Given a range [L, R], the task is to find all the numbers from the range which are composite as well as the eventual sum of their digits is 1.
Examples:
Input: L = 10, R = 100
Output: 10 28 46 55 64 82 91 100
10 = 1 + 0 = 1
28 = 2 + 8 = 10 = 1 + 0 = 1
…
91 = 9 + 1 = 10 = 1 + 0 = 1
100 = 1 + 0 + 0 = 1
Input: L = 250, R = 350
Output: 253 262 280 289 298 316 325 334 343
Approach: For every number in the range check if the number is composite i.e. it has a divisor other than 1 and the number itself. If the current number is a composite number then keep on calculating the sum of its digits until the number is reduced to a single digit, if this digit is 1 then the chosen number is a valid number.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <iostream>using namespace std;// Function that returns true if number n// is a composite numberbool isComposite(int n){ // Corner cases if (n <= 1) return false; if (n <= 3) return false; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return true; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return true; return false;}// Function that returns true if the eventual// digit sum of number nm is 1bool isDigitSumOne(int nm){ // Loop till the sum is not single digit number while (nm > 9) { // Initialize the sum as zero int sum_digit = 0; // Find the sum of digits while (nm > 0) { int digit = nm % 10; sum_digit = sum_digit + digit; nm = nm / 10; } nm = sum_digit; } // If sum is eventually 1 if (nm == 1) return true; else return false;}// Function to print the required numbers// from the given rangevoid printValidNums(int l, int r){ for (int i = l; i <= r; i++) { // If i is one of the required numbers if (isComposite(i) && isDigitSumOne(i)) cout << i << " "; }}// Driver codeint main(void){ int l = 10, r = 100; printValidNums(l, r); return 0;} |
Java
// Java implementation of the above approachpublic class GFG { // Function that returns true if number n // is a composite number static boolean isComposite(int n) { // Corner cases if (n <= 1) return false; if (n <= 3) return false; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return true; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return true; return false; } // Function that returns true if the eventual // digit sum of number nm is 1 static boolean isDigitSumOne(int nm) { // Loop till the sum is not single // digit number while (nm > 9) { // Initialize the sum as zero int sum_digit = 0; // Find the sum of digits while (nm > 0) { int digit = nm % 10; sum_digit = sum_digit + digit; nm = nm / 10; } nm = sum_digit; } // If sum is eventually 1 if (nm == 1) return true; else return false; } // Function to print the required numbers // from the given range static void printValidNums(int l, int r) { for (int i = l; i <= r; i++) { // If i is one of the required numbers if (isComposite(i) && isDigitSumOne(i)) System.out.print(i + " "); } } // Driver code public static void main(String arg[]) { int l = 10, r = 100; printValidNums(l, r); }} |
Python3
# Python3 implementation of the approach# Function that returns true if number n # is a composite numberdef isComposite(n): # Corner cases if (n <= 1): return False if (n <= 3): return False # This is checked so that we can skip # middle five numbers in below loop if (n % 2 == 0 or n % 3 == 0): return True i = 5 while(i * i <= n): if (n % i == 0 or n % (i + 2) == 0): return True i = i + 6 return False# Function that returns true if the eventual # digit sum of number nm is 1def isDigitSumOne(nm) : # Loop till the sum is not single # digit number while(nm>9) : # Initialize the sum as zero sum_digit = 0 # Find the sum of digits while(nm != 0) : digit = nm % 10 sum_digit = sum_digit + digit nm = nm // 10 nm = sum_digit # If sum is eventually 1 if(nm == 1): return True else: return False # Function to print the required numbers # from the given rangedef printValidNums(m, n ): for i in range(m, n + 1): # If i is one of the required numbers if(isComposite(i) and isDigitSumOne(i)) : print(i, end =" ")# Driver codel = 10r = 100printValidNums(m, n) |
C#
// C# implementation of the above approachusing System;class GFG{ // Function that returns true if number n // is a composite number static bool isComposite(int n) { // Corner cases if (n <= 1) return false; if (n <= 3) return false; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return true; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return true; return false; } // Function that returns true if the // eventual digit sum of number nm is 1 static bool isDigitSumOne(int nm) { // Loop till the sum is not single // digit number while (nm > 9) { // Initialize the sum as zero int sum_digit = 0; // Find the sum of digits while (nm > 0) { int digit = nm % 10; sum_digit = sum_digit + digit; nm = nm / 10; } nm = sum_digit; } // If sum is eventually 1 if (nm == 1) return true; else return false; } // Function to print the required numbers // from the given range static void printValidNums(int l, int r) { for (int i = l; i <= r; i++) { // If i is one of the required numbers if (isComposite(i) && isDigitSumOne(i)) Console.Write(i + " "); } } // Driver code static public void Main () { int l = 10, r = 100; printValidNums(l, r); }}// This code is contributed by jit_t |
PHP
<?php// PHP implementation of the above approach // Function that returns true if number n // is a composite number function isComposite($n) { // Corner cases if ($n <= 1) return false; if ($n <= 3) return false; // This is checked so that we can skip // middle five numbers in below loop if ($n % 2 == 0 || $n % 3 == 0) return true; for ($i = 5; $i * $i <= $n; $i = $i + 6) if ($n % $i == 0 || $n % ($i + 2) == 0) return true; return false; } // Function that returns true if the eventual // digit sum of number nm is 1 function isDigitSumOne($nm) { // Loop till the sum is not single // digit number while ($nm > 9) { // Initialize the sum as zero $sum_digit = 0; // Find the sum of digits while ($nm > 0) { $digit = $nm % 10; $sum_digit = $sum_digit + $digit; $nm = floor($nm / 10); } $nm = $sum_digit; } // If sum is eventually 1 if ($nm == 1) return true; else return false; } // Function to print the required numbers // from the given range function printValidNums($l, $r) { for ($i = $l; $i <= $r; $i++) { // If i is one of the required numbers if (isComposite($i) && isDigitSumOne($i)) echo $i, " "; } } // Driver code $l = 10; $r = 100; printValidNums($l, $r); // This code is contributed by Ryuga?> |
Javascript
<script>// Javascript implementation of the above approach // Function that returns true if number n // is a composite number function isComposite(n) { // Corner cases if (n <= 1) return false; if (n <= 3) return false; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return true; for (let i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return true; return false; } // Function that returns true if the eventual // digit sum of number nm is 1 function isDigitSumOne(nm) { // Loop till the sum is not single digit number while (nm > 9) { // Initialize the sum as zero let sum_digit = 0; // Find the sum of digits while (nm > 0) { let digit = nm % 10; sum_digit = sum_digit + digit; nm = Math.floor(nm / 10); } nm = sum_digit; } // If sum is eventually 1 if (nm == 1) return true; else return false; } // Function to print the required numbers // from the given range function printValidNums(l, r) { for (let i = l; i <= r; i++) { // If i is one of the required numbers if (isComposite(i) && isDigitSumOne(i)) document.write(i + " "); } } // Driver code let l = 10, r = 100; printValidNums(l, r); // This code is contributed by Mayank Tyagi</script> |
Output:
10 28 46 55 64 82 91 100
Time Complexity: O((r – l) * (sqrt(r – l) + log10(r – l)))
Auxiliary Space: O(1)
Optimizations : We can precompute composite numbers using Sieve Algorithms.
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