Suma máxima de subsecuencias obtenida al concatenar subarreglos disjuntos cuyas longitudes son primos

Dada una array arr[] de tamaño N, la tarea es encontrar la suma máxima de una subsecuencia formada por la concatenación de subarreglos disjuntos cuyas longitudes son números primos .

Ejemplos:

Entrada: arr[] = {10, 10, 7, 10, 10, 10}
Salida: 50
Explicación:
La subsecuencia con suma máxima se obtiene concatenando los siguientes dos subarreglos: 

1. {10, 10} de longitud 2, que es primo.
2. {10, 10, 10} de longitud 3, que es primo.

La subsecuencia resultante es {10, 10, 10, 10, 10}.
La suma de la subsecuencia es 50.

Entrada: arr[] = {11, 8, 12}
Salida: 31

Enfoque ingenuo: el enfoque más simple es usar la recursividad para calcular el valor de la suma máxima de la subsecuencia. En cada paso, realice varias llamadas recursivas para cada uno de los números primos menores que N . La relación de recurrencia está dada por:

sum(N) = (arr[N] + arr[N – 1] + … arr[N – P + 1]) + sum(N – P – 1)
donde, 
P es la longitud prima del subarreglo elegido, 
sum (N) es la función que encuentra la suma máxima de la subsecuencia resultante.

La relación de recurrencia anterior es para un solo número primo . Por lo tanto, puede haber más de una posible subsecuencia que se puede formar seleccionando diferentes subarreglos de diferente longitud principal. A continuación se muestra la relación de recurrencia resultante formada:

sum(N) = max(sum(a _N , …, a _N-P1+1 ) + sum(N – P1 – 1), sum(a _N , …, a _N-P2+1 ) + sum(N – P2 – 1), …, sum(a _N , …, a _N-Pk+1 ) + sum(N – P _k – 1))
donde, 
P1, P2, … Pk son números primos menores que N.

Complejidad Temporal: O(K ^N ) donde K es el número de losnúmeros primos menores que N. Aproximadamente K = (N / LogN) 
Espacio Auxiliar: O (1)

Programación dinámica usando el enfoque de abajo hacia arriba: las llamadas recursivas de arriba también se pueden reducir usando una array auxiliar dp[] y calcular el valor de cada estado en el enfoque de abajo hacia arriba . A continuación se muestran los pasos:

• Cree una array auxiliar prime[] para almacenar todos los números primos menores o iguales a N.
• Mantenga el Tamiz de Eratóstenes y atraviéselo para completar los valores de array prime[] .
• Cree una array auxiliar dp[] de tamaño N .
• Inicialice el estado 0 y 1 como dp[0] = 0 y dp[1] = 0 .
• Recorra la array dp[] sobre el rango [2, N] y actualice cada estado como:

 MSS(i) = max[sum(a _i …a _i-P1+1 ) + sum(i-P1-1), sum(a _i …a _i-P2+1 ) + sum(i-P2-1) , … sum(a _i …a _i-Pk+1 ) + sum(i-Pk-1) ] para todos los números primos P1, P2, … Pk menores que N.

• Inicialice la array pref[] para almacenar la suma del prefijo para calcular la suma (l, …, r) de manera eficiente.

a _i + a _i+1 + … a _j = sum(i … j) = pref[j] – pref[i – 1]

• Imprima el valor de dp[N] después de los pasos anteriores como resultado.

A continuación se muestra la implementación del enfoque anterior:

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
#define MAX 100005
 
// Function to return all prime numbers
// smaller than N
vector<int> SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..n]"
    bool sieve[MAX];
 
    // Initialize all its entries as true
    memset(sieve, true, sizeof(sieve));
 
    for (int p = 2; p * p < MAX; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (sieve[p] == true) {
 
            // Update all multiples of
            // p greater than or equal
            // to the square of it
            for (int i = p * p;
                 i < MAX; i += p) {
                sieve[i] = false;
            }
        }
    }
 
    // Stores all prime numbers
    // smaller than MAX
    vector<int> v;
 
    // Store all prime numbers
    for (int p = 2; p < MAX; p++) {
 
        // If p is prime
        if (sieve[p]) {
            v.push_back(p);
        }
    }
 
    return v;
}
 
// Function to build the auxiliary DP
// array from the start
void build(int dp[], int arr[], int N)
{
    // Base Case
    dp[0] = 0;
    dp[1] = 0;
 
    // Stores all prime numbers < N
    vector<int> prime
        = SieveOfEratosthenes();
 
    // Stores prefix sum
    int pref[N + 1];
    pref[0] = 0;
 
    // Update prefix sum
    for (int i = 1; i <= N; i++) {
        pref[i] = pref[i - 1]
                  + arr[i - 1];
    }
 
    // Iterate over range
    for (int i = 2; i <= N; i++) {
 
        // Update each state i.e.. when
        // current element is excluded
        dp[i] = dp[i - 1];
        for (int j = 0;
             j <= prime.size(); j++) {
 
            // Find start & end index
            // of subarrays when prime[i]
            // is taken
            int r = i - 1;
            int l = r - prime[j] + 1;
 
            // Check if starting point
            // lies in the array
            if (l < 0)
                break;
            int temp = 0;
 
            // Include the elements
            // al al+1 ... ar
            temp = pref[r + 1] - pref[l];
 
            // Check if element lies before
            // start of selected subarray
            if (l - 2 >= 0)
                temp += dp[l - 2 + 1];
 
            // Update value of dp[i]
            dp[i] = max(dp[i], temp);
        }
    }
}
 
// Function to find the maximum sum
// subsequence with prime length
void maxSumSubseq(int arr[], int N)
{
    // Auxiliary DP array
    int dp[N + 1];
 
    // Build DP array
    build(dp, arr, N);
 
    // Print the result
    cout << dp[N];
}
 
// Driver Code
int main()
{
    // Given arr[]
    int arr[] = { 10, 10, 7, 10, 10, 10 };
 
    // Size of array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    maxSumSubseq(arr, N);
 
    return 0;
}

// Java program for the above approach
import java.util.*;
 
class GFG{
     
static int MAX = 100005;
 
// Function to return all prime numbers
// smaller than N
static Vector<Integer> SieveOfEratosthenes()
{
     
    // Create a boolean array "prime[0..n]"
    boolean []sieve = new boolean[MAX];
 
    // Initialize all its entries as true
    Arrays.fill(sieve, true);
 
    for(int p = 2; p * p < MAX; p++)
    {
         
        // If prime[p] is not changed,
        // then it is a prime
        if (sieve[p] == true)
        {
             
            // Update all multiples of
            // p greater than or equal
            // to the square of it
            for(int i = p * p; i < MAX; i += p)
            {
                sieve[i] = false;
            }
        }
    }
 
    // Stores all prime numbers
    // smaller than MAX
    Vector<Integer> v = new Vector<Integer>();
 
    // Store all prime numbers
    for(int p = 2; p < MAX; p++)
    {
         
        // If p is prime
        if (sieve[p])
        {
            v.add(p);
        }
    }
    return v;
}
 
// Function to build the auxiliary DP
// array from the start
static void build(int dp[], int arr[], int N)
{
     
    // Base Case
    dp[0] = 0;
    dp[1] = 0;
 
    // Stores all prime numbers < N
    Vector<Integer> prime = SieveOfEratosthenes();
 
    // Stores prefix sum
    int []pref = new int[N + 1];
    pref[0] = 0;
 
    // Update prefix sum
    for(int i = 1; i <= N; i++)
    {
        pref[i] = pref[i - 1] + arr[i - 1];
    }
 
    // Iterate over range
    for(int i = 2; i <= N; i++)
    {
         
        // Update each state i.e.. when
        // current element is excluded
        dp[i] = dp[i - 1];
        for(int j = 0; j <= prime.size(); j++)
        {
             
            // Find start & end index
            // of subarrays when prime[i]
            // is taken
            int r = i - 1;
            int l = r - prime.get(j) + 1;
 
            // Check if starting point
            // lies in the array
            if (l < 0)
                break;
                 
            int temp = 0;
 
            // Include the elements
            // al al+1 ... ar
            temp = pref[r + 1] - pref[l];
 
            // Check if element lies before
            // start of selected subarray
            if (l - 2 >= 0)
                temp += dp[l - 2 + 1];
 
            // Update value of dp[i]
            dp[i] = Math.max(dp[i], temp);
        }
    }
}
 
// Function to find the maximum sum
// subsequence with prime length
static void maxSumSubseq(int arr[], int N)
{
     
    // Auxiliary DP array
    int []dp = new int[N + 1];
 
    // Build DP array
    build(dp, arr, N);
 
    // Print the result
    System.out.print(dp[N]);
}
 
// Driver Code
public static void main(String args[])
{
     
    // Given arr[]
    int arr[] = { 10, 10, 7, 10, 10, 10 };
 
    // Size of array
    int N = arr.length;
 
    // Function Call
    maxSumSubseq(arr, N);
}
}
 
// This code is contributed by ipg2016107

# Python3 program for the above approach
MAX = 100005
 
# Function to return all prime numbers
# smaller than N
def SieveOfEratosthenes():
     
    # Create a boolean array "prime[0..n]"
    sieve = [True for i in range(MAX)]
 
    # Initialize all its entries as true
    # memset(sieve, true, sizeof(sieve))
    for p in range(2, MAX):
        if p * p > MAX:
            break
         
        # If prime[p] is not changed,
        # then it is a prime
        if (sieve[p] == True):
 
            # Update all multiples of
            # p greater than or equal
            # to the square of it
            for i in range(p * p, MAX, p):
                sieve[i] = False
 
    # Stores all prime numbers
    # smaller than MAX
    v = []
 
    # Store all prime numbers
    for p in range(2, MAX):
         
        # If p is prime
        if (sieve[p]):
            v.append(p)
 
    return v
 
# Function to build the auxiliary DP
# array from the start
def build(dp, arr, N):
     
    # Base Case
    dp[0] = 0
    dp[1] = 0
 
    # Stores all prime numbers < N
    prime = SieveOfEratosthenes()
 
    # Stores prefix sum
    pref = [0 for i in range(N + 1)]
    pref[0] = 0
 
    # Update prefix sum
    for i in range(1, N + 1):
        pref[i] = pref[i - 1] + arr[i - 1]
 
    # Iterate over range
    for i in range(2, N + 1):
 
        # Update each state i.e.. when
        # current element is excluded
        dp[i] = dp[i - 1]
         
        for j in range(len(prime) + 1):
 
            # Find start & end index
            # of subarrays when prime[i]
            # is taken
            r = i - 1
            l = r - prime[j] + 1
 
            # Check if starting point
            # lies in the array
            if (l < 0):
                break
             
            temp = 0
 
            # Include the elements
            # al al+1 ... ar
            temp = pref[r + 1] - pref[l]
 
            # Check if element lies before
            # start of selected subarray
            if (l - 2 >= 0):
                temp += dp[l - 2 + 1]
 
            # Update value of dp[i]
            dp[i] = max(dp[i], temp)
 
# Function to find the maximum sum
# subsequence with prime length
def maxSumSubseq(arr, N):
     
    # Auxiliary DP array
    dp = [0 for i in range(N + 1)]
 
    # Build DP array
    build(dp, arr, N)
 
    # Print the result
    print(dp[N])
 
# Driver Code
if __name__ == '__main__':
     
    # Given arr[]
    arr = [ 10, 10, 7, 10, 10, 10 ]
 
    # Size of array
    N = len(arr)
 
    # Function Call
    maxSumSubseq(arr, N)
 
# This code is contributed by mohit kumar 29

// C# program for the
// above approach
using System;
using System.Collections.Generic;
class GFG{
     
static int MAX = 100005;
 
// Function to return all
// prime numbers smaller than N
static List<int> SieveOfEratosthenes()
{   
  // Create a bool array
  // "prime[0..n]"
  bool []sieve = new bool[MAX];
 
  // Initialize all its entries
  // as true
  for(int i = 0; i < MAX; i++)
    sieve[i] = true;
 
  for(int p = 2; p * p < MAX; p++)
  {
    // If prime[p] is not changed,
    // then it is a prime
    if (sieve[p] == true)
    {
      // Update all multiples of
      // p greater than or equal
      // to the square of it
      for(int i = p * p;
              i < MAX; i += p)
      {
        sieve[i] = false;
      }
    }
  }
 
  // Stores all prime numbers
  // smaller than MAX
  List<int> v = new List<int>();
 
  // Store all prime numbers
  for(int p = 2; p < MAX; p++)
  {
    // If p is prime
    if (sieve[p])
    {
      v.Add(p);
    }
  }
  return v;
}
 
// Function to build the auxiliary
// DP array from the start
static void build(int []dp,
                  int []arr, int N)
{   
  // Base Case
  dp[0] = 0;
  dp[1] = 0;
 
  // Stores all prime
  // numbers < N
  List<int> prime =
            SieveOfEratosthenes();
 
  // Stores prefix sum
  int []pref = new int[N + 1];
  pref[0] = 0;
 
  // Update prefix sum
  for(int i = 1; i <= N; i++)
  {
    pref[i] = pref[i - 1] +
              arr[i - 1];
  }
 
  // Iterate over range
  for(int i = 2; i <= N; i++)
  {
    // Update each state i.e..
    // when current element
    // is excluded
    dp[i] = dp[i - 1];
    for(int j = 0;
            j <= prime.Count; j++)
    {
      // Find start & end index
      // of subarrays when prime[i]
      // is taken
      int r = i - 1;
      int l = r - prime[j] + 1;
 
      // Check if starting point
      // lies in the array
      if (l < 0)
        break;
 
      int temp = 0;
 
      // Include the elements
      // al al+1 ... ar
      temp = pref[r + 1] - pref[l];
 
      // Check if element lies
      // before start of selected
      // subarray
      if (l - 2 >= 0)
        temp += dp[l - 2 + 1];
 
      // Update value of dp[i]
      dp[i] = Math.Max(dp[i],
                       temp);
    }
  }
}
 
// Function to find the maximum
// sum subsequence with prime
// length
static void maxSumSubseq(int []arr,
                         int N)
{
  // Auxiliary DP array
  int []dp = new int[N + 1];
 
  // Build DP array
  build(dp, arr, N);
 
  // Print the result
  Console.Write(dp[N]);
}
 
// Driver Code
public static void Main(String []args)
{
  // Given []arr
  int []arr = {10, 10, 7,
               10, 10, 10};
 
  // Size of array
  int N = arr.Length;
 
  // Function Call
  maxSumSubseq(arr, N);
}
}
 
// This code is contributed by shikhasingrajput

<script>
 
// Javascript program for the above approach
 
let MAX = 100005
 
// Function to return all prime numbers
// smaller than N
function SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..n]"
    let sieve = new Array(MAX);
 
    // Initialize all its entries as true
    sieve.fill(true)
 
    for (let p = 2; p * p < MAX; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (sieve[p] == true) {
 
            // Update all multiples of
            // p greater than or equal
            // to the square of it
            for (let i = p * p;
                 i < MAX; i += p) {
                sieve[i] = false;
            }
        }
    }
 
    // Stores all prime numbers
    // smaller than MAX
    let v = new Array();
 
    // Store all prime numbers
    for (let p = 2; p < MAX; p++) {
 
        // If p is prime
        if (sieve[p]) {
            v.push(p);
        }
    }
 
    return v;
}
 
// Function to build the auxiliary DP
// array from the start
function build(dp, arr, N)
{
    // Base Case
    dp[0] = 0;
    dp[1] = 0;
 
    // Stores all prime numbers < N
    let prime = SieveOfEratosthenes();
 
    // Stores prefix sum
    let pref = new Array(N + 1);
    pref[0] = 0;
 
    // Update prefix sum
    for (let i = 1; i <= N; i++) {
        pref[i] = pref[i - 1]
                  + arr[i - 1];
    }
 
    // Iterate over range
    for (let i = 2; i <= N; i++) {
 
        // Update each state i.e.. when
        // current element is excluded
        dp[i] = dp[i - 1];
        for (let j = 0;
             j <= prime.length; j++) {
 
            // Find start & end index
            // of subarrays when prime[i]
            // is taken
            let r = i - 1;
            let l = r - prime[j] + 1;
 
            // Check if starting point
            // lies in the array
            if (l < 0)
                break;
            let temp = 0;
 
            // Include the elements
            // al al+1 ... ar
            temp = pref[r + 1] - pref[l];
 
            // Check if element lies before
            // start of selected subarray
            if (l - 2 >= 0)
                temp += dp[l - 2 + 1];
 
            // Update value of dp[i]
            dp[i] = Math.max(dp[i], temp);
        }
    }
}
 
// Function to find the maximum sum
// subsequence with prime length
function maxSumSubseq(arr, N)
{
    // Auxiliary DP array
    let dp = new Array(N + 1);
 
    // Build DP array
    build(dp, arr, N);
 
    // Print the result
    document.write(dp[N]);
}
 
// Driver Code
 
    // Given arr[]
    let arr = [ 10, 10, 7, 10, 10, 10 ];
 
    // Size of array
    let N = arr.length;
 
    // Function Call
    maxSumSubseq(arr, N);
 
// This code is contributed by gfgking
 
</script>

50

Complejidad temporal: O(N * K)
Espacio auxiliar: O(N)