Merge branch 'master' of https://github.com/kedarj0shi/ScriptsDump

Author: kedarj0shi

Mathematical_Algorithms/src/Sieve_of_Eratosthenes.cpp

@@ -1,4 +1,4 @@
-// Prime Seive aka Sieve of Eratosthenes
+// Prime Sieve aka Sieve of Eratosthenes
 // Time complexity : O(n*log(log(n)))
 // Space complexity: O(n)
 // DESC : here we will discuss how to find prime numbers in a given range 
@@ -11,40 +11,45 @@ void sieve(vector<bool>& isprime, int n){
     isprime[0] = isprime[1] = false;
     // 1st method
     for(int i = 2; i * i <= n; i++){
-        if(isprime[i] == true){ //if we find i'th number to be a prime number, we will mark every multiple of that number as non-prime
-            for(int j = i*i ; j <= n; j += i){
+        if(isprime[i] == true){ // if we find i'th number to be a prime number, we will mark every multiple of that number as non-prime
+            for(int j = i*i ; j <= n; j += i){ // marking every multiple of i false
                 isprime[j] = false;
             }
         }
     }
     // here we are iterating every element from 2 to sqrt(n), but since we know that 2 is the only even prime number
-    // this can be further optimized to
-    // first we will mark every multiple of 2, and then in the next loop we will iterating only for odd numbers 
-    // 2. more optimized
+    // this can be further optimized
+    // first we will mark every multiple of 2, and then in the next loop we will be iterating only for odd numbers 
+    // 2. more optimized approach
     for(int i = 2; i < 3; i++){
-        if(isprime[i] == true){ //if we find i'th number to be a prime number, we will mark every multiple of that number as non-prime
-            for(int j = i*i ; j <= n; j += i){
+        if(isprime[i] == true){ // if we find i'th number to be a prime number, we will mark every multiple of that number as non-prime
+            for(int j = i*i ; j <= n; j += i){ // marking every multiple of i false
                 isprime[j] = false;
             }
         }
     }
-    for(int i = 3; i * i <= n; i += 2){ // will only check for odd number
-        if(isprime[i] == true){ //if we find i'th number to be a prime number, we will mark every multiple of that number as non-prime
-            for(int j = i*i ; j <= n; j += i){
+    for(int i = 3; i * i <= n; i += 2){ // will only check for odd numbers
+        if(isprime[i] == true){ // if we find i'th number to be a prime number, we will mark every multiple of that number as non-prime
+            for(int j = i*i ; j <= n; j += i){// marking every multiple of i false
                 isprime[j] = false;
             }
         }
     }
     // now in the isprime vector we have prime numbers from 0 to 1000000
-    // note that we are using isprime vector of boolean data type and not integer as it will reduce space (integer takes 4bytes, whereas bool takes 1byte)
+    // note that we are using isprime vector of boolean data type and not integer as it will reduce space (integer takes 4bytes, whereas bool takes only 1byte)
 }
 
 signed main(){
     int n=1e6;
-    //initialize a vector isprime such that if value at an index is true then it is a prime number, else a composite
-    //initially we consider every number as prime
+    // initialize a vector isprime such that if value at an index is true then it is a prime number, else a composite
+    // initially we consider every number as a prime number
     vector<bool> isprime(n+1,true);  
     sieve(v,n);
-
+    
+    // printing prime numbers from 1 to n
+    for( int i = 1; i <= n; i++) {
+        if( isprime[i] == true )
+            cout << i << " ";
+    }
     return 0;
-}
+}