Ya hemos discutido los subproblemas superpuestos y las propiedades de la subestructura óptima .
Ahora, analicemos el problema de la subsecuencia creciente más larga (LIS) como un problema de ejemplo que se puede resolver mediante la programación dinámica.
El problema de la subsecuencia creciente más larga (LIS) es encontrar la longitud de la subsecuencia más larga de una secuencia dada de modo que todos los elementos de la subsecuencia se clasifiquen en orden creciente. Por ejemplo, la longitud de LIS para {10, 22, 9, 33, 21, 50, 41, 60, 80} es 6 y LIS es {10, 22, 33, 50, 60, 80}.
Ejemplos:
Input: arr[] = {3, 10, 2, 1, 20}
Output: Length of LIS = 3
The longest increasing subsequence is 3, 10, 20
Input: arr[] = {3, 2}
Output: Length of LIS = 1
The longest increasing subsequences are {3} and {2}
Input: arr[] = {50, 3, 10, 7, 40, 80}
Output: Length of LIS = 4
The longest increasing subsequence is {3, 7, 40, 80}
Método 1 : Recursión.
Subestructura óptima: Sea arr[0..n-1] la array de entrada y L(i) la longitud del LIS que termina en el índice i, de modo que arr[i] es el último elemento del LIS.
Entonces, L(i) puede escribirse recursivamente como:
L(i) = 1 + max( L(j) ) where 0 < j < i and arr[j] < arr[i]; or L(i) = 1, if no such j exists.
Para encontrar el LIS para una array dada, necesitamos devolver max(L(i)) donde 0 < i < n.
Formalmente, la longitud de la subsecuencia creciente más larga que termina en el índice i será 1 mayor que la longitud máxima de todas las subsecuencias crecientes más largas que terminan en los índices anteriores a i, donde arr[j] < arr[i] (j < i).
Por lo tanto, vemos que el problema LIS satisface la propiedad de subestructura óptima ya que el problema principal se puede resolver utilizando soluciones a los subproblemas.
El árbol recursivo que se muestra a continuación hará que el enfoque sea más claro:
Input : arr[] = {3, 10, 2, 11}
f(i): Denotes LIS of subarray ending at index 'i'
(LIS(1)=1)
f(4) {f(4) = 1 + max(f(1), f(2), f(3))}
/ | \
f(1) f(2) f(3) {f(3) = 1, f(2) and f(1) are > f(3)}
| | \
f(1) f(2) f(1) {f(2) = 1 + max(f(1)}
|
f(1) {f(1) = 1}
A continuación se muestra la implementación del enfoque recursivo:
C++
/* A Naive C++ recursive implementation
of LIS problem */
#include <iostream>
using namespace std;
/* To make use of recursive calls, this
function must return two things:
1) Length of LIS ending with element arr[n-1].
We use max_ending_here for this purpose
2) Overall maximum as the LIS may end with
an element before arr[n-1] max_ref is
used this purpose.
The value of LIS of full array of size n
is stored in *max_ref which is our final result
*/
int _lis(int arr[], int n, int* max_ref)
{
/* Base case */
if (n == 1)
return 1;
// 'max_ending_here' is length of LIS
// ending with arr[n-1]
int res, max_ending_here = 1;
/* Recursively get all LIS ending with arr[0],
arr[1] ... arr[n-2]. If arr[i-1] is smaller
than arr[n-1], and max ending with arr[n-1]
needs to be updated, then update it */
for (int i = 1; i < n; i++) {
res = _lis(arr, i, max_ref);
if (arr[i - 1] < arr[n - 1]
&& res + 1 > max_ending_here)
max_ending_here = res + 1;
}
// Compare max_ending_here with the overall
// max. And update the overall max if needed
if (*max_ref < max_ending_here)
*max_ref = max_ending_here;
// Return length of LIS ending with arr[n-1]
return max_ending_here;
}
// The wrapper function for _lis()
int lis(int arr[], int n)
{
// The max variable holds the result
int max = 1;
// The function _lis() stores its result in max
_lis(arr, n, &max);
// returns max
return max;
}
/* Driver program to test above function */
int main()
{
int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
int n = sizeof(arr) / sizeof(arr[0]);
cout <<"Length of lis is "<< lis(arr, n);
return 0;
}
// This code is contributed by shivanisinghss2110
C
/* A Naive C recursive implementation
of LIS problem */
#include <stdio.h>
#include <stdlib.h>
/* To make use of recursive calls, this
function must return two things:
1) Length of LIS ending with element arr[n-1].
We use max_ending_here for this purpose
2) Overall maximum as the LIS may end with
an element before arr[n-1] max_ref is
used this purpose.
The value of LIS of full array of size n
is stored in *max_ref which is our final result
*/
int _lis(int arr[], int n, int* max_ref)
{
/* Base case */
if (n == 1)
return 1;
// 'max_ending_here' is length of LIS
// ending with arr[n-1]
int res, max_ending_here = 1;
/* Recursively get all LIS ending with arr[0],
arr[1] ... arr[n-2]. If arr[i-1] is smaller
than arr[n-1], and max ending with arr[n-1]
needs to be updated, then update it */
for (int i = 1; i < n; i++) {
res = _lis(arr, i, max_ref);
if (arr[i - 1] < arr[n - 1]
&& res + 1 > max_ending_here)
max_ending_here = res + 1;
}
// Compare max_ending_here with the overall
// max. And update the overall max if needed
if (*max_ref < max_ending_here)
*max_ref = max_ending_here;
// Return length of LIS ending with arr[n-1]
return max_ending_here;
}
// The wrapper function for _lis()
int lis(int arr[], int n)
{
// The max variable holds the result
int max = 1;
// The function _lis() stores its result in max
_lis(arr, n, &max);
// returns max
return max;
}
/* Driver program to test above function */
int main()
{
int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("Length of lis is %d", lis(arr, n));
return 0;
}
Java
/* A Naive Java Program for LIS Implementation */
class LIS {
static int max_ref; // stores the LIS
/* To make use of recursive calls, this function must
return two things: 1) Length of LIS ending with element
arr[n-1]. We use max_ending_here for this purpose 2)
Overall maximum as the LIS may end with an element
before arr[n-1] max_ref is used this purpose.
The value of LIS of full array of size n is stored in
*max_ref which is our final result */
static int _lis(int arr[], int n)
{
// base case
if (n == 1)
return 1;
// 'max_ending_here' is length of LIS ending with
// arr[n-1]
int res, max_ending_here = 1;
/* Recursively get all LIS ending with arr[0],
arr[1] ... arr[n-2]. If arr[i-1] is smaller
than arr[n-1], and max ending with arr[n-1] needs
to be updated, then update it */
for (int i = 1; i < n; i++) {
res = _lis(arr, i);
if (arr[i - 1] < arr[n - 1]
&& res + 1 > max_ending_here)
max_ending_here = res + 1;
}
// Compare max_ending_here with the overall max. And
// update the overall max if needed
if (max_ref < max_ending_here)
max_ref = max_ending_here;
// Return length of LIS ending with arr[n-1]
return max_ending_here;
}
// The wrapper function for _lis()
static int lis(int arr[], int n)
{
// The max variable holds the result
max_ref = 1;
// The function _lis() stores its result in max
_lis(arr, n);
// returns max
return max_ref;
}
// driver program to test above functions
public static void main(String args[])
{
int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
int n = arr.length;
System.out.println("Length of lis is " + lis(arr, n)
+ "\n");
}
}
/*This code is contributed by Rajat Mishra*/
Python3
# A naive Python implementation of LIS problem
""" To make use of recursive calls, this function must return
two things:
1) Length of LIS ending with element arr[n-1]. We use
max_ending_here for this purpose
2) Overall maximum as the LIS may end with an element
before arr[n-1] max_ref is used this purpose.
The value of LIS of full array of size n is stored in
*max_ref which is our final result """
# global variable to store the maximum
global maximum
def _lis(arr, n):
# to allow the access of global variable
global maximum
# Base Case
if n == 1:
return 1
# maxEndingHere is the length of LIS ending with arr[n-1]
maxEndingHere = 1
"""Recursively get all LIS ending with arr[0], arr[1]..arr[n-2]
IF arr[n-1] is smaller than arr[n-1], and max ending with
arr[n-1] needs to be updated, then update it"""
for i in range(1, n):
res = _lis(arr, i)
if arr[i-1] < arr[n-1] and res+1 > maxEndingHere:
maxEndingHere = res + 1
# Compare maxEndingHere with overall maximum. And
# update the overall maximum if needed
maximum = max(maximum, maxEndingHere)
return maxEndingHere
def lis(arr):
# to allow the access of global variable
global maximum
# length of arr
n = len(arr)
# maximum variable holds the result
maximum = 1
# The function _lis() stores its result in maximum
_lis(arr, n)
return maximum
# Driver program to test the above function
arr = [10, 22, 9, 33, 21, 50, 41, 60]
n = len(arr)
print ("Length of lis is ", lis(arr))
# This code is contributed by NIKHIL KUMAR SINGH
C#
using System;
/* A Naive C# Program for LIS Implementation */
class LIS {
static int max_ref; // stores the LIS
/* To make use of recursive calls, this function must
return two things: 1) Length of LIS ending with element
arr[n-1]. We use max_ending_here for this purpose 2)
Overall maximum as the LIS may end with an element
before arr[n-1] max_ref is used this purpose.
The value of LIS of full array of size n is stored in
*max_ref which is our final result */
static int _lis(int[] arr, int n)
{
// base case
if (n == 1)
return 1;
// 'max_ending_here' is length of LIS ending with
// arr[n-1]
int res, max_ending_here = 1;
/* Recursively get all LIS ending with arr[0],
arr[1] ... arr[n-2]. If arr[i-1] is smaller
than arr[n-1], and max ending with arr[n-1] needs
to be updated, then update it */
for (int i = 1; i < n; i++) {
res = _lis(arr, i);
if (arr[i - 1] < arr[n - 1]
&& res + 1 > max_ending_here)
max_ending_here = res + 1;
}
// Compare max_ending_here with the overall max. And
// update the overall max if needed
if (max_ref < max_ending_here)
max_ref = max_ending_here;
// Return length of LIS ending with arr[n-1]
return max_ending_here;
}
// The wrapper function for _lis()
static int lis(int[] arr, int n)
{
// The max variable holds the result
max_ref = 1;
// The function _lis() stores its result in max
_lis(arr, n);
// returns max
return max_ref;
}
// driver program to test above functions
public static void Main()
{
int[] arr = { 10, 22, 9, 33, 21, 50, 41, 60 };
int n = arr.Length;
Console.Write("Length of lis is " + lis(arr, n)
+ "\n");
}
}
Javascript
<script>
/* A Naive javascript Program for LIS Implementation */
let max_ref; // stores the LIS
/* To make use of recursive calls, this function must return
two things:
1) Length of LIS ending with element arr[n-1]. We use
max_ending_here for this purpose
2) Overall maximum as the LIS may end with an element
before arr[n-1] max_ref is used this purpose.
The value of LIS of full array of size n is stored in
*max_ref which is our final result */
function _lis(arr,n)
{
// base case
if (n == 1)
return 1;
// 'max_ending_here' is length of LIS ending with arr[n-1]
let res, max_ending_here = 1;
/* Recursively get all LIS ending with arr[0], arr[1] ...
arr[n-2]. If arr[i-1] is smaller than arr[n-1], and
max ending with arr[n-1] needs to be updated, then
update it */
for (let i = 1; i < n; i++)
{
res = _lis(arr, i);
if (arr[i-1] < arr[n-1] && res + 1 > max_ending_here)
max_ending_here = res + 1;
}
// Compare max_ending_here with the overall max. And
// update the overall max if needed
if (max_ref < max_ending_here)
max_ref = max_ending_here;
// Return length of LIS ending with arr[n-1]
return max_ending_here;
}
// The wrapper function for _lis()
function lis(arr,n)
{
// The max variable holds the result
max_ref = 1;
// The function _lis() stores its result in max
_lis( arr, n);
// returns max
return max_ref;
}
// driver program to test above functions
let arr=[10, 22, 9, 33, 21, 50, 41, 60 ]
let n = arr.length;
document.write("Length of lis is "
+ lis(arr, n) + "<br>");
// This code is contributed by avanitrachhadiya2155
</script>
Producción
Length of lis is 5
Análisis de Complejidad:
- Complejidad de tiempo: La complejidad de tiempo de este enfoque recursivo es exponencial ya que hay un caso de subproblemas superpuestos como se explica en el diagrama de árbol recursivo anterior.
- Espacio Auxiliar: O(1). No se utiliza espacio externo para almacenar valores aparte del espacio de pila interno.
Método 2 : Programación Dinámica.
Podemos ver que hay muchos subproblemas en la solución recursiva anterior que se resuelven una y otra vez. Por lo tanto, este problema tiene la propiedad de subestructura superpuesta y el recálculo de los mismos subproblemas se puede evitar mediante Memoización o Tabulación.
La simulación de enfoque aclarará las cosas:
Input : arr[] = {3, 10, 2, 11}
LIS[] = {1, 1, 1, 1} (initially)
Simulación iterativa:
- array[2] > array[1] {LIS[2] = max(LIS [2], LIS[1]+1)=2}
- arr[3] < arr[1] {Sin cambios}
- arr[3] < arr[2] {Sin cambios}
- array[4] > array[1] {LIS[4] = max(LIS[4], LIS[1]+1)=2}
- array[4] > array[2] {LIS[4] = max(LIS[4], LIS[2]+1)=3}
- array[4] > array[3] {LIS[4] = max(LIS [4], LIS[3]+1)=3}
Podemos evitar el recálculo de los subproblemas usando la tabulación como se muestra en el siguiente código:
A continuación se muestra la implementación del enfoque anterior:
C++
/* Dynamic Programming C++ implementation
of LIS problem */
#include <bits/stdc++.h>
using namespace std;
/* lis() returns the length of the longest
increasing subsequence in arr[] of size n */
int lis(int arr[], int n)
{
int lis[n];
lis[0] = 1;
/* Compute optimized LIS values in
bottom up manner */
for (int i = 1; i < n; i++) {
lis[i] = 1;
for (int j = 0; j < i; j++)
if (arr[i] > arr[j] && lis[i] < lis[j] + 1)
lis[i] = lis[j] + 1;
}
// Return maximum value in lis[]
return *max_element(lis, lis + n);
}
/* Driver program to test above function */
int main()
{
int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("Length of lis is %d\n", lis(arr, n));
return 0;
}
Java
/* Dynamic Programming Java implementation
of LIS problem */
class LIS {
/* lis() returns the length of the longest
increasing subsequence in arr[] of size n */
static int lis(int arr[], int n)
{
int lis[] = new int[n];
int i, j, max = 0;
/* Initialize LIS values for all indexes */
for (i = 0; i < n; i++)
lis[i] = 1;
/* Compute optimized LIS values in
bottom up manner */
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (arr[i] > arr[j] && lis[i] < lis[j] + 1)
lis[i] = lis[j] + 1;
/* Pick maximum of all LIS values */
for (i = 0; i < n; i++)
if (max < lis[i])
max = lis[i];
return max;
}
public static void main(String args[])
{
int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
int n = arr.length;
System.out.println("Length of lis is " + lis(arr, n)
+ "\n");
}
}
/*This code is contributed by Rajat Mishra*/
Python3
# Dynamic programming Python implementation
# of LIS problem
# lis returns length of the longest
# increasing subsequence in arr of size n
def lis(arr):
n = len(arr)
# Declare the list (array) for LIS and
# initialize LIS values for all indexes
lis = [1]*n
# Compute optimized LIS values in bottom up manner
for i in range(1, n):
for j in range(0, i):
if arr[i] > arr[j] and lis[i] < lis[j] + 1:
lis[i] = lis[j]+1
# Initialize maximum to 0 to get
# the maximum of all LIS
maximum = 0
# Pick maximum of all LIS values
for i in range(n):
maximum = max(maximum, lis[i])
return maximum
# end of lis function
# Driver program to test above function
arr = [10, 22, 9, 33, 21, 50, 41, 60]
print ("Length of lis is", lis(arr))
# This code is contributed by Nikhil Kumar Singh
C#
/* Dynamic Programming C# implementation of LIS problem */
using System;
class LIS {
/* lis() returns the length of the longest increasing
subsequence in arr[] of size n */
static int lis(int[] arr, int n)
{
int[] lis = new int[n];
int i, j, max = 0;
/* Initialize LIS values for all indexes */
for (i = 0; i < n; i++)
lis[i] = 1;
/* Compute optimized LIS values in bottom up manner
*/
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (arr[i] > arr[j] && lis[i] < lis[j] + 1)
lis[i] = lis[j] + 1;
/* Pick maximum of all LIS values */
for (i = 0; i < n; i++)
if (max < lis[i])
max = lis[i];
return max;
}
public static void Main()
{
int[] arr = { 10, 22, 9, 33, 21, 50, 41, 60 };
int n = arr.Length;
Console.WriteLine("Length of lis is " + lis(arr, n)
+ "\n");
}
// This code is contributed by Ryuga
}
Javascript
<script>
// Dynamic Programming Javascript implementation
// of LIS problem
// lis() returns the length of the longest
// increasing subsequence in arr[] of size n
function lis(arr, n)
{
let lis = Array(n).fill(0);
let i, j, max = 0;
// Initialize LIS values for all indexes
for(i = 0; i < n; i++)
lis[i] = 1;
// Compute optimized LIS values in
// bottom up manner
for(i = 1; i < n; i++)
for(j = 0; j < i; j++)
if (arr[i] > arr[j] && lis[i] < lis[j] + 1)
lis[i] = lis[j] + 1;
// Pick maximum of all LIS values
for(i = 0; i < n; i++)
if (max < lis[i])
max = lis[i];
return max;
}
// Driver code
let arr = [ 10, 22, 9, 33, 21, 50, 41, 60 ];
let n = arr.length;
document.write("Length of lis is " + lis(arr, n) + "\n");
// This code is contributed by avijitmondal1998
</script>
Producción
Length of lis is 5
Análisis de Complejidad:
- Complejidad Temporal: O(n 2 ).
Como se usa el bucle anidado. - Espacio Auxiliar: O(n).
Uso de cualquier array para almacenar valores LIS en cada índice.
Nota: La complejidad de tiempo de la solución de programación dinámica (DP) anterior es O (n ^ 2) y hay una solución O (N log N) para el problema LIS. No hemos discutido la solución O (N log N) aquí ya que el propósito de esta publicación es explicar la programación dinámica con un ejemplo simple. Consulte la publicación a continuación para obtener la solución O (N log N).
Tamaño de subsecuencia creciente más largo (N log N)
Método 3: Programación Dinámica
Si observamos de cerca el problema, podemos convertirlo en el Problema de subsecuencia común más largo. En primer lugar, crearemos otra array de elementos únicos de la array original y la ordenaremos. Ahora, la subsecuencia creciente más larga de nuestra array debe estar presente como una subsecuencia en nuestra array ordenada. Es por eso que nuestro problema ahora se reduce a encontrar la subsecuencia común entre las dos arrays.
Eg. arr =[50,3,10,7,40,80]
// Sorted array
arr1 = [3,7,10,40,50,80]
// LIS is longest common subsequence between the two arrays
ans = 4
The longest increasing subsequence is {3, 7, 40, 80}
C++
// Dynamic Programming Approach of Finding LIS by reducing
// the problem to longest common Subsequence
#include <bits/stdc++.h>
using namespace std;
/* lis() returns the length of the longest
increasing subsequence in arr[] of size n */
int lis(int arr[], int n)
{
vector<int> b;
set<int> s;
// setting iterator for set
set<int>::iterator it;
// storing unique elements
for (int i = 0; i < n; i++) {
s.insert(arr[i]);
}
// creating sorted vector
for (it = s.begin(); it != s.end(); it++) {
b.push_back(*it);
}
int m = b.size(), i, j;
int dp[m + 1][n + 1];
// storing -1 in dp multidimensional array
for (i = 0; i < m + 1; i++) {
for (j = 0; j < n + 1; j++) {
dp[i][j] = -1;
}
}
// Finding Longest common Subsequence of the two
// arrays
for (i = 0; i < m + 1; i++) {
for (j = 0; j < n + 1; j++) {
if (i == 0 || j == 0) {
dp[i][j] = 0;
}
else if (arr[j - 1] == b[i - 1]) {
dp[i][j] = 1 + dp[i - 1][j - 1];
}
else {
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[m][n];
}
/* Driver program to test above function */
int main()
{
int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("Length of lis is %d\n", lis(arr, n));
}
/* This code is contributed by Arun Bang */
Java
import static java.lang.Math.max;
import java.util.SortedSet;
import java.util.TreeSet;
// Dynamic Programming Approach of Finding LIS by reducing
// the problem to longest common Subsequence
public class Main {
/* lis() returns the length of the longest
increasing subsequence in arr[] of size n */
static int lis(int arr[], int n)
{
SortedSet<Integer> hs = new TreeSet<Integer>();
// Storing and Sorting unique elements.
for (int i = 0; i < n; i++)
hs.add(arr[i]);
int lis[] = new int[hs.size()];
int k = 0;
// Storing all the unique values in a sorted manner.
for (int val : hs) {
lis[k] = val;
k++;
}
int m = k, i, j;
int dp[][] = new int[m + 1][n + 1];
// Storing -1 in dp multidimensional array.
for (i = 0; i < m + 1; i++) {
for (j = 0; j < n + 1; j++) {
dp[i][j] = -1;
}
}
// Finding the Longest Common Subsequence of the two
// arrays
for (i = 0; i < m + 1; i++) {
for (j = 0; j < n + 1; j++) {
if (i == 0 || j == 0) {
dp[i][j] = 0;
}
else if (arr[j - 1] == lis[i - 1]) {
dp[i][j] = 1 + dp[i - 1][j - 1];
}
else {
dp[i][j]
= max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[m][n];
}
// Driver Program for the above test function.
public static void main(String[] args)
{
int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
int n = arr.length;
System.out.println("Length of lis is " + lis(arr, n)
+ "\n");
}
}
// This Code is Contributed by Omkar Subhash Ghongade
Python3
# Dynamic Programming Approach of Finding LIS by reducing the problem to longest common Subsequence
def lis(a):
n = len(a)
# Creating the sorted list
b = sorted(list(set(a)))
m = len(b)
# Creating dp table for storing the answers of sub problems
dp = [[-1 for i in range(m+1)] for j in range(n+1)]
# Finding Longest common Subsequence of the two arrays
for i in range(n+1):
for j in range(m+1):
if i == 0 or j == 0:
dp[i][j] = 0
elif a[i-1] == b[j-1]:
dp[i][j] = 1+dp[i-1][j-1]
else:
dp[i][j] = max(dp[i-1][j], dp[i][j-1])
return dp[-1][-1]
# Driver program to test above function
arr = [10, 22, 9, 33, 21, 50, 41, 60]
print("Length of lis is ", lis(arr))
# This code is Contributed by Dheeraj Khatri
Producción
Length of lis is 5
Análisis de Complejidad : O(n*n)
Como se usa el bucle anidado
Complejidad espacial : O(n*n)
Como array se utiliza para almacenar los valores.
Método 4: Memoización DP
Esta es una extensión del método recursivo.
Podemos ver que hay muchos subproblemas en la solución recursiva anterior que se resuelven una y otra vez. Por lo tanto, este problema tiene la propiedad de subestructura superpuesta y el recálculo de los mismos subproblemas se puede evitar utilizando Memoization
C++
/* A Naive C++ recursive implementation
of LIS problem */
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
/* To make use of recursive calls, this
function must return two things:
1) Length of LIS ending with element arr[n-1].
We use max_ending_here for this purpose
2) Overall maximum as the LIS may end with
an element before arr[n-1] max_ref is
used this purpose.
The value of LIS of full array of size n
is stored in *max_ref which is our final result
*/
int f(int idx, int prev_idx, int n, int a[],
vector<vector<int> >& dp)
{
if (idx == n) {
return 0;
}
if (dp[idx][prev_idx + 1] != -1) {
return dp[idx][prev_idx + 1];
}
int notTake = 0 + f(idx + 1, prev_idx, n, a, dp);
int take = INT_MIN;
if (prev_idx == -1 || a[idx] > a[prev_idx]) {
take = 1 + f(idx + 1, idx, n, a, dp);
}
return dp[idx][prev_idx + 1] = max(take, notTake);
}
// Function to find length of longest increasing
// subsequence.
int longestSubsequence(int n, int a[])
{
vector<vector<int> > dp(n + 1, vector<int>(n + 1, -1));
return f(0, -1, n, a, dp);
}
/* Driver program to test above function */
int main()
{
int a[] = { 3, 10, 2, 1, 20 };
int n = sizeof(a) / sizeof(a[0]);
cout << "Length of lis is " << longestSubsequence(n, a);
return 0;
}
Java
/* A Memoization Java Program for LIS Implementation */
import java.util.Arrays;
import java.lang.*;
class LIS {
/* To make use of recursive calls, this function must
return two things: 1) Length of LIS ending with element
arr[n-1]. We use max_ending_here for this purpose 2)
Overall maximum as the LIS may end with an element
before arr[n-1] max_ref is used this purpose.
The value of LIS of full array of size n is stored in
*max_ref which is our final result */
static int f(int idx, int prev_idx, int n, int a[],
int[][] dp)
{
if (idx == n) {
return 0;
}
if (dp[idx][prev_idx + 1] != -1) {
return dp[idx][prev_idx + 1];
}
int notTake = 0 + f(idx + 1, prev_idx, n, a, dp);
int take = Integer.MIN_VALUE;
if (prev_idx == -1 || a[idx] > a[prev_idx]) {
take = 1 + f(idx + 1, idx, n, a, dp);
}
return dp[idx][prev_idx + 1] = Math.max(take, notTake);
}
// The wrapper function for _lis()
static int lis(int arr[], int n)
{
// The function _lis() stores its result in max
int dp[][] = new int[n+1][n+1];
for (int row[] : dp)
Arrays.fill(row, -1);
return f(0, -1, n, arr, dp);
}
// driver program to test above functions
public static void main(String args[])
{
int a[] = { 3, 10, 2, 1, 20 };
int n = a.length;
System.out.println("Length of lis is " + lis(a, n)
+ "\n");
}
}
// This code is contributed by Sanskar.
Python3
# A Naive Python recursive implementation
# of LIS problem
# To make use of recursive calls, this
# function must return two things:
# 1) Length of LIS ending with element arr[n-1].
# We use max_ending_here for this purpose
# 2) Overall maximum as the LIS may end with
# an element before arr[n-1] max_ref is
# used this purpose.
# The value of LIS of full array of size n
# is stored in *max_ref which is our final result
import sys
def f(idx, prev_idx, n, a,dp):
if (idx == n):
return 0
if (dp[idx][prev_idx + 1] != -1):
return dp[idx][prev_idx + 1]
notTake = 0 + f(idx + 1, prev_idx, n, a, dp)
take = -sys.maxsize -1
if (prev_idx == -1 or a[idx] > a[prev_idx]):
take = 1 + f(idx + 1, idx, n, a, dp)
dp[idx][prev_idx + 1] = max(take, notTake)
return dp[idx][prev_idx + 1]
# Function to find length of longest increasing
# subsequence.
def longestSubsequence(n, a):
dp = [[-1 for i in range(n + 1)]for j in range(n + 1)]
return f(0, -1, n, a, dp)
# Driver program to test above function
a = [ 3, 10, 2, 1, 20 ]
n = len(a)
print("Length of lis is ",longestSubsequence(n, a))
# This code is contributed by shinjanpatra
Producción
Length of lis is 3
Análisis de Complejidad:
Complejidad Temporal: O(n 2 ).
Espacio Auxiliar: O(n 2 ).
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA