sachinvirgo
BAN USER
sort the array.
A rough algo can be:-
Now take mid = n/2
if (a[mid] == mid+2 && a[mid-1] = mid)
return mid+1;
if (a[mid] == mid+2)
search in subarray from start to mid-1
else
search in subarray from mid+1 to end
The only way is to redesign the application, keeping in mind the concurrency as a requirement.
- sachinvirgo February 24, 2008The array is the best data structure for this case. we only need to search the array by its index to check for ladder and snake. Once the array gets intialised we need not to insert or delete anthing, only search needs to be performed. This is what array is all about.
- sachinvirgo February 24, 2008here is the sample code(:-
public static void main(String args[])
{
String s= "C:\Dir1\Dir2\..\Dir3\.\Dir4\Dir5";
int i = s.indexOf("..",s);
while(i!=-1)
{
if(i==s.length()-2) \\incase .. is at end of string
s = s.substring(0,s.lastindexOf("\",i-2));
else
s = s.substring(0,s.lastindexOf("\",i-2)) + s.substring(i+2);
i = s.indexOf("..",s);
}
i = s.indexOf(".",s);
while(i!=-1)
{
if(i==s.length()-1) \\incase . is at end of string
s = s.substring(i-1);
else
s = s.substring(i-1) + s.substring(i+1);
i = s.indexOf(".",s);
}
System.out.println(s);
}
We can also use java strings with function lastindexOf() to attain the same.
-Sachin
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An idea: compare the last element of a list1(say a) with first element of other list2(say b). If a<b, just append the list2 to list1.
- sachinvirgo February 26, 2008We can further optimise it by comparing fisrt with first as well to proceed with correct order of merging