Given a number n. Print the minimum positive integer by which it should be divided so that a result is an odd number.
Examples :
Input : 36 Output : 4 36 must be divided by 4 or 12 to make it odd. We take minimum of 4 and 12 i.e. 4 Input : 8 Output : 8 8 must be divided by 8 to make it an odd number.
Method 1:
Find the minimum number that makes the given number odd by dividing it one by one from 2(i) to N If (N/i) is odd then return i.
C++
// C++ program to make a number odd #include <iostream> using namespace std; int makeOdd(int n) { // Return 1 if already odd if (n % 2 != 0) return 1; // Check on dividing with a number when // the result becomes odd Return that number for (int i = 2 ; i <= n ; i++) // If n is divided by i and n/i is odd // then return i if ((n % i == 0) && ((n / i) % 2 == 1)) return i; } // Driver code int main() { int n = 36; cout << makeOdd(n); return 0; } |
Java
// Java program to make a number odd import java.io.*; class GFG { static int makeOdd(int n) { // Return 1 if already odd if (n % 2 != 0) return 1; // Check on dividing with a number when // the result becomes odd Return that number int i; for (i = 2 ; i <= n ; i++) // If n is divided by i and n/i is odd // then return i if ((n % i == 0) && ((n / i) % 2 == 1)) break; return i; } // Driver code public static void main (String[] args) { int n = 36; int res = makeOdd(n); System.out.println(res); } } // This code is contributed by Pramod Kumar |
Python3
# Python3 program to # make a number odd def makeOdd(n): # Return 1 if # already odd if n % 2 != 0: return 1; # Check on dividing # with a number when # the result becomes # odd Return that number for i in range(2, n): # If n is divided by # i and n/i is odd # then return i if (n % i == 0 and (int)(n / i) % 2 == 1): return i; # Driver code n = 36; print(makeOdd(n)); # This code is contributed # by mits |
C#
// C# program to make a number odd using System; class GFG { static int makeOdd(int n) { // Return 1 if already odd if (n % 2 != 0) return 1; // Check on dividing with a number when // the result becomes odd Return that number int i; for (i = 2 ; i <= n ; i++) // If n is divided by i and n/i is odd // then return i if ((n % i == 0) && ((n / i) % 2 == 1)) break; return i; } // Driver code public static void Main () { int n = 36; int res = makeOdd(n); Console.Write(res); } } // This code is contributed by nitin mittal |
PHP
<?php // PHP program to make // a number odd function makeOdd($n) { // Return 1 if already odd if ($n % 2 != 0) return 1; // Check on dividing with // a number when the result // becomes odd Return that // number for ($i = 2 ; $i <= $n ; $i++) // If n is divided by i and // n/i is odd then return i if (($n % $i == 0) && (($n / $i) % 2 == 1)) return $i; } // Driver code $n = 36; echo makeOdd($n); // This code is contributed by nitin mittal. ?> |
Javascript
<script> // javascript program to make a number odd function makeOdd(n) { // Return 1 if already odd if (n % 2 != 0) return 1; // Check on dividing with a number when // the result becomes odd Return that number var i; for (i = 2 ; i <= n ; i++) // If n is divided by i and n/i is odd // then return i if ((n % i == 0) && ((n / i) % 2 == 1)) break; return i; } // Driver code var n = 36; var res = makeOdd(n); document.write(res); // This code is contributed by Amit Katiyar </script> |
Output
4
Time complexity: O(n) where n is a given number
Auxiliary space: O(1)
Method 2:
As there is only one case to make a number odd i.e. It is an even number i.e. divisible by 2. So, the point is to find the smallest multiple of 2 that can make the given number odd.
C++
// C++ program to make a number odd #include <iostream> using namespace std; // Function to find the value int makeOdd(int n) { // Return 1 if already odd if (n%2 != 0) return 1; // Check how many times it is divided by 2 int result = 1; while (n%2 == 0) { n /= 2; result *= 2; } return result; } // Driver code int main() { int n = 36; cout << makeOdd(n); return 0; } |
Java
// Java program to make a number odd import java.io.*; class GFG { // Function to find the value static int makeOdd(int n) { // Return 1 if already odd if (n % 2 != 0) return 1; // Check how many times it is divided by 2 int ans = 1; while (n % 2 == 0) { n /= 2; ans *= 2; } return ans; } // Driver code public static void main (String[] args) { int n = 36; int res = makeOdd(n); System.out.println(res); } } // This code is contributed by Pramod Kumar |
Python3
# Python3 program to # make a number odd # Function to find # the value def makeOdd(n): # Return 1 if # already odd if (n % 2 != 0): return 1; # Check how many times # it is divided by 2 result = 1; while (n % 2 == 0): n = n/ 2; result = result * 2; return result; # Driver code n = 36; print(makeOdd(n)); # This code is contributed # by mits |
C#
// C# program to make a number odd using System; class GFG { // Function to find the value static int makeOdd(int n) { // Return 1 if already odd if (n % 2 != 0) return 1; // Check how many times it // is divided by 2 int ans = 1; while (n % 2 == 0) { n /= 2; ans *= 2; } return ans; } // Driver code public static void Main () { int n = 36; int res = makeOdd(n); Console.Write(res); } } // This code is contributed by // nitin mittal. |
PHP
<?php // PHP program to make // a number odd // Function to find the value function makeOdd( $n) { // Return 1 if already odd if ($n % 2 != 0) return 1; // Check how many times // it is divided by 2 $result = 1; while ($n % 2 == 0) { $n /= 2; $resul *= 2; } return $resul; } // Driver code $n = 36; echo makeOdd($n); // This code is contributed by anuj_67. ?> |
Javascript
<script> // javascript program to make a number odd // Function to find the value function makeOdd( n) { // Return 1 if already odd if (n % 2 != 0) return 1; // Check how many times it is divided by 2 var ans = 1; while (n % 2 == 0) { n = parseInt(n/2); ans *= 2; } return ans; } // Driver code var n = 36; var res = makeOdd(n); document.write(res); // This code contributed by Princi Singh </script> |
Output
4
Time complexity: O(log2n) where n is a given number
Auxiliary space: O(1)
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