Dada una array de N enteros, la tarea es ordenarlos en un arreglo circular de tal manera que el elemento sea estrictamente menor que la suma de sus elementos adyacentes. En caso de que tal arreglo no sea posible, imprima -1 .
Tenga en cuenta que puede haber múltiples formas de organizar los elementos de modo que se cumpla la condición y la tarea es encontrar tal disposición.
Ejemplos:
Entrada: arr[] = {1, 4, 4, 3, 2}
Salida: 1 3 4 4 2
arr[0] = 1 < (2 + 3)
arr[1] = 4 < (1 + 4)
arr[ 2] = 4 < (4 + 3)
arr[3] = 3 < (4 + 2)
arr[4] = 2 < (3 + 1)
Entrada: arr[] = {8, 13, 5}
Salida: – 1
Enfoque: el problema se puede resolver utilizando un enfoque codicioso, primero ordenamos la array y luego colocamos el elemento más pequeño al principio, el segundo más pequeño al final, el tercero más pequeño en la segunda posición y el cuarto más pequeño en la penúltima posición en otra array. Una vez que se completa el arreglo, verifique si la condición dada se cumple o no.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the arrangement that
// satisfies the given condition
void printArrangement(int a[], int n)
{
// Sort the array initially
sort(a, a + n);
// Array that stores the arrangement
int b[n];
// Once the array is sorted
// Re-fill the array again in the
// mentioned way in the approach
int low = 0, high = n - 1;
for (int i = 0; i < n; i++) {
if (i % 2 == 0)
b[low++] = a[i];
else
b[high--] = a[i];
}
// Iterate in the array
// and check if the arrangement made
// satisfies the given condition or not
for (int i = 0; i < n; i++) {
// For the first element
// the adjacents will be a[1] and a[n-1]
if (i == 0) {
if (b[n - 1] + b[1] <= b[i]) {
cout << -1;
return;
}
}
// For the last element
// the adjacents will be a[0] and a[n-2]
else if (i == (n - 1)) {
if (b[n - 2] + b[0] <= b[i]) {
cout << -1;
return;
}
}
else {
if (b[i - 1] + b[i + 1] <= b[i]) {
cout << -1;
return;
}
}
}
// If we reach this position then
// the arrangement is possible
for (int i = 0; i < n; i++)
cout << b[i] << " ";
}
// Driver code
int main()
{
int a[] = { 1, 4, 4, 3, 2 };
int n = sizeof(a) / sizeof(a[0]);
printArrangement(a, n);
return 0;
}
Java
// Java implementation of the approach
import java.util.Arrays;
class GFG
{
// Function to print the arrangement that
// satisfies the given condition
static void printArrangement(int a[], int n)
{
// Sort the array initially
Arrays.sort(a);
// Array that stores the arrangement
int b[] = new int[n];
// Once the array is sorted
// Re-fill the array again in the
// mentioned way in the approach
int low = 0, high = n - 1;
for (int i = 0; i < n; i++)
{
if (i % 2 == 0)
b[low++] = a[i];
else
b[high--] = a[i];
}
// Iterate in the array
// and check if the arrangement made
// satisfies the given condition or not
for (int i = 0; i < n; i++)
{
// For the first element
// the adjacents will be a[1] and a[n-1]
if (i == 0)
{
if (b[n - 1] + b[1] <= b[i])
{
System.out.print(-1);
return;
}
}
// For the last element
// the adjacents will be a[0] and a[n-2]
else if (i == (n - 1))
{
if (b[n - 2] + b[0] <= b[i])
{
System.out.print(-1);
return;
}
}
else
{
if (b[i - 1] + b[i + 1] <= b[i])
{
System.out.print(-1);
return;
}
}
}
// If we reach this position then
// the arrangement is possible
for (int i = 0; i < n; i++)
System.out.print(b[i] + " ");
}
// Driver code
public static void main (String[] args)
{
int a[] = { 1, 4, 4, 3, 2 };
int n = a.length;
printArrangement(a, n);
}
}
// This code is contributed by anuj_67..
Python3
# Python3 implementation of the approach
# Function to print the arrangement that
# satisfies the given condition
def printArrangement(a, n):
# Sort the array initially
a = sorted(a)
# Array that stores the arrangement
b = [0 for i in range(n)]
# Once the array is sorted
# Re-fill the array again in the
# mentioned way in the approach
low = 0
high = n - 1
for i in range(n):
if (i % 2 == 0):
b[low] = a[i]
low += 1
else:
b[high] = a[i]
high -= 1
# Iterate in the array
# and check if the arrangement made
# satisfies the given condition or not
for i in range(n):
# For the first element
# the adjacents will be a[1] and a[n-1]
if (i == 0):
if (b[n - 1] + b[1] <= b[i]):
print("-1")
return
# For the last element
# the adjacents will be a[0] and a[n-2]
elif (i == (n - 1)) :
if (b[n - 2] + b[0] <= b[i]):
print("-1")
return
else:
if (b[i - 1] + b[i + 1] <= b[i]):
print("-1")
return
# If we reach this position then
# the arrangement is possible
for i in range(n):
print(b[i], end = " ")
# Driver code
a = [ 1, 4, 4, 3, 2 ]
n = len(a)
printArrangement(a, n)
# This code is contributed by Mohit Kumar
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to print the arrangement that
// satisfies the given condition
static void printArrangement(int []a, int n)
{
// Sort the array initially
Array.Sort(a);
// Array that stores the arrangement
int []b = new int[n];
// Once the array is sorted
// Re-fill the array again in the
// mentioned way in the approach
int low = 0, high = n - 1;
for (int i = 0; i < n; i++)
{
if (i % 2 == 0)
b[low++] = a[i];
else
b[high--] = a[i];
}
// Iterate in the array
// and check if the arrangement made
// satisfies the given condition or not
for (int i = 0; i < n; i++)
{
// For the first element
// the adjacents will be a[1] and a[n-1]
if (i == 0)
{
if (b[n - 1] + b[1] <= b[i])
{
Console.Write(-1);
return;
}
}
// For the last element
// the adjacents will be a[0] and a[n-2]
else if (i == (n - 1))
{
if (b[n - 2] + b[0] <= b[i])
{
Console.Write(-1);
return;
}
}
else
{
if (b[i - 1] + b[i + 1] <= b[i])
{
Console.Write(-1);
return;
}
}
}
// If we reach this position then
// the arrangement is possible
for (int i = 0; i < n; i++)
Console.Write(b[i] + " ");
}
// Driver code
public static void Main ()
{
int []a = { 1, 4, 4, 3, 2 };
int n = a.Length;
printArrangement(a, n);
}
}
// This code is contributed by anuj_67..
Javascript
<script>
// javascript implementation of the approach
// Function to print the arrangement that
// satisfies the given condition
function printArrangement(a, n)
{
// Sort the array initially
a.sort();
// Array that stores the arrangement
var b = Array(n).fill(0);
// Once the array is sorted
// Re-fill the array again in the
// mentioned way in the approach
var low = 0, high = n - 1;
for (i = 0; i < n; i++) {
if (i % 2 == 0)
b[low++] = a[i];
else
b[high--] = a[i];
}
// Iterate in the array
// and check if the arrangement made
// satisfies the given condition or not
for (i = 0; i < n; i++) {
// For the first element
// the adjacents will be a[1] and a[n-1]
if (i == 0) {
if (b[n - 1] + b[1] <= b[i]) {
document.write(-1);
return;
}
}
// For the last element
// the adjacents will be a[0] and a[n-2]
else if (i == (n - 1)) {
if (b[n - 2] + b[0] <= b[i]) {
document.write(-1);
return;
}
} else {
if (b[i - 1] + b[i + 1] <= b[i]) {
document.write(-1);
return;
}
}
}
// If we reach this position then
// the arrangement is possible
for (i = 0; i < n; i++)
document.write(b[i] + " ");
}
// Driver code
var a = [ 1, 4, 4, 3, 2 ];
var n = a.length;
printArrangement(a, n);
// This code is contributed by todaysgaurav
</script>
Producción:
1 3 4 4 2
Complejidad temporal: O(N log N)
Espacio auxiliar: O(n)