Create FindKthSmallestElementBetweenTwoSortedArraysOfUnequalLength

Author: nkatre

FindKthSmallestElementBetweenTwoSortedArraysOfUnequalLength

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+/*
+Question: Find Kth smallest element in two sorted arrays
+Source:	http://algorithmsandme.blogspot.in/2014/12/fins-kth-smallest-element-in-two-sorted.html#.VLNiz5_0_VN
+*/	
+
+package FindKthSmallestElementBetweenTwoSortedArraysOfUnequalLength;
+
+import java.util.Arrays;
+import java.util.Scanner;
+
+public class UsingRecursion {
+public static void main(String[] args) {
+	Scanner in = new Scanner(System.in);
+	try{
+		System.out.println("Enter the number of elements in the first SORTED array");
+		int n = in.nextInt();
+		int[] array1 = new int[n];
+		System.out.println("Enter the elements of the first SORTED array");
+		for(int i=0;i<n;i++)
+			array1[i]=in.nextInt();
+		System.out.println("Enter the number of elements in the second SORTED array");
+		int m = in.nextInt();
+		int[] array2 = new int[m];
+		System.out.println("Enter the elements of the second SORTED array");
+		for(int i=0;i<m;i++)
+			array2[i]=in.nextInt();
+		System.out.println("Enter the kth smallest element to be found");
+		int k = in.nextInt();
+		System.out.println("The kth smallest element of the two SORTED arrays is: "+findKthSmallestElement(array1,array2,array1.length,array2.length,k));
+	}
+	finally{
+		in.close();
+	}
+}
+
+private static int findKthSmallestElement(int[] a, int[] b,
+		int aLength, int bLength, int k) {                    // All the 5 parameters passed are VERY VERY IMP
+	
+	/* to maintain uniformity, we will assume that size_a is smaller than size_b
+    else we will swap array in call :) */
+	if(aLength>bLength)
+		return findKthSmallestElement(b, a, bLength, aLength, k);
+	
+	/* We have TWO BASE CASES
+	 * Now case when size of smaller array is 0 i.e there is no elemt in one array*/
+    //BASE CASE 1. If the smallest array length is 0
+	if(aLength == 0 && bLength > 0)
+            return b[k-1]; // due to zero based index
+    
+    /* case where k==1 that means we have hit limit */
+	//BASE CASE 2. If k==1
+	if(k==1)
+            return Math.min(a[0], b[0]);
+
+    /* Now the divide and conquer part */
+    int i =  Math.min(aLength, k/2) ; // k should be less than the size of array  
+    int j =  Math.min(bLength, k/2) ; // k should be less than the size of array  
+
+    if(a[i-1] > b[j-1])
+            // Now we need to find only K-j th element
+            return findKthSmallestElement(a, Arrays.copyOfRange(b, j, b.length), a.length, b.length -j, k-j);
+    else
+            return findKthSmallestElement(Arrays.copyOfRange(a, i, a.length), b, a.length-i, b.length,  k-i);
+}
+}
+/*
+Analysis:
+	Time Complexity = O(log(n+m))
+    Space Complexity = O(1)*/