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:
- {10, 10} of length 2, which is prime.
- {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++ 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
Time Complexity: O(N * K)
Auxiliary Space: O(N)