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Given an array of size N in which every number is between 1
and N, determine if there are any duplicates in it. You are
allowed to destroy the array if you like.
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Answer Posted / ashish
srry.....few mistakes in above soln.....
just add as many n(length of array) to an index as much it
has duplicates i.e. if array has size 7 and no. 5 is written
3 times then add 7 three times.so that after first loop
completion, you can just take division (a[5-1]/7 in this
case.so u got 3.also original data at a[5-1] cant be loss
bcoz u can just tak (a[5-1]%len) nd u get the original data)
//array is a[]
//len=length of array
for(i=0;i<len;i++)
{
b=a[i]%len;
a[b-1]=a[b-1]+len;
}
//printing output
for(i=0;i<len;i++)
{
cout<<i+1<<" occurs: "<<(a[i]/len)<<endl;
}
| Is This Answer Correct ? | 1 Yes | 1 No |
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