Given a string with lowercase English alphabets. The task is to check whether the frequency of the characters in the string can be arranged as a Fibonacci series. If yes, print “YES”, otherwise print “NO”.
Note:
- Frequencies can be arranged in any way to form Fibonacci Series.
- The Fibonacci Series starts from 1. That is the series is 1,1,2,3,5,…..
Examples:
Input : str = "abeeedd"
Output : YES
Frequency of 'a' => 1
Frequency of 'b' => 1
Frequency of 'e' => 3
Frequency of 'd' => 2
These frequencies are first 4 terms of
Fibonacci series => {1, 1, 2, 3}
Input : str = "dzzddz"
Output : NO
Frequencies are not in Fibonacci series
Approach:
- Store the frequencies of each character of the string in a map. Let the size of the map be
after storing frequencies. - Then, make a vector and insert first ‘n’ elements of the Fibonacci series in this vector.
- Then, compare each element of the vector with values of the map. If both elements of vector and values of the map are same, print ‘YES’, otherwise print ‘NO’.
after storing frequencies.Below is the implementation of the above approach:
C++
// C++ program to check whether frequency of// characters in a string makes// Fibonacci Sequence#include <bits/stdc++.h>using namespace std;int main() { string s = "abeedddccccc"; // create a map to store the frequency of each character map<char, int> freq; for (char c : s) { freq++; } // create an array to store the frequency values int f[freq.size()]; int i = 0; for (auto p : freq) { f[i] = p.second; i++; } // check if the frequency values form a Fibonacci sequence int n = freq.size(); bool isFibonacci = true; sort(f, f + freq.size()); if(f[0] != 1 || f[1] != 1) isFibonacci = false; else{ for (int j = 2; j < n; j++) { if (f[j] != f[j-1] + f[j-2]) { isFibonacci = false; break; } } } // print the result if (isFibonacci) { cout << "YES" << endl; } else { cout << "NO" << endl; } return 0;} |
Java
// Java program to check whether frequency of // characters in a string makes // Fibonacci Sequence import java.util.HashMap;import java.util.Vector;class GFG { // Function to check if the frequencies // are in Fibonacci series static String isFibonacci(String s) { // map to store the // frequencies of character HashMap<Character, Integer> m = new HashMap<>(); for (int i = 0; i < s.length(); i++) m.put(s.charAt(i), m.get(s.charAt(i)) == null ? 1 : m.get(s.charAt(i)) + 1); // Vector to store first n // fibonacci numbers Vector<Integer> v = new Vector<>(); // Get the size of the map int n = m.size(); // a and b are first and second terms of // fibonacci series int a = 1, b = 1; int c; v.add(a); v.add(b); // vector v contains elements of // fibonacci series for (int i = 0; i < n - 2; i++) { v.add(a + b); c = a + b; a = b; b = c; } int flag = 1; int i = 0; // Compare vector elements with values in Map for (HashMap.Entry<Character, Integer> entry : m.entrySet()) { if (entry.getValue() != v.elementAt(i)) { flag = 1; break; } i++; } if (flag == 1) return "YES"; else return "NO"; } // Driver Code public static void main(String[] args) { String s = "abeebbbccccc"; System.out.println(isFibonacci(s)); }}// This code is contributed by// sanjeev2552 |
Python3
# Python3 program to check whether the frequency# of characters in a string make Fibonacci Sequencefrom collections import defaultdict# Function to check if the frequencies# are in Fibonacci seriesdef isFibonacci(s): # map to store the frequencies of character m = defaultdict(lambda:0) for i in range(0, len(s)): m[s[i]] += 1 # Vector to store first n fibonacci numbers v = [] # Get the size of the map n = len(m) # a and b are first and second # terms of fibonacci series a = b = 1 v.append(a) v.append(b) # vector v contains elements of # fibonacci series for i in range(0, n - 2): v.append(a + b) c = a + b a, b = b, c flag, i = 1, 0 # Compare vector elements with values in Map for itr in sorted(m): if m[itr] != v[i]: flag = 0 break i += 1 if flag == 1: return "YES" else: return "NO"# Driver codeif __name__ == "__main__": s = "abeebbbccccc" print(isFibonacci(s))# This code is contributed by Rituraj Jain |
C#
// C# program to check whether frequency of // characters in a string makes // Fibonacci Sequence using System;using System.Collections.Generic; class GFG { // Function to check if the frequencies // are in Fibonacci series static String isFibonacci(String s) { // map to store the // frequencies of character int i = 0; Dictionary<int, int> mp = new Dictionary<int, int>(); for (i = 0; i < s.Length; i++) { if(mp.ContainsKey(s[i])) { var val = mp[s[i]]; mp.Remove(s[i]); mp.Add(s[i], val + 1); } else { mp.Add(s[i], 1); } } // List to store first n // fibonacci numbers List<int> v = new List<int>(); // Get the size of the map int n = mp.Count; // a and b are first and second terms of // fibonacci series int a = 1, b = 1; int c; v.Add(a); v.Add(b); // vector v contains elements of // fibonacci series for (i = 0; i < n - 2; i++) { v.Add(a + b); c = a + b; a = b; b = c; } int flag = 1; // Compare vector elements with values in Map foreach(KeyValuePair<int, int> entry in mp) { if (entry.Value != v[i]) { flag = 1; break; } i++; } if (flag == 1) return "YES"; else return "NO"; } // Driver Code public static void Main(String[] args) { String s = "abeebbbccccc"; Console.WriteLine(isFibonacci(s)); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to check whether frequency of// characters in a string makes// Fibonacci Sequence// Function to check if the frequencies// are in Fibonacci seriesfunction isFibonacci(s){ // map to store the // frequencies of character var m = new Map(); for (var i = 0; i < s.length; i++) { if(m.has(s[i])) { m.set(s[i], m.get(s[i])); } else { m.set(s[i], 1); } } // Vector to store first n // fibonacci numbers var v = []; // Get the size of the map var n = m.length; // a and b are first and second terms of // fibonacci series var a = 1, b = 1; var c; v.push(a); v.push(b); // vector v contains elements of fibonacci series for (var i = 0; i < n - 2; i++) { v.push(a + b); c = a + b; a = b; b = c; } var flag = 1; var i = 0; // Compare vector elements with values in Map m.forEach((value, key) => { if (value != v[i]) { flag = 0; } }); if (flag == 1) return "YES"; else return "NO";}// Driver codevar s = "abeeedd";document.write( isFibonacci(s));</script> |
Output
YES
Complexity Analysis:
- Time Complexity: O(n), where n is the length of the given string.
- Auxiliary Space: O(n)
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