Check if there exists any subarray with the given conditions

Given two integers N and X. Then the task is to return YES or NO by checking whether there exists a subarray in any permutation of length N such that it contains a subarray, where A*B is equal to the X. Here A and B denote the number of elements in sub-array and the first element of sorted subarray respectively.

Examples:

Input: N = 5, X = 3
Output: YES
Explanation: Considered the permutation is: {5, 2, 1, 3, 4}. Take the sub-array {A₄. . . .A₄} = { 3 }. Then A = 1 (As only 1 element is there), B = 3 (If the sub-array is sorted then first element of that sub-array will be 3). So, A * B = 1 * 3 = 3, Which is equal to Y. Therefore, output is YES.

Input: N = 7, X = 56
Output: NO
Explanation: It can be verified that no permutation of length N exists such that, It gives value of A * B as 56. Therefore, output is NO.

Approach: To solve the problem follow the below idea:

The problem is based on Greedy logic and observation based. It can be solved by implementing those observations by implementing them in a code. The observation is, there will surely exist a subarray if the condition (X % i == 0 && (X / i) ≥ 1 && (( X /  i) ≤ N - i + 1) successfully meet, where i is the current element.

Below are the steps for the above approach:

• Create a Boolean Flag and mark it as False initially.  
• Run a for loop from i = 1 to i ≤ N and follow the below-mentioned steps under the scope of the loop:
  •  If (X % i == 0 && (X / i) ≥ 1 && (( X /  i) ≤ N - i + 1)  is true then mark the flag as true and break the loop.
• Check if the flag is true, print YES else, print NO.

Below is the code to implement the approach:

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Method for checking whether
// SubArray exists or not
void SubArrayExists(int N, long X, bool Flag)
{

    // Loop for traversing
    // from 1 to N
    for (int i = 1; i <= N; i++) {

        // Testing the condition
        if (X % i == 0 && (X / i) >= 1
            && ((X / i) <= N - i + 1)) {

            // Marking the flag true and
            // breaking the loop
            Flag = true;
            break;
        }
    }

    cout << (Flag ? "YES" : "NO") << endl;
}

int main()
{

    // Inputs
    int N = 5;
    long X = 3;
    bool Flag = false;

    // Function call
    SubArrayExists(N, X, Flag);
}

// This code is contributed by Rahul Bhardwaj @rahulbhardwaj.rb

import java.util.*;

public class GFG {

    // Driver Function
    public static void main(String[] args)
    {

        // Inputs
        int N = 5;
        long X = 3;
        Boolean Flag = false;

        // Function call
        SubArrayExists(N, X, Flag);
    }

    // Method for checking whether
    // SubArray exists or not
    static void SubArrayExists(int N, long X, boolean Flag)
    {

        // Loop for traversing
        // from 1 to N
        for (int i = 1; i <= N; i++) {

            // Testing the condition
            if (X % i == 0 && (X / i) >= 1
                && ((X / i) <= N - i + 1)) {

                // Marking the flag true and
                // breaking the loop
                Flag = true;
                break;
            }
        }

        System.out.println(Flag ? "YES" : "NO");
    }
}

def subarray_exists(N, X):
    flag = False

    # Loop for traversing from 1 to N
    for i in range(1, N+1):
        # Testing the condition
        if X % i == 0 and (X // i) >= 1 and ((X // i) <= N - i + 1):
            # Marking the flag true and breaking the loop
            flag = True
            break

    print("YES" if flag else "NO")

# Inputs
N = 5
X = 3

# Function call
subarray_exists(N, X)

// C# code implementation

using System;

public class GFG {

    static public void Main()
    {

        // Inputs
        int N = 5;
        long X = 3;
        bool Flag = false;

        // Function call
        SubArrayExists(N, X, Flag);
    }

    // Method for checking whether SubArray exists or not
    static void SubArrayExists(int N, long X, bool Flag)
    {

        // Loop for traversing from 1 to N
        for (int i = 1; i <= N; i++) {

            // Testing the condition
            if (X % i == 0 && (X / i) >= 1
                && ((X / i) <= N - i + 1)) {

                // Marking the flag true and breaking the
                // loop
                Flag = true;
                break;
            }
        }

        Console.WriteLine(Flag ? "YES" : "NO");
    }
}

// This code is contributed by sankar.

// JavaScript code to implement the approach

function SubArrayExists(N, X) {

    let Flag = false;

    // Loop for traversing from 1 to N
    for (let i = 1; i <= N; i++) {
        // Testing the condition
        if (X % i == 0 && (X / i) >= 1
            && ((X / i) <= N - i + 1)) {

            // Marking the flag true and breaking the loop
            Flag = true;
            break;
        }
    }

    console.log(Flag ? "YES" : "NO");
}

// Inputs
let N = 5;
let X = 3;

// Function call
SubArrayExists(N, X);

YES

Time Complexity: O(N)
Auxiliary Space: O(1), As no extra space is used.