Question

I’m trying to fill an array of 20 ints with numbers from 1-20 in random sequence.
here’s my code:

 int lookup[20]={0}; 
 int array[20]={0};
 srand(time(NULL));
 for(int i=0;i<20;++i){ 
    bool done=false;
    while(!done){
      int n=rand()%20;
      if(lookup[n]==0){
          array[i]=n;
          lookup[n]=1;
          done=true;
      }
    }
 }

I’ve created a lookup array to check if the random number is not yet chosen and stored it in array. As you can see I’ve created 2 loops, one for traversing array and the while for choosing the random number. In every while loop iteration the number may reappear and causing another while loop. Is there faster way to do this?

Answer 1

You could fill the array in sequence and then shuffle it. That would prevent having to ever do more than 20 random number generations.

Fisher-Yates shuffle: can be done in O(n) time.

From wikipedia:

Properly implemented, the Fisher–Yates shuffle is unbiased, so that every permutation is equally likely. The modern version of the algorithm is also rather efficient, requiring only time proportional to the number of items being shuffled and no additional storage space.