An algorithm to find the nth largest number in two arrays of size n

Question

I have this question:

Given two sorted lists (stored in arrays) of size n, find an O(log n) algorithm that computes the nth largest element in the union

Answer 1

Evgeny Kluev gives a better answer - mine was O(n log n) since i didn't think about them as being sorted.

what i can add is give you a link to a very nice video explaining binary search, courtesy of MIT:

https://www.youtube.com/watch?v=UNHQ7CRsEtU

Answer 2

Compare A[n/2] and B[n/2]. If equal, any of them is our result. Other stopping condition for this algorithm is when both arrays are of size 1 (either initially or after several recursion steps). In this case we just choose the largest of A[n/2] and B[n/2].

If A[n/2] < B[n/2], repeat this procedure recursively for second half of A[] and first half of B[].

If A[n/2] > B[n/2], repeat this procedure recursively for second half of B[] and first half of A[].

Since on each step the problem size is (in worst case) halved, we'll get O(log n) algorithm.

---

Always dividing array size by two to get the index works properly only if n is a power of two. More correct way of choosing indexes (for arbitrary n) would be using the same strategy for one array but choosing complementing index: j=n-i for other one.

Answer 3

public static void main(String[] args) {  


int[] fred = { 60, 5, 7, 3, 20, 3, 44 };  

int[] tmp = new int[fred.length];  
go(fred, 1, tmp, 3);  
}  

public static void go(int[] fred, int cnt, int[] tmp, int whatPosition) {  
int max = 0;  
int currentPosition = 0;  
for (int i = 0; i < fred.length; i++) {  
if (i == 0)  
max = fred[i];  
else {  
if (fred[i] > max) {  
max = fred[i];  
currentPosition = i;  
}  
}  
}  
System.arraycopy(fred, 0, tmp, 0, fred.length);  
tmp[currentPosition] = 0;  
cnt++;  
if(cnt != whatPosition)  
go(tmp, cnt, tmp, whatPosition);  
else{  
for (int i = 0; i < tmp.length; i++) {  
if (i == 0)  
max = tmp[i];  
else {  
if (tmp[i] > max) {  
max = tmp[i];  
}  
}  
}  
System.out.println(max);  
}  




}