C/C++ Program to find Prime Numbers between given range

Given two numbers L and R, the task is to find the prime numbers between L and R.

Examples:

Input: L = 1, R = 10
Output: 2 3 5 7

Explanation:
Prime number between the 1 and 10 are 2, 3, 5, and 7

Input: L = 30, R = 40
Output: 31 37

Approach: The idea is to iterate from in the range [L, R] and check if any number in the given range is prime or not. If yes then print that number and check for the next number till we iterate all the numbers.

Below the implementation of the above approach:

C

// C program to find the prime numbers 
// between a given interval 
#include <stdio.h> 
  
// Function for print prime 
// number in given range 
void primeInRange(int L, int R) 
{ 
    int i, j, flag; 
  
    // Traverse each number in the 
    // interval with the help of for loop 
    for (i = L; i <= R; i++) { 
  
        // Skip 0 and 1 as they are 
        // niether prime nor composite 
        if (i == 1 || i == 0) 
            continue; 
  
        // flag variable to tell 
        // if i is prime or not 
        flag = 1; 
  
        // Iterate to check if i is prime 
        // or not 
        for (j = 2; j <= i / 2; ++j) { 
            if (i % j == 0) { 
                flag = 0; 
                break; 
            } 
        } 
  
        // flag = 1 means i is prime 
        // and flag = 0 means i is not prime 
        if (flag == 1) 
            printf("%d ", i); 
    } 
} 
  
// Driver Code 
int main() 
{ 
    // Given Range 
    int L = 1; 
    int R = 10; 
  
    // Function Call 
    primeInRange(L, R); 
  
    return 0; 
}

C++

// C++ program to find the prime numbers 
// between a given interval 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function for print prime 
// number in given range 
void primeInRange(int L, int R) 
{ 
    int flag; 
  
    // Traverse each number in the 
    // interval with the help of for loop 
    for (int i = L; i <= R; i++) { 
  
        // Skip 0 and 1 as they are 
        // niether prime nor composite 
        if (i == 1 || i == 0) 
            continue; 
  
        // flag variable to tell 
        // if i is prime or not 
        flag = 1; 
  
        // Iterate to check if i is prime 
        // or not 
        for (int j = 2; j <= i / 2; ++j) { 
            if (i % j == 0) { 
                flag = 0; 
                break; 
            } 
        } 
  
        // flag = 1 means i is prime 
        // and flag = 0 means i is not prime 
        if (flag == 1) 
            cout << i << " "; 
    } 
} 
  
// Driver Code 
int main() 
{ 
    // Given Range 
    int L = 1; 
    int R = 10; 
  
    // Function Call 
    primeInRange(L, R); 
  
    return 0; 
}

Output:

2 3 5 7

Time Complexity: O((R-L)*N), where N is the number, and L and R are the given range.
Auxiliary Space: O(1)

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My Personal Notes

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