Nearest prime less than given number n
You are given a number n ( 3 <= n < 10^6 ) and you have to find nearest prime less than n?
Examples:
Input : n = 10 Output: 7 Input : n = 17 Output: 13 Input : n = 30 Output: 29
A simple solution for this problem is to iterate from n-1 to 2, and for every number, check if it is a prime. If prime, then return it and break the loop. This solution looks fine if there is only one query. But not efficient if there are multiple queries for different values of n.
An efficient solution for this problem is to generate all primes less tha 10^6 using Sieve of Sundaram and store then in a array in increasing order. Now apply modified binary search to search nearest prime less than n. Time complexity of this solution is O(n log n + log n) = O(n log n).
C++
// C++ program to find the nearest prime to n. #include<bits/stdc++.h> #define MAX 1000000 using namespace std; // array to store all primes less than 10^6 vector<int> primes; // Utility function of Sieve of Sundaram void Sieve() { int n = MAX; // In general Sieve of Sundaram, produces primes // smaller than (2*x + 2) for a number given // number x int nNew = sqrt(n); // This array is used to separate numbers of the // form i+j+2ij from others where 1 <= i <= j int marked[n/2+500] = {0}; // eliminate indexes which does not produce primes for (int i=1; i<=(nNew-1)/2; i++) for (int j=(i*(i+1))<<1; j<=n/2; j=j+2*i+1) marked[j] = 1; // Since 2 is a prime number primes.push_back(2); // Remaining primes are of the form 2*i + 1 such // that marked[i] is false. for (int i=1; i<=n/2; i++) if (marked[i] == 0) primes.push_back(2*i + 1); } // modified binary search to find nearest prime less than N int binarySearch(int left,int right,int n) { if (left<=right) { int mid = (left + right)/2; // base condition is, if we are reaching at left // corner or right corner of primes[] array then // return that corner element because before or // after that we don't have any prime number in // primes array if (mid == 0 || mid == primes.size()-1) return primes[mid]; // now if n is itself a prime so it will be present // in primes array and here we have to find nearest // prime less than n so we will return primes[mid-1] if (primes[mid] == n) return primes[mid-1]; // now if primes[mid]<n and primes[mid+1]>n that // mean we reached at nearest prime if (primes[mid] < n && primes[mid+1] > n) return primes[mid]; if (n < primes[mid]) return binarySearch(left, mid-1, n); else return binarySearch(mid+1, right, n); } return 0; } // Driver program to run the case int main() { Sieve(); int n = 17; cout << binarySearch(0, primes.size()-1, n); return 0; } |
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Java
// Java program to find the nearest prime to n. import java.util.*; class GFG { static int MAX=1000000; // array to store all primes less than 10^6 static ArrayList<Integer> primes = new ArrayList<Integer>(); // Utility function of Sieve of Sundaram static void Sieve() { int n = MAX; // In general Sieve of Sundaram, produces primes // smaller than (2*x + 2) for a number given // number x int nNew = (int)Math.sqrt(n); // This array is used to separate numbers of the // form i+j+2ij from others where 1 <= i <= j int[] marked = new int[n / 2 + 500]; // eliminate indexes which does not produce primes for (int i = 1; i <= (nNew - 1) / 2; i++) for (int j = (i * (i + 1)) << 1; j <= n / 2; j = j + 2 * i + 1) marked[j] = 1; // Since 2 is a prime number primes.add(2); // Remaining primes are of the form 2*i + 1 such // that marked[i] is false. for (int i = 1; i <= n / 2; i++) if (marked[i] == 0) primes.add(2 * i + 1); } // modified binary search to find nearest prime less than N static int binarySearch(int left,int right,int n) { if (left <= right) { int mid = (left + right) / 2; // base condition is, if we are reaching at left // corner or right corner of primes[] array then // return that corner element because before or // after that we don't have any prime number in // primes array if (mid == 0 || mid == primes.size() - 1) return primes.get(mid); // now if n is itself a prime so it will be present // in primes array and here we have to find nearest // prime less than n so we will return primes[mid-1] if (primes.get(mid) == n) return primes.get(mid - 1); // now if primes[mid]<n and primes[mid+1]>n that // mean we reached at nearest prime if (primes.get(mid) < n && primes.get(mid + 1) > n) return primes.get(mid); if (n < primes.get(mid)) return binarySearch(left, mid - 1, n); else return binarySearch(mid + 1, right, n); } return 0; } // Driver code public static void main (String[] args) { Sieve(); int n = 17; System.out.println(binarySearch(0, primes.size() - 1, n)); } } // This code is contributed by mits |
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Python3
# Python3 program to find the nearest # prime to n. import math MAX = 10000; # array to store all primes less # than 10^6 primes = []; # Utility function of Sieve of Sundaram def Sieve(): n = MAX; # In general Sieve of Sundaram, produces # primes smaller than (2*x + 2) for a # number given number x nNew = int(math.sqrt(n)); # This array is used to separate numbers # of the form i+j+2ij from others where # 1 <= i <= j marked = [0] * (int(n / 2 + 500)); # eliminate indexes which does not # produce primes for i in range(1, int((nNew - 1) / 2) + 1): for j in range(((i * (i + 1)) << 1), (int(n / 2) + 1), (2 * i + 1)): marked[j] = 1; # Since 2 is a prime number primes.append(2); # Remaining primes are of the form # 2*i + 1 such that marked[i] is false. for i in range(1, int(n / 2) + 1): if (marked[i] == 0): primes.append(2 * i + 1); # modified binary search to find nearest # prime less than N def binarySearch(left, right, n): if (left <= right): mid = int((left + right) / 2); # base condition is, if we are reaching # at left corner or right corner of # primes[] array then return that corner # element because before or after that # we don't have any prime number in # primes array if (mid == 0 or mid == len(primes) - 1): return primes[mid]; # now if n is itself a prime so it will # be present in primes array and here # we have to find nearest prime less than # n so we will return primes[mid-1] if (primes[mid] == n): return primes[mid - 1]; # now if primes[mid]<n and primes[mid+1]>n # that means we reached at nearest prime if (primes[mid] < n and primes[mid + 1] > n): return primes[mid]; if (n < primes[mid]): return binarySearch(left, mid - 1, n); else: return binarySearch(mid + 1, right, n); return 0; # Driver Code Sieve(); n = 17; print(binarySearch(0, len(primes) - 1, n)); # This code is contributed by chandan_jnu |
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C#
// C# program to find the nearest prime to n. using System; using System.Collections; class GFG { static int MAX = 1000000; // array to store all primes less than 10^6 static ArrayList primes = new ArrayList(); // Utility function of Sieve of Sundaram static void Sieve() { int n = MAX; // In general Sieve of Sundaram, produces // primes smaller than (2*x + 2) for a // number given number x int nNew = (int)Math.Sqrt(n); // This array is used to separate numbers of the // form i+j+2ij from others where 1 <= i <= j int[] marked = new int[n / 2 + 500]; // eliminate indexes which does not produce primes for (int i = 1; i <= (nNew - 1) / 2; i++) for (int j = (i * (i + 1)) << 1; j <= n / 2; j = j + 2 * i + 1) marked[j] = 1; // Since 2 is a prime number primes.Add(2); // Remaining primes are of the form 2*i + 1 // such that marked[i] is false. for (int i = 1; i <= n / 2; i++) if (marked[i] == 0) primes.Add(2 * i + 1); } // modified binary search to find // nearest prime less than N static int binarySearch(int left, int right, int n) { if (left <= right) { int mid = (left + right) / 2; // base condition is, if we are reaching at left // corner or right corner of primes[] array then // return that corner element because before or // after that we don't have any prime number in // primes array if (mid == 0 || mid == primes.Count - 1) return (int)primes[mid]; // now if n is itself a prime so it will be // present in primes array and here we have // to find nearest prime less than n so we // will return primes[mid-1] if ((int)primes[mid] == n) return (int)primes[mid - 1]; // now if primes[mid]<n and primes[mid+1]>n // that mean we reached at nearest prime if ((int)primes[mid] < n && (int)primes[mid + 1] > n) return (int)primes[mid]; if (n < (int)primes[mid]) return binarySearch(left, mid - 1, n); else return binarySearch(mid + 1, right, n); } return 0; } // Driver code static void Main() { Sieve(); int n = 17; Console.WriteLine(binarySearch(0, primes.Count - 1, n)); } } // This code is contributed by chandan_jnu |
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PHP
<?php // PHP program to find the nearest // prime to n. $MAX = 10000; // array to store all primes less // than 10^6 $primes = array(); // Utility function of Sieve of Sundaram function Sieve() { global $MAX, $primes; $n = $MAX; // In general Sieve of Sundaram, produces // primes smaller than (2*x + 2) for a // number given number x $nNew = (int)(sqrt($n)); // This array is used to separate numbers // of the form i+j+2ij from others where // 1 <= i <= j $marked = array_fill(0, (int)($n / 2 + 500), 0); // eliminate indexes which does not // produce primes for ($i = 1; $i <= ($nNew - 1) / 2; $i++) for ($j = ($i * ($i + 1)) << 1; $j <= $n / 2; $j = $j + 2 * $i + 1) $marked[$j] = 1; // Since 2 is a prime number array_push($primes, 2); // Remaining primes are of the form // 2*i + 1 such that marked[i] is false. for ($i = 1; $i <= $n / 2; $i++) if ($marked[$i] == 0) array_push($primes, 2 * $i + 1); } // modified binary search to find nearest // prime less than N function binarySearch($left, $right, $n) { global $primes; if ($left <= $right) { $mid = (int)(($left + $right) / 2); // base condition is, if we are reaching // at left corner or right corner of // primes[] array then return that corner // element because before or after that // we don't have any prime number in // primes array if ($mid == 0 || $mid == count($primes) - 1) return $primes[$mid]; // now if n is itself a prime so it will // be present in primes array and here // we have to find nearest prime less than // n so we will return primes[mid-1] if ($primes[$mid] == $n) return $primes[$mid - 1]; // now if primes[mid]<n and primes[mid+1]>n // that means we reached at nearest prime if ($primes[$mid] < $n && $primes[$mid + 1] > $n) return $primes[$mid]; if ($n < $primes[$mid]) return binarySearch($left, $mid - 1, $n); else return binarySearch($mid + 1, $right, $n); } return 0; } // Driver Code Sieve(); $n = 17; echo binarySearch(0, count($primes) - 1, $n); // This code is contributed by chandan_jnu ?> |
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Output:
13
If you have another approach to solve this problem then please share in comments.
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