Maximum path sum for each position with jumps under divisibility condition

Given an array of n positive integers. Initially we are at first position. We can jump to position y (1 <= y <= n) from position x (1 <= x <= n) if x divides y and x < y. The task is to print maximum sum path ending at every position x.

Note : Since first element is at position 1, we can jump to any position from here as 1 divides all other position numbers.

Examples :

Input :  arr[] = {2, 3, 1, 4, 6, 5}
Output : 2 5 3 9 8 10
Maximum sum path ending with position 1 is 2.
For position 1, last position to visit is 1 only.
So maximum sum for position 1 = 2.

Maximum sum path ending with position 2 is 5.
For position 2, path can be jump from position 1 
to 2 as 1 divides 2.
So maximum sum for position 2 = 2 + 3 = 5.

For position 3, path can be jump from position 1 
to 3 as 1 divides 3.
So maximum sum for position 3 = 2 + 3 = 5.

For position 4, path can be jump from position 1
to 2 and 2 to 4.
So maximum sum for position 4 = 2 + 3 + 4 = 9.

For position 5, path can be jump from position 1 
to 5.
So maximum sum for position 5 = 2 + 6 = 8.

For position 6, path can be jump from position 
1 to 2 and 2 to 6 or 1 to 3 and 3 to 6.
But path 1 -> 2 -> 6 gives maximum sum for
position 6 = 2 + 3 + 5 = 10.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to use Dynamic Programming to solve this problem.

Create an 1-D array dp[] where each element dp[i] 
stores maximum sum path ending at index i (or 
position x where x = i+1) with divisible jumps.

The recurrence relation for dp[i] can be defined as:

dp[i] = max(dp[i], dp[divisor of i+1] + arr[i]) 

To find all the divisor of i+1, move from 1 
divisor to sqrt(i+1).

Below is the implementation of this approach:

C++

// C++ program to print maximum path sum ending with 
// each position x such that all path step positions 
// divide x. 
#include <bits/stdc++.h> 
using namespace std; 
  
void printMaxSum(int arr[], int n) 
{ 
    // Create an array such that dp[i] stores maximum 
    // path sum ending with i. 
    int dp[n]; 
    memset(dp, 0, sizeof dp); 
  
    // Calculating maximum sum path for each element. 
    for (int i = 0; i < n; i++) { 
        dp[i] = arr[i]; 
  
        // Finding previous step for arr[i] 
        // Moving from 1 to sqrt(i+1) since all the 
        // divisors are present from sqrt(i+1). 
        int maxi = 0; 
        for (int j = 1; j <= sqrt(i + 1); j++) { 
            // Checking if j is divisor of i+1. 
            if (((i + 1) % j == 0) && (i + 1) != j) { 
                // Checking which divisor will provide 
                // greater value. 
                if (dp[j - 1] > maxi) 
                    maxi = dp[j - 1]; 
                if (dp[(i + 1) / j - 1] > maxi && j != 1) 
                    maxi = dp[(i + 1) / j - 1]; 
            } 
        } 
  
        dp[i] += maxi; 
    } 
  
    // Printing the answer (Maximum path sum ending 
    // with every position i+1. 
    for (int i = 0; i < n; i++) 
        cout << dp[i] << " "; 
} 
  
// Driven Program 
int main() 
{ 
    int arr[] = { 2, 3, 1, 4, 6, 5 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    printMaxSum(arr, n); 
  
    return 0; 
} 

Java

// Java program to print maximum path 
// sum ending with each position x such 
// that all path step positions divide x. 
import java.util.*; 
  
class GFG { 
  
    static void printMaxSum(int arr[], int n) 
    { 
        // Create an array such that dp[i] 
        // stores maximum path sum ending with i. 
        int dp[] = new int[n]; 
        Arrays.fill(dp, 0); 
  
        // Calculating maximum sum 
        // path for each element. 
        for (int i = 0; i < n; i++) { 
            dp[i] = arr[i]; 
  
            // Finding previous step for arr[i] 
            // Moving from 1 to sqrt(i+1) since all the 
            // divisors are present from sqrt(i+1). 
            int maxi = 0; 
            for (int j = 1; j <= Math.sqrt(i + 1); j++) { 
                  
                // Checking if j is divisor of i+1. 
                if (((i + 1) % j == 0) && (i + 1) != j) { 
                      
                    // Checking which divisor will 
                    // provide greater value. 
                    if (dp[j - 1] > maxi) 
                        maxi = dp[j - 1]; 
                    if (dp[(i + 1) / j - 1] > maxi && j != 1) 
                        maxi = dp[(i + 1) / j - 1]; 
                } 
            } 
  
            dp[i] += maxi; 
        } 
  
        // Printing the answer (Maximum path sum  
        // ending with every position i+1.) 
        for (int i = 0; i < n; i++) 
            System.out.print(dp[i] + " "); 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        int arr[] = { 2, 3, 1, 4, 6, 5 }; 
        int n = arr.length; 
          
        // Function calling 
        printMaxSum(arr, n); 
    } 
} 
  
// This code is contributed by Anant Agarwal. 

Python3

# Python3 program to print maximum  
# path sum ending with each position 
# x such that all path step positions 
# divide x. 
  
def printMaxSum(arr, n): 
      
    # Create an array such that dp[i]  
    # stores maximum path sum ending with i. 
    dp = [0 for i in range(n)] 
  
    # Calculating maximum sum path  
    # for each element. 
    for i in range(n): 
        dp[i] = arr[i] 
  
        # Finding previous step for arr[i] 
        # Moving from 1 to sqrt(i + 1) since all the 
        # divisiors are present from sqrt(i + 1). 
        maxi = 0
        for j in range(1, int((i + 1) ** 0.5) + 1): 
  
            # Checking if j is divisior of i + 1. 
            if ((i + 1) % j == 0 and (i + 1) != j): 
  
                # Checking which divisor will provide 
                # greater value. 
                if (dp[j - 1] > maxi): 
                    maxi = dp[j - 1] 
                if (dp[(i + 1) // j - 1] > maxi and j != 1): 
                    maxi = dp[(i + 1) // j - 1] 
  
        dp[i] += maxi 
  
    # Printing the answer  
    # (Maximum path sum ending 
    # with every position i + 1). 
    for i in range(n): 
        print(dp[i], end = ' ') 
  
# Driver Program 
arr = [2, 3, 1, 4, 6, 5] 
n = len(arr) 
printMaxSum(arr, n) 
  
# This code is contributed by Soumen Ghosh. 

C#

// C# program to print maximum path 
// sum ending with each position x such 
// that all path step positions divide x. 
using System; 
  
class GFG { 
  
    static void printMaxSum(int[] arr, int n) 
    { 
        // Create an array such that dp[i] 
        // stores maximum path sum ending with i. 
        int[] dp = new int[n]; 
  
        // Calculating maximum sum 
        // path for each element. 
        for (int i = 0; i < n; i++) { 
            dp[i] = arr[i]; 
  
            // Finding previous step for arr[i] 
            // Moving from 1 to sqrt(i+1) since all the 
            // divisors are present from sqrt(i+1). 
            int maxi = 0; 
            for (int j = 1; j <= Math.Sqrt(i + 1); j++)  
            { 
                // Checking if j is divisor of i+1. 
                if (((i + 1) % j == 0) && (i + 1) != j) 
                { 
                    // Checking which divisor will 
                    // provide greater value. 
                    if (dp[j - 1] > maxi) 
                        maxi = dp[j - 1]; 
                    if (dp[(i + 1) / j - 1] > maxi && j != 1) 
                        maxi = dp[(i + 1) / j - 1]; 
                } 
            } 
  
            dp[i] += maxi; 
        } 
  
        // Printing the answer (Maximum path sum ending 
        // with every position i+1.) 
        for (int i = 0; i < n; i++) 
            Console.Write(dp[i] + " "); 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int[] arr = { 2, 3, 1, 4, 6, 5 }; 
        int n = arr.Length; 
          
        // Function calling 
        printMaxSum(arr, n); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program to print maximum path sum ending with  
// each position x such that all path step positions  
// divide x.  
  
function printMaxSum($arr, $n)  
{  
    // Create an array such that dp[i] stores maximum  
    // path sum ending with i.  
    $dp = array() ; 
  
    // Calculating maximum sum path for each element.  
    for ($i = 0; $i < $n; $i++) {  
        $dp[$i] = $arr[$i];  
  
        // Finding previous step for arr[i]  
        // Moving from 1 to sqrt(i+1) since all the  
        // divisors are present from sqrt(i+1).  
        $maxi = 0;  
        for ($j = 1; $j <= sqrt($i + 1); $j++) {  
            // Checking if j is divisor of i+1.  
            if ((($i + 1) % $j == 0) && ($i + 1) != $j) {  
                // Checking which divisor will provide  
                // greater value.  
                if ($dp[$j - 1] > $maxi)  
                    $maxi = $dp[$j - 1];  
                if ($dp[($i + 1) / $j - 1] > $maxi && $j != 1)  
                    $maxi = $dp[($i + 1) / $j - 1];  
            }  
        }  
  
        $dp[$i] += $maxi;  
    }  
  
    // Printing the answer (Maximum path sum ending  
    // with every position i+1.  
    for ($i = 0; $i < $n; $i++)  
        echo $dp[$i] , " " ;  
}  
  
// Driven Program  
$arr = array(2, 3, 1, 4, 6, 5 ); 
$n = sizeof($arr); 
  
printMaxSum($arr, $n);  
  
// This code is contributed by Ryuga 
?> 

Output:

2 5 3 9 8 10

Time Complexity: O(n*sqrt(n)).

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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