Simple coding.
Feb. 28th, 2010 04:54 pmAfter reading this: http://www.codinghorror.com/blog/2010/02/the-nonprogramming-programmer.html
I've got a bit scared: last year I do little coding, and when I do it's 90% of the time making other ppl's code more efficient/scalable, rather then developing my own. Or even Powerpoint coding, shame on me.
I've never been a fan of topcoder etc, because I am pretty slow thinker. But I was a bit scared that may be by now my brain is already dead, and I've became a non-programming programmer.
Went to http://codility.com/ (mentioned in the comments to the post above), and checked. Last time I've used Java was 3 years ago, but it took me 6 minutes to do the first simple task that passed all tests but BigInteger test (I knew in advance but was lazy) So I hope I am not hopeless, but still I would not pass interview
ivan_ghandhi style.
int equi ( int[] A ) {
int len = A.length;
int sum1 = 0; int sum2 = 0; int i;
for (i = 0; i < len; i++) sum2 += A[i];
for (i = 0; i < len; i++) {
sum2 -= A[i];
if (sum1 == sum2)
return i;
sum1 += A[i];
}
return -1;
}
I've got a bit scared: last year I do little coding, and when I do it's 90% of the time making other ppl's code more efficient/scalable, rather then developing my own. Or even Powerpoint coding, shame on me.
I've never been a fan of topcoder etc, because I am pretty slow thinker. But I was a bit scared that may be by now my brain is already dead, and I've became a non-programming programmer.
Went to http://codility.com/ (mentioned in the comments to the post above), and checked. Last time I've used Java was 3 years ago, but it took me 6 minutes to do the first simple task that passed all tests but BigInteger test (I knew in advance but was lazy) So I hope I am not hopeless, but still I would not pass interview
![[livejournal.com profile]](https://www.dreamwidth.org/img/external/lj-userinfo.gif)
int equi ( int[] A ) {
int len = A.length;
int sum1 = 0; int sum2 = 0; int i;
for (i = 0; i < len; i++) sum2 += A[i];
for (i = 0; i < len; i++) {
sum2 -= A[i];
if (sum1 == sum2)
return i;
sum1 += A[i];
}
return -1;
}