Check if X and Y elements can be selected from two arrays respectively such that the maximum in X is less than the minimum in Y

// Java program for the above approach
import java.io.*;
import java.util.*;

class GFG{

// Function to check if it is possible to
// choose X and Y elements from a[] and
// b[] such that maximum element among
// X element is less than minimum
// element among Y elements
static String check(int[] a, int[] b,
                    int Na, int Nb,
                    int k, int m)
{
    
    // Check if there are atleast X
    // elements in arr1[] and atleast
    // Y elements in arr2[]
    if (Na < k || Nb < m)
        return "No";

    // Sort arrays in ascending order
    Arrays.sort(a);
    Arrays.sort(b);

    // Check if (X - 1)-th element in arr1[]
    // is less than from M-Yth element
    // in arr2[]
    if (a[k - 1] < b[Nb - m])
    {
        return "Yes";
    }

    // Return false
    return "No";
}

// Driver Code
public static void main(String[] args)
{
    int[] arr1 = { 1, 2, 3 };
    int[] arr2 = { 3, 4, 5 };
    int N = arr1.length;
    int M = arr2.length;

    int X = 2, Y = 1;

    // Function Call
    System.out.println(check(
        arr1, arr2, N, M, X, Y));
}
}

// This code is contributed by Dharanendra L V

// Java program for the above approach
import java.util.*;

class GFG {
    // Function to check if it is possible to
    // choose X and Y elements from a[] and b[]
    // such that maximum element among X element
    // is less than minimum element among Y elements
    static boolean check(int[] a, int[] b, int n, int m,
                         int x, int y)
    {

        // Sort both arrays in ascending order
        Arrays.sort(a);
        Arrays.sort(b);

        // For each element in arr1[],
        // find the largest element in arr2[]
        // that is smaller than that element
        for (int i = 0; i < n; i++) {
            int j = Arrays.binarySearch(b, a[i]);

            // If binary search returns a negative
            // number, then the element is not found
            // in arr2[] and we need to convert the
            // index to its corresponding positive
            // value using the formula: -(index + 1)
            if (j < 0) {
                j = -(j + 1);
            }

            // If the index of the largest element
            // in arr2[] is greater than or equal
            // to M-Y, then it is possible to choose
            // X elements from arr1[] and Y elements
            // from arr2[] such that the maximum
            // element among X is less than the
            // minimum element among Y
            if (j <= m - y && a[i + x - 1] < b[j + y - 1]) {
                return true;
            }
        }

        // If none of the elements in arr1[]
        // satisfy the above condition, then
        // it is not possible to choose any
        // such combinations
        return false;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[] arr1 = { 1, 2, 3 };
        int[] arr2 = { 3, 4, 5 };
        int n = arr1.length;
        int m = arr2.length;
        int x = 2, y = 1;

        // Function Call
        if (check(arr1, arr2, n, m, x, y)) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }
}