 = -1; for (k = 0; k <= (n1 + n2) / 2; k++) { if (i < n1 && j < n2) { if (a[i] < b[j]) { m2 = m1; m1 = a[i]; i++; } else { m2 = m1; m1 = b[j]; j++; } } /* Below is to handle the case where all elements of a[] are smaller than smallest(or first) element of b[] or a[] is empty*/ else if (i == n1) { m2 = m1; m1 = b[j]; j++; } /* Below is to handle case where all elements of b[] are smaller than smallest(or first) element of a[] or b[] is empty*/ else if (j == n2) { m2 = m1; m1 = a[i]; i++; } } /*Below is to handle the case where sum of number of elements of the arrays is even */ if ((n1 + n2) % 2 == 0) { return (m1 + m2) * (float)1.0 / 2; } /* Below is to handle the case where sum of number of elements of the arrays is odd */ return m1; } // Driver program to test above functions public static void main(String[] args) { int a[] = { 1, 12, 15, 26, 38 }; int b[] = { 2, 13, 24 }; int n1 = a.length; int n2 = b.length; System.out.println(findmedian(a, n1, b, n2)); }}// This code has been contributed by inder_verma. |
Python3
# Python3 program to find median of # two sorted arrays of unequal sizes# A utility function to find median # of two integers''' This function returns median of a[] and b[]. Assumptions in this function: Both a[] and b[] are sortedarrays '''def findmedian(a, n1, b, n2): i = 0 # Current index of i / p array a[] j = 0 # Current index of i / p array b[] m1 = -1 m2 = -1 for k in range(((n1 + n2) // 2) + 1) : if (i < n1 and j < n2) : if (a[i] < b[j]) : m2 = m1 m1 = a[i] i += 1 else : m2 = m1 m1 = b[j] j += 1 # Below is to handle the case where # all elements of a[] are # smaller than smallest(or first) # element of b[] or a[] is empty elif(i == n1) : m2 = m1 m1 = b[j] j += 1 # Below is to handle case where # all elements of b[] are # smaller than smallest(or first) # element of a[] or b[] is empty elif (j == n2) : m2 = m1 m1 = a[i] i += 1 '''Below is to handle the case where sum of number of elements of the arrays is even ''' if ((n1 + n2) % 2 == 0): return (m1 + m2) * 1.0 / 2 ''' Below is to handle the case where sum of number of elements of the arrays is odd ''' return m1# Driver Codeif __name__ == "__main__": a = [ 1, 12, 15, 26, 38 ] b = [ 2, 13, 24 ] n1 = len(a) n2 = len(b) print(findmedian(a, n1, b, n2))# This code is contributed # by ChitraNayal |
C#
// A C# program to find median// of two sorted arrays of// unequal sizesusing System;class GFG { // A utility function to find // median of two integers /* This function returns median of a[] and b[]. Assumptions in this function: Both a[] and b[] are sorted arrays */ static float findmedian(int[] a, int n1, int[] b, int n2) { int i = 0; /* Current index of i/p array a[] */ int j = 0; /* Current index of i/p array b[] */ int k; int m1 = -1, m2 = -1; for (k = 0; k <= (n1 + n2) / 2; k++) { if (i < n1 && j < n2) { if (a[i] < b[j]) { m2 = m1; m1 = a[i]; i++; } else { m2 = m1; m1 = b[j]; j++; } } /* Below is to handle the case where all elements of a[] are smaller than smallest(or first) element of b[] or a[] is empty*/ else if (i == n1) { m2 = m1; m1 = b[j]; j++; } /* Below is to handle case where all elements of b[] are smaller than smallest(or first) element of a[] or b[] is empty*/ else if (j == n2) { m2 = m1; m1 = a[i]; i++; } } /*Below is to handle the case wheresum of number of elements of the arrays is even */ if ((n1 + n2) % 2 == 0) { return (m1 + m2) * (float)1.0 / 2; } /* Below is to handle the case where sum of number of elements of the arrays is odd */ return m1; } // Driver Code public static void Main() { int[] a = { 1, 12, 15, 26, 38 }; int[] b = { 2, 13, 24 }; int n1 = a.Length; int n2 = b.Length; Console.WriteLine(findmedian(a, n1, b, n2)); }}// This code is contributed// by Subhadeep Gupta |
PHP
<?php// A PHP program to find median of // two sorted arrays of unequal sizes// A utility function to find // median of two integers/* This function returns median of a[] and b[]. Assumptions in thisfunction: Both a[] and b[] are sorted arrays */function findmedian($a, $n1, $b, $n2){ $i = 0; /* Current index of i/p array a[] */ $j = 0; /* Current index of i/p array b[] */ $k; $m1 = -1; $m2 = -1; for ($k = 0; $k <= ($n1 + $n2) / 2; $k++) { if ($i < $n1 and $j < $n2) { if ($a[$i] < $b[$j]) { $m2 = $m1; $m1 = $a[$i]; $i++; } else { $m2 = $m1; $m1 = $b[$j]; $j++; } } /* Below is to handle the case where all elements of a[] are smaller than smallest(or first) element of b[] or a[] is empty*/ else if (i == n1) { $m2 = $m1; $m1 = $b[j]; $j++; } /* Below is to handle case where all elements of b[] are smaller than smallest(or first) element of a[] or b[] is empty*/ else if ($j == $n2) { $m2 = $m1; $m1 = $a[$i]; $i++; } } /*Below is to handle the case where sum of number of elements of the arrays is even */ if (($n1 + $n2) % 2 == 0) return ($m1 + $m2) * 1.0 / 2; /* Below is to handle the case where sum of number of elements of the arrays is odd */ return m1;}// Driver Code$a = array( 1, 12, 15, 26, 38 );$b = array( 2, 13, 24 );$n1 = count($a);$n2 = count($b);echo(findmedian($a, $n1, $b, $n2));// This code is contributed // by inder_verma.?> |
Javascript
<script>// A Javascript program to find median// of two sorted arrays of unequal sizes// A utility function to find median of two integers// This function returns median of a[] and b[].// Assumptions in this function: Both a[] and b[]// are sorted arrays function findmedian(a, n1, b, n2){ let i = 0; /* Current index of i/p array a[] */ let j = 0; /* Current index of i/p array b[] */ let k; let m1 = -1, m2 = -1; for(k = 0; k <= (n1 + n2) / 2; k++) { if (i < n1 && j < n2) { if (a[i] < b[j]) { m2 = m1; m1 = a[i]; i++; } else { m2 = m1; m1 = b[j]; j++; } } /* Below is to handle the case where all elements of a[] are smaller than smallest(or first) element of b[] or a[] is empty*/ else if (i == n1) { m2 = m1; m1 = b[j]; j++; } /* Below is to handle case where all elements of b[] are smaller than smallest(or first) element of a[] or b[] is empty*/ else if (j == n2) { m2 = m1; m1 = a[i]; i++; } } /*Below is to handle the case where sum of number of elements of the arrays is even */ if ((n1 + n2) % 2 == 0) { return (m1 + m2) * 1.0 / 2; } /* Below is to handle the case where sum of number of elements of the arrays is odd */ return m1;}// Driver codelet a = [ 1, 12, 15, 26, 38 ];let b = [ 2, 13, 24 ];let n1 = a.length;let n2 = b.length;document.write(findmedian(a, n1, b, n2));// This code is contributed by rag2127</script> |
Output:
14.000000
Complexity Analysis:
- Time Complexity: O(n), Both the lists need to be traversed so the time complexity is O(n).
- Auxiliary Space: O(1), no extra space is required.
More Efficient (Log time complexity methods)
- Median of two sorted arrays of different sizes.
- Median of two sorted arrays of different sizes in min(Log (n1, n2))
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