Dado un número N , la tarea es verificar si los dígitos del número son crecientes y decrecientes alternativamente. Si el número de dígitos es inferior a 3 , devuelve falso . Por ejemplo, si d1, d2 y d3 son los dígitos de N, d1 < d2 > d3 debe cumplirse .
Ejemplos :
Entrada : N = 6834
Salida : verdadero
Explicación : Los dígitos de N aumentan y disminuyen alternativamente, es decir, 6 < 8 > 3 < 4
Entrada : N = 32
Salida : falso
Enfoque : la tarea se puede resolver convirtiendo el entero dado en una string. Itere sobre la string y verifique si los caracteres adyacentes siguen las condiciones dadas o no.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation for the above approach
#include <bits/stdc++.h>
using namespace std;
// Utility function to check if
// the digits of the current
// integer forms a wave pattern
bool check(int N)
{
// Convert the number to a string
string S = to_string(N);
// Loop to iterate over digits
for (int i = 0; i < S.size(); i++) {
if (i == 0) {
// Next character of
// the number
int next = i + 1;
// Current character is
// not a local minimum
if (next < S.size()) {
if (S[i] >= S[next]) {
return false;
}
}
}
else if (i == S.size() - 1) {
// Previous character of
// the number
int prev = i - 1;
if (prev >= 0) {
// Character is a
// local maximum
if (i & 1) {
// Character is not
// a local maximum
if (S[i] <= S[prev]) {
return false;
}
}
else {
// Character is a
// local minimum
if (S[i] >= S[prev]) {
return false;
}
}
}
}
else {
int prev = i - 1;
int next = i + 1;
if (i & 1) {
// Character is a
// local maximum
if ((S[i] > S[prev])
&& (S[i] > S[next])) {
}
else {
return false;
}
}
else {
// Character is a
// local minimum
if ((S[i] < S[prev])
&& (S[i] < S[next])) {
}
else {
return false;
}
}
}
}
return true;
}
// Driver Code
int main()
{
int N = 64;
cout << (check(N) ? "true" : "false");
}
Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Utility function to check if
// the digits of the current
// integer forms a wave pattern
static Boolean check(int N)
{
// Convert the number to a string
String S = Integer.toString(N);
// Loop to iterate over digits
for (int i = 0; i < S.length(); i++) {
if (i == 0) {
// Next character of
// the number
int next = i + 1;
// Current character is
// not a local minimum
if (next < S.length()) {
if (S.charAt(i) >= S.charAt(next)) {
return false;
}
}
}
else if (i == S.length() - 1) {
// Previous character of
// the number
int prev = i - 1;
if (prev >= 0) {
// Character is a
// local maximum
if (i % 2 == 1) {
// Character is not
// a local maximum
if (S.charAt(i) <= S.charAt(prev)) {
return false;
}
}
else {
// Character is a
// local minimum
if (S.charAt(i) >= S.charAt(prev)) {
return false;
}
}
}
}
else {
int prev = i - 1;
int next = i + 1;
if (i % 2 == 1) {
// Character is a
// local maximum
if ((S.charAt(i) > S.charAt(prev))
&& (S.charAt(i) > S.charAt(next))) {
}
else {
return false;
}
}
else
{
// Character is a
// local minimum
if ((S.charAt(i) < S.charAt(prev))
&& (S.charAt(i) < S.charAt(next))) {
}
else {
return false;
}
}
}
}
return true;
}
// Driver Code
public static void main (String[] args) {
int N = 64;
if(check(N) == true)
System.out.print("true");
else
System.out.print("false");
}
}
// This code is contributed by hrithikgarg03188
Python3
# Python implementation for the above approach
# Utility function to check if
# the digits of the current
# integer forms a wave pattern
def check(N):
# Convert the number to a string
S = str(N)
# Loop to iterate over digits
for i in range(len(S)):
if (i == 0):
# Next character of
# the number
next = i + 1
# Current character is
# not a local minimum
if (next < len(S)):
if (S[i] >= S[next]):
return False
elif (i == len(S) - 1):
# Previous character of
# the number
prev = i - 1
if (prev >= 0):
# Character is a
# local maximum
if (i & 1):
# Character is not
# a local maximum
if (S[i] <= S[prev]):
return False
else:
# Character is a
# local minimum
if (S[i] >= S[prev]):
return False
else:
prev = i - 1
next = i + 1
if (i & 1):
# Character is a
# local maximum
if ((S[i] > S[prev]) and (S[i] > S[next])):
print("", end="")
else:
return False
else:
# Character is a
# local minimum
if ((S[i] < S[prev]) and (S[i] < S[next])):
print("", end="")
else:
return False
return True
# Driver Code
N = 64
print("true" if check(N) else "false")
# This code is contributed by Saurabh Jaiswal
C#
// C# program for the above approach
using System;
class GFG {
// Utility function to check if
// the digits of the current
// integer forms a wave pattern
static bool check(int N)
{
// Convert the number to a string
string S = N.ToString();
// Loop to iterate over digits
for (int i = 0; i < S.Length; i++) {
if (i == 0) {
// Next character of
// the number
int next = i + 1;
// Current character is
// not a local minimum
if (next < S.Length) {
if (S[i] >= S[next]) {
return false;
}
}
}
else if (i == S.Length - 1) {
// Previous character of
// the number
int prev = i - 1;
if (prev >= 0) {
// Character is a
// local maximum
if (i % 2 == 1) {
// Character is not
// a local maximum
if (S[i] <= S[prev]) {
return false;
}
}
else {
// Character is a
// local minimum
if (S[i] >= S[prev]) {
return false;
}
}
}
}
else {
int prev = i - 1;
int next = i + 1;
if (i % 2 == 1) {
// Character is a
// local maximum
if ((S[i] > S[prev])
&& (S[i] > S[next])) {
}
else {
return false;
}
}
else
{
// Character is a
// local minimum
if ((S[i] < S[prev])
&& (S[i] < S[next])) {
}
else {
return false;
}
}
}
}
return true;
}
// Driver Code
public static void Main () {
int N = 64;
if(check(N) == true)
Console.Write("true");
else
Console.Write("false");
}
}
// This code is contributed by Samim Hossain Mondal.
Javascript
<script>
// Javascript implementation for the above approach
// Utility function to check if
// the digits of the current
// integer forms a wave pattern
function check(N){
// Convert the number to a string
let S = new String(N);
// Loop to iterate over digits
for (let i = 0; i < S.length; i++) {
if (i == 0) {
// Next character of
// the number
let next = i + 1;
// Current character is
// not a local minimum
if (next < S.length) {
if (S[i] >= S[next]) {
return false;
}
}
}
else if (i == S.length - 1) {
// Previous character of
// the number
let prev = i - 1;
if (prev >= 0) {
// Character is a
// local maximum
if (i & 1) {
// Character is not
// a local maximum
if (S[i] <= S[prev]) {
return false;
}
}
else {
// Character is a
// local minimum
if (S[i] >= S[prev]) {
return false;
}
}
}
}
else {
let prev = i - 1;
let next = i + 1;
if (i & 1) {
// Character is a
// local maximum
if ((S[i] > S[prev])
&& (S[i] > S[next])) {
}
else {
return false;
}
}
else {
// Character is a
// local minimum
if ((S[i] < S[prev])
&& (S[i] < S[next])) {
}
else {
return false;
}
}
}
}
return true;
}
// Driver Code
let N = 64;
document.write(check(N) ? "true" : "false");
// This code is contributed by gfgking.
</script>
Producción
false
Complejidad de tiempo : O(N), N es el número de dígitos
Espacio auxiliar : O(N)