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:  

1. It starts from the 0th index of an array.
2. It ends at (N-1)th index of an array.
3. 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.
4. If the path involves selecting the next (i + 1)th element from the adjacent array, instead of the 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 
 

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 the 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 Programming Approach: The DP can be used in two steps – Recursion and Memoization.  

• Recursion: The problem could be broken down using the 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));

• Similarly, 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));

• Similarly, 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)));
 

• Memoization: The complexity of the above recursion solution can be reduced with the help of memoization. 
  • 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:  

67

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

Another approach : Dp tabulation (iterative)

In previous approach we solve the problem using recursion + memoization and use where we use DP to store the computation and also check whether we compute the particular problem previously or not but in this approach we solve the problem iteratively by using only DP and no recursion.

Implementation Steps:

• Create a 3d Dp to store computation previous of previous subproblems.
• Initialize DP with base cases.
• Here dp[i][j][x] stores the maximum sum using at most x jumps till index j of the i-th array.
• Initialize 3 nested loops to compute every subpreoblems.
• Noe initialize a variable nextMax with 0  and Calculate maximum sum possible after making a jump from the current array.
• At last initialize the variable maxSum the contains maximum sum is the maximum value in the dp table.
• Finally print maxSum.

Implementation:

Output:

67

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