# 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
Select C[2]: 13 => sum = 34
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:

## 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 memoization     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 memoization         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

## Python

 #Python3 program to maximum path sum in #the given arrays with at most K jumps M = 3 N = 5 K = 2 dp=[[[-1 for i in range(K)]       for i in range(N)]       for i in range(M)] def initializeDp():     for i in range(M):         for j in range(N):             for k in range(K):                 dp[i][j][k] = -1   #Function to calculate maximum path sum def pathSum(a, b, c, i, n, k, on):       #Base Case     if (i == n):         return 0       if (dp[on][i][k] != -1):         return dp[on][i][k]     current, sum = 0, 0     if on == 0:         current = a[i]         #break     if on == 1:         current = b[i]         #break     if on == 2:         current = c[i]         #break       #No jumps available.     #Hence pathSum can be     #from current array only     if (k == 0):         dp[on][i][k] = current +                        pathSum(a, b, c,                                i + 1, n, k, on)         return dp[on][i][k]       #Since jumps are available     #pathSum can be from current     #or adjacent array     if on == 0:         sum = current + max(pathSum(a, b, c, i + 1,                                     n, k - 1, 1),                             pathSum(a, b, c,                                     i + 1, n, k, 0))               #break     if on == 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     if on == 2:         sum = current + max(pathSum(a, b, c, i + 1,                                     n, k - 1, 1),                             pathSum(a, b, c, i + 1,                                     n, k, 2))               #break     dp[on][i][k] = sum       return sum   def findMaxSum(a, b, c, n, k):     sum = 0       #Creating the DP array for memoization     initializeDp()       #Find the pathSum using recursive approach     for i in range(3):           #Maximise the sum         sum = max(sum, pathSum(a, b, c, 0, n, k, i))       print(sum)   #Driver Code if __name__ == '__main__':     n = 5     k = 1     A = [4, 5, 1, 2, 10]     B = [9, 7, 3, 20, 16]     C = [6, 12, 13, 9, 8]     findMaxSum(A, B, C, n, k)   #This code is contributed by Mohit Kumar 29

## 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 memoization     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

## Javascript



Output

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:

## C++

 #include #include using namespace std;   #define M 3 #define N 5 #define K 2   // Dp to store computation of previous subproblems int dp[M][N][K];   // Function to calculate maximum path sum void findMaxSum(int* a, int* b,                 int* c, int n, int k) {     // Initializing dp table with base case values     for (int i = 0; i < M; i++)         for (int j = 0; j < N; j++)             for (int x = 0; x <= k; x++)                 dp[i][j][x] = 0;       // dp[i][j][x] stores the maximum sum     // using at most x jumps till index j     // of the i-th array     for (int x = 0; x <= k; x++) {         for (int j = n - 1; j >= 0; j--) {             for (int i = 0; i < M; i++) {                 int current;                 switch (i) {                 case 0:                     current = a[j];                     break;                 case 1:                     current = b[j];                     break;                 case 2:                     current = c[j];                     break;                 }                   if (j == n - 1) {                     // Base case when at the last index                     dp[i][j][x] = current;                 }                 else if (x == 0) {                     // If no jumps available, path can only                     // continue from current array                     dp[i][j][x] = current + dp[i][j + 1][x];                 }                 else {                     int nextMax = 0;                     // Calculating maximum sum possible after                     // making a jump from the current array                     for (int next = 0; next < M; next++) {                         if (next != i) {                             nextMax = max(nextMax, dp[next][j + 1][x - 1]);                         }                         else {                             nextMax = max(nextMax, dp[next][j + 1][x]);                         }                     }                     dp[i][j][x] = current + nextMax;                 }             }         }     }       // Maximum sum is the maximum value in the dp table     int maxSum = 0;     for (int i = 0; i < M; i++) {         maxSum = max(maxSum, dp[i][0][k]);     }       cout << maxSum << endl; }     // 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 };             // function call     findMaxSum(A, B, C, n, k);       return 0; }

## Java

 import java.util.*;   public class Main {   static final int M = 3;   static final int N = 5;   static final int K = 2;     // Dp to store computation of previous subproblems   static int[][][] dp = new int[M][N][K + 1];     // Function to calculate maximum path sum   static void findMaxSum(int[] a, int[] b, int[] c, int n, int k)   {       // Initializing dp table with base case values     for (int i = 0; i < M; i++) {       for (int j = 0; j < N; j++) {         for (int x = 0; x <= k; x++) {           dp[i][j][x] = 0;         }       }     }       // dp[i][j][x] stores the maximum sum     // using at most x jumps till index j     // of the i-th array     for (int x = 0; x <= k; x++) {       for (int j = n - 1; j >= 0; j--) {         for (int i = 0; i < M; i++) {           int current;           switch (i) {             case 0:               current = a[j];               break;             case 1:               current = b[j];               break;             case 2:               current = c[j];               break;             default:               current = 0;               break;           }             if (j == n - 1) {             // Base case when at the last index             dp[i][j][x] = current;           } else if (x == 0) {             // If no jumps available, path can only             // continue from current array             dp[i][j][x] = current + dp[i][j + 1][x];           } else {             int nextMax = 0;             // Calculating maximum sum possible after             // making a jump from the current array             for (int next = 0; next < M; next++) {               if (next != i) {                 nextMax = Math.max(nextMax, dp[next][j + 1][x - 1]);               } else {                 nextMax = Math.max(nextMax, dp[next][j + 1][x]);               }             }             dp[i][j][x] = current + nextMax;           }         }       }     }       // Maximum sum is the maximum value in the dp table     int maxSum = 0;     for (int i = 0; i < M; i++) {       maxSum = Math.max(maxSum, dp[i][0][k]);     }       System.out.println(maxSum);   }     // 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 };       // function call     findMaxSum(A, B, C, n, k);   } }

## Python3

 # Function to calculate maximum path sum def findMaxSum(a, b, c, n, k):     # Initialize a 3D DP table with zeros     dp = [[[0 for _ in range(k + 1)] for _ in range(n)] for _ in range(3)]       # dp[i][j][x] stores the maximum sum using at most x jumps till index j of the i-th array     for x in range(k + 1):         for j in range(n - 1, -1, -1):             for i in range(3):                 current = 0                 if i == 0:                     current = a[j]                 elif i == 1:                     current = b[j]                 elif i == 2:                     current = c[j]                   if j == n - 1:                     # Base case when at the last index                     dp[i][j][x] = current                 elif x == 0:                     # If no jumps available, path can only continue from the current array                     dp[i][j][x] = current + dp[i][j + 1][x]                 else:                     nextMax = 0                     # Calculating maximum sum possible after making a jump from the current array                     for nextArr in range(3):                         if nextArr != i:                             nextMax = max(nextMax, dp[nextArr][j + 1][x - 1])                         else:                             nextMax = max(nextMax, dp[nextArr][j + 1][x])                     dp[i][j][x] = current + nextMax       # Maximum sum is the maximum value in the dp table     maxSum = 0     for i in range(3):         maxSum = max(maxSum, dp[i][0][k])       print(maxSum)   # Driver Code if __name__ == "__main__":     n = 5     k = 1     A = [4, 5, 1, 2, 10]     B = [9, 7, 3, 20, 16]     C = [6, 12, 13, 9, 8]           # Function call     findMaxSum(A, B, C, n, k)       # This code is contributed by Dwaipayan Bandyopadhyay

## C#

 using System;   class Program {     const int M = 3;     const int N = 5;     const int K = 2;       // Dp to store computation of previous subproblems     static int[,,] dp = new int[M, N, K + 1];       // Function to calculate maximum path sum     static void FindMaxSum(int[] a, int[] b, int[] c, int n, int k)     {         // Initializing dp table with base case values         for (int i = 0; i < M; i++)         {             for (int j = 0; j < N; j++)             {                 for (int x = 0; x <= k; x++)                 {                     dp[i, j, x] = 0;                 }             }         }           // dp[i, j, x] stores the maximum sum         // using at most x jumps till index j         // of the i-th array         for (int x = 0; x <= k; x++)         {             for (int j = n - 1; j >= 0; j--)             {                 for (int i = 0; i < M; i++)                 {                     int current = 0; // Initialize current here                       switch (i)                     {                         case 0:                             current = a[j];                             break;                         case 1:                             current = b[j];                             break;                         case 2:                             current = c[j];                             break;                     }                       if (j == n - 1)                     {                         // Base case when at the last index                         dp[i, j, x] = current;                     }                     else if (x == 0)                     {                         // If no jumps available, path can only                         // continue from the current array                         dp[i, j, x] = current + dp[i, j + 1, x];                     }                     else                     {                         int nextMax = 0;                         // Calculating maximum sum possible after                         // making a jump from the current array                         for (int next = 0; next < M; next++)                         {                             if (next != i)                             {                                 nextMax = Math.Max(nextMax, dp[next, j + 1, x - 1]);                             }                             else                             {                                 nextMax = Math.Max(nextMax, dp[next, j + 1, x]);                             }                         }                         dp[i, j, x] = current + nextMax;                     }                 }             }         }           // Maximum sum is the maximum value in the dp table         int maxSum = 0;         for (int i = 0; i < M; i++)         {             maxSum = Math.Max(maxSum, dp[i, 0, k]);         }           Console.WriteLine(maxSum);     }       // Driver Code     static void Main()     {         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 };           // Function call         FindMaxSum(A, B, C, n, k);     } }     // This code is contributed by rambabuguphka

## Javascript

 const M = 3; const N = 5; const K = 2;   // Dp to store computation of previous subproblems const dp = new Array(M).fill(0).map(() => new Array(N).fill(0).map(() => new Array(K + 1).fill(0)));   // Function to calculate maximum path sum function findMaxSum(a, b, c, n, k) {   // Initializing dp table with base case values   for (let i = 0; i < M; i++) {     for (let j = 0; j < N; j++) {       for (let x = 0; x <= k; x++) {         dp[i][j][x] = 0;       }     }   }     // dp[i][j][x] stores the maximum sum   // using at most x jumps till index j   // of the i-th array   for (let x = 0; x <= k; x++) {     for (let j = n - 1; j >= 0; j--) {       for (let i = 0; i < M; i++) {         let current;         switch (i) {           case 0:             current = a[j];             break;           case 1:             current = b[j];             break;           case 2:             current = c[j];             break;         }           if (j == n - 1) {           // Base case when at the last index           dp[i][j][x] = current;         } else if (x == 0) {           // If no jumps available, path can only           // continue from current array           dp[i][j][x] = current + dp[i][j + 1][x];         } else {           let nextMax = 0;           // Calculating maximum sum possible after           // making a jump from the current array           for (let next = 0; next < M; next++) {             if (next != i) {               nextMax = Math.max(nextMax, dp[next][j + 1][x - 1]);             } else {               nextMax = Math.max(nextMax, dp[next][j + 1][x]);             }           }           dp[i][j][x] = current + nextMax;         }       }     }   }     // Maximum sum is the maximum value in the dp table   let maxSum = 0;   for (let i = 0; i < M; i++) {     maxSum = Math.max(maxSum, dp[i][0][k]);   }     console.log(maxSum); }   // Driver Code const n = 5; const k = 1; const A = [4, 5, 1, 2, 10]; const B = [9, 7, 3, 20, 16]; const C = [6, 12, 13, 9, 8];   // function call findMaxSum(A, B, C, n, k);

Output:

67

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

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