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K-th Element of Two Sorted Arrays

  • Difficulty Level : Hard
  • Last Updated : 25 Aug, 2021

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.

Basic Approach 
Since we are given two sorted arrays, we can use the 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];
}
 
// 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 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];
    }
     
    // Driver Code
    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
?>

Javascript




<script>
 
// JavaScript Program to find kth element
// from two sorted arrays
 
    function kth(arr1 , arr2 , m , n , k) {
        var sorted1 = Array(m + n).fill(0);
        var 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
     
        var arr1 = [ 2, 3, 6, 7, 9 ];
        var arr2 = [ 1, 4, 8, 10 ];
        var k = 5;
        document.write(kth(arr1, arr2, 5, 4, k));
 
// This code contributed by umadevi9616
 
</script>
Output
6

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



Space Optimized Version of above approach: We can avoid the use of extra array.

C++




// C++ program to find kth element
// from two sorted arrays
#include <bits/stdc++.h>
using namespace std;
 
int find(int A[], int B[], int m,
         int n, int k_req)
{
    int k = 0, i = 0, j = 0;
 
    // Keep taking smaller of the current
    // elements of two sorted arrays and
    // keep incrementing k
    while(i < m && j < n)
    {
        if(A[i] < B[j])
        {
            k++;
            if(k == k_req)
                return A[i];
            i++;
        }
        else
        {
            k++;
            if(k == k_req)
                return B[j];
            j++;
        }
    }
 
    // If array B[] is completely traversed
    while(i < m)
    {
        k++;
        if(k == k_req)
            return A[i];
        i++;
    }
 
    // If array A[] is completely traversed
    while(j < n)
    {
        k++;
        if(k == k_req)
            return B[j];
        j++;
    }
}
 
// Driver Code
int main()
{
    int A[5] = { 2, 3, 6, 7, 9 };
    int B[4] = { 1, 4, 8, 10 };
    int k = 5;
     
    cout << find(A, B, 5, 4, k);
     
    return 0;
}
 
// This code is contributed by Sreejith S

Java




import java.io.*;
 
class GFG {
    public static int find(int A[], int B[], int m, int n,
                           int k_req)
    {
        int k = 0, i = 0, j = 0;
 
        // Keep taking smaller of the current
        // elements of two sorted arrays and
        // keep incrementing k
        while (i < m && j < n) {
            if (A[i] < B[j]) {
                k++;
                if (k == k_req)
                    return A[i];
                i++;
            }
            else {
                k++;
                if (k == k_req)
                    return B[j];
                j++;
            }
        }
 
        // If array B[] is completely traversed
        while (i < m) {
            k++;
            if (k == k_req)
                return A[i];
            i++;
        }
 
        // If array A[] is completely traversed
        while (j < n) {
            k++;
            if (k == k_req)
                return B[j];
            j++;
        }
        return -1;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] A = { 2, 3, 6, 7, 9 };
        int[] B = { 1, 4, 8, 10 };
        int k = 5;
 
        System.out.println(find(A, B, 5, 4, k));
    }
}

Python3




# Python3 Program to find kth element
# from two sorted arrays
 
def find(A, B, m, n, k_req):   
    i, j, k = 0, 0, 0
 
    # Keep taking smaller of the current
    # elements of two sorted arrays and
    # keep incrementing k
    while i < len(A) and j < len(B):
        if A[i] < B[j]:
            k += 1
            if k == k_req:
                return A[i]
            i += 1
        else:
            k += 1
            if k == k_req:
                return B[j]       
            j += 1
 
    # If array B[] is completely traversed
    while i < len(A):
        k += 1
        if k == k_req:
                return A[i]
        i += 1
 
 
    # If array A[] is completely traversed
    while j < len(B):
        k += 1
        if k == k_req:
                return B[j]
        j += 1
 
# driver code
A = [2, 3, 6, 7, 9]
B = [1, 4, 8, 10]
k = 5;
print(find(A, B, 5, 4, k))
# time complexity of O(k)

C#




// C# program to find kth element
// from two sorted arrays
using System;
public class GFG
{
 
  public static int find(int[] A, int[] B,
                         int m, int n,int k_req)
  {
    int k = 0, i = 0, j = 0;
 
    // Keep taking smaller of the current
    // elements of two sorted arrays and
    // keep incrementing k
    while (i < m && j < n) {
      if (A[i] < B[j]) {
        k++;
        if (k == k_req)
          return A[i];
        i++;
      }
      else {
        k++;
        if (k == k_req)
          return B[j];
        j++;
      }
    }
 
    // If array B[] is completely traversed
    while (i < m)
    {
      k++;
      if (k == k_req)
        return A[i];
      i++;
    }
 
    // If array A[] is completely traversed
    while (j < n)
    {
      k++;
      if (k == k_req)
        return B[j];
      j++;
    }
    return -1;
  }
 
  // Driver Code
  static public void Main (){
    int[] A = { 2, 3, 6, 7, 9 };
    int[] B = { 1, 4, 8, 10 };
    int k = 5;
    Console.WriteLine(find(A, B, 5, 4, k));
  }
}
 
// This code is contributed by rag2127
Output
6

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

  1. 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.
  2. 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. Set the last element of arr2 to be arr2[mid2].
  3. In this way, 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;
}

Python3




# Python program to find k-th element from two sorted arrays
def kth(arr1, arr2, n, m, k):
 
    if n == 1 or m == 1:
        if m == 1:
            arr2, arr1 = arr1, arr2
            m = n
        if k == 1:
            return min(arr1[0], arr2[0])
        elif k == m + 1:
            return max(arr1[0], arr2[0])
        else:
            if arr2[k - 1] < arr1[0]:
                return arr2[k - 1]
            else:
                return max(arr1[0], arr2[k - 2])
 
    mid1 = (n - 1)//2
    mid2 = (m - 1)//2
 
    if mid1+mid2+1 < k:
        if arr1[mid1] < arr2[mid2]:
            return kth(arr1[mid1 + 1:], arr2, n - mid1 - 1, m, k - mid1 - 1)
        else:
            return kth(arr1, arr2[mid2 + 1:], n, m - mid2 - 1, k - mid2 - 1)
    else:
        if arr1[mid1] < arr2[mid2]:
            return kth(arr1, arr2[:mid2 + 1], n, mid2 + 1, k)
        else:
            return kth(arr1[:mid1 + 1], arr2, mid1 + 1, m, k)
 
 
if __name__ == "__main__":
    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 harshitkap00r
Output
6

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. The below implementation displays this. 

Explanation:
Instead of comparing the middle element of the arrays,
we compare the k / 2nd 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++




// 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

Python3




# Python3 program to find kth element from
# two sorted arrays
def kth(arr1, arr2, m, n, k, st1 = 0, 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 or k > (m - st1) + (n - st2)):
        return -1
         
    # Compare first elements of arrays and return
    if (k == 1):
        if(arr1[st1] < arr2[st2]):
            return arr1[st1]
        else:
            return arr2[st2]
 
    curr = int(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
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 avanitrachhadiya2155

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
6

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);
 
    // Check if arr1 is empty returning
    // k-th element of arr2
    if (m == 0)
        return arr2[k - 1];
 
    // Check 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);
 
        // Check if arr1 is empty returning
        // k-th element of arr2
        if (m == 0)
            return arr2[k - 1];
 
        // Check 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
Output
6

Time Complexity:O(log k)

Another Approach: (Using Min Heap)

  1. Push the elements of both arrays to a priority queue (min-heap).
  2. Pop-out k-1 elements from the front.
  3. Element at the front of the priority queue is the required answer.

Below is the implementation of the above approach:

C++




// C++ Program to find kth
// element from two sorted arrays
#include <bits/stdc++.h>
using namespace std;
 
// Function to find K-th min
int kth(int* a, int* b, int n, int m, int k)
{
      // Declaring a min heap
    priority_queue<int, vector<int>,
                   greater<int> > pq;
       
      // Pushing elements for
    // array a to min-heap
    for (int i = 0; i < n; i++) {
        pq.push(a[i]);
    }
   
      // Pushing elements for
    // array b to min-heap
    for (int i = 0; i < m; i++) {
        pq.push(b[i]);
    }
   
      // Poping-out K-1 elements
    while (k-- > 1) {
        pq.pop();
    }
    return pq.top();
}
 
//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;
}
 
// This code is contributed by yashbeersingh42

Java




// Java Program to find kth element
// from two sorted arrays
import java.util.*;
 
class GFG {
   
    // Function to find K-th min
    static int kth(int a[], int b[],
                   int n, int m, int k)
    {
         
        // Declaring a min heap
        PriorityQueue<Integer> pq =
                        new PriorityQueue<>();
 
        // Pushing elements for
        // array a to min-heap
        for (int i = 0; i < n; i++) {
            pq.offer(a[i]);
        }
 
        // Pushing elements for
        // array b to min-heap
        for (int i = 0; i < m; i++) {
            pq.offer(b[i]);
        }
 
        // Poping-out K-1 elements
        while (k-- > 1) {
            pq.remove();
        }
        return pq.peek();
    }
   
    // Driver Code
    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 yashbeersingh42
Output
6

Time Complexity: O(NlogN)

Space Complexity: O(m+n)

Another Approach : (Using Upper Bound STL)

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, The 1st element of this array is 1. The thing to notice here is upper_bound(6) gives 5, upper_bound(4) gives 4 that is number of element equal to or less than the number we are giving as input to upper_bound().

Here is another example

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.

Observation required :

The simplest method to solve this question is using upper_bound to check what is the position of a element in the sorted array. The upper_bound function return the pointer to element which is greater than the element we searched.

So to find the kth element we need to just find the element whose upper_bound() is 4. So again now we now what upper_bound() gives us we need 1 last observation to solve this question. If we have been given 2 arrays, We just need to the sum of upper_bound for the 2 arrays

Input : Array 1 – 2 3 6 7 9



     Array 2 – 1 4 8 10

     k = 5

Value of upper_bound for value(6) in array1 is 3 and for array 2 is 2. This give us a total of 5. which is the answer.

Algorithm :

  • We take a mid between [L,R] using the formula mid = (L+R)/2.
  • Check if the middle can be the kth element using upper_bound() function
  • Find the sum of upper_bound() for both the arrays and if the sum is >= K, It’s a possible value of kth element.
  • If sum is >= K then we assign R = mid – 1.
  • else if sum <k then the current mid is too small and we assign L = mid+1.
  • Repeat from top
  • Return the smallest value found.

Here is the implementation for the optimized method :

C++




// C++ program to find the kth element
#include <bits/stdc++.h>
using namespace std;
 
long long int maxN
    = 1e10; // the maximum value in the array possible.
 
long long int kthElement(int arr1[], int arr2[], int n,
                         int m, int k)
{
    long long int left = 1,
                  right
                  = maxN; // The range of where ans can lie.
    long long int ans = 1e15; // We have to find min of all
                              // the ans so take .
 
    // using binary search to check all possible values of
    // kth element
    while (left <= right) {
        long long int mid = (left + right) / 2;
        long long int up_cnt
            = upper_bound(arr1, arr1 + n, mid) - arr1;
        up_cnt += upper_bound(arr2, arr2 + m, mid) - arr2;
 
        if (up_cnt >= k) {
            ans = min(ans,
                      mid); // find the min of all answers.
            right
                = mid - 1; // Try to find a smaller answer.
        }
        else
            left = mid + 1; // Current mid is too small so
                            // shift right.
    }
 
    return ans;
}
 
// Driver code
int main()
{
    // Example 1
    int n = 5, m = 7, k = 7;
    int arr1[n] = { 100, 112, 256, 349, 770 };
    int arr2[m] = { 72, 86, 113, 119, 265, 445, 892 };
    cout << kthElement(arr1, arr2, n, m, k) << endl;
    return 0;
}
Output
256

Time Complexity : O( Log( maxN ).log( N+M ) )
Auxiliary Space : O( 1 )

This article is contributed by Aditya Kamath. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.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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