Dada una array arr[] de N enteros, la tarea es encontrar la amplitud y el número de ondas para la array dada. Si la array no es una array de ondas , imprima -1 .
Conjunto de ondas: un conjunto es un conjunto de ondas si aumenta y disminuye estrictamente de forma continua o viceversa.
La amplitud se define como la diferencia máxima de números consecutivos.
Ejemplos:
Entrada: arr[] = {1, 2, 1, 5, 0, 7, -6}
Salida: Amplitud = 13, Ondas = 3
Explicación:
Para la array observe el patrón 1->2 (aumento), 2-> 1 (disminuir), 1->5 (aumentar), 5->0 (disminuir), 0->7 (aumentar), 7->-6 (disminuir). Amplitud = 13 (entre 7 y -6) y ondas totales = 3
Entrada: arr[] = {1, 2, 1, 5, 0, 7, 7}
Salida: -1
Explicación:
El arreglo no es un arreglo ondulado como el los dos últimos elementos de la array son iguales, por lo que la respuesta es -1.
Enfoque:
la idea es verificar los elementos adyacentes de ambos lados donde ambos deben ser menores o mayores que el elemento actual. Si se cumple esta condición, cuente el número de ondas; de lo contrario, imprima -1 , donde el número de ondas es (n – 1)/2 . Mientras atraviesa la array, siga actualizando la diferencia máxima entre los elementos consecutivos para obtener la amplitud de la array de ondas dada .
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 find the amplitude and
// number of waves for the given array
bool check(int a[], int n)
{
int ma = a[1] - a[0];
// Check for both sides adjacent
// elements that both must be less
// or both must be greater
// than current element
for (int i = 1; i < n - 1; i++) {
if ((a[i] > a[i - 1]
&& a[i + 1] < a[i])
|| (a[i] < a[i - 1]
&& a[i + 1] > a[i]))
// Update amplitude with max value
ma = max(ma, abs(a[i] - a[i + 1]));
else
return false;
}
// Print the Amplitude
cout << "Amplitude = " << ma;
cout << endl;
return true;
}
// Driver Code
int main()
{
// Given array a[]
int a[] = { 1, 2, 1, 5, 0, 7, -6 };
int n = sizeof a / sizeof a[0];
// Calculate number of waves
int wave = (n - 1) / 2;
// Function Call
if (check(a, n))
cout << "Waves = " << wave;
else
cout << "-1";
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the amplitude and
// number of waves for the given array
static boolean check(int a[], int n)
{
int ma = a[1] - a[0];
// Check for both sides adjacent
// elements that both must be less
// or both must be greater
// than current element
for (int i = 1; i < n - 1; i++)
{
if ((a[i] > a[i - 1] &&
a[i + 1] < a[i]) ||
(a[i] < a[i - 1] &&
a[i + 1] > a[i]))
// Update amplitude with max value
ma = Math.max(ma, Math.abs(a[i] - a[i + 1]));
else
return false;
}
// Print the Amplitude
System.out.print("Amplitude = " + ma);
System.out.println();
return true;
}
// Driver Code
public static void main(String[] args)
{
// Given array a[]
int a[] = { 1, 2, 1, 5, 0, 7, -6 };
int n = a.length;
// Calculate number of waves
int wave = (n - 1) / 2;
// Function Call
if (check(a, n))
System.out.print("Waves = " + wave);
else
System.out.print("-1");
}
}
// This code is contributed by sapnasingh4991
Python3
# Python3 program for the above approach
# Function to find the amplitude and
# number of waves for the given array
def check(a, n):
ma = a[1] - a[0]
# Check for both sides adjacent
# elements that both must be less
# or both must be greater
# than current element
for i in range(1, n - 1):
if ((a[i] > a[i - 1] and
a[i + 1] < a[i]) or
(a[i] < a[i - 1] and
a[i + 1] > a[i])):
# Update amplitude with max value
ma = max(ma, abs(a[i] - a[i + 1]))
else:
return False
# Print the Amplitude
print("Amplitude = ", ma)
return True
# Driver Code
if __name__ == '__main__':
# Given array a[]
a = [1, 2, 1, 5, 0, 7, -6]
n = len(a)
# Calculate number of waves
wave = (n - 1) // 2
# Function Call
if (check(a, n)):
print("Waves = ",wave)
else:
print("-1")
# This code is contributed by Mohit Kumar
C#
// C# program for the above approach
using System;
class GFG{
// Function to find the amplitude and
// number of waves for the given array
static bool check(int []a, int n)
{
int ma = a[1] - a[0];
// Check for both sides adjacent
// elements that both must be less
// or both must be greater
// than current element
for (int i = 1; i < n - 1; i++)
{
if ((a[i] > a[i - 1] &&
a[i + 1] < a[i]) ||
(a[i] < a[i - 1] &&
a[i + 1] > a[i]))
// Update amplitude with max value
ma = Math.Max(ma, Math.Abs(a[i] - a[i + 1]));
else
return false;
}
// Print the Amplitude
Console.Write("Amplitude = " + ma);
Console.WriteLine();
return true;
}
// Driver Code
public static void Main(String[] args)
{
// Given array []a
int []a = { 1, 2, 1, 5, 0, 7, -6 };
int n = a.Length;
// Calculate number of waves
int wave = (n - 1) / 2;
// Function Call
if (check(a, n))
Console.Write("Waves = " + wave);
else
Console.Write("-1");
}
}
// This code is contributed by sapnasingh4991
Javascript
<script>
// JavaScript program for the above approach
// Function to find the amplitude and
// number of waves for the given array
function check(a, n)
{
let ma = a[1] - a[0];
// Check for both sides adjacent
// elements that both must be less
// or both must be greater
// than current element
for (let i = 1; i < n - 1; i++)
{
if ((a[i] > a[i - 1] &&
a[i + 1] < a[i]) ||
(a[i] < a[i - 1] &&
a[i + 1] > a[i]))
// Update amplitude with max value
ma = Math.max(ma, Math.abs(a[i] - a[i + 1]));
else
return false;
}
// Print the Amplitude
document.write("Amplitude = " + ma);
document.write("<br/>");
return true;
}
// Driver Code
// Given array a[]
let a = [ 1, 2, 1, 5, 0, 7, -6 ];
let n = a.length;
// Calculate number of waves
let wave = (n - 1) / 2;
// Function Call
if (check(a, n))
document.write("Waves = " + wave);
else
document.write("-1");
</script>
Amplitude = 13 Waves = 3
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por grand_master y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA