m.khooryani's blog

Faster way to generate these numbers?
By m.khooryani, history, 9 years ago, In English 
        int limit = 1 << n;
        for (int i = 0; i < limit; i++) {
            if (Integer.bitCount(i) >= 3) {
                System.out.println(i);
            }
        }
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9 years ago, # |
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There's technically no asymptotically better way. Note that the number of integers in that range that have bitCount < 3 is So the total number of integers you would be printing out is O(2n) - O(n2) = O(2n),  which is at least necessary asymptotically.

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