Given n queries of the form range [L, R]. The task is to find the maximum difference between two prime numbers in the range for each query. If there are no prime in the range then print 0. All ranges are below 100005.
Examples:
Input : Q = 3
query1 = [2, 5]
query2 = [2, 2]
query3 = [24, 28]
Output : 3
0
0
In first query, 2 and 5 are prime number
in the range with maximum difference which
is 3. In second there
is only 1 prime number in range, so output
is 0. And in third query, there is no prime number in the given range so the output is 0.
The idea is to compute Prime numbers using Sieve of Eratosthenes along with some precomputing.
Below are the step to solve the question:
Step 1: Find the prime numbers using Sieve of Eratosthenes algorithm.
Step 2: Make an array, let say prefix[], where prefix[i] represents largest prime number smaller or equal to i.
Step 3: Make an array, let say suffix[], where suffix[i] represents smallest prime number greater or equal to i.
Step 4: Now for each query having [L, R], do the following:
if (prefix[R] R)
return 0;
else
return prefix[R] - suffix[L];
Below is the implementation of this approach:
C++
// CPP program to find maximum differences between// two prime numbers in given ranges#include <bits/stdc++.h>using namespace std;#define MAX 100005// Declare global variables to assign heap memory and avoid// stack overflowbool prime[MAX];int prefix[MAX], suffix[MAX];// Precompute Sieve, Prefix array, Suffix arrayvoid precompute(int prefix[], int suffix[]){ memset(prime, true, sizeof(prime)); // Sieve of Eratosthenes for (int i = 2; i * i < MAX; i++) { if (prime[i]) { for (int j = i * i; j < MAX; j += i) prime[j] = false; } } prefix[1] = 1; suffix[MAX - 1] = 1e9 + 7; // Precomputing Prefix array. for (int i = 2; i < MAX; i++) { if (prime[i]) prefix[i] = i; else prefix[i] = prefix[i - 1]; } // Precompute Suffix array. for (int i = MAX - 1; i > 1; i--) { if (prime[i]) suffix[i] = i; else suffix[i] = suffix[i + 1]; }}// Function to solve each queryint query(int prefix[], int suffix[], int L, int R){ if (prefix[R] < L || suffix[L] > R) return 0; else return prefix[R] - suffix[L];}// Driven Programint main(){ int q = 3; int L[] = { 2, 2, 24 }; int R[] = { 5, 2, 28 }; precompute(prefix, suffix); for (int i = 0; i < q; i++) cout << query(prefix, suffix, L[i], R[i]) << endl; return 0;} |
Java
// Java program to find maximum differences between// two prime numbers in given rangespublic class GFG { final static int MAX = 100005; // Precompute Sieve, Prefix array, Suffix array static void precompute(int prefix[], int suffix[]) { boolean prime[] = new boolean[MAX]; for (int i = 0; i < MAX; i++) { prime[i] = true; } // Sieve of Eratosthenes for (int i = 2; i * i < MAX; i++) { if (prime[i]) { for (int j = i * i; j < MAX; j += i) { prime[j] = false; } } } prefix[1] = 1; suffix[MAX - 1] = (int)1e9 + 7; // Precomputing Prefix array. for (int i = 2; i < MAX; i++) { if (prime[i]) { prefix[i] = i; } else { prefix[i] = prefix[i - 1]; } } // Precompute Suffix array. for (int i = MAX - 2; i > 1; i--) { if (prime[i]) { suffix[i] = i; } else { suffix[i] = suffix[i + 1]; } } } // Function to solve each query static int query(int prefix[], int suffix[], int L, int R) { if (prefix[R] < L || suffix[L] > R) { return 0; } else { return prefix[R] - suffix[L]; } } // Driven Program public static void main(String[] args) { int q = 3; int L[] = { 2, 2, 24 }; int R[] = { 5, 2, 28 }; int prefix[] = new int[MAX], suffix[] = new int[MAX]; precompute(prefix, suffix); for (int i = 0; i < q; i++) { System.out.println( query(prefix, suffix, L[i], R[i])); } }}/*This code is contributed by Rajput-Ji*/ |
Python3
# Python 3 program to find maximum# differences between two prime numbers# in given rangesfrom math import sqrtMAX = 100005# Precompute Sieve, Prefix array, Suffix arraydef precompute(prefix, suffix): prime = [True for i in range(MAX)] # Sieve of Eratosthenes k = int(sqrt(MAX)) for i in range(2, k, 1): if (prime[i]): for j in range(i * i, MAX, i): prime[j] = False prefix[1] = 1 suffix[MAX - 1] = int(1e9 + 7) # Precomputing Prefix array. for i in range(2, MAX, 1): if (prime[i]): prefix[i] = i else: prefix[i] = prefix[i - 1] # Precompute Suffix array. i = MAX - 2 while(i > 1): if (prime[i]): suffix[i] = i else: suffix[i] = suffix[i + 1] i -= 1# Function to solve each querydef query(prefix, suffix, L, R): if (prefix[R] < L or suffix[L] > R): return 0 else: return prefix[R] - suffix[L]# Driver Codeif __name__ == '__main__': q = 3 L = [2, 2, 24] R = [5, 2, 28] prefix = [0 for i in range(MAX)] suffix = [0 for i in range(MAX)] precompute(prefix, suffix) for i in range(0, q, 1): print(query(prefix, suffix, L[i], R[i]))# This code is contributed by# Surendra_Gangwar |
C#
// C# program to find maximum differences between// two prime numbers in given rangesusing System;public class GFG { static readonly int MAX = 100005; // Precompute Sieve, Prefix array, Suffix array static void precompute(int[] prefix, int[] suffix) { bool[] prime = new bool[MAX]; for (int i = 0; i < MAX; i++) { prime[i] = true; } // Sieve of Eratosthenes for (int i = 2; i * i < MAX; i++) { if (prime[i]) { for (int j = i * i; j < MAX; j += i) { prime[j] = false; } } } prefix[1] = 1; suffix[MAX - 1] = (int)1e9 + 7; // Precomputing Prefix array. for (int i = 2; i < MAX; i++) { if (prime[i]) { prefix[i] = i; } else { prefix[i] = prefix[i - 1]; } } // Precompute Suffix array. for (int i = MAX - 2; i > 1; i--) { if (prime[i]) { suffix[i] = i; } else { suffix[i] = suffix[i + 1]; } } } // Function to solve each query static int query(int[] prefix, int[] suffix, int L, int R) { if (prefix[R] < L || suffix[L] > R) { return 0; } else { return prefix[R] - suffix[L]; } } // Driven Program public static void Main() { int q = 3; int[] L = { 2, 2, 24 }; int[] R = { 5, 2, 28 }; int[] prefix = new int[MAX]; int[] suffix = new int[MAX]; precompute(prefix, suffix); for (int i = 0; i < q; i++) { Console.WriteLine( query(prefix, suffix, L[i], R[i])); } }}/*This code is contributed by 29AjayKumar*/ |
PHP
<?php// PHP program to find maximum differences// between two prime numbers in given ranges$MAX = 100005;// Precompute Sieve, Prefix array, // Suffix arrayfunction precompute(&$prefix, &$suffix){ global $MAX; $prime = array_fill(0, $MAX, true); // Sieve of Eratosthenes for ($i = 2; $i * $i < $MAX; $i++) { if ($prime[$i]) { for ($j = $i * $i; $j < $MAX; $j += $i) $prime[$j] = false; } } $prefix[1] = 1; $suffix[$MAX - 1] = 1e9 + 7; // Precomputing Prefix array. for ($i = 2; $i < $MAX; $i++) { if ($prime[$i]) $prefix[$i] = $i; else $prefix[$i] = $prefix[$i - 1]; } // Precompute Suffix array. for ($i = $MAX - 1; $i > 1; $i--) { if ($prime[$i]) $suffix[$i] = $i; else $suffix[$i] = $suffix[$i + 1]; }}// Function to solve each queryfunction query($prefix, $suffix, $L, $R){ if ($prefix[$R] < $L || $suffix[$L] > $R) return 0; else return $prefix[$R] - $suffix[$L];}// Driver Code$q = 3;$L = array( 2, 2, 24 );$R = array( 5, 2, 28 );$prefix = array_fill(0, $MAX + 1, 0);$suffix = array_fill(0, $MAX + 1, 0);precompute($prefix, $suffix);for ($i = 0; $i < $q; $i++) echo query($prefix, $suffix, $L[$i], $R[$i]) . "\n";// This code is contributed by mits?> |
Javascript
<script>// JavaScript program to find maximum // differences between two prime // numbers in given rangeslet MAX = 100005; // Precompute Sieve, Prefix array, Suffix arrayfunction precompute(prefix, suffix){ let prime = []; for(let i = 0; i < MAX; i++) { prime[i] = true; } // Sieve of Eratosthenes for(let i = 2; i * i < MAX; i++) { if (prime[i]) { for(let j = i * i; j < MAX; j += i) { prime[j] = false; } } } prefix[1] = 1; suffix[MAX - 1] = 1e9 + 7; // Precomputing Prefix array. for(let i = 2; i < MAX; i++) { if (prime[i]) { prefix[i] = i; } else { prefix[i] = prefix[i - 1]; } } // Precompute Suffix array. for(let i = MAX - 2; i > 1; i--) { if (prime[i]) { suffix[i] = i; } else { suffix[i] = suffix[i + 1]; } }}// Function to solve each queryfunction query(prefix, suffix, L, R){ if (prefix[R] < L || suffix[L] > R) { return 0; } else { return prefix[R] - suffix[L]; }}// Driver Codelet q = 3;let L = [ 2, 2, 24 ];let R = [ 5, 2, 28 ];let prefix = [], suffix = [];precompute(prefix, suffix);for(let i = 0; i < q; i++){ document.write(query(prefix, suffix, L[i], R[i]) + "<br/>");}// This code is contributed by sanjoy_62</script> |
Output:
3 0 0
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