Count subarrays with Prime sum
Given an array A[] of integers. The task is to count total subarrays whose sum is prime with ( size > 1 ).
Examples:
Input : A[] = { 1, 2, 3, 4, 5 }
Output : 3
Subarrays are -> {1, 2}, {2, 3}, {3, 4}
Input : A = { 22, 33, 4, 1, 10 };
Output : 4
Approach: Generate all possible subarrays and store their sum in a vector. Iterate the vector and check whether a sum is prime or not. It YES increment the count.
You can use sieve-of-eratosthenes to check whether a sum is prime in O(1).
Below is the implementation of above approach:
C++
// C++ program to count subarrays // with Prime sum #include <bits/stdc++.h> using namespace std; // Function to count subarrays // with Prime sum int primeSubarrays(int A[], int n) { int max_val = int(pow(10, 7)); // USE SIEVE TO FIND ALL PRIME NUMBERS LESS // THAN OR EQUAL TO max_val // Create a boolean array "prime[0..n]". A // value in prime[i] will finally be false // if i is Not a prime, else true. vector<bool> prime(max_val + 1, true); // Remaining part of SIEVE prime[0] = false; prime[1] = false; for (int p = 2; p * p <= max_val; p++) { // If prime[p] is not changed, then // it is a prime if (prime[p] == true) { // Update all multiples of p for (int i = p * 2; i <= max_val; i += p) prime[i] = false; } } int cnt = 0; // Initialize result // Traverse through the array for (int i = 0; i < n - 1; ++i) { int val = A[i]; for (int j = i + 1; j < n; ++j) { val += A[j]; if (prime[val]) ++cnt; } } // return answer return cnt; } // Driver program int main() { int A[] = { 1, 2, 3, 4, 5 }; int n = sizeof(A) / sizeof(A[0]); cout << primeSubarrays(A, n); return 0; } |
Java
// Java program to count subarrays // with Prime sum import java.util.*; class Solution { // Function to count subarrays // with Prime sum static int primeSubarrays(int A[], int n) { int max_val = (int)(Math.pow(10, 7)); // USE SIEVE TO FIND ALL PRIME NUMBERS LESS // THAN OR EQUAL TO max_val // Create a boolean array "prime[0..n]". A // value in prime[i] will finally be false // if i is Not a prime, else true. Vector<Boolean> prime=new Vector<Boolean>(max_val + 1); //initilize initial value for (int p = 0; p <max_val + 1; p++) prime.add(p,true); // Remaining part of SIEVE prime.set(0, false); prime.set(1, false); for (int p = 2; p * p <= max_val; p++) { // If prime[p] is not changed, then // it is a prime if (prime.get(p) == true) { // Update all multiples of p for (int i = p * 2; i <= max_val; i += p) prime.set(i, false); } } int cnt = 0; // Initialize result // Traverse through the array for (int i = 0; i < n - 1; ++i) { int val = A[i]; for (int j = i + 1; j < n; ++j) { val += A[j]; if (prime.get(val)) ++cnt; } } // return answer return cnt; } // Driver program public static void main(String args[]) { int A[] = { 1, 2, 3, 4, 5 }; int n = A.length; System.out.print( primeSubarrays(A, n)); } } //contributed by Arnab Kundu |
Python3
# Python3 program to count subarrays # with Prime sum # Function to count subarrays # with Prime sum def primeSubarrays(A, n): max_val = 10**7 # USE SIEVE TO FIND ALL PRIME NUMBERS # LESS THAN OR EQUAL TO max_val # Create a boolean array "prime[0..n]". A # value in prime[i] will finally be false # if i is Not a prime, else true. prime = [True] * (max_val + 1) # Remaining part of SIEVE prime[0] = False prime[1] = False for p in range(2, int(max_val**(0.5)) + 1): # If prime[p] is not changed, then # it is a prime if prime[p] == True: # Update all multiples of p for i in range(2 * p, max_val + 1, p): prime[i] = False cnt = 0 # Initialize result # Traverse through the array for i in range(0, n - 1): val = A[i] for j in range(i + 1, n): val += A[j] if prime[val] == True: cnt += 1 # return answer return cnt # Driver Code if __name__ == "__main__": A = [1, 2, 3, 4, 5] n = len(A) print(primeSubarrays(A, n)) # This code is contributed by Rituraj Jain |
C#
// C# program to count subarrays // with Prime sum class Solution { // Function to count subarrays // with Prime sum static int primeSubarrays(int[] A, int n) { int max_val = (int)(System.Math.Pow(10, 7)); // USE SIEVE TO FIND ALL PRIME NUMBERS LESS // THAN OR EQUAL TO max_val // Create a boolean array "prime[0..n]". A // value in prime[i] will finally be false // if i is Not a prime, else true. bool[] prime=new bool[max_val + 1]; //initilize initial value for (int p = 0; p <max_val + 1; p++) prime[p]=true; // Remaining part of SIEVE prime[0]=false; prime[1]=false; for (int p = 2; p * p <= max_val; p++) { // If prime[p] is not changed, then // it is a prime if (prime[p] == true) { // Update all multiples of p for (int i = p * 2; i <= max_val; i += p) prime[i]=false; } } int cnt = 0; // Initialize result // Traverse through the array for (int i = 0; i < n - 1; ++i) { int val = A[i]; for (int j = i + 1; j < n; ++j) { val += A[j]; if (prime[val]) ++cnt; } } // return answer return cnt; } // Driver program static void Main() { int[] A = { 1, 2, 3, 4, 5 }; int n = A.Length; System.Console.WriteLine( primeSubarrays(A, n)); } } //contributed by mits |
PHP
<?php // PHP program to count subarrays // with Prime sum // Function to count subarrays // with Prime sum function primeSubarrays($A, $n) { $max_val = pow(10, 5); // USE SIEVE TO FIND ALL PRIME NUMBERS LESS // THAN OR EQUAL TO max_val // Create a boolean array "prime[0..n]". A // value in prime[i] will finally be false // if i is Not a prime, else true. $prime=array_fill(0,$max_val + 1,true); // Remaining part of SIEVE $prime[0] = false; $prime[1] = false; for ($p = 2; $p * $p <= $max_val; $p++) { // If prime[p] is not changed, then // it is a prime if ($prime[$p] == true) { // Update all multiples of p for ($i = $p * 2; $i <= $max_val; $i += $p) $prime[$i] = false; } } $cnt = 0; // Initialize result // Traverse through the array for ($i = 0; $i < $n - 1; ++$i) { $val = $A[$i]; for ($j = $i + 1; $j < $n; ++$j) { $val += $A[$j]; if ($prime[$val]) ++$cnt; } } // return answer return $cnt; } // Driver program $A = array( 1, 2, 3, 4, 5 ); $n = count($A); echo primeSubarrays($A, $n); // This code is contributed by mits ?> |
Output:
3
Time Complexity: O(N2)
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