Yahoo Interview Question for Software Engineer / Developers






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No replies ????

- Aryan October 02, 2008 | Flag Reply
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brutal force solution. Start from sqroot of the number down to 2. Check if it is a prime number, if then done. Otherwise, find next smaller prime number. Needs O(N) though.

- Anonymous October 27, 2008 | Flag Reply
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it fails for 143= 11*13
sqrt(143) =12.

- venkat September 10, 2011 | Flag
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who told you sqrt(143) ==12

- Fatty August 25, 2013 | Flag
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void largestPrime(int N)
{
int i;
for(i=2; i<=(int)sqrt((double)N); i++)
{
if(N % i == 0)
{

largestPrime(N/i);
return;
}

}
printf("%d ", N);

}

- Anonymous November 11, 2008 | Flag Reply
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void prime_factors(int n)
{
int t = sqrt(n) ,i;

for(i = 2 ; i<= t && n > 1 ;i++)
while(n % i == 0){
printf("%d\t",i);
n/= i;
}
if(n > 1)
printf("%d", n);
}

- venkat September 10, 2011 | Flag
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i<=(int)sqrt((double)N) should be i<=(int)2*sqrt((double)N) because this condition fails when N = 15, sqrt(15) is about 3.7, but 5 is still a factor.

- Anonymous August 29, 2011 | Flag Reply


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