Given a string that contains digits of a number. The number may contain many same continuous digits in it. The task is to count number of ways to spell the number.
For example, consider 8884441100, one can spell it simply as triple eight triple four double two and double zero. One can also spell as double eight, eight, four, double four, two, two, double zero.
Examples :
Input : num = 100 Output : 2 The number 100 has only 2 possibilities, 1) one zero zero 2) one double zero. Input : num = 11112 Output: 8 1 1 1 1 2, 11 1 1 2, 1 1 11 2, 1 11 1 2 11 11 2, 1 111 2, 111 1 2, 1111 2 Input : num = 8884441100 Output: 64 Input : num = 12345 Output: 1 Input : num = 11111 Output: 16
This is a simple problem of permutation and combination. If we take example test case given in the question, 11112. The answer depends on the number of possible substrings of 1111. The number of possible substrings of “1111” is 2^3 = 8 because it is the number of combinations of 4 – 1 = 3 separators ‘|’ between two characters of the string (digits of number represented by the string) : “1|1|1|1”. As our combinations will depend on whether we choose a particular 1 and for “2” there will be only one possibility 2^0 = 1, so answer for “11112” will be 8*1 = 8.
So, the approach is to count the particular continuous digit in string and multiply 2^(count-1) with previous result.
C++
// C++ program to count number of ways we// can spell a number#include<bits/stdc++.h>using namespace std;typedef long long int ll;// Function to calculate all possible spells of// a number with repeated digits// num --> string which is favourite numberll spellsCount(string num){ int n = num.length(); // final count of total possible spells ll result = 1; // iterate through complete number for (int i=0; i<n; i++) { // count contiguous frequency of particular // digit num[i] int count = 1; while (i < n-1 && num[i+1] == num[i]) { count++; i++; } // Compute 2^(count-1) and multiply with result result = result * pow(2, count-1); } return result;}// Driver program to run the caseint main(){ string num = "11112"; cout << spellsCount(num); return 0;} |
Java
// Java program to count number of ways we// can spell a numberimport java.io.*;class GFG { // Function to calculate all possible // spells of a number with repeated digits // num --> string which is favourite number static long spellsCount(String num) { int n = num.length(); // final count of total possible spells long result = 1; // iterate through complete number for (int i = 0; i < n; i++) { // count contiguous frequency of // particular digit num[i] int count = 1; while (i < n - 1 && num.charAt(i + 1) == num.charAt(i)) { count++; i++; } // Compute 2^(count-1) and multiply // with result result = result * (long)Math.pow(2, count - 1); } return result; } public static void main(String[] args) { String num = "11112"; System.out.print(spellsCount(num)); }}// This code is contributed by Anant Agarwal. |
Python3
# Python3 program to count number of# ways we can spell a number# Function to calculate all possible # spells of a number with repeated # digits num --> string which is # favourite numberdef spellsCount(num): n = len(num); # final count of total # possible spells result = 1; # iterate through complete # number i = 0; while(i<n): # count contiguous frequency # of particular digit num[i] count = 1; while (i < n - 1 and num[i + 1] == num[i]): count += 1; i += 1; # Compute 2^(count-1) and # multiply with result result = result * int(pow(2, count - 1)); i += 1; return result;# Driver Codenum = "11112";print(spellsCount(num));# This code is contributed# by mits |
C#
// C# program to count number of ways we// can spell a numberusing System;class GFG { // Function to calculate all possible // spells of a number with repeated // digits num --> string which is // favourite number static long spellsCount(String num) { int n = num.Length; // final count of total possible // spells long result = 1; // iterate through complete number for (int i = 0; i < n; i++) { // count contiguous frequency of // particular digit num[i] int count = 1; while (i < n - 1 && num[i + 1] == num[i]) { count++; i++; } // Compute 2^(count-1) and multiply // with result result = result * (long)Math.Pow(2, count - 1); } return result; } // Driver code public static void Main() { String num = "11112"; Console.Write(spellsCount(num)); }}// This code is contributed by nitin mittal. |
PHP
<?php// PHP program to count // number of ways we// can spell a number// Function to calculate // all possible spells of// a number with repeated // digits num --> string// which is favourite numberfunction spellsCount($num){ $n = strlen($num); // final count of total // possible spells $result = 1; // iterate through // complete number for ($i = 0; $i < $n; $i++) { // count contiguous frequency // of particular digit num[i] $count = 1; while ($i < $n - 1 && $num[$i + 1] == $num[$i]) { $count++; $i++; } // Compute 2^(count-1) and // multiply with result $result = $result * pow(2, $count - 1); } return $result;}// Driver Code$num = "11112";echo spellsCount($num);// This code is contributed// by nitin mittal. ?> |
Javascript
<script>// Javascript program to count number of // ways we can spell a number// Function to calculate all possible // spells of a number with repeated // digits num --> string which is// favourite numberfunction spellsCount(num){ let n = num.length; // Final count of total possible // spells let result = 1; // Iterate through complete number for (let i = 0; i < n; i++) { // Count contiguous frequency of // particular digit num[i] let count = 1; while (i < n - 1 && num[i + 1] == num[i]) { count++; i++; } // Compute 2^(count-1) and multiply // with result result = result * Math.pow(2, count - 1); } return result;} // Driver codelet num = "11112";document.write(spellsCount(num));// This code is contributed by code_hunt </script> |
8
Time Complexity: O(n*log(n))
Auxiliary Space: O(1)
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