Dadas dos arrays arr1[] y arr2[] , ambas de tamaño N y un número entero X , la tarea es verificar si la suma de los elementos del mismo índice de ambas arrays en los índices correspondientes se puede hacer como máximo K después de reorganizar la segunda formación. Si es posible, escriba «Sí» , de lo contrario, escriba «No» .
Ejemplos:
Entrada: arr1[] = {1, 2, 3}, arr2[] = {1, 1, 2}, X = 4
Salida: Sí
Explicación:
Reorganizar la array B[] como {1, 2, 1}. Ahora la suma de los índices correspondientes son:
A[0] + B[0] = 1 + 1 = 2 ≤ 4
A[1] + B[1] = 2 + 2 = 4 ≤ 4
A[2] + B[2 ] = 3 + 1 = 4 ≤ 4Entrada: arr1[] = {1, 2, 3, 4}, arr2[] = {1, 2, 3, 4}, X = 4
Salida: No
Explicación: No hay manera de que la array B[] pueda ser reorganizado de tal manera que se satisfaga la condición A[i] + B[i] <= X.
Enfoque ingenuo: el enfoque más simple es generar todas las permutaciones posibles de la array B[] y, si alguna permutación satisface la condición dada, imprimir Yes . De lo contrario , imprima No.
Complejidad temporal: O(N!)
Espacio auxiliar: O(1)
Enfoque eficiente: para optimizar el enfoque anterior, la idea es usar Sorting . Siga los pasos a continuación para resolver el problema:
- Ordene la array arr1[] en orden ascendente y arr2[] en orden descendente.
- Ahora, recorra ambas arrays simultáneamente y si existe algún par tal que la suma de arr1[i] y arr2[] exceda X , entonces imprima «No» . De lo contrario, escriba «Sí» .
A continuación se muestra la implementación del enfoque anterior:
C++
// 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
// 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
# 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#
// 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
Javascript
<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
Complejidad de tiempo: O(N*log N)
Espacio auxiliar: O(1)