diff --git a/algorithms/Java/README.md b/algorithms/Java/README.md
index 9f71e23e..5ce8e900 100644
--- a/algorithms/Java/README.md
+++ b/algorithms/Java/README.md
@@ -8,6 +8,7 @@
 4. [Unique Digits of Large Number](arrays/unique-digits-of-large-number.java)
 5. [Majority Element](arrays/majority-element.java)
 6. [Longest Consecutive Subsequence](arrays/longest-consecutive-subsequence.java)
+7. [K-th Element of Two Sorted Arrays](arrays/kth-element-2sorted-array.java)
 
 ## Graphs
 
diff --git a/algorithms/Java/arrays/kth-element-2sorted-array.java b/algorithms/Java/arrays/kth-element-2sorted-array.java
new file mode 100644
index 00000000..3893a571
--- /dev/null
+++ b/algorithms/Java/arrays/kth-element-2sorted-array.java
@@ -0,0 +1,76 @@
+//Given two sorted arrays arr1 and arr2 of size M and N respectively and an element K. 
+//The task is to find the element that would be at the k’th position of the final sorted array.
+
+// Time Complexity: O(k) 
+// Auxiliary Space: O(1)
+
+import java.util.*;
+import java.lang.*;
+import java.io.*;
+
+class Solution {
+    public long kthElement( int arr1[], int arr2[], int n, int m, int k) {
+        int i=0, j=0, curr_k=0;
+        while(i<n && j<m){
+            if(arr1[i]<arr2[j]){
+                curr_k++;
+                if(k==curr_k) return arr1[i];
+                i++;
+            } else {
+                curr_k++;
+                if(k==curr_k) return arr2[j];
+                j++;
+            }
+        }
+        while (i < n) {
+            curr_k++;
+            if (k == curr_k)
+                return arr1[i];
+            i++;
+        }
+        while (j < m) {
+            curr_k++;
+            if (k == curr_k)
+                return arr2[j];
+            j++;
+        }
+        return -1;
+    }
+}
+
+class Main {
+	  public static void main(String[] args) throws IOException	{
+        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
+        int t = Integer.parseInt(br.readLine().trim());
+        while(t-->0) {
+            StringTokenizer stt = new StringTokenizer(br.readLine());
+            int n = Integer.parseInt(stt.nextToken());
+            int m = Integer.parseInt(stt.nextToken());
+            int k = Integer.parseInt(stt.nextToken());
+            int a[] = new int[(int)(n)];
+            int b[] = new int[(int)(m)];
+            String inputLine[] = br.readLine().trim().split(" ");
+            for (int i = 0; i < n; i++) {
+                a[i] = Integer.parseInt(inputLine[i]);
+            }
+            String inputLine1[] = br.readLine().trim().split(" ");
+            for (int i = 0; i < m; i++) {
+                b[i] = Integer.parseInt(inputLine1[i]);
+            }
+            Solution obj = new Solution();
+            System.out.println(obj.kthElement( a, b, n, m, k)); 
+        }
+	  }
+}
+
+/* Test Case:
+Input:
+arr1[] = {2, 3, 6, 7, 9}
+arr2[] = {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. */
\ No newline at end of file