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

**Examples:**

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

Explanation:

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

Input:L = 30, R = 40Output: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; ` `}` |

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## 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; ` `}` |

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**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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