Dada una array de enteros arr[] de tamaño N y un entero K . La tarea es encontrar la longitud de la subsecuencia más larga cuya suma sea menor o igual a K .
Ejemplo:
Entrada: arr[] = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15} K = 40
Salida: 5
Explicación:
Si seleccionamos subsecuencia {0, 1, 3, 7, 15} entonces la suma total será 26 , que es menor que 40 . Por lo tanto, la longitud de subsecuencia creciente más larga posible es 5 .Entrada: arr[] = {5, 8, 3, 7, 9, 1} K = 4
Salida: 1
Acercarse:
- El problema anterior se puede resolver mediante recursividad.
- Elija el elemento en esa posición si la suma total es menor que K y explore los demás elementos.
- Deje el elemento en esa posición y explore el resto.
La relación de recurrencia se dará como:
Relación de recurrencia:
T(N) = max(solve(arr, N, arr[i], i+1, K-arr[i])+1, solve(arr, N, prevele, i+1, K)) ;
Condiciones base: if(i >= N || K <= 0) devuelve 0
Aquí está la implementación del enfoque anterior:
C++
// C++ program to find the Longest
// Increasing Subsequence having sum
// value atmost K
#include <bits/stdc++.h>
using namespace std;
int solve(int arr[], int N,
int prevele, int i, int K)
{
// check for base cases
if (i >= N || K <= 0)
return 0;
// check if it is possible to take
// current elements
if (arr[i] <= prevele
|| (K - arr[i] < 0)) {
return solve(arr, N, prevele,
i + 1, K);
}
// if current element is ignored
else {
int ans = max(
solve(arr, N, arr[i],
i + 1, K - arr[i])
+ 1,
solve(arr, N, prevele,
i + 1, K));
return ans;
}
}
// Driver Code
int main()
{
int N = 16;
int arr[N]
= { 0, 8, 4, 12,
2, 10, 6, 14,
1, 9, 5, 13,
3, 11, 7, 15 };
int K = 40;
cout << solve(arr, N,
INT_MIN, 0, K)
<< endl;
}
Java
// Java program to find the Longest
// Increasing Subsequence having sum
// value atmost K
import java.io.*;
class GFG{
static int solve(int arr[], int N,
int prevele, int i, int K)
{
// Check for base cases
if (i >= N || K <= 0)
return 0;
// Check if it is possible to take
// current elements
if (arr[i] <= prevele ||
(K - arr[i] < 0))
{
return solve(arr, N, prevele,
i + 1, K);
}
// If current element is ignored
else
{
int ans = Math.max(solve(arr, N, arr[i],
i + 1, K - arr[i]) + 1,
solve(arr, N, prevele,
i + 1, K));
return ans;
}
}
// Driver code
public static void main (String[] args)
{
int N = 16;
int arr[] = new int[]{ 0, 8, 4, 12,
2, 10, 6, 14,
1, 9, 5, 13,
3, 11, 7, 15 };
int K = 40;
System.out.print(solve(arr, N,
Integer.MIN_VALUE, 0, K));
}
}
// This code is contributed by Pratima Pandey
Python3
# Python3 program to find the Longest # Increasing Subsequence having sum # value atmost K import sys def solve(arr, N, prevele, i, K): # Check for base cases if (i >= N or K <= 0): return 0; # Check if it is possible to take # current elements if (arr[i] <= prevele or (K - arr[i] < 0)): return solve(arr, N, prevele, i + 1, K); # If current element is ignored else: ans = max(solve(arr, N, arr[i], i + 1, K - arr[i]) + 1, solve(arr, N, prevele, i + 1, K)); return ans; # Driver code if __name__ == '__main__': N = 16; arr = [ 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15 ]; K = 40; print(solve(arr, N, -sys.maxsize, 0, K)); # This code is contributed by 29AjayKumar
C#
// C# program to find the Longest
// Increasing Subsequence having sum
// value atmost K
using System;
class GFG{
static int solve(int[] arr, int N,
int prevele, int i, int K)
{
// Check for base cases
if (i >= N || K <= 0)
return 0;
// Check if it is possible to take
// current elements
if (arr[i] <= prevele ||
(K - arr[i] < 0))
{
return solve(arr, N, prevele,
i + 1, K);
}
// If current element is ignored
else
{
int ans = Math.Max(solve(arr, N, arr[i],
i + 1, K - arr[i]) + 1,
solve(arr, N, prevele,
i + 1, K));
return ans;
}
}
// Driver code
public static void Main ()
{
int N = 16;
int[] arr = new int[]{ 0, 8, 4, 12,
2, 10, 6, 14,
1, 9, 5, 13,
3, 11, 7, 15 };
int K = 40;
Console.Write(solve(arr, N,
Int32.MinValue, 0, K));
}
}
// This code is contributed by sanjoy_62
Javascript
<script>
// Javascript program to find the Longest
// Increasing Subsequence having sum
// value atmost K
function solve(arr, N, prevele, i, K)
{
// check for base cases
if (i >= N || K <= 0)
return 0;
// check if it is possible to take
// current elements
if (arr[i] <= prevele
|| (K - arr[i] < 0)) {
return solve(arr, N, prevele,
i + 1, K);
}
// if current element is ignored
else {
var ans = Math.max(
solve(arr, N, arr[i],
i + 1, K - arr[i])
+ 1,
solve(arr, N, prevele,
i + 1, K));
return ans;
}
}
// Driver Code
var N = 16;
var arr
= [0, 8, 4, 12,
2, 10, 6, 14,
1, 9, 5, 13,
3, 11, 7, 15];
var K = 40;
document.write( solve(arr, N,
-1000000000, 0, K));
</script>
5
Tiempo Complejidad: O (2 N )
Espacio Auxiliar: O (1)
Publicación traducida automáticamente
Artículo escrito por sunilkannur98 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA