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Given two integers N and X. Then the duty is to return YES or NO by checking whether or not there exists a subarray in any permutation of size N such that it incorporates a subarray, the place A*B is the same as the X. Right here A and B denote the variety of components in sub-array and the primary factor of sorted subarray respectively.
Examples:
Enter: N = 5, X = 3Output: YESExplanation: Thought-about the permutation is: {5, 2, 1, 3, 4}. Take the sub-array {A4. . . .A4} = { 3 }. Then A = 1 (As only one factor is there), B = 3 (If the sub-array is sorted then first factor of that sub-array can be 3). So, A * B = 1 * 3 = 3, Which is the same as Y. Subsequently, output is YES.
Enter: N = 7, X = 56Output: NOExplanation: It may be verified that no permutation of size N exists such that, It offers worth of A * B as 56. Subsequently, output is NO.
Strategy: To unravel the issue observe the beneath thought:
The issue is predicated on Grasping logic and commentary primarily based. It may be solved by implementing these observations by implementing them in a code. The commentary is, there’ll absolutely exist a subarray if the situation (X % i == 0 && (X / i) ≥ 1 && (( X / i) ≤ N – i + 1) efficiently meet, the place i is the present factor.
Beneath are the steps for the above method:
Create a Boolean Flag and mark it as False initially. Run a for loop from i = 1 to i ≤ N and observe the below-mentioned steps underneath 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.Examine if the flag is true, print YES else, print NO.
Beneath is the code to implement the method:
Java
import java.util.*;
public class GFG {
public static void important(String[] args)
{
int N = 5;
lengthy X = 3;
Boolean Flag = false;
SubArrayExists(N, X, Flag);
}
static void SubArrayExists(int N, lengthy X, boolean Flag)
{
for (int i = 1; i <= N; i++) {
if (X % i == 0 && (X / i) >= 1
&& ((X / i) <= N – i + 1)) {
Flag = true;
break;
}
}
System.out.println(Flag ? “YES” : “NO”);
}
}
Time Complexity: O(N)Auxiliary House: O(1), As no further area is used.
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