Length of largest sub-array having primes strictly greater than non-primes

Given an array ‘arr’ of length ‘n’. The task is to find the largest contiguous sub-array having the count of prime numbers strictly greater than the count of non-prime numbers.

Examples:

Input: arr[] = {4, 7, 4, 7, 11, 5, 4, 4, 4, 5}
Output: 9

Input:  arr[] = { 1, 9, 3, 4, 5, 6, 7, 8 }
Output: 5


Approach: To find the largest subarray in which count of prime is strictly greater than the count of non-prime:
First of all, use sieve to find the prime number.
Replace all primes with 1 in the array and all non-primes with -1. Now this problem is similar to Longest subarray having count of 1s one more than count of 0s
Below is the implementation of above approach:

C++

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// C++ implementation of above approach
  
#include <bits/stdc++.h>
using namespace std;
  
bool prime[1000000 + 5];
  
void findPrime()
{
    memset(prime, true, sizeof(prime));
    prime[1] = false;
  
    for (int p = 2; p * p <= 1000000; 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 <= 1000000; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to find the length of longest
// subarray having count of primes more
// than count of non-primes
int lenOfLongSubarr(int arr[], int n)
{
    // unordered_map 'um' implemented as
    // hash table
    unordered_map<int, int> um;
    int sum = 0, maxLen = 0;
  
    // traverse the given array
    for (int i = 0; i < n; i++) {
  
        // consider -1 as non primes and 1 as primes
        sum += prime[arr[i]] == 0 ? -1 : 1;
  
        // when subarray starts form index '0'
        if (sum == 1)
            maxLen = i + 1;
  
        // make an entry for 'sum' if it is
        // not present in 'um'
        else if (um.find(sum) == um.end())
            um[sum] = i;
  
        // check if 'sum-1' is present in 'um'
        // or not
        if (um.find(sum - 1) != um.end()) {
  
            // update maxLength
            if (maxLen < (i - um[sum - 1]))
                maxLen = i - um[sum - 1];
        }
    }
  
    // required maximum length
    return maxLen;
}
  
// Driver code
int main()
{
    findPrime();
  
    int arr[] = { 1, 9, 3, 4, 5, 6, 7, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << lenOfLongSubarr(arr, n) << endl;
    return 0;
}

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Java

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// Java implementation of above approach
  
import java.util.*;
class GfG {
    static boolean prime[] = new boolean[1000000 + 5];
  
    static void findPrime()
    {
        Arrays.fill(prime, true);
        prime[1] = false;
  
        for (int p = 2; p * p <= 1000000; 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 <= 1000000; i += p)
                    prime[i] = false;
            }
        }
    }
  
    // Function to find the length of longest
    // subarray having count of primes more
    // than count of non-primes
    static int lenOfLongSubarr(int arr[], int n)
    {
        // unordered_map 'um' implemented as
        // hash table
        Map<Integer, Integer> um = new HashMap<Integer, Integer>();
        int sum = 0, maxLen = 0;
  
        // traverse the given array
        for (int i = 0; i < n; i++) {
  
            // consider -1 as non primes and 1 as primes
            sum += prime[arr[i]] == false ? -1 : 1;
  
            // when subarray starts form index '0'
            if (sum == 1)
                maxLen = i + 1;
  
            // make an entry for 'sum' if it is
            // not present in 'um'
            else if (!um.containsKey(sum))
                um.put(sum, i);
  
            // check if 'sum-1' is present in 'um'
            // or not
            if (um.containsKey(sum - 1)) {
  
                // update maxLength
                if (maxLen < (i - um.get(sum - 1)))
                    maxLen = i - um.get(sum - 1);
            }
        }
  
        // required maximum length
        return maxLen;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        findPrime();
  
        int arr[] = { 1, 9, 3, 4, 5, 6, 7, 8 };
        int n = arr.length;
  
        System.out.println(lenOfLongSubarr(arr, n));
    }
}

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Python3

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# Python3 implementation of above approach 
  
prime = [True] * (1000000 + 5
  
def findPrime(): 
  
    prime[0], prime[1] = False, False
  
    for p in range(2, 1001): 
  
        # If prime[p] is not changed, 
        # then it is a prime 
        if prime[p] == True
  
            # Update all multiples of p 
            for i in range(p * 2, 1000001, p):
                prime[i] = False
  
# Function to find the length of longest 
# subarray having count of primes more 
# than count of non-primes 
def lenOfLongSubarr(arr, n): 
  
    # unordered_map 'um' implemented as 
    # hash table 
    um = {} 
    Sum, maxLen = 0, 0
  
    # traverse the given array 
    for i in range(0, n): 
  
        # consider -1 as non primes and 1 as primes 
        Sum = Sum-1 if prime[arr[i]] == False else Sum + 1
          
        # when subarray starts form index '0' 
        if Sum == 1
            maxLen = i + 1
  
        # make an entry for 'sum' if 
        # it is not present in 'um' 
        elif Sum not in um: 
            um[Sum] =
  
        # check if 'sum-1' is present 
        # in 'um' or not 
        if (Sum - 1) in um: 
  
            # update maxLength 
            if maxLen < (i - um[Sum - 1]): 
                maxLen = i - um[Sum - 1
  
    # required maximum length 
    return maxLen 
  
# Driver Code
if __name__ == "__main__":
  
    findPrime() 
  
    arr = [1, 9, 3, 4, 5, 6, 7, 8
    n = len(arr) 
  
    print(lenOfLongSubarr(arr, n))
  
# This code is contributed 
# by Rituraj Jain

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C#

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// C# implementation of above approach
using System;
using System.Collections.Generic;
  
class GfG {
  
    static bool[] prime = new bool[1000000 + 5];
  
    static void findPrime()
    {
        Array.Fill(prime, true);
        prime[1] = false;
  
        for (int p = 2; p * p <= 1000000; 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 <= 1000000; i += p)
                    prime[i] = false;
            }
        }
    }
  
    // Function to find the length of longest
    // subarray having count of primes more
    // than count of non-primes
    static int lenOfLongSubarr(int[] arr, int n)
    {
        // unordered_map 'um' implemented as
        // hash table
        Dictionary<int, int> um = new Dictionary<int, int>();
        int sum = 0, maxLen = 0;
  
        // traverse the given array
        for (int i = 0; i < n; i++) {
  
            // consider -1 as non primes and 1 as primes
            sum += prime[arr[i]] == false ? -1 : 1;
  
            // when subarray starts form index '0'
            if (sum == 1)
                maxLen = i + 1;
  
            // make an entry for 'sum' if it is
            // not present in 'um'
            else if (!um.ContainsKey(sum))
                um[sum] = i;
  
            // check if 'sum-1' is present in 'um'
            // or not
            if (um.ContainsKey(sum - 1)) {
  
                // update maxLength
                if (maxLen < (i - um[sum - 1]))
                    maxLen = i - um[sum - 1];
            }
        }
  
        // required maximum length
        return maxLen;
    }
  
    // Driver code
    public static void Main()
    {
        findPrime();
  
        int[] arr = { 1, 9, 3, 4, 5, 6, 7, 8 };
        int n = arr.Length;
  
        Console.WriteLine(lenOfLongSubarr(arr, n));
    }
}
  
// This code is contributed by Code_Mech.

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Output:

5


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