Merge pull request #102 from Shoyeb45/feature/next-prime

Feature: added solution of next prime

Author: ks-iitjmu

7_mathematics/next_prime.cpp

@@ -0,0 +1,116 @@
+#include <bits/stdc++.h>
+using namespace std;
+typedef long long ll;
+
+/*
+
+# Problem: Next Prime (https://cses.fi/problemset/task/3396/)
+
+# Prerequisite:
+-> Prime Numbers, Divisors, Sieve
+
+# Main Idea:
+-> There is not a large gap between prime numbers. The number of prime numbers from 1 to n is given by π(n) - prime counting function.
+-> And π(n) = n / log(n), since there are many prime number and their count increases as we go further. So we can start from n + 1
+   and keep going until we found the prime number(p > n). 
+-> For checking prime number we can check all numbers from 1 to sqrt(n) or we can use sieve of Erastosthenes to get all the prime numbers from 1 to 1e6
+   and since n <= 1e12, so we can efficiently check if the number is prime or not just by checking the divisibility with these primes from 1 to 1e6. 
+
+- is_prime_1 : This function is implemented using square root solution
+- precompute : it computes all the primes from 1 to 1e6 and return the prime vector
+- is_prime_2 : Uses sieve to check if the number is prime or not.
+
+Complexties:
+If we use square root method -
+    1. Time Complexity: O(t * sqrt(n) * k)
+    2. Space Commplexity: O(1)
+If we use sieve method:
+    1. Time Complexity: O(N(log(log(N))) + t * (N / log(N) * k)), where N = 1e6
+    2. Space Commplexity: O(N / log(N))
+
+And approximately k can be upto 600 for numbers <= 1e6
+*/
+
+/// Checks if the given number is prime or not. Uses sqaure root method to check.
+/// TC: O(sqrt(n))
+/// @return true if number is prime or false  
+bool is_prime_1(ll number) {
+    if (number == 2 || number == 3) {
+        return true;
+    }
+    if (number < 2 || number % 2 == 0 || number % 6 % 4 != 1) {
+        return false;
+    }
+ 
+    for (ll i = 3; 1LL * i * i <= number; i += 2) {
+        if (number % i == 0) {
+            return false;
+        }
+    }
+    return true;
+}
+
+/// Function to generate prime numbers.
+/// @return All prime numbers from 1 to 1e6
+vector<ll> precompute() {
+    const int N = 1e6 + 1;
+    vector<ll> primes;
+    bool is_prime[N];
+    fill(is_prime, is_prime + N, true);
+    is_prime[0] = is_prime[1] = false;
+
+    for (int i = 2; i < N; i++) {
+        if (is_prime[i]) {
+            primes.push_back((ll) i);
+            for (ll j = 1LL * i * i; j < N; j += i) {
+                is_prime[j] = false;
+            }
+        }
+    }
+    return primes;
+}
+
+/// Checks if the number is prime or not. Uses generated primes to check the primality.
+/// @return true if number is prime or false
+bool is_prime_2(ll number, vector<ll>& primes) {
+    if (number == 2 || number == 3) {
+        return true;
+    }
+    if (number < 2 || number % 2 == 0 || number % 6 % 4 != 1) {
+        return false;
+    }
+
+    for (ll prime: primes) {
+        if (prime == number) {
+            return true;
+        }
+        if (number % prime == 0) {
+            return false;
+        }
+    }
+    return true;
+}
+
+/// Function to solve one testcase
+void solve() {
+    ll n;
+    cin >> n;
+
+    ll next_prime = n + 1;
+    while (!is_prime_1(next_prime)) {
+        next_prime++;
+    }
+    cout << next_prime << "\n";
+}
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+    
+    while (t--) {
+        solve();
+    }
+}

README.md

@@ -71,6 +71,7 @@ The solutions are organized by problem categories:
 - **Exponentiation 2** - Fast exponentiation with modular arithmetic
 - **Sum of Divisors** - Mathematical series calculation
 - **Creating Strings** - Combinatorics with factorial calculations
+- **Next Prime** - Finding next prime
 
 ## 🔧 Implementation Details