Given two arrays arr1[] and arr2[], both of size N and an integer X, the task is to check if the sum of same-indexed elements of both the arrays at corresponding indices can be made at most K after rearranging the second array. If it is possible then print “Yes” else print “No”.
Examples:
Input: arr1[] = {1, 2, 3}, arr2[] = {1, 1, 2}, X = 4
Output: Yes
Explanation:
Rearranging the array B[] as {1, 2, 1}. Now the sum of corresponding indices are:
A[0] + B[0] = 1 + 1 = 2 ? 4
A[1] + B[1] = 2 + 2 = 4 ? 4
A[2] + B[2] = 3 + 1 = 4 ? 4Input: arr1[] = {1, 2, 3, 4}, arr2[] = {1, 2, 3, 4}, X = 4
Output: No
Explanation: There is no way that the array B[] can be rearranged such that the condition A[i] + B[i] <= X is satisfied.
Naive Approach: The simplest approach is to generate all possible permutations of the array B[] and if any permutation satisfies the given condition, then print Yes. Otherwise, print No.
Time Complexity: O(N!)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use Sorting. Follow the steps below to solve the problem:
- Sort the array arr1[] in ascending order and arr2[] in descending order.
- Now, traverse both the arrays simultaneously and if there exists any pair such that the sum of arr1[i] and arr2[] exceeds X, then print “No”. Otherwise, print “Yes”.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to check if elements // of B[] can be rearranged // such that A[i] + B[i] <= X void rearrange( int A[], int B[],
int N, int X)
{ // Checks the given condition
bool flag = true ;
// Sort A[] in ascending order
sort(A, A + N);
// Sort B[] in descending order
sort(B, B + N, greater< int >());
// Traverse the arrays A[] and B[]
for ( int i = 0; i < N; i++) {
// If A[i] + B[i] exceeds X
if (A[i] + B[i] > X) {
// Rearrangement not possible,
// set flag to false
flag = false ;
break ;
}
}
// If flag is true
if (flag)
cout << "Yes" ;
// Otherwise
else
cout << "No" ;
} // Driver Code int main()
{ int A[] = { 1, 2, 3 };
int B[] = { 1, 1, 2 };
int X = 4;
int N = sizeof (A) / sizeof (A[0]);
// Function Call
rearrange(A, B, N, X);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG
{ // Function to check if elements
// of B[] can be rearranged
// such that A[i] + B[i] <= X
static void rearrange( int A[], int B[],
int N, int X)
{
// Checks the given condition
boolean flag = true ;
// Sort A[] in ascending order
Arrays.sort(A);
// Sort B[] in descending order
Arrays.sort(B);
// Traverse the arrays A[] and B[]
for ( int i = 0 ; i < N; i++)
{
// If A[i] + B[i] exceeds X
if (A[i] + B[N - 1 - i] > X)
{
// Rearrangement not possible,
// set flag to false
flag = false ;
break ;
}
}
// If flag is true
if (flag == true )
System.out.print( "Yes" );
// Otherwise
else
System.out.print( "No" );
}
// Driver Code
public static void main (String[] args)
{
int A[] = { 1 , 2 , 3 };
int B[] = { 1 , 1 , 2 };
int X = 4 ;
int N = A.length;
// Function Call
rearrange(A, B, N, X);
}
} // This code is contributed by AnkThon |
# Python3 program for the above approach # Function to check if elements # of B can be rearranged # such that A[i] + B[i] <= X def rearrange(A, B, N, X):
# Checks the given condition
flag = True
# Sort A in ascending order
A = sorted (A)
# Sort B in descending order
B = sorted (B)[:: - 1 ]
# Traverse the arrays A and B
for i in range (N):
# If A[i] + B[i] exceeds X
if (A[i] + B[i] > X):
# Rearrangement not possible,
# set flag to false
flag = False
break
# If flag is true
if (flag):
print ( "Yes" )
# Otherwise
else :
print ( "No" )
# Driver Code if __name__ = = '__main__' :
A = [ 1 , 2 , 3 ]
B = [ 1 , 1 , 2 ]
X = 4
N = len (A)
# Function Call
rearrange(A, B, N, X)
# This code is contributed by mohit kumar 29 |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG
{ // Function to check if elements
// of B[] can be rearranged
// such that A[i] + B[i] <= X
static void rearrange( int [] A, int [] B,
int N, int X)
{
// Checks the given condition
bool flag = true ;
// Sort A[] in ascending order
Array.Sort(A);
// Sort B[] in descending order
Array.Sort(B);
// Traverse the arrays A[] and B[]
for ( int i = 0; i < N; i++)
{
// If A[i] + B[i] exceeds X
if (A[i] + B[N - 1 - i] > X)
{
// Rearrangement not possible,
// set flag to false
flag = false ;
break ;
}
}
// If flag is true
if (flag == true )
Console.WriteLine( "Yes" );
// Otherwise
else
Console.WriteLine( "No" );
}
// Driver Code
public static void Main()
{
int []A = { 1, 2, 3 };
int []B = { 1, 1, 2 };
int X = 4;
int N = A.Length;
// Function Call
rearrange(A, B, N, X);
}
} // This code is contributed by AnkThon |
<script> // Javascript program of the above approach // Function to check if elements // of B[] can be rearranged // such that A[i] + B[i] <= X function rearrange(A, B, N, X)
{ // Checks the given condition
let flag = true ;
// Sort A[] in ascending order
A.sort();
// Sort B[] in descending order
B.sort();
// Traverse the arrays A[] and B[]
for (let i = 0; i < N; i++)
{
// If A[i] + B[i] exceeds X
if (A[i] + B[N - 1 - i] > X)
{
// Rearrangement not possible,
// set flag to false
flag = false ;
break ;
}
}
// If flag is true
if (flag == true )
document.write( "Yes" );
// Otherwise
else
document.write( "No" );
} // Driver Code let A = [ 1, 2, 3 ]; let B = [ 1, 1, 2 ]; let X = 4; let N = A.length; // Function Call rearrange(A, B, N, X); // This code is contributed by avijitmondal1998 </script> |
Yes
Time Complexity: O(N*log N)
Auxiliary Space: O(1)