Examine if there exists any subarray with the specified situations

Given two integers N and X. Then the job is to return Yes or NO by examining irrespective of whether there exists a subarray in any permutation of duration N such that it contains a subarray, the place A*B is equivalent to the X. In this article A and B denote the range of aspects in sub-array and the to start with element of sorted subarray respectively.

Illustrations:

Enter: N = 5, X = 3
Output: Indeed
Rationalization: Viewed as the permutation is: 5, 2, 1, 3, 4. Just take the sub-array A₄. . . .A₄ = 3 . Then A = 1 (As only 1 component is there), B = 3 (If the sub-array is sorted then initially component of that sub-array will be 3). So, A * B = 1 * 3 = 3, Which is equal to Y. Consequently, output is Of course.

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

Approach: To fix the problem abide by the beneath plan:

The problem is centered on Greedy logic and observation primarily based. It can be solved by utilizing all those observations by applying them in a code. The observation is, there will surely exist a subarray if the ailment (X % i == && (X / i) ≥ 1 && (( X /  i) ≤ N – i + 1) correctly fulfill, where by i is the present element.

Below are the measures for the over technique:

• Produce a Boolean Flag and mark it as Untrue originally.  
• Operate a for loop from i = 1 to i ≤ N and follow the down below-described ways underneath the scope of the loop:
  •  If (X % i == && (X / i) ≥ 1 && (( X /  i) ≤ N – i + 1)  is correct then mark the flag as true and split the loop.
• Verify if the flag is true, print Indeed else, print NO.

Down below is the code to apply the solution:

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