Count Odd and Even numbers in a range from L to R
Given two numbers L and R, the task is to count the number of odd numbers in the range L to R.
Examples:
Input: l = 3, r = 7
Output: 3 2
Count of odd numbers is 3 i.e. 3, 5, 7
Count of even numbers is 2 i.e. 4, 6
Input: l = 4, r = 8
Output: 2
Approach: Total numbers in the range will be (R – L + 1) i.e. N.
- If N is even then the count of both odd and even numbers will be N/2.
- If N is odd,
- If L or R is odd, then the count of odd number will be N/2 + 1 and even numbers = N – countofOdd.
- Else, count of odd numbers will be N/2 and even numbers = N – countofOdd.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Return the number of odd numbers // in the range [L, R] int countOdd(int L, int R){ int N = (R - L) / 2; // if either R or L is odd if (R % 2 != 0 || L % 2 != 0) N += 1; return N; } // Driver code int main() { int L = 3, R = 7; int odds = countOdd(L, R); int evens = (R - L + 1) - odds; cout << "Count of odd numbers is " << odds << endl; cout << "Count of even numbers is " << evens << endl; return 0; } // This code is contributed by Rituraj Jain |
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Java
// Java implementation of the above approach class GFG { // Return the number of odd numbers // in the range [L, R] static int countOdd(int L, int R) { int N = (R - L) / 2; // if either R or L is odd if (R % 2 != 0 || L % 2 != 0) N++; return N; } // Driver code public static void main(String[] args) { int L = 3, R = 7; int odds = countOdd(L, R); int evens = (R - L + 1) - odds; System.out.println("Count of odd numbers is " + odds); System.out.println("Count of even numbers is " + evens); } } |
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Python 3
# Python 3 implementation of the # above approach # Return the number of odd numbers # in the range [L, R] def countOdd(L, R): N = (R - L) // 2 # if either R or L is odd if (R % 2 != 0 or L % 2 != 0): N += 1 return N # Driver code if __name__ == "__main__": L = 3 R = 7 odds = countOdd(L, R) evens = (R - L + 1) - odds print("Count of odd numbers is", odds) print("Count of even numbers is", evens) # This code is contributed by ita_c |
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C#
// C# implementation of the above approach using System; class GFG { // Return the number of odd numbers // in the range [L, R] static int countOdd(int L, int R) { int N = (R - L) / 2; // if either R or L is odd if (R % 2 != 0 || L % 2 != 0) N++; return N; } // Driver code public static void Main() { int L = 3, R = 7; int odds = countOdd(L, R); int evens = (R - L + 1) - odds; Console.WriteLine("Count of odd numbers is " + odds); Console.WriteLine("Count of even numbers is " + evens); } } // This code is contributed by Ryuga |
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PHP
<?php // PHP implementation of the above approach // Return the number of odd numbers // in the range [L, R] function countOdd($L, $R) { $N = ($R - $L) / 2; // if either R or L is odd if ($R % 2 != 0 || $L % 2 != 0) $N++; return $N; } // Driver code $L = 3; $R = 7; $odds = countOdd($L, $R); $evens = ($R - $L + 1) - $odds; echo "Count of odd numbers is " . $odds . "\n"; echo "Count of even numbers is " . $evens; // This code is contributed // by Akanksha Rai ?> |
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Output:
Count of odd numbers is 3 Count of even numbers is 2
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