Dada una array arr[] de enteros positivos, la tarea es encontrar si es posible hacer que esta array sea estrictamente decreciente modificando como máximo un elemento.
Ejemplos:
Entrada: arr[] = {12, 9, 10, 5, 2}
Salida: Sí
{12, 11, 10, 5, 2} es una de las soluciones válidas.
Entrada: arr[] = {1, 2, 3, 4}
Salida: No
Enfoque: Para cada elemento arr[i] , si es mayor que arr[i – 1] y arr[i + 1] o es más pequeño que arr[i – 1] y arr[i + 1] entonces arr [i] necesita ser modificado.
es decir , arr[i] = (arr[i – 1] + arr[i + 1]) / 2 . Si después de la modificación, arr[i] = arr[i – 1] o arr[i + 1] , entonces la array no se puede hacer estrictamente decreciente sin afectar como máximo a un elemento; de lo contrario, cuente todas esas modificaciones, si el recuento de modificaciones al final es menor o igual a 1 , entonces imprima Sí , de lo contrario imprima No.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns true if the array
// can be made strictly decreasing
// with at most one change
bool check(vector<int> arr, int n)
{
// To store the number of modifications
// required to make the array
// strictly decreasing
int modify = 0;
// Check whether the last element needs
// to be modify or not
if (arr[n - 1] >= arr[n - 2]) {
arr[n - 1] = arr[n - 2] - 1;
modify++;
}
// Check whether the first element needs
// to be modify or not
if (arr[0] <= arr[1]) {
arr[0] = arr[1] + 1;
modify++;
}
// Loop from 2nd element to the 2nd last element
for (int i = n - 2; i > 0; i--) {
// Check whether arr[i] needs to be modified
if ((arr[i - 1] <= arr[i] && arr[i + 1] <= arr[i])
|| (arr[i - 1] >= arr[i] && arr[i + 1] >= arr[i])) {
// Modifying arr[i]
arr[i] = (arr[i - 1] + arr[i + 1]) / 2;
modify++;
// Check if arr[i] is equal to any of
// arr[i-1] or arr[i+1]
if (arr[i] == arr[i - 1] || arr[i] == arr[i + 1])
return false;
}
}
// If more than 1 modification is required
if (modify > 1)
return false;
return true;
}
// Driver code
int main()
{
vector<int> arr = { 10, 5, 11, 2 };
int n = arr.size();
if (check(arr, n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java implementation of the approach
class GFG {
// Function that returns true if the array
// can be made strictly decreasing
// with at most one change
public static boolean check(int[] arr, int n)
{
// To store the number of modifications
// required to make the array
// strictly decreasing
int modify = 0;
// Check whether the last element needs
// to be modify or not
if (arr[n - 1] >= arr[n - 2]) {
arr[n - 1] = arr[n - 2] - 1;
modify++;
}
// Check whether the first element needs
// to be modify or not
if (arr[0] <= arr[1]) {
arr[0] = arr[1] + 1;
modify++;
}
// Loop from 2nd element to the 2nd last element
for (int i = n - 2; i > 0; i--) {
// Check whether arr[i] needs to be modified
if ((arr[i - 1] <= arr[i] && arr[i + 1] <= arr[i])
|| (arr[i - 1] >= arr[i] && arr[i + 1] >= arr[i])) {
// Modifying arr[i]
arr[i] = (arr[i - 1] + arr[i + 1]) / 2;
modify++;
// Check if arr[i] is equal to any of
// arr[i-1] or arr[i+1]
if (arr[i] == arr[i - 1] || arr[i] == arr[i + 1])
return false;
}
}
// If more than 1 modification is required
if (modify > 1)
return false;
return true;
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 10, 5, 11, 3 };
int n = arr.length;
if (check(arr, n))
System.out.print("Yes");
else
System.out.print("No");
}
}
Python3
# Python3 implementation of the approach
# Function that returns true if the array
# can be made strictly decreasing
# with at most one change
def check(arr, n):
modify = 0
# Check whether the last element needs
# to be modify or not
if (arr[n - 1] >= arr[n - 2]):
arr[n-1] = arr[n-2] - 1
modify += 1
# Check whether the first element needs
# to be modify or not
if (arr[0] <= arr[1]):
arr[0] = arr[1] + 1
modify += 1
# Loop from 2nd element to the 2nd last element
for i in range(n-2, 0, -1):
# Check whether arr[i] needs to be modified
if (arr[i - 1] <= arr[i] and arr[i + 1] <= arr[i]) or \
(arr[i - 1] >= arr[i] and arr[i + 1] >= arr[i]):
# Modifying arr[i]
arr[i] = (arr[i - 1] + arr[i + 1]) // 2
modify += 1
# Check if arr[i] is equal to any of
# arr[i-1] or arr[i + 1]
if (arr[i] == arr[i - 1] or arr[i] == arr[i + 1]):
return False
# If more than 1 modification is required
if (modify > 1):
return False
return True
# Driver code
if __name__ == "__main__":
arr = [10, 5, 11, 3]
n = len(arr)
if (check(arr, n)):
print("Yes")
else:
print("No")
C#
// C# implementation of the approach
using System;
class GFG
{
// Function that returns true if the array
// can be made strictly decreasing
// with at most one change
public static bool check(int[]arr, int n)
{
// To store the number of modifications
// required to make the array
// strictly decreasing
int modify = 0;
// Check whether the last element needs
// to be modify or not
if (arr[n - 1] >= arr[n - 2])
{
arr[n - 1] = arr[n - 2] - 1;
modify++;
}
// Check whether the first element needs
// to be modify or not
if (arr[0] <= arr[1])
{
arr[0] = arr[1] + 1;
modify++;
}
// Loop from 2nd element to the 2nd last element
for (int i = n - 2; i > 0; i--)
{
// Check whether arr[i] needs to be modified
if ((arr[i - 1] <= arr[i] && arr[i + 1] <= arr[i])
|| (arr[i - 1] >= arr[i] && arr[i + 1] >= arr[i]))
{
// Modifying arr[i]
arr[i] = (arr[i - 1] + arr[i + 1]) / 2;
modify++;
// Check if arr[i] is equal to any of
// arr[i-1] or arr[i+1]
if (arr[i] == arr[i - 1] || arr[i] == arr[i + 1])
return false;
}
}
// If more than 1 modification is required
if (modify > 1)
return false;
return true;
}
// Driver code
static public void Main ()
{
int[]arr = { 10, 5, 11, 3 };
int n = arr.Length;
if (check(arr, n))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by ajit.
Javascript
<script>
// Javascript implementation of the approach
// Function that returns true if the array
// can be made strictly decreasing
// with at most one change
function check(arr, n)
{
// To store the number of modifications
// required to make the array
// strictly decreasing
let modify = 0;
// Check whether the last element needs
// to be modify or not
if (arr[n - 1] >= arr[n - 2])
{
arr[n - 1] = arr[n - 2] - 1;
modify++;
}
// Check whether the first element needs
// to be modify or not
if (arr[0] <= arr[1])
{
arr[0] = arr[1] + 1;
modify++;
}
// Loop from 2nd element to the 2nd last element
for (let i = n - 2; i > 0; i--)
{
// Check whether arr[i] needs to be modified
if ((arr[i - 1] <= arr[i] && arr[i + 1] <= arr[i])
|| (arr[i - 1] >= arr[i] && arr[i + 1] >= arr[i]))
{
// Modifying arr[i]
arr[i] = parseInt((arr[i - 1] + arr[i + 1]) / 2, 10);
modify++;
// Check if arr[i] is equal to any of
// arr[i-1] or arr[i+1]
if (arr[i] == arr[i - 1] || arr[i] == arr[i + 1])
return false;
}
}
// If more than 1 modification is required
if (modify > 1)
return false;
return true;
}
let arr = [ 10, 5, 11, 3 ];
let n = arr.length;
if (check(arr, n))
document.write("Yes");
else
document.write("No");
</script>
Producción:
Yes
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por sanjeev2552 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA