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Code Optimization Technique (logical AND and logical OR)

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  • Difficulty Level : Medium
  • Last Updated : 23 Jan, 2023
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Logical AND (&&)

While using && (logical AND), we must put the condition first whose probability of getting false is high so that compiler doesn’t need to check the second condition if the first condition is false.
 

C++14




#include <bits/stdc++.h>
using namespace std;
// Function to check whether n is odd
bool isOdd(int n) { return (n & 1); }
 
// Function to check whether n is prime
bool isPrime(int n)
{
    if (n <= 1)
        return false;
    for (int i = 2; i <= sqrt(n); i++)
        if ((n % i) == 0)
            return false;
    return true;
}
 
int main()
{
    int cnt = 0, n = 10;
 
    // Implementation 1
    for (int i = 2; i <= n; i++) {
        if (isOdd(i) && isPrime(i))
            cnt++;
    }
 
    cnt = 0;
    n = 10;
 
    // Implementation 2
    for (int i = 2; i <= n; i++) {
        if (isPrime(i) && isOdd(i))
            cnt++;
    }
}

Python3




# Function to check whether n is odd
def isOdd(n):
    pass
 
# Function to check whether n is prime
def isPrime(n):
    pass
 
if __name__ == '__main__':
    cnt = 0; n = 10
 
    # Implementation 1
    for i in range(2,n) :
        if isOdd(i) and isPrime(i):
            cnt+=1
     
 
    cnt = 0
    n = 10
 
    # Implementation 2
    for i in range(2,n):
        if isPrime(i) and isOdd(i):
            cnt+=1
    

Javascript




<script>
    // Function to check whether n is odd
    function isOdd(n) { return (n & 1); }
     
    // Function to check whether n is prime
    function isPrime(n)
    {
        if (n <= 1)
            return false;
        for (let i = 2; i <= Math.floor(Math.sqrt(n)); i++)
            if ((n % i) == 0)
                return false;
        return true;
    }
     
     
        let cnt = 0, n = 10;
     
        // Implementation 1
        for (let i = 2; i <= n; i++) {
            if (isOdd(i) && isPrime(i))
                cnt++;
        }
     
        cnt = 0;
        n = 10;
     
        // Implementation 2
        for (let i = 2; i <= n; i++) {
            if (isPrime(i) && isOdd(i))
                cnt++;
        }
     
    // This code is contributed by Pushpesh Raj.
</script>

Consider the above implementation: 
 

In implementation 1, we avoid checking even numbers whether they are prime or not as primality test requires more computation than checking a number for even/odd. 
Probability of a number getting odd is more than of it being a prime that’s why we first check whether the number is odd before checking it for prime. 
 

 

On the other hand in implementation 2, we are checking whether the number is prime or not before checking whether it is odd which makes unnecessary computation as all even numbers other than 2 are not prime but the implementation still checks them for prime. 
 

 

Logical OR (||)

While using || (logical OR), we must put the condition first whose probability of getting true is high so that compiler doesn’t need to check the second condition if the first condition is true.
 

C




#include <iostream.h>
 
// Function to check whether n is odd
bool isEven(int n);
 
// Function to check whether n is prime
bool isPrime(int n);
 
int main()
{
    int cnt = 0, n = 10;
 
    // Implementation 1
    for (int i = 3; i <= n; i++) {
        if (isEven(i) || !isPrime(i))
            cnt++;
    }
}

Java




import java.io.*;
 
class GFG {
// Function to check whether n is odd
boolean isEven(int n);
  
// Function to check whether n is prime
boolean isPrime(int n);
  
public static void main (String[] args)
{
    int cnt = 0, n = 10;
  
    // Implementation 1
    for (int i = 3; i <= n; i++) {
        if (isEven(i) || !isPrime(i))
            cnt++;
    }
}
}
 
// This code is contributed by Utkarsh

Python3




# Function to check whether n is even
def isEven(n):
    pass
 
# Function to check whether n is prime
def isPrime(n):
    pass
 
if __name__ == '__main__':
    cnt = 0; n = 10
 
    # Implementation 1
    for i in range(3,n) :
        if isOdd(i) or not isPrime(i):
            cnt+=1

Javascript




// Function to check whether n is odd
function isEven(n);
 
// Function to check whether n is prime
function isPrime(n);
 
 
    let cnt = 0, n = 10;
 
    // Implementation 1
    for (let i = 3; i <= n; i++) {
        if (isEven(i) || !isPrime(i))
            cnt++;
    }
 
// This code is contributed by Aman Kumar

As described earlier that the probability of a number being even is more than that of it being a non-prime. The current order of execution of the statements doesn’t allow even numbers greater than 2 to be checked whether they are non-prime (as they are all non-primes). 
 

Note: For larger inputs, the order of the execution of statements can affect the overall execution time for the program.
 


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