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 the 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 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 Programing 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 <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 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
<script> // Javascript program to maximum path sum in // the given arrays with at most K jumps let M = 3; let N = 5; let K = 2; let dp= new Array(M); function initializeDp() { for (let i = 0; i < M; i++) { dp[i]= new Array(N); for (let j = 0; j < N; j++) { dp[i][j]= new Array(K); for (let k = 0; k < K; k++) dp[i][j][k] = -1; } } } // Function to calculate maximum path sum function 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]; let 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; } function findMaxSum(a,b,c,n,k) { let sum = 0; // Creating the DP array for memoization initializeDp(); // Find the pathSum using recursive approach for (let i = 0; i < 3; i++) { // Maximise the sum sum = Math.max(sum, pathSum(a, b, c, 0, n, k, i)); } document.write(sum); } // Driver Code let n = 5, k = 1; let A=[4, 5, 1, 2, 10]; let B=[9, 7, 3, 20, 16]; let C=[6, 12, 13, 9, 8 ]; findMaxSum(A, B, C, n, k); // This code is contributed by avanitrachhadiya2155 </script> |
67
Time Complexity: O(N * K)
Auxiliary Space: O(N * K)