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# Program to find Prime Numbers Between given Interval

Given two numbers  a and b as interval range, the task is to find the prime numbers in between this interval.

Examples:

`Input : a = 1, b = 10Output : 2, 3, 5, 7Input : a = 10, b = 20Output : 11, 13, 17, 19`

In the below program, the range of numbers is taken as input and stored in the variables ‘a’ and ‘b’. Then using a for-loop, the numbers between the interval of a and b are traversed. For each number in the for loop, it is checked if this number is prime or not. If found prime, print the number. Then the next number in the loop is checked, till all numbers are checked.

Program:

## C++

 `// C++ program to find the prime numbers``// between a given interval` `#include ``using` `namespace` `std;` `int` `main()``{``    ``// Declare the variables``    ``int` `a, b, i, j, flag;` `    ``// Ask user to enter lower value of interval``    ``cout << ``"Enter lower bound of the interval: "``;``    ``cin >> a; ``// Take input` `    ``// Ask user to enter upper value of interval``    ``cout << ``"\nEnter upper bound of the interval: "``;``    ``cin >> b; ``// Take input` `    ``// Print display message``    ``cout << ``"\nPrime numbers between "``         ``<< a << ``" and "` `<< b << ``" are: "``;` `    ``// Traverse each number in the interval``    ``// with the help of for loop``    ``for` `(i = a; i <= b; i++) {``        ``// Skip 0 and 1 as they are``        ``// neither prime nor composite``        ``if` `(i == 1 || i == 0)``            ``continue``;` `        ``// flag variable to tell``        ``// if i is prime or not``        ``flag = 1;` `        ``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)``            ``cout << i << ``" "``;``    ``}` `    ``return` `0;``}` `// This code is contributed by Akanksha Rai`

## C

 `// C program to find the prime numbers``// between a given interval` `#include ` `int` `main()``{``    ``// Declare the variables``    ``int` `a, b, i, j, flag;` `    ``// Ask user to enter lower value of interval``    ``printf``(``"Enter lower bound of the interval: "``);``    ``scanf``(``"%d"``, &a); ``// Take input` `    ``// Ask user to enter upper value of interval``    ``printf``(``"\nEnter upper bound of the interval: "``);``    ``scanf``(``"%d"``, &b); ``// Take input` `    ``// Print display message``    ``printf``(``"\nPrime numbers between %d and %d are: "``, a, b);` `    ``// Traverse each number in the interval``    ``// with the help of for loop``    ``for` `(i = a; i <= b; i++) {``        ``// Skip 0 and 1 as they are``        ``// neither prime nor composite``        ``if` `(i == 1 || i == 0)``            ``continue``;` `        ``// flag variable to tell``        ``// if i is prime or not``        ``flag = 1;` `        ``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);``    ``}` `    ``return` `0;``}`

## Java

 `import` `java.util.Scanner;` `// Java program to find the prime numbers``// between a given interval``public` `class` `GFG {` `    ``// driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``Scanner sc = ``new` `Scanner(System.in);``        ``// Declare the variables``        ``int` `a, b, i, j, flag;` `        ``// Ask user to enter lower value of interval``        ``System.out.printf(``"Enter lower bound of the interval: "``);``        ``a = sc.nextInt(); ``// Take input` `        ``// Ask user to enter upper value of interval``        ``System.out.printf(``"\nEnter upper bound of the interval: "``);``        ``b = sc.nextInt(); ``// Take input` `        ``// Print display message``        ``System.out.printf(``"\nPrime numbers between %d and %d are: "``, a, b);` `        ``// Traverse each number in the interval``        ``// with the help of for loop``        ``for` `(i = a; i <= b; i++) {` `            ``// Skip 0 and 1 as they are``            ``// neither prime nor composite``            ``if` `(i == ``1` `|| i == ``0``)``                ``continue``;` `            ``// flag variable to tell``            ``// if i is prime or not``            ``flag = ``1``;` `            ``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``)``                ``System.out.println(i);``        ``}``    ``}``}`

## Python3

 `# Python3 program to find the prime``# numbers between a given interval` `if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Declare the variables``    ``a, b, i, j, flag ``=` `0``, ``0``, ``0``, ``0``, ``0` `    ``# Ask user to enter lower value of interval``    ``print``(``"Enter lower bound of the interval:"``,``                                      ``end ``=` `"")``    ``a ``=` `int``(``input``()) ``# Take input``    ``print``(a)``    ` `    ``# Ask user to enter upper value of interval``    ``print``(``"Enter upper bound of the interval:"``,``                                      ``end ``=` `"")``    ``b ``=` `int``(``input``()) ``# Take input``    ``print``(b)``    ` `    ``# Print display message``    ``print``(``"Prime numbers between"``, a, ``"and"``,``                        ``b, ``"are:"``, end ``=` `"")` `    ``# Traverse each number in the interval``    ``# with the help of for loop``    ``for` `i ``in` `range``(a, b ``+` `1``):` `        ``# Skip 1 as1 is neither``        ``# prime nor composite``        ``if` `(i ``=``=` `1``):``            ``continue` `        ``# flag variable to tell``        ``# if i is prime or not``        ``flag ``=` `1``        ` `        ``for` `j ``in` `range``(``2``, i ``/``/` `2` `+` `1``):``            ``if` `(i ``%` `j ``=``=` `0``):``                ``flag ``=` `0``                ``break``            ` `        ``# flag = 1 means i is prime``        ``# and flag = 0 means i is not prime``        ``if` `(flag ``=``=` `1``):``            ``print``(i, end ``=` `" "``)``            ` `# This code is contributed``# by Mohit kumar 29`

## C#

 `// C# program to find the prime numbers``// between a given interval``using` `System;` `class` `GFG{` `// Driver code   ``public` `static` `void` `Main(``string``[] args)``{``    ` `    ``// Declare the variables``    ``int` `a, b, i, j, flag;` `    ``// Ask user to enter lower value of interval``    ``Console.WriteLine(``"Enter lower bound of "` `+``                      ``"the interval: "``);``                      ` `    ``// Take input``    ``a = ``int``.Parse(Console.ReadLine());` `    ``// Ask user to enter upper value of interval``    ``Console.WriteLine(``"\nEnter upper bound "` `+``                      ``"of the interval: "``);``                      ` `    ``// Take input``    ``b = ``int``.Parse(Console.ReadLine());` `    ``// Print display message``    ``Console.WriteLine(``"\nPrime numbers between "` `+ ``                      ``"{0} and {1} are: "``, a, b);` `    ``// Traverse each number in the interval``    ``// with the help of for loop``    ``for``(i = a; i <= b; i++)``    ``{``        ` `        ``// Skip 0 and 1 as they are``        ``// neither prime nor composite``        ``if` `(i == 1 || i == 0)``            ``continue``;` `        ``// flag variable to tell``        ``// if i is prime or not``        ``flag = 1;` `        ``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)``            ``Console.WriteLine(i);``    ``}``}``}` `// This code is contributed by jana_sayantan`

## Javascript

 ``

Output:

`Enter lower bound of the interval: 1Enter upper bound of the interval: 10Prime numbers between 1 and 10 are: 2 3 5 7 `

Time Complexity: O(N2), Where N is the difference between the range
Auxiliary Space: O(1)

### Optimized Solution :

The idea is to use the fact that even numbers (except 2) are not primes.

## C++

 `// C++ program to find the prime numbers``// between a given interval` `#include ``using` `namespace` `std;` `int` `main()``{``    ``// Declare the variables``    ``int` `a, b, i, j;` `    ``// Ask user to enter lower value of interval``    ``// interval < 0 cannot be prime numbers``    ``cout << ``"Enter lower bound of the interval: "``;``    ``cin >> a; ``// Take input` `    ``// Ask user to enter upper value of interval``    ``cout << ``"\nEnter upper bound of the interval: "``;``    ``cin >> b; ``// Take input` `    ``// Print display message``    ``cout << ``"\nPrime numbers between "` `<< a << ``" and "` `<< b``         ``<< ``" are: "``;` `    ``// Explicitly handling the cases when a is less than 2``    ``// since 0 and 1 are not prime numbers``    ``if` `(a <= 2) {``        ``a = 2;``        ``if` `(b >= 2) {``            ``cout << a << ``" "``;``            ``a++;``        ``}``    ``}` `    ``// MAKING SURE THAT a IS ODD BEFORE WE BEGIN``    ``// THE LOOP``    ``if` `(a % 2 == 0)``        ``a++;` `    ``// NOTE : WE TRAVERSE THROUGH ODD NUMBERS ONLY``    ``for` `(i = a; i <= b; i = i + 2) {` `        ``// flag variable to tell``        ``// if i is prime or not``        ``bool` `flag = 1;` `        ``// WE TRAVERSE TILL SQUARE ROOT OF j only.``        ``// (LARGEST POSSIBLE VALUE OF A PRIME FACTOR)``        ``for` `(j = 2; j * j <= i; ++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){``          ``if``(i==1)``            ``continue``;``          ``else``            ``cout << i << ``" "``;``        ``}``    ``}` `    ``return` `0;``}`

## Java

 `import` `java.util.*;` `public` `class` `PrimeNumbersInRange {` `    ``// function which checks whether a number is Prime or Not``    ``// If the number is prime, it returns true. Else, it returns false.``    ``public` `static` `boolean` `isPrime(``int` `n) {``        ``// 0 and 1 are neither prime nor composite numbers``        ``if` `(n == ``0` `|| n == ``1``) {``            ``return` `false``;``        ``}``        ``// 2 is a prime number``        ``if` `(n == ``2``) {``            ``return` `true``;``        ``}``        ``// every composite number has a prime factor``        ``// less than or equal to its square root.``        ``for` `(``int` `i = ``2``; i * i <= n; i++) {``            ``if` `(n % i == ``0``) {``                ``return` `false``;``            ``}``        ``}``        ``return` `true``;` `    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args) {``        ``Scanner sc = ``new` `Scanner(System.in);` `        ``// Ask user to enter lower value of interval``        ``System.out.print(``"Enter lower bound of the interval: "``);``        ``int` `lower = sc.nextInt(); ``// Take input` `        ``// Ask user to enter upper value of interval``        ``System.out.print(``"\nEnter upper bound of the interval: "``);``        ``int` `upper = sc.nextInt(); ``// Take input` `        ``// Print display message``        ``System.out.printf(``"\nPrime numbers between %d and %d are: "``, lower, upper);` `        ``for` `(``int` `i = lower; i <= upper; i++) {``            ``if` `(isPrime(i)) {``                ``System.out.print(i + ``" "``);``            ``}``        ``}``        ``sc.close();``    ``}``}`

## Python3

 `# Python3 program to find the prime``# numbers between a given interval`` ` `if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Declare the variables``    ``a, b, i, j ``=` `0``, ``0``, ``0``, ``0``    ` `    ``# Ask user to enter lower value of interval``    ``print``(``"Enter lower bound of the interval:"``,end ``=` `"")``    ` `    ``a ``=` `int``(``input``()) ``# Take input``    ``print``(a)``    ` `    ``# Ask user to enter upper value of interval``    ``print``(``"Enter upper bound of the interval:"``,end ``=` `"")``    ` `    ``b ``=` `int``(``input``()) ``# Take input``    ``print``(b)``    ` `    ``# Print display message``    ``print``(``"Prime numbers between"``, a, ``"and"``,b, ``"are:"``, end ``=` `"")``    ` `    ` `    ``# Explicitly handling the cases when a is less than 2``    ``if` `(a ``=``=` `1``):``        ``print``(a,end``=``" "``)``        ``a``+``=``1``        ``if` `(b >``=` `2``):``            ``print``(a,end``=``" "``)``            ``a``+``=``1``    ``if` `(a ``=``=` `2``):``        ``print``(a,end``=``" "``)``    ` `    ``# MAKING SURE THAT a IS ODD BEFORE WE BEGIN``    ``# THE LOOP``    ``if` `(a ``%` `2` `=``=` `0``):``        ``a``+``=``1``    ``# NOTE : WE TRAVERSE THROUGH ODD NUMBERS ONLY``    ``for` `i ``in` `range``(a,b``+``1``,``2``):``        ` `        ``# flag variable to tell``        ``# if i is prime or not``        ``flag ``=` `1``        ``# WE TRAVERSE TILL SQUARE ROOT OF j only.``        ``# (LARGEST POSSIBLE VALUE OF A PRIME FACTOR)``        ``j ``=` `2``        ``while``(j ``*` `j <``=` `i):``            ``if` `(i ``%` `j ``=``=` `0``):``                ``flag ``=` `0``                ``break``            ``j``+``=``1``        ` `        ``# flag = 1 means i is prime``        ``# and flag = 0 means i is not prime``        ``if` `(flag ``=``=` `1``):``            ``print``(i,end``=``" "``)` `# This code is contributed by shubhamsingh10`

## C#

 `// C# program to find the prime numbers``// between a given interval``using` `System;``class` `GFG``{``  ``// Driver code``  ``static` `public` `void` `Main()``  ``{` `    ``// Declare the variables``    ``int` `a, b, i, j, flag;` `    ``// Ask user to enter lower value of interval``    ``Console.Write(``      ``"Enter lower bound of the interval: "``);``    ``a = Convert.ToInt32(``      ``Console.ReadLine()); ``// Take input` `    ``// Ask user to enter upper value of interval``    ``Console.Write(``      ``"\nEnter upper bound of the interval: "``);``    ``b = Convert.ToInt32(``      ``Console.ReadLine()); ``// Take input` `    ``// Print display message``    ``Console.Write(``"\nPrime numbers between "` `+ a``                  ``+ ``" and "` `+ b + ``" are: "``);` `    ``// Explicitly handling the cases when a is less than``    ``// 2``    ``if` `(a == 1)``    ``{``      ``Console.Write(a + ``" "``);``      ``a++;``      ``if` `(b >= 2)``      ``{``        ``Console.Write(a + ``" "``);``        ``a++;``      ``}``    ``}``    ``if` `(a == 2)``    ``{``      ``Console.Write(a + ``" "``);``    ``}` `    ``// MAKING SURE THAT a IS ODD BEFORE WE BEGIN``    ``// THE LOOP``    ``if` `(a % 2 == 0)``    ``{``      ``a++;``    ``}` `    ``// NOTE : WE TRAVERSE THROUGH ODD NUMBERS ONLY``    ``for` `(i = a; i <= b; i = i + 2)``    ``{` `      ``// flag variable to tell``      ``// if i is prime or not``      ``flag = 1;` `      ``// WE TRAVERSE TILL SQUARE ROOT OF j only.``      ``// (LARGEST POSSIBLE VALUE OF A PRIME FACTOR)``      ``for` `(j = 2; j * j <= i; ++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)``      ``{``        ``Console.Write(i + ``" "``);``      ``}``    ``}``  ``}``}` `// This code is contributed by rag2127`

## Javascript

 ``

Output:

`Enter lower bound of the interval: 1Enter upper bound of the interval: 10Prime numbers between 1 and 10 are: 2 3 5 7`

Time Complexity: O(N*sqrt(N)), Where N is the difference between the range
Auxiliary Space: O(1)

### Sieve of Eratosthenes Method:

The sieve of Eratosthenes is one of the most efficient ways to find all primes smaller than N when N is smaller than 10 million or so.

#### Steps:

1. Create a boolean array prime[srt to n] and initialize all its entries as true.
2. Mark prime[0] and prime[1] as false because they are not prime numbers.
3. Starting from p = srt, for each prime number p less than or equal to the square root of n, mark all multiples of p greater than or equal to p*p as composite by setting prime[i] to false.
4. Finally, print all prime numbers between srt and n.

Below is the implementation:

## C++

 `// C++ program to find the prime numbers``// between a given interval using Sieve of Eratosthenes``#include ``using` `namespace` `std;` `void` `SieveOfEratosthenes(``int` `srt, ``int` `n)``{``    ``// Create a boolean array "prime[srt to n]" and``    ``// initialize all entries it as true. A value in``    ``// prime[i] will finally be false if i is Not a prime,``    ``// else true.``    ``bool` `prime[n + 2];``    ``memset``(prime, ``true``, ``sizeof``(prime));``    ``prime[0] = ``false``;``    ``prime[1] = ``false``;` `    ``for` `(``int` `p = 2; p * p <= n; p++) {``        ``// If prime[p] is not changed, then it is a prime``        ``if` `(prime[p] == ``true``) {``            ``// Update all multiples of p greater than or``            ``// equal to the square of it numbers which are``            ``// multiple of p and are less than p^2 are``            ``// already been marked.``            ``for` `(``int` `i = p * p; i <= n; i += p)``                ``prime[i] = ``false``;``        ``}``    ``}` `    ``// Print all prime numbers``    ``for` `(``int` `p = srt; p <= n; p++)``        ``if` `(prime[p])``            ``cout << p << ``" "``;``}` `// Driver Code``int` `main()``{``    ``int` `srt = 1;``    ``int` `end = 10;``    ``SieveOfEratosthenes(srt, end);``    ``return` `0;``}` `// This code is contributed by Susobhan Akhuli`

## Java

 `// java program to find the prime numbers``// between a given interval using Sieve of Eratosthenes``import` `java.util.*;` `public` `class` `GFG {``    ``public` `static` `void` `SieveOfEratosthenes(``int` `srt, ``int` `n)``    ``{``        ``// Create a boolean array "prime[srt to n]" and``        ``// initialize all entries it as true. A value in``        ``// prime[i] will finally be false if i is Not a``        ``// prime, else true.``        ``boolean``[] prime = ``new` `boolean``[n + ``2``];``        ``Arrays.fill(prime, ``true``);``        ``prime[``0``] = ``false``;``        ``prime[``1``] = ``false``;` `        ``for` `(``int` `p = ``2``; p * p <= n; p++) {``            ``// If prime[p] is not changed, then it is a``            ``// prime``            ``if` `(prime[p] == ``true``) {``                ``// Update all multiples of p greater than or``                ``// equal to the square of it numbers which``                ``// are multiple of p and are less than p^2``                ``// are already been marked.``                ``for` `(``int` `i = p * p; i <= n; i += p) {``                    ``prime[i] = ``false``;``                ``}``            ``}``        ``}` `        ``// Print all prime numbers``        ``for` `(``int` `p = srt; p <= n; p++) {``            ``if` `(prime[p]) {``                ``System.out.print(p + ``" "``);``            ``}``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `srt = ``1``;``        ``int` `end = ``10``;``        ``SieveOfEratosthenes(srt, end);``    ``}``}` `// This code is contributed by Susobhan Akhuli`

## Python3

 `# Python program to find the prime numbers``# between a given interval using Sieve of Eratosthenes``import` `math` `def` `SieveOfEratosthenes(srt, n):``    ``# Create a boolean array "prime[srt to n]" and``    ``# initialize all entries it as true. A value in``    ``# prime[i] will finally be false if i is Not a prime,``    ``# else true.``    ``prime ``=` `[``True` `for` `i ``in` `range``(n ``+` `2` `)]``    ``prime[``0``] ``=` `False``    ``prime[``1``] ``=` `False` `    ``for` `p ``in` `range``(``2``, ``int``(math.sqrt(n))``+``1``):``        ``# If prime[p] is not changed, then it is a prime``        ``if` `prime[p] ``=``=` `True``:``            ``# Update all multiples of p greater than or``            ``# equal to the square of it numbers which are``            ``# multiple of p and are less than p^2 are``            ``# already been marked.``            ``for` `i ``in` `range``(p``*``p, n``+``1``, p):``                ``prime[i] ``=` `False` `    ``# Print all prime numbers``    ``for` `p ``in` `range``(srt, n``+``1``):``        ``if` `prime[p]:``            ``print``(p, end``=``" "``)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``srt ``=` `1``    ``end ``=` `10``    ``SieveOfEratosthenes(srt, end)` `# This code is contributed by Susobhan Akhuli`

## C#

 `// C# program to find the prime numbers``// between a given interval using Sieve of Eratosthenes``using` `System;` `public` `class` `GFG``{``    ``public` `static` `void` `SieveOfEratosthenes(``int` `srt, ``int` `n)``    ``{``        ``// Create a boolean array "prime[srt to n]" and``        ``// initialize all entries it as true. A value in``        ``// prime[i] will finally be false if i is Not a``        ``// prime, else true.``        ``bool``[] prime = ``new` `bool``[n + 2];``        ``Array.Fill(prime, ``true``);``        ``prime[0] = ``false``;``        ``prime[1] = ``false``;` `        ``for` `(``int` `p = 2; p * p <= n; p++)``        ``{``            ``// If prime[p] is not changed, then it is a``            ``// prime``            ``if` `(prime[p] == ``true``)``            ``{``                ``// Update all multiples of p greater than or``                ``// equal to the square of it. Numbers which``                ``// are multiples of p and are less than p^2``                ``// are already marked.``                ``for` `(``int` `i = p * p; i <= n; i += p)``                ``{``                    ``prime[i] = ``false``;``                ``}``            ``}``        ``}` `        ``// Print all prime numbers``        ``for` `(``int` `p = srt; p <= n; p++)``        ``{``            ``if` `(prime[p])``            ``{``                ``Console.Write(p + ``" "``);``            ``}``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``int` `srt = 1;``        ``int` `end = 10;``        ``SieveOfEratosthenes(srt, end);``    ``}``}` `// This code is contributed by Pushpesh Raj`

## Javascript

 `function` `SieveOfEratosthenes(srt, n) {``    ``// Create a boolean array "prime[srt to n]" and``    ``// initialize all entries it as true. A value in``    ``// prime[i] will finally be false if i is Not a prime,``    ``// else true.``    ``let prime = ``new` `Array(n + 2 ).fill(``true``);``    ``prime[0] = ``false``;``    ``prime[1] = ``false``;` `    ``for` `(let p = 2; p * p <= n; p++) {``        ``// If prime[p] is not changed, then it is a prime``        ``if` `(prime[p] == ``true``) {``            ``// Update all multiples of p greater than or``            ``// equal to the square of it numbers which are``            ``// multiple of p and are less than p^2 are``            ``// already been marked.``            ``for` `(let i = p * p; i <= n; i += p)``                ``prime[i] = ``false``;``        ``}``    ``}` `    ``// Print all prime numbers``    ``for` `(let p = srt; p <= n; p++)``        ``if` `(prime[p])``            ``console.log(p + ``" "``);``}` `let srt = 1;``let end = 10;``SieveOfEratosthenes(srt, end);`

Output

```2 3 5 7

```

Time Complexity:
The time complexity of the Sieve of Eratosthenes algorithm is O(n*log(log(n))) as it iterates over all numbers from 2 to n and removes all the multiples of each prime number.

In this code, the algorithm is applied only to a range [srt, n], which reduces the time complexity to O((n-srt+1)*log(log(n))) for the range [srt, n]. The loop for marking the multiples of each prime number iterates from p*p to n, which takes O((n-srt+1)*log(log(n))) time. Therefore, the overall time complexity of this code is O((n-srt+1)*log(log(n))).

Auxiliary Space:
The space complexity of the algorithm is O(n), which is the size of the boolean array prime. However, the size of the array is reduced to n+2-srt, which is the size of the array required for the given range [srt, n]. Therefore, the auxiliary space complexity of the code is O(n-srt+1).