Given two arrays A[] and B[] and two integers X and Y, the task is to choose X elements from A[] and Y elements from B[] such that all the chosen elements from A[] are less than all the chosen elements from B[]
Examples:
Input: A[] = {1, 1, 1, 1, 1}, B[] = {3, 1}, X = 3, Y = 1
Output: Yes
Choose {1, 1, 1} from A[] and {3} from B[].
Input: A[] = {5, 4}, B[] = {2, 3, 2, 2}, X = 2, Y = 1
Output: No
Approach: In order to satisfy the given conditions, the minimum X elements have to be chosen from A[] and the maximum Y elements have to be chosen from B[]. This can be done by sorting both the arrays and then choosing the Xth smallest element from A[] say xSmall and Yth largest element from B[] say yLarge.
This is because if xSmall is smaller than yLarge then all the elements smaller than it will definitely be smaller than yLarge and all the elements larger than yLarge will definitely be greater than xSmall.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to that returns true if // it possible to choose the elements bool isPossible( int A[], int B[], int n,
int m, int x, int y)
{ // If elements can't be chosen
if (x > n || y > m)
return false ;
// Sort both the arrays
sort(A, A + n);
sort(B, B + m);
// If xth smallest element of A[]
// is smaller than the yth
// greatest element of B[]
if (A[x - 1] < B[m - y])
return true ;
else
return false ;
} // Driver code int main()
{ int A[] = { 1, 1, 1, 1, 1 };
int B[] = { 2, 2 };
int n = sizeof (A) / sizeof ( int );
int m = sizeof (B) / sizeof ( int );
int x = 3, y = 1;
if (isPossible(A, B, n, m, x, y))
cout << "Yes" ;
else
cout << "No" ;
return 0;
} |
// Java implementation of the above approach import java.util.*;
class GFG
{ // Function to that returns true if
// it possible to choose the elements
static boolean isPossible( int A[], int B[], int n,
int m, int x, int y)
{
// If elements can't be chosen
if (x > n || y > m)
return false ;
// Sort both the arrays
Arrays.sort(A);
Arrays.sort(B);
// If xth smallest element of A[]
// is smaller than the yth
// greatest element of B[]
if (A[x - 1 ] < B[m - y])
return true ;
else
return false ;
}
// Driver code
public static void main (String[] args)
{
int A[] = { 1 , 1 , 1 , 1 , 1 };
int B[] = { 2 , 2 };
int n = A.length;
int m = B.length;;
int x = 3 , y = 1 ;
if (isPossible(A, B, n, m, x, y))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
} // This code is contributed by AnkitRai01 |
# Python3 implementation of the approach # Function to that returns true if # it possible to choose the elements def isPossible(A, B, n, m, x, y) :
# If elements can't be chosen
if (x > n or y > m) :
return False
# Sort both the arrays
A.sort()
B.sort()
# If xth smallest element of A[]
# is smaller than the yth
# greatest element of B[]
if (A[x - 1 ] < B[m - y]) :
return True
else :
return False
# Driver code A = [ 1 , 1 , 1 , 1 , 1 ]
B = [ 2 , 2 ]
n = len (A)
m = len (B)
x = 3
y = 1
if (isPossible(A, B, n, m, x, y)):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed by # divyamohan123 |
// C# implementation of the above approach using System;
class GFG
{ // Function to that returns true if
// it possible to choose the elements
static bool isPossible( int []A, int []B, int n,
int m, int x, int y)
{
// If elements can't be chosen
if (x > n || y > m)
return false ;
// Sort both the arrays
Array.Sort(A);
Array.Sort(B);
// If xth smallest element of A[]
// is smaller than the yth
// greatest element of B[]
if (A[x - 1] < B[m - y])
return true ;
else
return false ;
}
// Driver code
public static void Main (String[] args)
{
int []A = { 1, 1, 1, 1, 1 };
int []B = { 2, 2 };
int n = A.Length;
int m = B.Length;;
int x = 3, y = 1;
if (isPossible(A, B, n, m, x, y))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
} // This code is contributed by Rajput-Ji |
<script> // javascript implementation of the above approach // Function to that returns true if // it possible to choose the elements function isPossible(A , B , n, m , x , y)
{ // If elements can't be chosen
if (x > n || y > m)
return false ;
// Sort both the arrays
A.sort();
B.sort();
// If xth smallest element of A
// is smaller than the yth
// greatest element of B
if (A[x - 1] < B[m - y])
return true ;
else
return false ;
} // Driver code var A = [ 1, 1, 1, 1, 1 ];
var B = [ 2, 2 ];
var n = A.length;
var m = B.length;;
var x = 3, y = 1;
if (isPossible(A, B, n, m, x, y))
document.write( "Yes" );
else document.write( "No" );
// This code is contributed by 29AjayKumar </script> |
Yes
Time Complexity: O(N*log(N))
Auxiliary Space: O(1)