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:
- {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++
#include <bits/stdc++.h>
using namespace std;
#define MAX 100005
vector< int > SieveOfEratosthenes()
{
bool sieve[MAX];
memset (sieve, true , sizeof (sieve));
for ( int p = 2; p * p < MAX; p++) {
if (sieve[p] == true ) {
for ( int i = p * p;
i < MAX; i += p) {
sieve[i] = false ;
}
}
}
vector< int > v;
for ( int p = 2; p < MAX; p++) {
if (sieve[p]) {
v.push_back(p);
}
}
return v;
}
void build( int dp[], int arr[], int N)
{
dp[0] = 0;
dp[1] = 0;
vector< int > prime
= SieveOfEratosthenes();
int pref[N + 1];
pref[0] = 0;
for ( int i = 1; i <= N; i++) {
pref[i] = pref[i - 1]
+ arr[i - 1];
}
for ( int i = 2; i <= N; i++) {
dp[i] = dp[i - 1];
for ( int j = 0;
j <= prime.size(); j++) {
int r = i - 1;
int l = r - prime[j] + 1;
if (l < 0)
break ;
int temp = 0;
temp = pref[r + 1] - pref[l];
if (l - 2 >= 0)
temp += dp[l - 2 + 1];
dp[i] = max(dp[i], temp);
}
}
}
void maxSumSubseq( int arr[], int N)
{
int dp[N + 1];
build(dp, arr, N);
cout << dp[N];
}
int main()
{
int arr[] = { 10, 10, 7, 10, 10, 10 };
int N = sizeof (arr) / sizeof (arr[0]);
maxSumSubseq(arr, N);
return 0;
}
|
Java
import java.util.*;
class GFG{
static int MAX = 100005 ;
static Vector<Integer> SieveOfEratosthenes()
{
boolean []sieve = new boolean [MAX];
Arrays.fill(sieve, true );
for ( int p = 2 ; p * p < MAX; p++)
{
if (sieve[p] == true )
{
for ( int i = p * p; i < MAX; i += p)
{
sieve[i] = false ;
}
}
}
Vector<Integer> v = new Vector<Integer>();
for ( int p = 2 ; p < MAX; p++)
{
if (sieve[p])
{
v.add(p);
}
}
return v;
}
static void build( int dp[], int arr[], int N)
{
dp[ 0 ] = 0 ;
dp[ 1 ] = 0 ;
Vector<Integer> prime = SieveOfEratosthenes();
int []pref = new int [N + 1 ];
pref[ 0 ] = 0 ;
for ( int i = 1 ; i <= N; i++)
{
pref[i] = pref[i - 1 ] + arr[i - 1 ];
}
for ( int i = 2 ; i <= N; i++)
{
dp[i] = dp[i - 1 ];
for ( int j = 0 ; j <= prime.size(); j++)
{
int r = i - 1 ;
int l = r - prime.get(j) + 1 ;
if (l < 0 )
break ;
int temp = 0 ;
temp = pref[r + 1 ] - pref[l];
if (l - 2 >= 0 )
temp += dp[l - 2 + 1 ];
dp[i] = Math.max(dp[i], temp);
}
}
}
static void maxSumSubseq( int arr[], int N)
{
int []dp = new int [N + 1 ];
build(dp, arr, N);
System.out.print(dp[N]);
}
public static void main(String args[])
{
int arr[] = { 10 , 10 , 7 , 10 , 10 , 10 };
int N = arr.length;
maxSumSubseq(arr, N);
}
}
|
Python3
MAX = 100005
def SieveOfEratosthenes():
sieve = [ True for i in range ( MAX )]
for p in range ( 2 , MAX ):
if p * p > MAX :
break
if (sieve[p] = = True ):
for i in range (p * p, MAX , p):
sieve[i] = False
v = []
for p in range ( 2 , MAX ):
if (sieve[p]):
v.append(p)
return v
def build(dp, arr, N):
dp[ 0 ] = 0
dp[ 1 ] = 0
prime = SieveOfEratosthenes()
pref = [ 0 for i in range (N + 1 )]
pref[ 0 ] = 0
for i in range ( 1 , N + 1 ):
pref[i] = pref[i - 1 ] + arr[i - 1 ]
for i in range ( 2 , N + 1 ):
dp[i] = dp[i - 1 ]
for j in range ( len (prime) + 1 ):
r = i - 1
l = r - prime[j] + 1
if (l < 0 ):
break
temp = 0
temp = pref[r + 1 ] - pref[l]
if (l - 2 > = 0 ):
temp + = dp[l - 2 + 1 ]
dp[i] = max (dp[i], temp)
def maxSumSubseq(arr, N):
dp = [ 0 for i in range (N + 1 )]
build(dp, arr, N)
print (dp[N])
if __name__ = = '__main__' :
arr = [ 10 , 10 , 7 , 10 , 10 , 10 ]
N = len (arr)
maxSumSubseq(arr, N)
|
C#
using System;
using System.Collections.Generic;
class GFG{
static int MAX = 100005;
static List< int > SieveOfEratosthenes()
{
bool []sieve = new bool [MAX];
for ( int i = 0; i < MAX; i++)
sieve[i] = true ;
for ( int p = 2; p * p < MAX; p++)
{
if (sieve[p] == true )
{
for ( int i = p * p;
i < MAX; i += p)
{
sieve[i] = false ;
}
}
}
List< int > v = new List< int >();
for ( int p = 2; p < MAX; p++)
{
if (sieve[p])
{
v.Add(p);
}
}
return v;
}
static void build( int []dp,
int []arr, int N)
{
dp[0] = 0;
dp[1] = 0;
List< int > prime =
SieveOfEratosthenes();
int []pref = new int [N + 1];
pref[0] = 0;
for ( int i = 1; i <= N; i++)
{
pref[i] = pref[i - 1] +
arr[i - 1];
}
for ( int i = 2; i <= N; i++)
{
dp[i] = dp[i - 1];
for ( int j = 0;
j <= prime.Count; j++)
{
int r = i - 1;
int l = r - prime[j] + 1;
if (l < 0)
break ;
int temp = 0;
temp = pref[r + 1] - pref[l];
if (l - 2 >= 0)
temp += dp[l - 2 + 1];
dp[i] = Math.Max(dp[i],
temp);
}
}
}
static void maxSumSubseq( int []arr,
int N)
{
int []dp = new int [N + 1];
build(dp, arr, N);
Console.Write(dp[N]);
}
public static void Main(String []args)
{
int []arr = {10, 10, 7,
10, 10, 10};
int N = arr.Length;
maxSumSubseq(arr, N);
}
}
|
Javascript
<script>
let MAX = 100005
function SieveOfEratosthenes()
{
let sieve = new Array(MAX);
sieve.fill( true )
for (let p = 2; p * p < MAX; p++) {
if (sieve[p] == true ) {
for (let i = p * p;
i < MAX; i += p) {
sieve[i] = false ;
}
}
}
let v = new Array();
for (let p = 2; p < MAX; p++) {
if (sieve[p]) {
v.push(p);
}
}
return v;
}
function build(dp, arr, N)
{
dp[0] = 0;
dp[1] = 0;
let prime = SieveOfEratosthenes();
let pref = new Array(N + 1);
pref[0] = 0;
for (let i = 1; i <= N; i++) {
pref[i] = pref[i - 1]
+ arr[i - 1];
}
for (let i = 2; i <= N; i++) {
dp[i] = dp[i - 1];
for (let j = 0;
j <= prime.length; j++) {
let r = i - 1;
let l = r - prime[j] + 1;
if (l < 0)
break ;
let temp = 0;
temp = pref[r + 1] - pref[l];
if (l - 2 >= 0)
temp += dp[l - 2 + 1];
dp[i] = Math.max(dp[i], temp);
}
}
}
function maxSumSubseq(arr, N)
{
let dp = new Array(N + 1);
build(dp, arr, N);
document.write(dp[N]);
}
let arr = [ 10, 10, 7, 10, 10, 10 ];
let N = arr.length;
maxSumSubseq(arr, N);
</script>
|
Time Complexity: O(N * K)
Auxiliary Space: O(N)
Last Updated :
15 Jan, 2022
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