I need help for Time Complexity of this???

[lulzim]

Does anyone knws what is the time complexity of this implementation I need O(?)

for(int i= 0,j= 0; i< n; i++,j++ )
{

  for(int a= 0,b= 0; a< n; a++,b++ )
   {			
           1 statement goes   here...  
			  
   }
		 
}

Is this O(n2) or O(n4) im doubting on this.

Does anybody of u can convince me in this??

[sunde887]

Why not set the value of n to some small number and do this in your head? This will help you convince yourself, how many times will it execute if n is 3?

Beware of shadowing variables in the outer loop, it can cause some confusion. After trying this for 3, if you can't find the pattern, try it with n=2,4,5 and see if a pattern appears.

[lulzim]

so i was asking from u guys to make me sure for that bc i said im doubting is it O(N2) or ???

[sunde887]

Why do you doubt it? After testing for n=2,3, and 4, this should be obvious.

[lulzim]

hehe would u have mentioned O(?) :P

[lulzim]

I mean can u explain step by step how it comes to O(?) actually if i knew i wouldnt ask about this :S.

[sunde887]

Add this to your inner loop: 'System.out.println("Count");' then run the program and count how many times 'Count' is displayed: this number should show a pattern if you change n around a bit(try n=2,3,4).

int count = 0;
for(int n = 0; n < 3;++n){
  count=0;
  for(int i = 0; i < n; ++i)
    for(int j = 0; j < n; ++j)
      System.out.println(++count);
}

This may help.

[lulzim]

for(int SmallMatrixStartRow = 0,MainMatrixStartROW = startROW ; SmallMatrixStartRow < sizeOfMatrixPart; SmallMatrixStartRow++,MainMatrixStartROW++ ){
			  for(int SmallMatrixStartColumn = 0, MainMatrixStartColumn = startColumn ; SmallMatrixStartColumn < sizeOfMatrixPart ; MainMatrixStartColumn++, SmallMatrixStartColumn++){
				 
			//     Bpart.setElementAt(B.getElementAt(MainMatrixStartROW, MainMatrixStartColumn), SmallMatrixStartRow, SmallMatrixStartColumn);
			     
			     DestinationMatrix.setElementAt(SourcePart.getElementAt(SmallMatrixStartRow, SmallMatrixStartColumn), MainMatrixStartROW, MainMatrixStartColumn);
			       }
		  }

So this is O(n2) right???

[sunde887]

Yes O(n^2), since the inner and outer loop has n as the condition it will go until. Since the inner loop will loop n times for every cycle of the outer loop, it makes sense to readon that it does n*n, or n^2.

[doWhile]

Crossposted I need help for this time complexity measurment?