给定长度为’n’的数组’arr’。任务是找到最大素数计数严格大于非素数计数的连续子数组。
例子：
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
方法：要找到素数严格大于非素数的最大子数组：
首先，使用筛子查找素数。
将数组中的所有素数替换为1，并将所有非素数替换为-1。现在，这个问题类似于最长子数组，其计数为1s比计数为0s多一
下面是上述方法的实现：
C++
// C++ implementation of above approach
#include
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 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;
} Java
// 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 um = new HashMap();
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));
}
} Python3
# 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] = i
# 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 JainC#
// 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 um = new Dictionary();
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. 输出：
5
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