Maximum path sum in the given arrays with at most K jumps

Given three arrays A, B and C each having N elements, the task is to find the maximum sum that can be obtained along any valid path with at most K jumps.

A path is valid if it follows the following properties:

It starts from 0th index of an array.
It ends at (N-1)th index of an array.
For any element in the path at index i, the next element should be on the index i+1 of either current or adjacent array only.
If the path involves selecting the next (i + 1)th element from the adjacent array, instead of current one, then it is said to be 1 jump

Examples:

Input: A[] = {4, 5, 1, 2, 10}, B[] = {9, 7, 3, 20, 16}, C[] = {6, 12, 13, 9, 8}, K = 2
Output: 70
Explanation:
Starting from array B and selecting the elements as follows:
Select B[0]: 9 => sum = 9
Jump to C[1]: 12 => sum = 21
Select C[2]: 13 => sum = 34
Jump to B[3]: 20 => sum = 54
Select B[4]: 16 => sum = 70
Therefore maximum sum with at most 2 jumps = 70

Input: A[] = {10, 4, 1, 8}, B[] = {9, 0, 2, 5}, C[] = {6, 2, 7, 3}, K = 2
Output: 24

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

Intuitive Greedy Approach (Not Correct): One possible idea to solve the problem could be to pick the maximum element at the current index and move to the next index having maximum value either from the current array or the adjacent array if jumps are left.

For example:

Given,
A[] = {4, 5, 1, 2, 10},
B[] = {9, 7, 3, 20, 16},
C[] = {6, 12, 13, 9, 8},
K = 2

Finding the solution using Greedy approach:

Current maximum: 9, K = 2, sum = 9
Next maximum: 12, K = 1, sum = 12
Next maximum: 13, K = 1, sum = 25
Next maximum: 20, K = 0, sum = 45
Adding rest of elements: 16, K = 0, sum = 61

Clearly, this is not the maximum sum.

Hence this approach is incorrect.

Dynamic Programing Approach: The DP can be used in two steps – Recursion and Memorization.

Recursion: The problem could be break down using following recursive relation:
- On Array A, index i with K jumps
  
  pathSum(A, i, k) = A[i] + max(pathSum(A, i+1, k), pathSum(B, i+1, k-1));
- Similiarly on Array B,
  
  pathSum(B, i, k) = B[i] + max(pathSum(B, i+1, k), max(pathSum(A, i+1, k-1), pathSum(C, i+1, k-1));
- Similiarly on Array C,
  
  pathSum(C, i, k) = C[i] + max(pathSum(C, i+1, k), pathSum(B, i+1, k-1));
- Therefore the maximum sum can be found as:
  
  maxSum = max(pathSum(A, i, k), max(pathSum(B, i, k), pathSum(C, i, k)));
Memorisation: The complexity of the above recursion solution can be reduce with help of memorisation.
- Store the results after calculating in a 3-dimensional array (dp) of size [3][N][K].
- The value of any element of dp array stores the maximum sum on ith index with x jumps left in an array

Below is the implementation of the approach:

C++

// C++ program to maximum path sum in 
// the given arrays with at most K jumps 
  
#include <iostream> 
using namespace std; 
  
#define M 3 
#define N 5 
#define K 2 
  
int dp[M][N][K]; 
  
void initializeDp() 
{ 
    for (int i = 0; i < M; i++) 
        for (int j = 0; j < N; j++) 
            for (int k = 0; k < K; k++) 
                dp[i][j][k] = -1; 
} 
  
// Function to calculate maximum path sum 
int pathSum(int* a, int* b, int* c, 
            int i, int n, 
            int k, int on) 
{ 
    // Base Case 
    if (i == n) 
        return 0; 
  
    if (dp[on][i][k] != -1) 
        return dp[on][i][k]; 
  
    int current, sum; 
    switch (on) { 
    case 0: 
        current = a[i]; 
        break; 
    case 1: 
        current = b[i]; 
        break; 
    case 2: 
        current = c[i]; 
        break; 
    } 
  
    // No jumps available. 
    // Hence pathSum can be 
    // from current array only 
    if (k == 0) { 
        return dp[on][i][k] 
               = current 
                 + pathSum(a, b, c, i + 1, 
                           n, k, on); 
    } 
  
    // Since jumps are available 
    // pathSum can be from current 
    // or adjacent array 
    switch (on) { 
    case 0: 
        sum = current 
              + max(pathSum(a, b, c, i + 1, 
                            n, k - 1, 1), 
                    pathSum(a, b, c, i + 1, 
                            n, k, 0)); 
        break; 
    case 1: 
        sum = current 
              + max(pathSum(a, b, c, i + 1, 
                            n, k - 1, 0), 
                    max(pathSum(a, b, c, i + 1, 
                                n, k, 1), 
                        pathSum(a, b, c, i + 1, 
                                n, k - 1, 2))); 
        break; 
    case 2: 
        sum = current 
              + max(pathSum(a, b, c, i + 1, 
                            n, k - 1, 1), 
                    pathSum(a, b, c, i + 1, 
                            n, k, 2)); 
        break; 
    } 
  
    return dp[on][i][k] = sum; 
} 
  
void findMaxSum(int* a, int* b, 
                int* c, int n, int k) 
{ 
    int sum = 0; 
  
    // Creating the DP array for memorisation 
    initializeDp(); 
  
    // Find the pathSum using recursive approach 
    for (int i = 0; i < 3; i++) { 
  
        // Maximise the sum 
        sum = max(sum, 
                  pathSum(a, b, c, 0, 
                          n, k, i)); 
    } 
  
    cout << sum; 
} 
  
// Driver Code 
int main() 
{ 
    int n = 5, k = 1; 
    int A[n] = { 4, 5, 1, 2, 10 }; 
    int B[n] = { 9, 7, 3, 20, 16 }; 
    int C[n] = { 6, 12, 13, 9, 8 }; 
  
    findMaxSum(A, B, C, n, k); 
  
    return 0; 
} 

Java

// Java program to maximum path sum in 
// the given arrays with at most K jumps 
import java.util.*; 
class GFG 
{ 
    static int M = 3; 
    static int N = 5; 
    static int K = 2; 
      
    static int dp[][][] = new int[M][N][K]; 
      
    static void initializeDp() 
    { 
        for (int i = 0; i < M; i++) 
            for (int j = 0; j < N; j++) 
                for (int k = 0; k < K; k++) 
                    dp[i][j][k] = -1; 
    } 
      
    // Function to calculate maximum path sum 
    static int pathSum(int a[], int b[], int c[], 
                int i, int n, 
                int k, int on) 
    { 
        // Base Case 
        if (i == n) 
            return 0; 
      
        if (dp[on][i][k] != -1) 
            return dp[on][i][k]; 
      
        int current = 0, sum = 0; 
          
        switch (on) { 
        case 0: 
            current = a[i]; 
            break; 
        case 1: 
            current = b[i]; 
            break; 
        case 2: 
            current = c[i]; 
            break; 
        } 
      
        // No jumps available. 
        // Hence pathSum can be 
        // from current array only 
        if (k == 0) { 
            return dp[on][i][k] 
                = current 
                    + pathSum(a, b, c, i + 1, 
                            n, k, on); 
        } 
      
        // Since jumps are available 
        // pathSum can be from current 
        // or adjacent array 
        switch (on) { 
        case 0: 
            sum = current 
                + Math.max(pathSum(a, b, c, i + 1, 
                                n, k - 1, 1), 
                        pathSum(a, b, c, i + 1, 
                                n, k, 0)); 
            break; 
        case 1: 
            sum = current 
                + Math.max(pathSum(a, b, c, i + 1, 
                                n, k - 1, 0), 
                        Math.max(pathSum(a, b, c, i + 1, 
                                    n, k, 1), 
                            pathSum(a, b, c, i + 1, 
                                    n, k - 1, 2))); 
            break; 
        case 2: 
            sum = current 
                + Math.max(pathSum(a, b, c, i + 1, 
                                n, k - 1, 1), 
                        pathSum(a, b, c, i + 1, 
                                n, k, 2)); 
            break; 
        } 
      
        return dp[on][i][k] = sum; 
    } 
      
    static void findMaxSum(int a[], int b[], 
                    int c[], int n, int k) 
    { 
        int sum = 0; 
      
        // Creating the DP array for memorisation 
        initializeDp(); 
      
        // Find the pathSum using recursive approach 
        for (int i = 0; i < 3; i++) { 
      
            // Maximise the sum 
            sum = Math.max(sum, 
                    pathSum(a, b, c, 0, 
                            n, k, i)); 
        } 
      
        System.out.print(sum); 
    } 
      
    // Driver Code 
    public static void main(String []args) 
    { 
        int n = 5, k = 1; 
        int A[] = { 4, 5, 1, 2, 10 }; 
        int B[] = { 9, 7, 3, 20, 16 }; 
        int C[] = { 6, 12, 13, 9, 8 }; 
      
        findMaxSum(A, B, C, n, k); 
    } 
} 
  
// This code is contributed by chitranayal 

C#

// C# program to maximum path sum in 
// the given arrays with at most K jumps 
using System; 
  
class GFG{ 
      
static int M = 3; 
static int N = 5; 
static int K = 2; 
      
static int [,,]dp = new int[M, N, K]; 
      
static void initializeDp() 
{ 
    for(int i = 0; i < M; i++) 
       for(int j = 0; j < N; j++) 
          for(int k = 0; k < K; k++) 
             dp[i, j, k] = -1; 
} 
      
// Function to calculate maximum path sum 
static int pathSum(int []a, int []b, int []c, 
                   int i, int n, 
                   int k, int on) 
{ 
      
    // Base Case 
    if (i == n) 
        return 0; 
      
    if (dp[on, i, k] != -1) 
        return dp[on, i, k]; 
      
    int current = 0, sum = 0; 
          
    switch (on) 
    { 
    case 0: 
        current = a[i]; 
        break; 
    case 1: 
        current = b[i]; 
        break; 
    case 2: 
        current = c[i]; 
        break; 
    } 
      
    // No jumps available. 
    // Hence pathSum can be 
    // from current array only 
    if (k == 0) 
    { 
        return dp[on, i, k] = current +  
                              pathSum(a, b, c, i + 1, 
                                      n, k, on); 
    } 
      
    // Since jumps are available 
    // pathSum can be from current 
    // or adjacent array 
    switch (on) 
    { 
    case 0: 
        sum = current + Math.Max(pathSum(a, b, c, i + 1, 
                                         n, k - 1, 1), 
                                 pathSum(a, b, c, i + 1, 
                                         n, k, 0)); 
        break; 
    case 1: 
        sum = current + Math.Max(pathSum(a, b, c, i + 1, 
                                         n, k - 1, 0), 
                        Math.Max(pathSum(a, b, c, i + 1, 
                                          n, k, 1), 
                                 pathSum(a, b, c, i + 1, 
                                         n, k - 1, 2))); 
        break; 
    case 2: 
        sum = current + Math.Max(pathSum(a, b, c, i + 1, 
                                         n, k - 1, 1), 
                                 pathSum(a, b, c, i + 1, 
                                         n, k, 2)); 
        break; 
    } 
      
    return dp[on, i, k] = sum; 
} 
      
static void findMaxSum(int []a, int []b, 
                       int []c, int n, int k) 
{ 
    int sum = 0; 
      
    // Creating the DP array for memorisation 
    initializeDp(); 
      
    // Find the pathSum using recursive approach 
    for(int i = 0; i < 3; i++) 
    { 
         
       // Maximise the sum 
       sum = Math.Max(sum, pathSum(a, b, c, 0, 
                                   n, k, i)); 
    } 
    Console.Write(sum); 
} 
      
// Driver Code 
public static void Main(String []args) 
{ 
    int n = 5, k = 1; 
    int []A = { 4, 5, 1, 2, 10 }; 
    int []B = { 9, 7, 3, 20, 16 }; 
    int []C = { 6, 12, 13, 9, 8 }; 
      
    findMaxSum(A, B, C, n, k); 
} 
} 
  
// This code is contributed by gauravrajput1 

Output:

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

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My Personal Notes

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

C++

Java

C#

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