Se le da una array de n elementos. Un extremo es un elemento que es mayor que sus dos vecinos o menor que sus dos vecinos. Tienes que calcular el número de extremos locales en una array dada.
Nota: los elementos primero y último no son extremos.
Ejemplos:
Input : a[] = {1, 5, 2, 5}
Output : 2
Input : a[] = {1, 2, 3}
Output : 0
Enfoque: para calcular el número de extremos, debemos verificar si un elemento es máximo o mínimo, es decir, si es mayor que sus dos vecinos o menor que ambos vecinos. Para esto, simplemente itere sobre la array y para cada elemento verifique su posibilidad de ser un extremo.
Nota: a[0] y a[n-1] tienen exactamente un vecino cada uno, no son mínimos ni máximos.
Implementación:
C++
// CPP to find number
// of extrema
#include <bits/stdc++.h>
using namespace std;
// function to find
// local extremum
int extrema(int a[], int n)
{
int count = 0;
// start loop from position 1
// till n-1
for (int i = 1; i < n - 1; i++)
{
// only one condition
// will be true at a
// time either a[i]
// will be greater than
// neighbours or less
// than neighbours
// check if a[i] is greater
// than both its neighbours
// then add 1 to x
count += (a[i] > a[i - 1] && a[i] > a[i + 1]);
// check if a[i] is
// less than both its
// neighbours, then
// add 1 to x
count += (a[i] < a[i - 1] && a[i] < a[i + 1]);
}
return count;
}
// driver program
int main()
{
int a[] = { 1, 0, 2, 1 };
int n = sizeof(a) / sizeof(a[0]);
cout << extrema(a, n);
return 0;
}
Java
// Java to find
// number of extrema
import java.io.*;
class GFG {
// function to find
// local extremum
static int extrema(int a[], int n)
{
int count = 0;
// start loop from
// position 1 till n-1
for (int i = 1; i < n - 1; i++)
{
// only one condition
// will be true at a
// time either a[i]
// will be greater than
// neighbours or less
// than neighbours
// check if a[i] is greater
// than both its neighbours
// then add 1 to x
if(a[i] > a[i - 1] && a[i] > a[i + 1])
count += 1;
// check if a[i] is
// less than both its
// neighbours, then
// add 1 to x
if(a[i] < a[i - 1] && a[i] < a[i + 1])
count += 1;
}
return count;
}
// driver program
public static void main(String args[])
throws IOException
{
int a[] = { 1, 0, 2, 1 };
int n = a.length;
System.out.println(extrema(a, n));
}
}
/* This code is contributed by Nikita Tiwari.*/
Python3
# Python 3 to find # number of extrema # function to find # local extremum def extrema(a, n): count = 0 # start loop from # position 1 till n-1 for i in range(1, n - 1) : # only one condition # will be true # at a time either # a[i] will be greater # than neighbours or # less than neighbours # check if a[i] if # greater than both its # neighbours, then add # 1 to x count += (a[i] > a[i - 1] and a[i] > a[i + 1]); # check if a[i] if # less than both its # neighbours, then # add 1 to x count += (a[i] < a[i - 1] and a[i] < a[i + 1]); return count # driver program a = [1, 0, 2, 1 ] n = len(a) print(extrema(a, n)) # This code is contributed by Smitha Dinesh Semwal
C#
// C# to find
// number of extrema
using System;
class GFG {
// function to find
// local extremum
static int extrema(int []a, int n)
{
int count = 0;
// start loop from
// position 1 till n-1
for (int i = 1; i < n - 1; i++)
{
// only one condition
// will be true at a
// time either a[i]
// will be greater than
// neighbours or less
// than neighbours
// check if a[i] is greater
// than both its neighbours
// then add 1 to x
if(a[i] > a[i - 1] && a[i] > a[i + 1])
count += 1;
// check if a[i] is
// less than both its
// neighbours, then
// add 1 to x
if(a[i] < a[i - 1] && a[i] < a[i + 1])
count += 1;
}
return count;
}
// Driver program
public static void Main()
{
int []a = { 1, 0, 2, 1 };
int n = a.Length;
Console.WriteLine(extrema(a, n));
}
}
/* This code is contributed by vt_m.*/
PHP
<?php
// PHP to find number
// of extrema
// function to find
// local extremum
function extrema($a, $n)
{
$count = 0;
// start loop from position 1
// till n-1
for ($i = 1; $i < $n - 1; $i++)
{
// only one condition
// will be true at a
// time either a[i]
// will be greater than
// neighbours or less
// than neighbours
// check if a[i] is greater
// than both its neighbours
// then add 1 to x
$count += ($a[$i] > $a[$i - 1] and
$a[$i] > $a[$i + 1]);
// check if a[i] is
// less than both its
// neighbours, then
// add 1 to x
$count += ($a[$i] < $a[$i - 1] and
$a[$i] < $a[$i + 1]);
}
return $count;
}
// Driver Code
$a = array( 1, 0, 2, 1 );
$n = count($a);
echo extrema($a, $n);
// This code is contributed by anuj_67.
?>
Javascript
<script>
// JavaScript program to find
// number of extrema
// function to find
// local extremum
function extrema(a, n)
{
let count = 0;
// start loop from
// position 1 till n-1
for (let i = 1; i < n - 1; i++)
{
// only one condition
// will be true at a
// time either a[i]
// will be greater than
// neighbours or less
// than neighbours
// check if a[i] is greater
// than both its neighbours
// then add 1 to x
if(a[i] > a[i - 1] && a[i] > a[i + 1])
count += 1;
// check if a[i] is
// less than both its
// neighbours, then
// add 1 to x
if(a[i] < a[i - 1] && a[i] < a[i + 1])
count += 1;
}
return count;
}
// Driver code
let a = [ 1, 0, 2, 1 ];
let n = a.length;
document.write(extrema(a, n));
</script>
Producción
2
Complejidad temporal: O(n)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por Shivam.Pradhan y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA