Check for Subarray splitting possibility in Prime number

Given a positive integer array arr[] of size N and a prime number p, the task is to form an N number of non-empty subarrays by performing a series of steps. In each step, you can select an existing subarray (which may be the result of previous steps) with a length of at least two and split it into two sub-subarrays, if, for each resulting sub-subarray, at least one of the following holds:

• The length of the sub-subarray is one.
• The sum of elements of any one of the two sub-subarrays is greater than or equal to the given prime number.

Examples:

Input: arr = [2, 5, 8, 3] p = 11
Output: Possible to Split
Explanation: We can split the array into [2] and [5, 8, 3] in the first step. Then, in the second step, we can split [5, 8, 3] into [5] and [8, 3]. and in the last step, we can split [8, 3] into [8] and [3]. As a result, the answer is possible.

Input: arr = [4, 2, 1] p = 7
Output: Not possible to split
Explanation: We cannot split the array into two valid subarrays, as none of the valid splitting conditions can satisfy any combination of splitting steps.

Input: arr = [2, 4, 5, 7, 4] p = 13
Output: Not possible to split

Approach: To solve the problem follow the below Intuition:

• We just need subarray of length 2 whose sum >= p, through this subarray we can reduce all elements which are present before and after the subarray one by one.
  • If N == 2, can split once.
  • If N > 2, we must have subarray of length 2 whose arr[i] + arr[i + 1] >= p. Otherwise it is not possible to split the array into N subarrays.

Follow the steps to solve the problem:

• When splitting an array arr[] if we find a case where arr[i] + arr[i+1] >= p we can split the whole array into N parts.
• Since while splitting the array into subarrays we may reach a stage where size of array arr[] is 3 and only way to split the array is when we have sum of 2 consecutive numbers is greater than or equal to p.

Below is the implementation for the above approach:

// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;

// Function to check if it is
// possible to split the array
bool isPossibleToSplitArray(vector<int>& arr, int p)
{
    int n = arr.size();

    // If the array has 2 or fewer elements,
    // it is always possible to split it
    if (n <= 2)
        return true;

    // Iterate through the array to find a pair
    // of adjacent elements whose sum is >= 'p'
    for (int i = 1; i < n; i++) {
        int sum = arr[i - 1] + arr[i];
        if (sum >= p)
            return true;
    }

    // If no such pair is found,
    // it is impossible to split
    // the array as required
    return false;
}

// Driver Code
int main()
{
    vector<int> arr = { 2, 5, 8, 3 };
    int p = 11;

    // Call the function to check if it
    // is possible to split the array
    bool ans = isPossibleToSplitArray(arr, p);

    if (ans == true)
        cout << "Possible to Split" << endl;
    else
        cout << "Not Possible to Split" << endl;

    return 0;
}

// Java Code

import java.io.*;
import java.util.ArrayList;
import java.util.List;

public class Main {
    // Function to check if it is possible to split the array
    public static boolean isPossibleToSplitArray(List<Integer> arr, int p) {
        int n = arr.size();

        // If the array has 2 or fewer elements,
        // it is always possible to split it
        if (n <= 2) {
            return true;
        }

        // Iterate through the array to find a pair
        // of adjacent elements whose sum is >= 'p'
        for (int i = 1; i < n; i++) {
            int sum = arr.get(i - 1) + arr.get(i);
            if (sum >= p) {
                return true;
            }
        }

        // If no such pair is found,
        // it is impossible to split
        // the array as required
        return false;
    }

    public static void main(String[] args) {
        List<Integer> arr = new ArrayList<>();
        arr.add(2);
        arr.add(5);
        arr.add(8);
        arr.add(3);
        int p = 11;

        // Call the function to check if it
        // is possible to split the array
        boolean ans = isPossibleToSplitArray(arr, p);

        if (ans) {
            System.out.println("Possible to Split");
        } else {
            System.out.println("Not Possible to Split");
        }
    }
}

# Function to check if it is
# possible to split the array
def is_possible_to_split_array(arr, p):
    n = len(arr)

    # If the array has 2 or fewer elements,
    # it is always possible to split it
    if n <= 2:
        return True

    # Iterate through the array to find a pair
    # of adjacent elements whose sum is >= 'p'
    for i in range(1, n):
        sum_val = arr[i - 1] + arr[i]
        if sum_val >= p:
            return True

    # If no such pair is found,
    # it is impossible to split
    # the array as required
    return False

# Driver Code
if __name__ == "__main__":
    arr = [2, 5, 8, 3]
    p = 11

    # Call the function to check if it
    # is possible to split the array
    ans = is_possible_to_split_array(arr, p)

    if ans:
        print("Possible to Split")
    else:
        print("Not Possible to Split")

using System;
using System.Collections.Generic;

class Program
{
    // Function to check if it is possible to split the list
    public static bool IsPossibleToSplitList(List<int> list, int p)
    {
        int n = list.Count;

        // If the list has 2 or fewer elements,
        // it is always possible to split it
        if (n <= 2)
        {
            return true;
        }

        // Iterate through the list to find a pair
        // of adjacent elements whose sum is >= 'p'
        for (int i = 1; i < n; i++)
        {
            int sum = list[i - 1] + list[i];
            if (sum >= p)
            {
                return true;
            }
        }

        // If no such pair is found,
        // it is impossible to split
        // the list as required
        return false;
    }

    public static void Main(string[] args)
    {
        List<int> list = new List<int> { 2, 5, 8, 3 };
        int p = 11;

        // Call the function to check if it
        // is possible to split the list
        bool isPossible = IsPossibleToSplitList(list, p);

        if (isPossible)
        {
            Console.WriteLine("Possible to Split");
        }
        else
        {
            Console.WriteLine("Not Possible to Split");
        }
    }
}

// Function to check if it is possible to split the array
function isPossibleToSplitArray(arr, p) {
    const n = arr.length;

    // If the array has 2 or fewer elements,
    // it is always possible to split it
    if (n <= 2) {
        return true;
    }

    // Iterate through the array to find a pair
    // of adjacent elements whose sum is >= 'p'
    for (let i = 1; i < n; i++) {
        const sum = arr[i - 1] + arr[i];
        if (sum >= p) {
            return true;
        }
    }

    // If no such pair is found,
    // it is impossible to split
    // the array as required
    return false;
}

// Driver Code
function main() {
    const arr = [2, 5, 8, 3];
    const p = 11;

    // Call the function to check if it
    // is possible to split the array
    const ans = isPossibleToSplitArray(arr, p);

    if (ans) {
        console.log("Possible to Split");
    } else {
        console.log("Not Possible to Split");
    }
}

main();  // Call the main function to execute the code

Possible to Split

Time Complexity: O(n), where n is the length of the array
Auxiliary Space: O(1), no extra space is required, so it is a constant.