Given two arrays arr1[] and arr2[]. The task is to count the distinct sums that can be obtained while choosing a prime element from arr1[] and another prime element from arr2[].
Examples:
Input: arr1[] = {2, 3}, arr2[] = {2, 2, 4, 7}
Output: 4
All possible prime pairs are (2, 2), (2, 2), (2, 7), (3, 2), (3, 2)
and (3, 7) with sums 4, 4, 9, 5, 5 and 10 respectively.
Input: arr1[] = {3, 1, 4, 2, 5}, arr2[] = {8, 7, 10, 6, 5}
Output: 5
Approach: Use Sieve of Eratosthenes to check whether a number is prime or not then for every prime pair, store it’s sum in a set in order to avoid duplicates. The size of the final set will be the required answer.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
#define MAX 1000000
using namespace std;
bool prime[MAX];
void sieve()
{
memset(prime, true, sizeof(prime));
prime[0] = prime[1] = false;
for (int p = 2; p * p <= MAX; p++) {
if (prime[p] == true) {
for (int i = p * p; i <= MAX; i += p)
prime[i] = false;
}
}
}
int distinctSum(int arr1[], int arr2[], int m, int n)
{
sieve();
set<int, greater<int> > sumSet;
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (prime[arr1[i]] && prime[arr2[j]])
sumSet.insert(arr1[i] + arr2[j]);
return sumSet.size();
}
int main()
{
int arr1[] = { 2, 3 };
int arr2[] = { 2, 2, 4, 7 };
int m = sizeof(arr1) / sizeof(arr1[0]);
int n = sizeof(arr2) / sizeof(arr2[0]);
cout << distinctSum(arr1, arr2, m, n);
return 0;
}
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Java
import java.util.*;
class GFG
{
static int MAX = 1000000;
static boolean []prime = new boolean[MAX + 1];
static void sieve()
{
Arrays.fill(prime, true);
prime[0] = prime[1] = false;
for (int p = 2; p * p <= MAX; p++)
{
if (prime[p] == true)
{
for (int i = p * p;
i <= MAX; i += p)
prime[i] = false;
}
}
}
static int distinctSum(int arr1[],
int arr2[],
int m, int n)
{
sieve();
Set<Integer> sumSet = new HashSet<Integer>();
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (prime[arr1[i]] && prime[arr2[j]])
sumSet.add(arr1[i] + arr2[j]);
return sumSet.size();
}
public static void main(String[] args)
{
int arr1[] = { 2, 3 };
int arr2[] = { 2, 2, 4, 7 };
int m = arr1.length;
int n = arr2.length;
System.out.println(distinctSum(arr1, arr2, m, n));
}
}
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Python3
MAX = 1000000
prime = [True for i in range(MAX + 1)]
def sieve():
prime[0], prime[1] = False, False
for p in range(2, MAX + 1):
if p * p > MAX:
break
if (prime[p] == True):
for i in range(2 * p, MAX + 1, p):
prime[i] = False
def distinctSum(arr1, arr2, m, n):
sieve()
sumSet = dict()
for i in range(m):
for j in range(n):
if (prime[arr1[i]] and
prime[arr2[j]]):
sumSet[arr1[i] + arr2[j]] = 1
return len(sumSet)
arr1 = [2, 3 ]
arr2 = [2, 2, 4, 7 ]
m = len(arr1)
n = len(arr2)
print(distinctSum(arr1, arr2, m, n))
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C#
using System;
using System.Collections.Generic;
class GFG
{
static int MAX = 1000000;
static bool []prime = new bool[MAX + 1];
static void sieve()
{
for (int i = 0; i < MAX + 1; i++)
prime[i] = true;
prime[0] = prime[1] = false;
for (int p = 2; p * p <= MAX; p++)
{
if (prime[p] == true)
{
for (int i = p * p;
i <= MAX; i += p)
prime[i] = false;
}
}
}
static int distinctSum(int []arr1,
int []arr2,
int m, int n)
{
sieve();
HashSet<int> sumSet = new HashSet<int>();
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (prime[arr1[i]] && prime[arr2[j]])
sumSet.Add(arr1[i] + arr2[j]);
return sumSet.Count;
}
public static void Main(String[] args)
{
int []arr1 = { 2, 3 };
int []arr2 = { 2, 2, 4, 7 };
int m = arr1.Length;
int n = arr2.Length;
Console.WriteLine(distinctSum(arr1, arr2, m, n));
}
}
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Javascript
<script>
let MAX = 1000000
let prime = new Array(MAX);
function sieve() {
prime.fill(true)
prime[0] = prime[1] = false;
for (let p = 2; p * p <= MAX; p++) {
if (prime[p] == true) {
for (let i = p * p; i <= MAX; i += p)
prime[i] = false;
}
}
}
function distinctSum(arr1, arr2, m, n) {
sieve();
let sumSet = new Set();
for (let i = 0; i < m; i++)
for (let j = 0; j < n; j++)
if (prime[arr1[i]] && prime[arr2[j]])
sumSet.add(arr1[i] + arr2[j]);
return sumSet.size;
}
let arr1 = [2, 3];
let arr2 = [2, 2, 4, 7];
let m = arr1.length;
let n = arr2.length;
document.write(distinctSum(arr1, arr2, m, n));
</script>
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Time Complexity : O(N * M * log (N * M) + MAX * log(MAX) )
Auxiliary Space: O(MAX + N * M)