Dada una array arr[] de tamaño N , la tarea es encontrar el número mínimo de elementos de la array necesarios para eliminar de la array, de modo que la array dada se convierta en una array bitónica .
Ejemplos:
Entrada: arr[] = { 2, 1, 1, 5, 6, 2, 3, 1 }
Salida: 3
Explicación:
Eliminar arr[0], arr[1] y arr[5] modifica arr[] a { 1 , 5, 6, 3, 1 }
Dado que los elementos del arreglo siguen un orden creciente seguido de un orden decreciente, la salida requerida es 3.Entrada: arr[] = { 1, 3, 1 }
Salida: 0
Explicación:
La array dada ya es una array bitónica. Por lo tanto, la salida requerida es 3.
Enfoque: El problema se puede resolver con base en el concepto del problema de la subsecuencia creciente más larga . Siga los pasos a continuación para resolver el problema:
- Inicialice una variable, digamos left[] , tal que left[i] almacene la longitud de la subsecuencia creciente más larga hasta el i -ésimo índice.
- Inicialice una variable, digamos right[] , de modo que right[i] almacene la longitud de la subsecuencia decreciente más larga en el rango [i, N] .
- Recorra la array izquierda[] y derecha[] usando la variable i y encuentre el valor máximo de (izquierda[i] + derecha[i] – 1) y guárdelo en una variable, digamos maxLen .
- Finalmente, imprima el valor de N – maxLen .
A continuación se muestra la implementación del enfoque anterior:
C++14
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count minimum array elements
// required to be removed to make an array bitonic
void min_element_removal(int arr[], int N)
{
// left[i]: Stores the length
// of LIS up to i-th index
vector<int> left(N, 1);
// right[i]: Stores the length
// of decreasing subsequence
// over the range [i, N]
vector<int> right(N, 1);
// Calculate the length of LIS
// up to i-th index
for (int i = 1; i < N; i++) {
// Traverse the array
// upto i-th index
for (int j = 0; j < i; j++) {
// If arr[j] is less than arr[i]
if (arr[j] < arr[i]) {
// Update left[i]
left[i] = max(left[i],
left[j] + 1);
}
}
}
// Calculate the length of decreasing
// subsequence over the range [i, N]
for (int i = N - 2; i >= 0;
i--) {
// Traverse right[] array
for (int j = N - 1; j > i;
j--) {
// If arr[i] is greater
// than arr[j]
if (arr[i] > arr[j]) {
// Update right[i]
right[i] = max(right[i],
right[j] + 1);
}
}
}
// Stores length of the
// longest bitonic array
int maxLen = 0;
// Traverse left[] and right[] array
for (int i = 1; i < N - 1; i++) {
// Update maxLen
maxLen = max(maxLen, left[i] + right[i] - 1);
}
cout << (N - maxLen) << "\n";
}
// Function to print minimum removals
// required to make given array bitonic
void makeBitonic(int arr[], int N)
{
if (N == 1) {
cout << "0" << endl;
return;
}
if (N == 2) {
if (arr[0] != arr[1])
cout << "0" << endl;
else
cout << "1" << endl;
return;
}
min_element_removal(arr, N);
}
// Driver Code
int main()
{
int arr[] = { 2, 1, 1, 5, 6, 2, 3, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
makeBitonic(arr, N);
return 0;
}
C
// C program to implement
// the above approach
#include <stdio.h>
#define max(a,b) ((a) > (b) ? (a) : (b)) //defining max
// Function to count minimum array elements
// required to be removed to make an array bitonic
void min_element_removal(int arr[], int N)
{
// left[i]: Stores the length
// of LIS up to i-th index
int left[N];
for (int i = 0; i < N; i++)
left[i] = 1;
// right[i]: Stores the length
// of decreasing subsequence
// over the range [i, N]
int right[N];
for (int i = 0; i < N; i++)
right[i] = 1;
// Calculate the length of LIS
// up to i-th index
for (int i = 1; i < N; i++) {
// Traverse the array
// upto i-th index
for (int j = 0; j < i; j++) {
// If arr[j] is less than arr[i]
if (arr[j] < arr[i]) {
// Update left[i]
left[i] = max(left[i], left[j] + 1);
}
}
}
// Calculate the length of decreasing
// subsequence over the range [i, N]
for (int i = N - 2; i >= 0;
i--) {
// Traverse right[] array
for (int j = N - 1; j > i;
j--) {
// If arr[i] is greater
// than arr[j]
if (arr[i] > arr[j]) {
// Update right[i]
right[i] = max(right[i], right[j] + 1);
}
}
}
// Stores length of the
// longest bitonic array
int maxLen = 0;
// Traverse left[] and right[] array
for (int i = 1; i < N - 1; i++) {
// Update maxLen
maxLen = max(maxLen, left[i] + right[i] - 1);
}
printf("%d\n", (N - maxLen));
}
// Function to print minimum removals
// required to make given array bitonic
void makeBitonic(int arr[], int N)
{
if (N == 1) {
printf("0\n");
return;
}
if (N == 2) {
if (arr[0] != arr[1])
printf("0\n");
else
printf("1\n");
return;
}
min_element_removal(arr, N);
}
// Driver Code
int main()
{
int arr[] = { 2, 1, 1, 5, 6, 2, 3, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
makeBitonic(arr, N);
return 0;
}
// This code is contributed by phalashi.
Java
// Java program to implement
// the above approach
class GFG {
// Function to count minimum array elements
// required to be removed to make an array bitonic
static void min_element_removal(int arr[], int N)
{
// left[i]: Stores the length
// of LIS up to i-th index
int left[] = new int[N];
for(int i = 0; i < N; i++)
left[i] = 1;
// right[i]: Stores the length
// of decreasing subsequence
// over the range [i, N]
int right[] = new int[N];
for(int i = 0; i < N; i++)
right[i] = 1;
// Calculate the length of LIS
// up to i-th index
for (int i = 1; i < N; i++) {
// Traverse the array
// upto i-th index
for (int j = 0; j < i; j++) {
// If arr[j] is less than arr[i]
if (arr[j] < arr[i]) {
// Update left[i]
left[i] = Math.max(left[i],
left[j] + 1);
}
}
}
// Calculate the length of decreasing
// subsequence over the range [i, N]
for (int i = N - 2; i >= 0;
i--) {
// Traverse right[] array
for (int j = N - 1; j > i;
j--) {
// If arr[i] is greater
// than arr[j]
if (arr[i] > arr[j]) {
// Update right[i]
right[i] = Math.max(right[i],
right[j] + 1);
}
}
}
// Stores length of the
// longest bitonic array
int maxLen = 0;
// Traverse left[] and right[] array
for (int i = 1; i < N - 1; i++) {
// Update maxLen
maxLen = Math.max(maxLen, left[i] + right[i] - 1);
}
System.out.println(N - maxLen);
}
// Function to print minimum removals
// required to make given array bitonic
static void makeBitonic(int arr[], int N)
{
if (N == 1) {
System.out.println("0");
return;
}
if (N == 2) {
if (arr[0] != arr[1])
System.out.println("0");
else
System.out.println("1");
return;
}
min_element_removal(arr, N);
}
// Driver Code
public static void main (String[] args) {
int arr[] = { 2, 1, 1, 5, 6, 2, 3, 1 };
int N = arr.length;
makeBitonic(arr, N);
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 program to implement
# the above approach
# Function to count minimum array elements
# required to be removed to make an array bitonic
def min_element_removal(arr, N):
# left[i]: Stores the length
# of LIS up to i-th index
left = [1] * N
# right[i]: Stores the length
# of decreasing subsequence
# over the range [i, N]
right = [1] * (N)
#Calculate the length of LIS
#up to i-th index
for i in range(1, N):
#Traverse the array
#upto i-th index
for j in range(i):
#If arr[j] is less than arr[i]
if (arr[j] < arr[i]):
#Update left[i]
left[i] = max(left[i], left[j] + 1)
# Calculate the length of decreasing
# subsequence over the range [i, N]
for i in range(N - 2, -1, -1):
# Traverse right[] array
for j in range(N - 1, i, -1):
# If arr[i] is greater
# than arr[j]
if (arr[i] > arr[j]):
# Update right[i]
right[i] = max(right[i], right[j] + 1)
# Stores length of the
# longest bitonic array
maxLen = 0
# Traverse left[] and right[] array
for i in range(1, N - 1):
# Update maxLen
maxLen = max(maxLen, left[i] + right[i] - 1)
print((N - maxLen))
# Function to print minimum removals
# required to make given array bitonic
def makeBitonic(arr, N):
if (N == 1):
print("0")
return
if (N == 2):
if (arr[0] != arr[1]):
print("0")
else:
print("1")
return
min_element_removal(arr, N)
# Driver Code
if __name__ == '__main__':
arr=[2, 1, 1, 5, 6, 2, 3, 1]
N = len(arr)
makeBitonic(arr, N)
# This code is contributed by mohit kumar 29
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to count minimum array elements
// required to be removed to make an array bitonic
static void min_element_removal(int []arr, int N)
{
// left[i]: Stores the length
// of LIS up to i-th index
int []left = new int[N];
for(int i = 0; i < N; i++)
left[i] = 1;
// right[i]: Stores the length
// of decreasing subsequence
// over the range [i, N]
int []right = new int[N];
for(int i = 0; i < N; i++)
right[i] = 1;
// Calculate the length of LIS
// up to i-th index
for(int i = 1; i < N; i++)
{
// Traverse the array
// upto i-th index
for(int j = 0; j < i; j++)
{
// If arr[j] is less than arr[i]
if (arr[j] < arr[i])
{
// Update left[i]
left[i] = Math.Max(left[i],
left[j] + 1);
}
}
}
// Calculate the length of decreasing
// subsequence over the range [i, N]
for(int i = N - 2; i >= 0; i--)
{
// Traverse right[] array
for(int j = N - 1; j > i; j--)
{
// If arr[i] is greater
// than arr[j]
if (arr[i] > arr[j])
{
// Update right[i]
right[i] = Math.Max(right[i],
right[j] + 1);
}
}
}
// Stores length of the
// longest bitonic array
int maxLen = 0;
// Traverse left[] and right[] array
for(int i = 1; i < N - 1; i++)
{
// Update maxLen
maxLen = Math.Max(maxLen, left[i] +
right[i] - 1);
}
Console.WriteLine(N - maxLen);
}
// Function to print minimum removals
// required to make given array bitonic
static void makeBitonic(int []arr, int N)
{
if (N == 1)
{
Console.WriteLine("0");
return;
}
if (N == 2)
{
if (arr[0] != arr[1])
Console.WriteLine("0");
else
Console.WriteLine("1");
return;
}
min_element_removal(arr, N);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 2, 1, 1, 5, 6, 2, 3, 1 };
int N = arr.Length;
makeBitonic(arr, N);
}
}
// This code is contributed by AnkThon
Javascript
<script>
// Javascript program to implement
// the above approach
// Function to count minimum array elements
// required to be removed to make an array bitonic
function min_element_removal(arr, N)
{
// left[i]: Stores the length
// of LIS up to i-th index
var left = Array(N).fill(1);
// right[i]: Stores the length
// of decreasing subsequence
// over the range [i, N]
var right = Array(N).fill(1);
// Calculate the length of LIS
// up to i-th index
for (var i = 1; i < N; i++) {
// Traverse the array
// upto i-th index
for (var j = 0; j < i; j++) {
// If arr[j] is less than arr[i]
if (arr[j] < arr[i]) {
// Update left[i]
left[i] = Math.max(left[i],
left[j] + 1);
}
}
}
// Calculate the length of decreasing
// subsequence over the range [i, N]
for (var i = N - 2; i >= 0;
i--) {
// Traverse right[] array
for (var j = N - 1; j > i;
j--) {
// If arr[i] is greater
// than arr[j]
if (arr[i] > arr[j]) {
// Update right[i]
right[i] = Math.max(right[i],
right[j] + 1);
}
}
}
// Stores length of the
// longest bitonic array
var maxLen = 0;
// Traverse left[] and right[] array
for (var i = 1; i < N - 1; i++) {
// Update maxLen
maxLen = Math.max(maxLen, left[i] + right[i] - 1);
}
document.write((N - maxLen) + "<br>");
}
// Function to print minimum removals
// required to make given array bitonic
function makeBitonic(arr, N)
{
if (N == 1) {
document.write( "0" + "<br>");
return;
}
if (N == 2) {
if (arr[0] != arr[1])
document.write( "0" + "<br>");
else
document.write( "1" + "<br>");
return;
}
min_element_removal(arr, N);
}
// Driver Code
var arr = [2, 1, 1, 5, 6, 2, 3, 1];
var N = arr.length;
makeBitonic(arr, N);
// This code is contributed by rutvik_56.
</script>
Producción:
3
Tiempo Complejidad: O(N 2 )
Espacio Auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por saikumarkudikala y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA