mirror of
https://github.com/TheAlgorithms/Ruby
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102 lines
1.8 KiB
Ruby
102 lines
1.8 KiB
Ruby
# Write a program that outputs the string representation of numbers
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# from 1 to n. But for multiples of three it should output “Fizz”
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# instead of the number and for the multiples of five output “Buzz”.
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# For numbers which are multiples of both three and five output
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# “FizzBuzz”.
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#
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# Approach 1: Naive Approach
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#
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# Complexity Analysis
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# Time Complexity: O(N)
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# Space Complexity: O(1)
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# @param {Integer} n
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# @return {String[]}
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def fizz_buzz(n)
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str = []
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n.times do |i|
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i += 1
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if i % 5 == 0 && i % 3 == 0
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str.push('FizzBuzz')
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elsif i % 3 == 0
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str.push('Fizz')
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elsif i % 5 == 0
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str.push('Buzz')
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else
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str.push(i.to_s)
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end
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end
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str
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end
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n = 15
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fizz_buzz(n)
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# => [
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# "1",
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# "2",
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# "Fizz",
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# "4",
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# "Buzz",
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# "Fizz",
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# "7",
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# "8",
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# "Fizz",
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# "Buzz",
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# "11",
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# "Fizz",
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# "13",
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# "14",
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# "FizzBuzz"
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# ]
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#
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# Approach 2: String Concatenation
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#
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# Algorithm
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#
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# Instead of checking for every combination of these conditions,
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# check for divisibility by given numbers i.e. 3, 5 as given in the
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# problem. If the number is divisible, concatenate the corresponding
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# string mapping Fizz or Buzz to the current answer string.
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#
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# For eg. If we are checking for the number 15, the steps would be:
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#
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# Condition 1: 15 % 3 == 0 , num_ans_str = "Fizz"
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# Condition 2: 15 % 5 == 0 , num_ans_str += "Buzz"
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# => num_ans_str = "FizzBuzz"
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#
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# Complexity Analysis
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#
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# Time Complexity: O(N)
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# Space Complexity: O(1)
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# @param {Integer} n
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# @return {String[]}
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def fizz_buzz(n)
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str = []
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n.times do |i|
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i += 1
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num_str = ''
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num_str += 'Fizz' if i % 3 == 0
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num_str += 'Buzz' if i % 5 == 0
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num_str = i.to_s if num_str == ''
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str.push(num_str)
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end
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str
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end
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n = 15
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puts(fizz_buzz(n))
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