Maximum subsequence sum obtained by concatenating disjoint subarrays whose lengths are prime

Given an array arr[] of size N, the task is to find the maximum sum of a subsequence formed by concatenating disjoint subarrays whose lengths are prime numbers.

Examples:

Input: arr[] = {10, 10, 7, 10, 10, 10}
Output: 50
Explanation:
Subsequence with maximum sum is obtained by concatenating following two subarrays: 

  1. {10, 10} of length 2, which is prime.
  2. {10, 10, 10} of length 3, which is prime.

The resulting subsequence is {10, 10, 10, 10, 10}.
Sum of the subsequence is 50.

Input: arr[] = {11, 8, 12}
Output: 31



Naive Approach: The simplest approach is to use recursion to calculate the value of the maximum sum of subsequence. In each step, call multiple recursive calls for each of the prime numbers smaller than N. The recurrence relation is given by:

sum(N) = (arr[N] + arr[N – 1] + … arr[N – P + 1]) + sum(N – P – 1)
where, 
P is the prime length of the subarray chosen, 
sum(N) is the function that finds the maximum sum of resulting subsequence.

The above recurrence relation is for only one Prime Number. Therefore, there can be more than one possible subsequence can be formed by selecting different subarrays of different prime length. Below is the resulting recurrence relation formed:

sum(N) = max(sum(aN, …, aN-P1+1) + sum(N – P1 – 1), sum(aN, …, aN-P2+1) + sum(N – P2 – 1), …, sum(aN, …, aN-Pk+1) + sum(N – Pk – 1))
where, 
P1, P2, … Pk are prime numbers smaller than N.

Time Complexity: O(KN) where K is the number of the prime numbers smaller than N. Approximately K = (N / LogN) 
Auxiliary Space: O(1)

Dynamic Programming using Bottom-up Approach: The recursive calls in the above can also be reduced using an auxiliary array dp[] and calculate the value of each state in the bottom-up approach. Below are the steps:

  • Create an auxiliary array prime[] to store all the prime numbers smaller or equal to N.
  • Maintain the Sieve of Eratosthenes and traverse it to populate the values of array prime[].
  • Create an auxiliary array dp[] of size N.
  • Initialize the state 0 and 1 as dp[0] = 0 and dp[1] = 0.
  • Traverse the array dp[] over the range [2, N] and update each state as:

 MSS(i) = max[sum(ai…ai-P1+1) + sum(i-P1-1), sum(ai…ai-P2+1) + sum(i-P2-1), … sum(ai…ai-Pk+1) + sum(i-Pk-1) ] for all prime numbers P1, P2, … Pk smaller than N.

  • Initialize pref[] array to store prefix sum to calculate sum(l, …, r) efficiently.

ai + ai+1 + … aj = sum(i … j) = pref[j] – pref[i – 1]

  • Print the value of dp[N] after the above steps as the result.

Below is the implementation of the above approach:

C++

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// 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 seive[MAX];
 
    // Initialize all its entries as true
    memset(seive, true, sizeof(seive));
 
    for (int p = 2; p * p < MAX; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (seive[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) {
                seive[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 (seive[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;
}

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Java

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// 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 []seive = new boolean[MAX];
 
    // Initialize all its entries as true
    Arrays.fill(seive, true);
 
    for(int p = 2; p * p < MAX; p++)
    {
         
        // If prime[p] is not changed,
        // then it is a prime
        if (seive[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)
            {
                seive[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 (seive[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

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Python3

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# 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]"
    seive = [True for i in range(MAX)]
 
    # Initialize all its entries as true
    # memset(seive, true, sizeof(seive))
    for p in range(2, MAX):
        if p * p > MAX:
            break
         
        # If prime[p] is not changed,
        # then it is a prime
        if (seive[p] == True):
 
            # Update all multiples of
            # p greater than or equal
            # to the square of it
            for i in range(p * p, MAX, p):
                seive[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 (seive[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

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C#

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// 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 []seive = new bool[MAX];
 
  // Initialize all its entries
  // as true
  for(int i = 0; i < MAX; i++)
    seive[i] = true;
 
  for(int p = 2; p * p < MAX; p++)
  {
    // If prime[p] is not changed,
    // then it is a prime
    if (seive[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)
      {
        seive[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 (seive[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

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Output

50

Time Complexity: O(N * K)
Auxiliary Space: O(N)

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