Check if frequency of characters are in Recaman Series in Python

PythonServer Side ProgrammingProgramming

Suppose we have a lowercase string s. We have to check whether the occurrences of alphabets in s, after rearranging in any possible manner, generates the Recaman’s Sequence (ignoring the first term).

The Recaman’s sequence is as follows −

$$a_{n}=\begin{cases}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:0(if\:n=0) & \\a_{n-1}-n(if\:a_{n}-n>0\wedge not\:present\in sequence) & \\\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:a_{n-1}+n(otherwise)\end{cases}$$

Some of the items of Recaman's Sequence are [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24,...] (The first term is ignored as this is 0)

So, if the input is like s = "pppuvuuqquuu", then the output will be True as the characters and frequencies are (p, 3), (u, 6), (v, 1) and (q, 2). So the frequencies are forming first few terms of Recaman's Sequence [1, 3, 6, 2].

To solve this, we will follow these steps −

  • freq := a map holding all characters in s and their frequencies
  • n := size of freq
  • array := first n Recaman's sequence terms
  • f := 1
  • for each char in freq, do
    • is_found := 0
    • for j in range 1 to n, do
      • if freq[keys] is same as array[j], then
        • is_found := 1
        • come out from the loop
    • if is_found is false, then
      • f := 0
      • come out from the loop
  • return True when f is set otherwise False

Example

Let us see the following implementation to get better understanding −

 Live Demo

from collections import defaultdict
def recaman(array, n) :
   array[0] = 0
   for i in range(1, n + 1):
      temp = array[i - 1] - i
      for j in range(i):
         if array[j] == temp or temp < 0:
            temp = array[i - 1] + i
            break
      array[i] = temp
def solve(s) :
   freq = defaultdict(int)
   for i in range(len(s)) :
      freq[s[i]] += 1
   n = len(freq)
   array = [0] * (n + 1)
   recaman(array, n)
   f = 1
   for keys in freq.keys() :
      is_found = 0
      for j in range(1, n + 1) :
         if freq[keys] == array[j]:
            is_found = 1
            break;
      if not is_found:
         f = 0
         break
   return True if f else False
s = "pppuvuuqquuu"
print(solve(s))

Input

"pppuvuuqquuu"

Output

True
raja
Published on 18-Jan-2021 17:06:06
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