Doubt in answer

#1

1 bit for a;2 bits forb;3 bits for c;4 bits for d; 5 bits for e, 5 bits for f; total bits=416 bits.

therefore saved bits =800-416=384

#2

Hello @akshatkaush,

HUFFMAN :

-First calculating the frequencies of a – 3, d-5, c-6, b-7 for message string that is given.

-Then adding in min priority queue with respect to their frequencies

-Dequeue until queue is not empty and make a huffman tree…

  • Huffman tree is :

  • Now four charecters ,so 4 combinations therefore we required 2 bits.
  • This 4 characters repeat 21 times,so total bits are required is 21 * 2 = 42.

Hope, this would help.
Give a like if you are satisfied.

#3

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