K-th Element of Two Sorted Arrays

Given two sorted arrays of size m and n respectively, you are tasked with finding the element that would be at the k’th position of the final sorted array.

Examples:

Input : Array 1 - 2 3 6 7 9
        Array 2 - 1 4 8 10
        k = 5
Output : 6
Explanation: The final sorted array would be -
1, 2, 3, 4, 6, 7, 8, 9, 10
The 5th element of this array is 6.
Input : Array 1 - 100 112 256 349 770
        Array 2 - 72 86 113 119 265 445 892
        k = 7
Output : 256
Explanation: Final sorted array is -
72, 86, 100, 112, 113, 119, 256, 265, 349, 445, 770, 892
7th element of this array is 256.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Basic Approach
Since we are given two sorted arrays, we can use merging technique to get the final merged array. From this, we simply go to the k’th index.

C++

// Program to find kth element from two sorted arrays 
#include <iostream> 
using namespace std; 
  
int kth(int arr1[], int arr2[], int m, int n, int k) 
{ 
    int sorted1[m + n]; 
    int i = 0, j = 0, d = 0; 
    while (i < m && j < n) 
    { 
        if (arr1[i] < arr2[j]) 
            sorted1[d++] = arr1[i++]; 
        else
            sorted1[d++] = arr2[j++]; 
    } 
    while (i < m) 
        sorted1[d++] = arr1[i++]; 
    while (j < n) 
        sorted1[d++] = arr2[j++]; 
    return sorted1[k - 1]; 
} 
  
int main() 
{ 
    int arr1[5] = {2, 3, 6, 7, 9}; 
    int arr2[4] = {1, 4, 8, 10}; 
    int k = 5; 
    cout << kth(arr1, arr2, 5, 4, k); 
    return 0; 
}  

Java

// Java Program to find kth element  
// from two sorted arrays 
  
class Main 
{ 
    static int kth(int arr1[], int arr2[], int m, int n, int k) 
    { 
        int[] sorted1 = new int[m + n]; 
        int i = 0, j = 0, d = 0; 
        while (i < m && j < n) 
        { 
            if (arr1[i] < arr2[j]) 
                sorted1[d++] = arr1[i++]; 
            else
                sorted1[d++] = arr2[j++]; 
        } 
        while (i < m) 
            sorted1[d++] = arr1[i++]; 
        while (j < n) 
            sorted1[d++] = arr2[j++]; 
        return sorted1[k - 1]; 
    } 
      
    // main function 
    public static void main (String[] args)  
    { 
        int arr1[] = {2, 3, 6, 7, 9}; 
        int arr2[] = {1, 4, 8, 10}; 
        int k = 5; 
        System.out.print(kth(arr1, arr2, 5, 4, k)); 
    } 
} 
  
/* This code is contributed by Harsh Agarwal */

Python3

# Program to find kth element 
# from two sorted arrays 
  
def kth(arr1, arr2, m, n, k): 
      
    sorted1 = [0] * (m + n) 
    i = 0
    j = 0
    d = 0
    while (i < m and j < n): 
      
        if (arr1[i] < arr2[j]): 
            sorted1[d] = arr1[i] 
            i += 1
        else: 
            sorted1[d] = arr2[j] 
            j += 1
        d += 1
  
    while (i < m): 
        sorted1[d] = arr1[i] 
        d += 1
        i += 1
    while (j < n): 
        sorted1[d] = arr2[j] 
        d += 1
        j += 1
    return sorted1[k - 1] 
  
# driver code 
arr1 = [2, 3, 6, 7, 9] 
arr2 = [1, 4, 8, 10] 
k = 5; 
print(kth(arr1, arr2, 5, 4, k)) 
  
# This code is contributed by Smitha Dinesh Semwal 

C#

// C# Program to find kth element  
// from two sorted arrays 
class GFG 
{ 
static int kth(int[] arr1, int[] arr2,  
               int m, int n, int k) 
{ 
    int[] sorted1 = new int[m + n]; 
    int i = 0, j = 0, d = 0; 
    while (i < m && j < n) 
    { 
        if (arr1[i] < arr2[j]) 
            sorted1[d++] = arr1[i++]; 
        else
            sorted1[d++] = arr2[j++]; 
    } 
    while (i < m) 
        sorted1[d++] = arr1[i++]; 
    while (j < n) 
        sorted1[d++] = arr2[j++]; 
    return sorted1[k - 1]; 
} 
  
// Driver Code 
static void Main()  
{ 
    int[] arr1 = {2, 3, 6, 7, 9}; 
    int[] arr2 = {1, 4, 8, 10}; 
    int k = 5; 
    System.Console.WriteLine(kth(arr1, arr2, 
                                 5, 4, k)); 
} 
} 
  
// This code is contributed by mits  

PHP

<?php  
// Program to find kth 
// element from two  
// sorted arrays 
  
function kth($arr1, $arr2, 
             $m, $n, $k) 
{ 
    $sorted1[$m + $n] = 0; 
    $i = 0; 
    $j = 0; 
    $d = 0; 
    while ($i < $m && $j < $n) 
    { 
        if ($arr1[$i] < $arr2[$j]) 
            $sorted1[$d++] = $arr1[$i++]; 
        else
            $sorted1[$d++] = $arr2[$j++]; 
    } 
    while ($i < $m) 
        $sorted1[$d++] = $arr1[$i++]; 
    while ($j < $n) 
        $sorted1[$d++] = $arr2[$j++]; 
    return $sorted1[$k - 1]; 
} 
  
// Driver Code 
$arr1 = array(2, 3, 6, 7, 9); 
$arr2 = array(1, 4, 8, 10); 
$k = 5; 
echo kth($arr1, $arr2, 5, 4, $k); 
  
// This code is contributed  
// by ChitraNayal 
?> 

Output:

Time Complexity: O(n)
Auxiliary Space : O(m + n)

Divide And Conquer Approach 1
While the previous method works, can we make our algorithm more efficient? The answer is yes. By using a divide and conquer approach, similar to the one used in binary search, we can attempt to find the k’th element in a more efficient way.

Explanation:
We compare the middle elements of arrays arr1 and arr2,
let us call these indices mid1 and mid2 respectively.

Let us assume arr1[mid1]  k, then clearly the elements after 
mid2 cannot be the required element. We then set the last
element of arr2 to be arr2[mid2].

In this way, we define a new subproblem with half the size
of one of the arrays.

C++

// Program to find k-th element from two sorted arrays 
#include <iostream> 
using namespace std; 
  
int kth(int *arr1, int *arr2, int *end1, int *end2, int k) 
{ 
    if (arr1 == end1) 
        return arr2[k]; 
    if (arr2 == end2) 
        return arr1[k]; 
    int mid1 = (end1 - arr1) / 2; 
    int mid2 = (end2 - arr2) / 2; 
    if (mid1 + mid2 < k) 
    { 
        if (arr1[mid1] > arr2[mid2]) 
            return kth(arr1, arr2 + mid2 + 1, end1, end2, 
                k - mid2 - 1); 
        else
            return kth(arr1 + mid1 + 1, arr2, end1, end2, 
                k - mid1 - 1); 
    } 
    else
    { 
        if (arr1[mid1] > arr2[mid2]) 
            return kth(arr1, arr2, arr1 + mid1, end2, k); 
        else
            return kth(arr1, arr2, end1, arr2 + mid2, k); 
    } 
} 
  
int main() 
{ 
    int arr1[5] = {2, 3, 6, 7, 9}; 
    int arr2[4] = {1, 4, 8, 10}; 
  
    int k = 5; 
    cout << kth(arr1, arr2, arr1 + 5, arr2 + 4,  k - 1); 
    return 0; 
}  

Output:

Note that in the above code, k is 0 indexed, which means if we want a k that’s 1 indexed, we have to subtract 1 when passing it to the function.
Time Complexity: O(log n + log m)

Divide And Conquer Approach 2
While the above implementation is very efficient, we can still get away with making it more efficient. Instead of dividing the array into segments of n / 2 and m / 2 then recursing, we can divide them both by k / 2 and recurse. Below implementation displays this.

Explanation:
Instead of comparing the middle element of the arrays,
we compare the k / 2th element.
Let arr1 and arr2 be the arrays.
Now, if arr1[k / 2]  arr1[1]

New subproblem:
Array 1 - 6 7 9
Array 2 - 1 4 8 10
k = 5 - 2 = 3

floor(k / 2) = 1
arr1[1] = 6
arr2[1] = 1
arr1[1] > arr2[1]

New subproblem:
Array 1 - 6 7 9
Array 2 - 4 8 10
k = 3 - 1 = 2

floor(k / 2) = 1
arr1[1] = 6
arr2[1] = 4
arr1[1] > arr2[1]

New subproblem:
Array 1 - 6 7 9
Array 2 - 8 10
k = 2 - 1 = 1

Now, we directly compare first elements,
since k = 1. 
arr1[1] < arr2[1]
Hence, arr1[1] = 6 is the answer.

C++

// Program to find kth element from two sorted arrays 
#include <iostream> 
using namespace std; 
  
int kth(int arr1[], int arr2[], int m, int n, int k, 
                           int st1 = 0, int st2 = 0) 
{ 
    // In case we have reached end of array 1 
    if (st1 == m) 
        return arr2[st2 + k - 1]; 
  
    // In case we have reached end of array 2 
    if (st2 == n) 
        return arr1[st1 + k - 1]; 
  
    // k should never reach 0 or exceed sizes 
    // of arrays 
    if (k == 0 || k > (m - st1) + (n - st2)) 
        return -1; 
  
    // Compare first elements of arrays and return 
    if (k == 1) 
        return (arr1[st1] < arr2[st2]) ? 
            arr1[st1] : arr2[st2]; 
    int curr = k / 2; 
  
    // Size of array 1 is less than k / 2 
    if (curr - 1 >= m - st1) 
    { 
        // Last element of array 1 is not kth 
        // We can directly return the (k - m)th 
        // element in array 2 
        if (arr1[m - 1] < arr2[st2 + curr - 1]) 
            return arr2[st2 + (k - (m - st1) - 1)]; 
        else
            return kth(arr1, arr2, m, n, k - curr, 
                st1, st2 + curr); 
    } 
  
    // Size of array 2 is less than k / 2 
    if (curr-1 >= n-st2) 
    { 
        if (arr2[n - 1] < arr1[st1 + curr - 1]) 
            return arr1[st1 + (k - (n - st2) - 1)]; 
        else
            return kth(arr1, arr2, m, n, k - curr, 
                st1 + curr, st2); 
    } 
    else
    { 
        // Normal comparison, move starting index 
        // of one array k / 2 to the right 
        if (arr1[curr + st1 - 1] < arr2[curr + st2 - 1]) 
            return kth(arr1, arr2, m, n, k - curr, 
                st1 + curr, st2); 
        else
            return kth(arr1, arr2, m, n, k - curr, 
                st1, st2 + curr); 
    } 
} 
  
// Driver code 
int main() 
{ 
    int arr1[5] = {2, 3, 6, 7, 9}; 
    int arr2[4] = {1, 4, 8, 10}; 
  
    int k = 5; 
    cout << kth(arr1, arr2, 5, 4,  k); 
    return 0; 
} 

Java

// Java Program to find kth element from two sorted arrays 
class GFG  
{ 
  
    static int kth(int arr1[], int arr2[], int m, 
                   int n, int k, int st1, int st2)  
    { 
        // In case we have reached end of array 1 
        if (st1 == m)  
        { 
            return arr2[st2 + k - 1]; 
        } 
  
        // In case we have reached end of array 2 
        if (st2 == n)  
        { 
            return arr1[st1 + k - 1]; 
        } 
  
        // k should never reach 0 or exceed sizes 
        // of arrays 
        if (k == 0 || k > (m - st1) + (n - st2))  
        { 
            return -1; 
        } 
  
        // Compare first elements of arrays and return 
        if (k == 1)  
        { 
            return (arr1[st1] < arr2[st2]) 
                    ? arr1[st1] : arr2[st2]; 
        } 
        int curr = k / 2; 
  
        // Size of array 1 is less than k / 2 
        if (curr - 1 >= m - st1) 
        { 
              
            // Last element of array 1 is not kth 
            // We can directly return the (k - m)th 
            // element in array 2 
            if (arr1[m - 1] < arr2[st2 + curr - 1])  
            { 
                return arr2[st2 + (k - (m - st1) - 1)]; 
            }  
            else 
            { 
                return kth(arr1, arr2, m, n, k - curr, 
                        st1, st2 + curr); 
            } 
        } 
  
        // Size of array 2 is less than k / 2 
        if (curr - 1 >= n - st2) 
        { 
            if (arr2[n - 1] < arr1[st1 + curr - 1])  
            { 
                return arr1[st1 + (k - (n - st2) - 1)]; 
            } 
            else 
            { 
                return kth(arr1, arr2, m, n, k - curr, 
                        st1 + curr, st2); 
            } 
        }  
        else
          
        // Normal comparison, move starting index 
        // of one array k / 2 to the right 
        if (arr1[curr + st1 - 1] < arr2[curr + st2 - 1]) 
        { 
            return kth(arr1, arr2, m, n, k - curr, 
                    st1 + curr, st2); 
        }  
        else 
        { 
            return kth(arr1, arr2, m, n, k - curr, 
                    st1, st2 + curr); 
        } 
    } 
  
    // Driver code 
    public static void main(String[] args)  
    { 
        int arr1[] = {2, 3, 6, 7, 9}; 
        int arr2[] = {1, 4, 8, 10}; 
        int k = 5; 
        int st1 = 0, st2 = 0; 
        System.out.println(kth(arr1, arr2, 5, 4, k, st1, st2)); 
    } 
}  
  
// This code is contributed by 29AjayKumar 

C#

// C# Program to find kth element from two sorted arrays 
using System; 
  
class GFG  
{ 
  
    static int kth(int []arr1, int []arr2, int m, 
                int n, int k, int st1, int st2)  
    { 
        // In case we have reached end of array 1 
        if (st1 == m)  
        { 
            return arr2[st2 + k - 1]; 
        } 
  
        // In case we have reached end of array 2 
        if (st2 == n)  
        { 
            return arr1[st1 + k - 1]; 
        } 
  
        // k should never reach 0 or exceed sizes 
        // of arrays 
        if (k == 0 || k > (m - st1) + (n - st2))  
        { 
            return -1; 
        } 
  
        // Compare first elements of arrays and return 
        if (k == 1)  
        { 
            return (arr1[st1] < arr2[st2]) 
                    ? arr1[st1] : arr2[st2]; 
        } 
        int curr = k / 2; 
  
        // Size of array 1 is less than k / 2 
        if (curr - 1 >= m - st1) 
        { 
              
            // Last element of array 1 is not kth 
            // We can directly return the (k - m)th 
            // element in array 2 
            if (arr1[m - 1] < arr2[st2 + curr - 1])  
            { 
                return arr2[st2 + (k - (m - st1) - 1)]; 
            }  
            else
            { 
                return kth(arr1, arr2, m, n, k - curr, 
                        st1, st2 + curr); 
            } 
        } 
  
        // Size of array 2 is less than k / 2 
        if (curr - 1 >= n - st2) 
        { 
            if (arr2[n - 1] < arr1[st1 + curr - 1])  
            { 
                return arr1[st1 + (k - (n - st2) - 1)]; 
            } 
            else
            { 
                return kth(arr1, arr2, m, n, k - curr, 
                        st1 + curr, st2); 
            } 
        }  
        else
          
        // Normal comparison, move starting index 
        // of one array k / 2 to the right 
        if (arr1[curr + st1 - 1] < arr2[curr + st2 - 1]) 
        { 
            return kth(arr1, arr2, m, n, k - curr, 
                    st1 + curr, st2); 
        }  
        else
        { 
            return kth(arr1, arr2, m, n, k - curr, 
                    st1, st2 + curr); 
        } 
    } 
  
    // Driver code 
    public static void Main(String[] args)  
    { 
        int []arr1 = {2, 3, 6, 7, 9}; 
        int []arr2 = {1, 4, 8, 10}; 
        int k = 5; 
        int st1 = 0, st2 = 0; 
        Console.WriteLine(kth(arr1, arr2, 5, 4, k, st1, st2)); 
    } 
} 
  
// This code is contributed by PrinciRaj1992 

Output:

Time Complexity: O(log k)

Now, k can take a maximum value of m + n. This means that log k can be in the worst case, log(m + n). Logm + logn = log(mn) by properties of logarithms, and when m, n > 2, log(m + n) < log(mn). Thus this algorithm slightly outperforms the previous algorithm.Also see another simple implemented log k approach suggested by Raj Kumar.

C++

// C++ Program to find kth element from two sorted arrays 
// Time Complexity: O(log k) 
  
#include <iostream> 
using namespace std; 
  
int kth(int arr1[], int m, int arr2[], int n, int k) 
{ 
      
  if (k > (m+n) || k < 1) return -1; 
    
  // let m <= n 
  if (m > n) return kth(arr2, n, arr1, m, k); 
    
  // if arr1 is empty returning k-th element of arr2 
  if (m == 0) return arr2[k - 1]; 
    
  // if k = 1 return minimum of first two elements of both arrays  
  if (k == 1) return min(arr1[0], arr2[0]); 
    
  // now the divide and conquer part 
  int i = min(m, k / 2), j = min(n, k / 2); 
    
  if (arr1[i - 1] > arr2[j - 1] )  
    // Now we need to find only k-j th element since we have found out the lowest j 
    return kth(arr1, m, arr2 + j, n - j, k - j); 
  else 
    // Now we need to find only k-i th element since we have found out the lowest i 
    return kth(arr1 + i, m - i, arr2, n, k - i); 
} 
  
// Driver code 
int main() 
{ 
    int arr1[5] = {2, 3, 6, 7, 9}; 
    int arr2[4] = {1, 4, 8, 10}; 
    int m = sizeof(arr1)/sizeof(arr1[0]); 
    int n = sizeof(arr2)/sizeof(arr2[0]); 
    int k = 5; 
      
    int ans = kth(arr1,m,arr2, n, k); 
      
    if(ans == -1) cout<<"Invalid query"; 
    else cout<<ans; 
      
    return 0; 
} 
// This code is contributed by Raj Kumar 

Java

// Java Program to find kth element from two sorted arrays  
// Time Complexity: O(log k)  
import java.util.Arrays; 
  
class Gfg  
{ 
    static int kth(int arr1[], int m, int arr2[], 
                                    int n, int k)  
    { 
        if (k > (m + n) || k < 1) 
            return -1; 
  
        // let m > n 
        if (m > n) 
            return kth(arr2, n, arr1, m, k); 
  
        // if arr1 is empty returning k-th element of arr2 
        if (m == 0) 
            return arr2[k - 1]; 
  
        // if k = 1 return minimum of first  
        // two elements of both arrays 
        if (k == 1) 
            return Math.min(arr1[0], arr2[0]); 
  
        // now the divide and conquer part 
        int i = Math.min(m, k / 2); 
        int j = Math.min(n, k / 2); 
  
        if (arr1[i - 1] > arr2[j - 1])  
        { 
            // Now we need to find only k-j th element 
            // since we have found out the lowest j 
            int temp[] = Arrays.copyOfRange(arr2, j, n); 
            return kth(arr1, m, temp, n - j, k - j); 
        } 
  
        // Now we need to find only k-i th element  
        // since we have found out the lowest i 
        int temp[] = Arrays.copyOfRange(arr1, i, m); 
        return kth(temp, m - i, arr2, n, k - i); 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        int arr1[] = { 2, 3, 6, 7, 9 }; 
        int arr2[] = { 1, 4, 8, 10 }; 
        int m = arr1.length; 
        int n = arr2.length; 
  
        int k = 5; 
        int ans = kth(arr1, m, arr2, n, k); 
        if (ans == -1) 
            System.out.println("Invalid query"); 
        else
            System.out.println(ans); 
    } 
} 
  
// This code is contributed by Vivek Kumar Singh 

Time Complexity:O(log k)

This article is contributed by Aditya Kamath. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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