Last Updated: February 25, 2016
·
3.813K
· saurabhksingh

# Algorithm : You are given two sorted arrays of lengths m and n.Give a O(log(m)+log(n)) time algorithm for computing the k-th smallest element in the union of the two arrays.

Problem asks for O(log(m) + log(n)) solution here. However I am giving here O(logK) solution.

public class KthSmallestInTwoSortedArrays {

/**

• Problem : Find the kth smallest from two sorted arrays (length m and n) combined

• Solution : Find the kth smallest from among first k elements of each array.

• We will use the binary search here.

*/

/**

*/

public static void main(String[] args) {

int [] A = {4,5,7, 45, 67, 77, 86, 94, 101};

int [] B = {1,3,6,8,23,24,34,44,51};

int k = 6;

getKthSmallestInTwoArrays(A, B, k);

}

private static void getKthSmallestInTwoArrays(int[] a, int[] b, int k) {

int lowA = 0, lowB = 0, highA = k – 1, highB = k – 1;

if (a.length < (k – 1))

{

highA = a.length – 1;

}

if (b.length < (k -1))

{

highB = b.length – 1;

}

if ((highA + highB) < k)

{

//insufficient elements

return;

}

int midA = 0, midB = 0;

boolean canContinue = false;

int result = 0;

while(k >= 0)

{

midA = lowA + (highA – lowA)/2;

midB = lowB + (highB – lowB)/2;

if (a[midA] >= b[midB])

{

//it means the first midA elements of A are all greater than first midB elements of B

//it means that the kth smallest lies in second half of B OR first half of A

k = k – (midB – lowB + 1);

result = b[midB];

highA = midA – 1;

lowB = midB + 1;

}

else if (a[midA] < b[midB])

{

//it means the first midB elements of B are all greater than first midA elements of A

//it means that the kth smallest lies in second half of A OR first half of B

k = k – (midA – lowA + 1);

result = a[midA];

highB = midB – 1;

lowA = midA + 1;

}

if (k == 0)

break;

}

System.out.println(result);

}

}

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#### 1 Response

while loop (k>=0) and inside loop if (k==0) break;

redundant logic

just while (k>0) is enough

over 1 year ago ·

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