# 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 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 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: 9 => sum = 9
Select C: 13 => sum = 34
Select B: 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.

1. 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)));

2. 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 [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 ` `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:

```67
```

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

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Improved By : chitranayal, GauravRajput1