Here at NML, one of our senior developers (and chief instigators of office humour), Charl, likes to keep the team on its toes with a regular coding challenge. They’ve become so popular – with prizes and instant office-cred involved – we’ve dubbed it the NML Charllenge.
In the spirit of open-source inspiration, here’s the latest one. If you’d like to tackle it, email your solution to us at firstname.lastname@example.org with the subject line: Coding Challenge.
Write a Run-Length encoding algorithm to encode a piece of text and be able to decode the result back to the original.
You can use any language of your choice, and send as many versions as you want. Your submission will be evaluated on the following:
- Correct output for encoding and decoding (non-negotiable)
To recap, a run-length encoder takes a stream and encodes it by counting repeating bytes/characters and then outputting it as the number of the byte/characters that was repeated, plus the byte/character. So “aabbcde” will be “2a2b1c1d1e”.
Below is the runtime encode string you should be able to decode. A special prize for the first person to send in the decode text and tell us the name of the author. (We’ll know if you cheated by trying to figure out the latter before the former.)
(Note that the results of the any encode should be able to serve as the input for the next decode and that again for the next encode, without any loss of fidelity.)
1`1T1w1a1s1 1b1r1i2l1i1g1,1 1a1n1d1 1t1h1e1 1s1l1i1t1h1y1 1t1o1v1e1s2 1D1i1d1 1g1y1r1e1 1a1n1d1 1g1i1m1b1l1e1 1i1n1 1t1h1e1 1w1a1b1e1:1A2l1 1m1i1m1s1y1 1w1e1r1e1 1t1h1e1 1b1o1r1o1g1o1v1e1s1,2 1A1n1d1 1t1h1e1 1m1o1m1e1 1r1a1t1h1s1 o1u1t1g1r1a1b1e1.2n1’1B1e1w1a1r1e1 1t1h1e1 1J1a2b1e1r1w1o1c1k1,1 1m1y1 1s1o1n1!2 1T1h1e1 1j1a1w1s1 1t1h1a1t1 1b1i1t1e1,1 1t1h1e1 1c1l1a1w1s1 1t1h1a1t1 1c1a1t1c1h1!1B1e1w1a1r1e1 1t1h1e1 1J1u1b1j1u1b1 1b1i1r1d1,1 1a1n1d1 1s1h1u1n2 1T1h1e1 1f1r1u1m1i1o1u1s1 1B1a1n1d1e1r1s1n1a1t1c1h1!1’2n1H1e1 1t2o1k1 1h1i1s1 1v1o1r1p1a1l1 1s1w1o1r1d1 1i1n1 1h1a1n1d1:2 1L1o1n1g1 1t1i1m1e1 1t1h1e1 1m1a1n1x1o1m1e1 1f1o1e1 1h1e1 1s1o1u1g1h1t1 2-1S1o1 1r1e1s1t1e1d1 1h1e1 1b1y1 1t1h1e1 1T1u1m1t1u1m1 1t1r2e1,2 1A1n1d1 1s1t2o1d1 1a1w1h1i1l1e1 1i1n1 1t1h1o1u1g1h1t1.2n1A1n1d1,1 1a1s1 1i1n1 1u2f1i1s1h1 1t1h1o1u1g1h1t1 1h1e1 1s1t2o1d1,2 1T1h1e1 1J1a2b1e1r1w1o1c1k1,1 1w1i1t1h1 1e1y1e1s1 1o1f1 1f1l1a1m1e1,1C1a1m1e1 1w1h1i2f1l1i1n1g1 1t1h1r1o1u1g1h1 1t1h1e1 1t1u1l1g1e1y1 1w2o1d1,2 1A1n1d1 1b1u1r1b1l1e1d1 1a1s1 1i1t1 1c1a1m1e1!2n1O1n1e1,1 1t1w1o1!1 1O1n1e1,1 1t1w1o1!1 1A1n1d1 1t1h1r1o1u1g1h1 1a1n1d1 1t1h1r1o1u1g1h2 1T1h1e1 1v1o1r1p1a1l1 1b1l1a1d1e1 1w1e1n1t1 1s1n1i1c1k1e1r1-1s1n1a1c1k1!1H1e1 1l1e1f1t1 1i1t1 1d1e1a1d1,1 1a1n1d1 1w1i1t1h1 1i1t1s1 1h1e1a1d2 1H1e1 1w1e1n1t1 1g1a1l1u1m1p1h1i1n1g1 1b1a1c1k1.2n1’1A1n1d1,1 1h1a1s1 1t1h1o1u1 1s1l1a1i1n1 1t1h1e1 1J1a2b1e1r1w1o1c1k1?2 1C1o1m1e1 1t1o1 1m1y1 1a1r1m1s1,1 1m1y1 1b1e1a1m1i1s1h1 1b1o1y1!1O1 1f1r1a1b1j1o1u1s1 1d1a1y1!1 1C1a2l2o1h1!1 1C1a2l1a1y1!1’2 1H1e1 1c1h1o1r1t1l1e1d1 1i1n1 1h1i1s1 1j1o1y1.2n1`1T1w1a1s1 1b1r1i2l1i1g1,1 1a1n1d1 1t1h1e1 1s1l1i1t1h1y1 1t1o1v1e1s2 1D1i1d1 1g1y1r1e1 1a1n1d1 1g1i1m1b1l1e1 1i1n1 1t1h1e1 1w1a1b1e1;1A2l1 1m1i1m1s1y1 1w1e1r1e1 1t1h1e1 1b1o1r1o1g1o1v1e1s1,2 1A1n1d1 1t1h1e1 1m1o1m1e1 1r1a1t1h1s1 1o1u1t1g1r1a1b1e1.