Given a number n, print all primes smaller than n.
Input: N = 10
Output: 2, 3, 5, 7
Explanation : The output “2, 3, 5, 7” for input N = 10 represents the list of the prime numbers less than or equal to 10.
Input: N = 5
Output: 2, 3, 5
Explanation : The output “2, 3, 5” for input N = 5 represents the list of the prime numbers less than or equal to 5.
A Naive approach is to run a loop from 0 to n-1 and check each number for primeness. A Better Approach is to use Simple Sieve of Eratosthenes.
C++
#include <iostream>
#include <vector>
void simpleSieve(int limit) {
// Create a boolean array "mark[0..limit-1]" and
// initialize all entries of it as true. A value
// in mark[p] will finally be false if 'p' is Not
// a prime, else true.
std::vector<bool> mark(limit, true);
// One by one traverse all numbers so that their
// multiples can be marked as composite.
for (int p = 2; p * p < limit; p++) {
// If p is not changed, then it is a prime
if (mark[p] == true) {
// Update all multiples of p
for (int i = p * p; i < limit; i += p)
mark[i] = false;
}
}
// Print all prime numbers and store them in prime
for (int p = 2; p < limit; p++)
if (mark[p] == true)
std::cout << p << " ";
}
int main() {
int limit = 100;
simpleSieve(limit);
return 0;
}
C
// This functions finds all primes smaller than 'limit'
// using simple sieve of eratosthenes.
void simpleSieve(int limit)
{
// Create a boolean array "mark[0..limit-1]" and
// initialize all entries of it as true. A value
// in mark[p] will finally be false if 'p' is Not
// a prime, else true.
bool mark[limit];
for(int i = 0; i<limit; i++) {
mark[i] = true;
}
// One by one traverse all numbers so that their
// multiples can be marked as composite.
for (int p=2; p*p<limit; p++)
{
// If p is not changed, then it is a prime
if (mark[p] == true)
{
// Update all multiples of p
for (int i=p*p; i<limit; i+=p)
mark[i] = false;
}
}
// Print all prime numbers and store them in prime
for (int p=2; p<limit; p++)
if (mark[p] == true)
cout << p << " ";
}
Java
import java.util.Arrays;
public class Main {
public static void main(String[] args) {
int limit = 100;
simpleSieve(limit);
}
// This function finds all primes smaller than 'limit'
// using simple sieve of eratosthenes.
static void simpleSieve(int limit) {
// Create a boolean array "mark[0..limit-1]" and
// initialize all entries of it as true. A value
// in mark[p] will finally be false if 'p' is Not
// a prime, else true.
boolean []mark = new boolean[limit];
Arrays.fill(mark, true);
// One by one traverse all numbers so that their
// multiples can be marked as composite.
for (int p = 2; p * p < limit; p++) {
// If p is not changed, then it is a prime
if (mark[p] == true) {
// Update all multiples of p
for (int i = p * p; i < limit; i += p)
mark[i] = false;
}
}
// Print all prime numbers and store them in prime
for (int p = 2; p < limit; p++)
if (mark[p] == true)
System.out.print(p + " ");
}
}
Python
def simple_sieve(limit):
# Create a boolean array "mark[0..limit-1]" and
# initialize all entries of it as true. A value
# in mark[p] will finally be false if 'p' is Not
# a prime, else true.
mark = [True for _ in range(limit)]
# One by one traverse all numbers so that their
# multiples can be marked as composite.
for p in range(2, int(limit**0.5) + 1):
# If p is not changed, then it is a prime
if mark[p] == True:
# Update all multiples of p
for i in range(p * p, limit, p):
mark[i] = False
# Print all prime numbers and store them in prime
for p in range(2, limit):
if mark[p] == True:
print(p, end=" ")
limit = 100
simple_sieve(limit)
C#
// This functions finds all primes smaller than 'limit'
// using simple sieve of eratosthenes.
static void simpleSieve(int limit)
{
// Create a boolean array "mark[0..limit-1]" and
// initialize all entries of it as true. A value
// in mark[p] will finally be false if 'p' is Not
// a prime, else true.
bool []mark = new bool[limit];
Array.Fill(mark, true);
// One by one traverse all numbers so that their
// multiples can be marked as composite.
for (int p = 2; p * p < limit; p++)
{
// If p is not changed, then it is a prime
if (mark[p] == true)
{
// Update all multiples of p
for (int i = p * p; i < limit; i += p)
mark[i] = false;
}
}
// Print all prime numbers and store them in prime
for (int p = 2; p < limit; p++)
if (mark[p] == true)
Console.Write(p + " ");
}
// This code is contributed by pratham76.
JavaScript
function simpleSieve(limit) {
// Create a boolean array "mark[0..limit-1]" and
// initialize all entries of it as true. A value
// in mark[p] will finally be false if 'p' is Not
// a prime, else true.
let mark = new Array(limit).fill(true);
// One by one traverse all numbers so that their
// multiples can be marked as composite.
for (let p = 2; p * p < limit; p++) {
// If p is not changed, then it is a prime
if (mark[p] === true) {
// Update all multiples of p
for (let i = p * p; i < limit; i += p)
mark[i] = false;
}
}
// Print all prime numbers and store them in prime
for (let p = 2; p < limit; p++)
if (mark[p] === true)
console.log(p + " ");
}
let limit = 100;
simpleSieve(limit);
Problems with Simple Sieve:
The Sieve of Eratosthenes looks good, but consider the situation when n is large, the Simple Sieve faces the following issues.
- An array of size ?(n) may not fit in memory
- The simple Sieve is not cached friendly even for slightly bigger n. The algorithm traverses the array without locality of reference
Segmented Sieve
The idea of a segmented sieve is to divide the range [0..n-1] in different segments and compute primes in all segments one by one. This algorithm first uses Simple Sieve to find primes smaller than or equal to ?(n). Below are steps used in Segmented Sieve.
- Use Simple Sieve to find all primes up to the square root of ‘n’ and store these primes in an array “prime[]”. Store the found primes in an array ‘prime[]’.
- We need all primes in the range [0..n-1]. We divide this range into different segments such that the size of every segment is at-most ?n
- Do following for every segment [low..high]
- Create an array mark[high-low+1]. Here we need only O(x) space where x is a number of elements in a given range.
- Iterate through all primes found in step 1. For every prime, mark its multiples in the given range [low..high].
In Simple Sieve, we needed O(n) space which may not be feasible for large n. Here we need O(?n) space and we process smaller ranges at a time (locality of reference)
Below is the implementation of the above idea.
C++
// C++ program to print all primes smaller than
// n using segmented sieve
#include <bits/stdc++.h>
using namespace std;
// This functions finds all primes smaller than 'limit'
// using simple sieve of eratosthenes. It also stores
// found primes in vector prime[]
void simpleSieve(int limit, vector<int> &prime)
{
// Create a boolean array "mark[0..n-1]" and initialize
// all entries of it as true. A value in mark[p] will
// finally be false if 'p' is Not a prime, else true.
vector<bool> mark(limit + 1, true);
for (int p=2; p*p<limit; p++)
{
// If p is not changed, then it is a prime
if (mark[p] == true)
{
// Update all multiples of p
for (int i=p*p; i<limit; i+=p)
mark[i] = false;
}
}
// Print all prime numbers and store them in prime
for (int p=2; p<limit; p++)
{
if (mark[p] == true)
{
prime.push_back(p);
cout << p << " ";
}
}
}
// Prints all prime numbers smaller than 'n'
void segmentedSieve(int n)
{
// Compute all primes smaller than or equal
// to square root of n using simple sieve
int limit = floor(sqrt(n))+1;
vector<int> prime;
prime.reserve(limit);
simpleSieve(limit, prime);
// Divide the range [0..n-1] in different segments
// We have chosen segment size as sqrt(n).
int low = limit;
int high = 2*limit;
// While all segments of range [0..n-1] are not processed,
// process one segment at a time
while (low < n)
{
if (high >= n)
high = n;
// To mark primes in current range. A value in mark[i]
// will finally be false if 'i-low' is Not a prime,
// else true.
bool mark[limit+1];
memset(mark, true, sizeof(mark));
// Use the found primes by simpleSieve() to find
// primes in current range
for (int i = 0; i < prime.size(); i++)
{
// Find the minimum number in [low..high] that is
// a multiple of prime[i] (divisible by prime[i])
// For example, if low is 31 and prime[i] is 3,
// we start with 33.
int loLim = floor(low/prime[i]) * prime[i];
if (loLim < low)
loLim += prime[i];
/* Mark multiples of prime[i] in [low..high]:
We are marking j - low for j, i.e. each number
in range [low, high] is mapped to [0, high-low]
so if range is [50, 100] marking 50 corresponds
to marking 0, marking 51 corresponds to 1 and
so on. In this way we need to allocate space only
for range */
for (int j=loLim; j<high; j+=prime[i])
mark[j-low] = false;
}
// Numbers which are not marked as false are prime
for (int i = low; i<high; i++)
if (mark[i - low] == true)
cout << i << " ";
// Update low and high for next segment
low = low + limit;
high = high + limit;
}
}
// Driver program to test above function
int main()
{
int n = 100;
cout << "Primes smaller than " << n << ":\n";
segmentedSieve(n);
return 0;
}
Java
// Java program to print all primes smaller than
// n using segmented sieve
import java.util.Vector;
import static java.lang.Math.sqrt;
import static java.lang.Math.floor;
class Test
{
// This method finds all primes smaller than 'limit'
// using simple sieve of eratosthenes. It also stores
// found primes in vector prime[]
static void simpleSieve(int limit, Vector<Integer> prime)
{
// Create a boolean array "mark[0..n-1]" and initialize
// all entries of it as true. A value in mark[p] will
// finally be false if 'p' is Not a prime, else true.
boolean mark[] = new boolean[limit+1];
for (int i = 0; i < mark.length; i++)
mark[i] = true;
for (int p=2; p*p<limit; p++)
{
// If p is not changed, then it is a prime
if (mark[p] == true)
{
// Update all multiples of p
for (int i=p*p; i<limit; i+=p)
mark[i] = false;
}
}
// Print all prime numbers and store them in prime
for (int p=2; p<limit; p++)
{
if (mark[p] == true)
{
prime.add(p);
System.out.print(p + " ");
}
}
}
// Prints all prime numbers smaller than 'n'
static void segmentedSieve(int n)
{
// Compute all primes smaller than or equal
// to square root of n using simple sieve
int limit = (int) (floor(sqrt(n))+1);
Vector<Integer> prime = new Vector<>();
simpleSieve(limit, prime);
// Divide the range [0..n-1] in different segments
// We have chosen segment size as sqrt(n).
int low = limit;
int high = 2*limit;
// While all segments of range [0..n-1] are not processed,
// process one segment at a time
while (low < n)
{
if (high >= n)
high = n;
// To mark primes in current range. A value in mark[i]
// will finally be false if 'i-low' is Not a prime,
// else true.
boolean mark[] = new boolean[limit+1];
for (int i = 0; i < mark.length; i++)
mark[i] = true;
// Use the found primes by simpleSieve() to find
// primes in current range
for (int i = 0; i < prime.size(); i++)
{
// Find the minimum number in [low..high] that is
// a multiple of prime.get(i) (divisible by prime.get(i))
// For example, if low is 31 and prime.get(i) is 3,
// we start with 33.
int loLim = (int) (floor(low/prime.get(i)) * prime.get(i));
if (loLim < low)
loLim += prime.get(i);
/* Mark multiples of prime.get(i) in [low..high]:
We are marking j - low for j, i.e. each number
in range [low, high] is mapped to [0, high-low]
so if range is [50, 100] marking 50 corresponds
to marking 0, marking 51 corresponds to 1 and
so on. In this way we need to allocate space only
for range */
for (int j=loLim; j<high; j+=prime.get(i))
mark[j-low] = false;
}
// Numbers which are not marked as false are prime
for (int i = low; i<high; i++)
if (mark[i - low] == true)
System.out.print(i + " ");
// Update low and high for next segment
low = low + limit;
high = high + limit;
}
}
// Driver method
public static void main(String args[])
{
int n = 100;
System.out.println("Primes smaller than " + n + ":");
segmentedSieve(n);
}
}
Python
# Python3 program to print all primes
# smaller than n, using segmented sieve
import math
prime = []
# This method finds all primes
# smaller than 'limit' using
# simple sieve of eratosthenes.
# It also stores found primes in list prime
def simpleSieve(limit):
# Create a boolean list "mark[0..n-1]" and
# initialize all entries of it as True.
# A value in mark[p] will finally be False
# if 'p' is Not a prime, else True.
mark = [True for i in range(limit + 1)]
p = 2
while (p * p <= limit):
# If p is not changed, then it is a prime
if (mark[p] == True):
# Update all multiples of p
for i in range(p * p, limit + 1, p):
mark[i] = False
p += 1
# Print all prime numbers
# and store them in prime
for p in range(2, limit):
if mark[p]:
prime.append(p)
print(p,end = " ")
# Prints all prime numbers smaller than 'n'
def segmentedSieve(n):
# Compute all primes smaller than or equal
# to square root of n using simple sieve
limit = int(math.floor(math.sqrt(n)) + 1)
simpleSieve(limit)
# Divide the range [0..n-1] in different segments
# We have chosen segment size as sqrt(n).
low = limit
high = limit * 2
# While all segments of range [0..n-1] are not processed,
# process one segment at a time
while low < n:
if high >= n:
high = n
# To mark primes in current range. A value in mark[i]
# will finally be False if 'i-low' is Not a prime,
# else True.
mark = [True for i in range(limit + 1)]
# Use the found primes by simpleSieve()
# to find primes in current range
for i in range(len(prime)):
# Find the minimum number in [low..high]
# that is a multiple of prime[i]
# (divisible by prime[i])
# For example, if low is 31 and prime[i] is 3,
# we start with 33.
loLim = int(math.floor(low / prime[i]) *
prime[i])
if loLim < low:
loLim += prime[i]
# Mark multiples of prime[i] in [low..high]:
# We are marking j - low for j, i.e. each number
# in range [low, high] is mapped to [0, high-low]
# so if range is [50, 100] marking 50 corresponds
# to marking 0, marking 51 corresponds to 1 and
# so on. In this way we need to allocate space
# only for range
for j in range(loLim, high, prime[i]):
mark[j - low] = False
# Numbers which are not marked as False are prime
for i in range(low, high):
if mark[i - low]:
print(i, end = " ")
# Update low and high for next segment
low = low + limit
high = high + limit
# Driver Code
n = 100
print("Primes smaller than", n, ":")
segmentedSieve(100)
# This code is contributed by bhavyadeep
C#
// C# program to print
// all primes smaller than
// n using segmented sieve
using System;
using System.Collections;
class GFG
{
// This method finds all primes
// smaller than 'limit' using simple
// sieve of eratosthenes. It also stores
// found primes in vector prime[]
static void simpleSieve(int limit,
ArrayList prime)
{
// Create a boolean array "mark[0..n-1]"
// and initialize all entries of it as
// true. A value in mark[p] will finally be
// false if 'p' is Not a prime, else true.
bool[] mark = new bool[limit + 1];
for (int i = 0; i < mark.Length; i++)
mark[i] = true;
for (int p = 2; p * p < limit; p++)
{
// If p is not changed, then it is a prime
if (mark[p] == true)
{
// Update all multiples of p
for (int i = p * p; i < limit; i += p)
mark[i] = false;
}
}
// Print all prime numbers and store them in prime
for (int p = 2; p < limit; p++)
{
if (mark[p] == true)
{
prime.Add(p);
Console.Write(p + " ");
}
}
}
// Prints all prime numbers smaller than 'n'
static void segmentedSieve(int n)
{
// Compute all primes smaller than or equal
// to square root of n using simple sieve
int limit = (int) (Math.Floor(Math.Sqrt(n)) + 1);
ArrayList prime = new ArrayList();
simpleSieve(limit, prime);
// Divide the range [0..n-1] in
// different segments We have chosen
// segment size as sqrt(n).
int low = limit;
int high = 2*limit;
// While all segments of range
// [0..n-1] are not processed,
// process one segment at a time
while (low < n)
{
if (high >= n)
high = n;
// To mark primes in current range.
// A value in mark[i] will finally
// be false if 'i-low' is Not a prime,
// else true.
bool[] mark = new bool[limit + 1];
for (int i = 0; i < mark.Length; i++)
mark[i] = true;
// Use the found primes by
// simpleSieve() to find
// primes in current range
for (int i = 0; i < prime.Count; i++)
{
// Find the minimum number in
// [low..high] that is a multiple
// of prime.get(i) (divisible by
// prime.get(i)) For example,
// if low is 31 and prime.get(i)
// is 3, we start with 33.
int loLim = ((int)Math.Floor((double)(low /
(int)prime[i])) * (int)prime[i]);
if (loLim < low)
loLim += (int)prime[i];
/* Mark multiples of prime.get(i) in [low..high]:
We are marking j - low for j, i.e. each number
in range [low, high] is mapped to [0, high-low]
so if range is [50, 100] marking 50 corresponds
to marking 0, marking 51 corresponds to 1 and
so on. In this way we need to allocate space only
for range */
for (int j = loLim; j < high; j += (int)prime[i])
mark[j-low] = false;
}
// Numbers which are not marked as false are prime
for (int i = low; i < high; i++)
if (mark[i - low] == true)
Console.Write(i + " ");
// Update low and high for next segment
low = low + limit;
high = high + limit;
}
}
// Driver code
static void Main()
{
int n = 100;
Console.WriteLine("Primes smaller than " + n + ":");
segmentedSieve(n);
}
}
// This code is contributed by mits
JavaScript
// JavaSCript program to print all primes smaller than
// n using segmented sieve
// This functions finds all primes smaller than 'limit'
// using simple sieve of eratosthenes. It also stores
// found primes in vector prime[]
let res = "";
function simpleSieve(limit, prime)
{
// Create a boolean array "mark[0..n-1]" and initialize
// all entries of it as true. A value in mark[p] will
// finally be false if 'p' is Not a prime, else true.
let mark = new Array(limit+1).fill(true);
for (let p=2; p*p<limit; p++)
{
// If p is not changed, then it is a prime
if (mark[p] === true)
{
// Update all multiples of p
for (let i=p*p; i<limit; i+=p){
mark[i] = false;
}
}
}
// Print all prime numbers and store them in prime
for (let p=2; p<limit; p++)
{
if (mark[p] === true)
{
prime.push(p);
res = res + p + " ";
}
}
}
// Prints all prime numbers smaller than 'n'
function segmentedSieve(n)
{
// Compute all primes smaller than or equal
// to square root of n using simple sieve
let limit = Math.floor(Math.sqrt(n))+1;
let prime = new Array(limit);
simpleSieve(limit, prime);
// Divide the range [0..n-1] in different segments
// We have chosen segment size as sqrt(n).
let low = limit;
let high = 2*limit;
// While all segments of range [0..n-1] are not processed,
// process one segment at a time
while (low < n)
{
if (high >= n){
high = n;
}
// To mark primes in current range. A value in mark[i]
// will finally be false if 'i-low' is Not a prime,
// else true.
let mark = new Array(limit+1).fill(true);
// Use the found primes by simpleSieve() to find
// primes in current range
for (let i = 0; i < prime.length; i++)
{
// Find the minimum number in [low..high] that is
// a multiple of prime[i] (divisible by prime[i])
// For example, if low is 31 and prime[i] is 3,
// we start with 33.
let loLim = Math.floor(low/prime[i]) * prime[i];
if (loLim < low){
loLim += prime[i];
}
/* Mark multiples of prime[i] in [low..high]:
We are marking j - low for j, i.e. each number
in range [low, high] is mapped to [0, high-low]
so if range is [50, 100] marking 50 corresponds
to marking 0, marking 51 corresponds to 1 and
so on. In this way we need to allocate space only
for range */
for (let j=loLim; j<high; j+=prime[i]){
mark[j-low] = false;
}
}
// Numbers which are not marked as false are prime
for (let i = low; i<high; i++){
if (mark[i - low] == true){
res = res + i + " ";
}
}
// Update low and high for next segment
low = low + limit;
high = high + limit;
}
console.log(res);
}
// Driver program to test above function
let n = 100;
console.log("Primes smaller than", n);
segmentedSieve(n);
// The code is contributed by Gautam goel (gautamgoel962)
OutputPrimes smaller than 100:
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
Time Complexity : O(n * ln(sqrt(n)))
Auxiliary Space: O(sqrt(n))
Note that time complexity (or a number of operations) by Segmented Sieve is the same as Simple Sieve. It has advantages for large 'n' as it has better locality of reference thus allowing better caching by the CPU and also requires less memory space.
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Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
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Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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