Anil Singh
BAN USER
We can use combination of HashMap and customized Stack.
Stack here should maintained in such a way that we should see LFU object as peek,
and in case memory is full remove object from map same as peek of stack.
I always disagree with solutions coming out of naive iterations !!!
This can be solved by using Binary Search.
Perform a binary search for SUM - arr[i] in array.
Time complexity is O(n) here.
- Anil Singh April 20, 2013How come time coplexuty is O(n) here ?
- Anil Singh April 20, 2013public class ArrayMaxSum {
/**
* @param args
*/
public static void main(String[] args) {
String string = "1,2,-10,3,4,-20,5,6,4,-50,9,0,-1";
String[] numbers = string.split(",");
LinkedList<String> lst = new LinkedList<String>();
LinkedList<String> tmp = new LinkedList<String>();
int max_so_far = 0;
int max_ending_here = 0;
for (int i = 0; i < numbers.length-1; i++)
{
max_ending_here = max_ending_here + Integer.parseInt(numbers[i]);
lst.add(numbers[i]);
if (max_ending_here < 0)
{
max_ending_here = 0;
lst.clear();
}
if (max_ending_here > max_so_far)
{
tmp.clear();
max_so_far = max_ending_here;
tmp.addAll(lst);
}
}
for(String str : tmp)
{
System.out.print(str+", ");
}
System.out.print(" = "+max_so_far);
}
}
After for loop firstKth points to first Kth element in list.
After completion of while loop lastKth will point to Kth from last in list.
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Looks like an excellent solution driven by mathematics !!!
- Anil Singh January 16, 2014