Given a matrix of integers where every element represents the weight of the cell. Find the path having the maximum weight in matrix [N X N]. Path Traversal Rules are:
- It should begin from top left element.
- The path can end at any element of last row.
- We can move to follow two cells from a cell (i, j).
- Down Move : (i+1, j)
- Diagonal Move : (i+1, j+1)
Examples:
Input : N = 5
mat[5][5] = {{ 4, 2 ,3 ,4 ,1 },
{ 2 , 9 ,1 ,10 ,5 },
{15, 1 ,3 , 0 ,20 },
{16 ,92, 41, 44 ,1},
{8, 142, 6, 4, 8} };
Output : 255
Path with max weight : 4 + 2 +15 + 92 + 142 = 255
The above problem can be recursively defined.
Let maxCost(i, j) be the cost maximum cost to
reach mat[i][j]. Since end point can be any point
in last row, we finally return maximum of all values
maxCost(N-1, j) where j varies from 0 to N-1.
If i == 0 and j == 0
maxCost(0, 0) = mat[0][0]
// We can traverse through first column only by
// down move
Else if j = 0
maxCost(i, 0) = maxCost(i-1, 0) + mat[i][0]
// In other cases, a cell mat[i][j] can be reached
// through previous two cells ma[i-1][j] and
// mat[i-1][j-1]
Else
maxCost(i, j) = mat[i][j] + max(maxCost(i-1, j),
maxCost(i-1, j-1)),
If we draw the recursion tree of above recursive solution, we can observe overlapping subproblems. Since the problem has overlapping subproblems, we can solve it efficiently using Dynamic Programming. Below is Dynamic Programming based solution.
Implementation:
C++
// C++ program to find the path having the// maximum weight in matrix#include<bits/stdc++.h>using namespace std;const int MAX = 1000;/* Function which return the maximum weight path sum */int maxCost(int mat[][MAX], int N){ // create 2D matrix to store the sum of the path int dp[N][N]; memset(dp, 0, sizeof(dp)); dp[0][0] = mat[0][0]; // Initialize first column of total weight // array (dp[i to N][0]) for (int i=1; i<N; i++) dp[i][0] = mat[i][0] + dp[i-1][0]; // Calculate rest path sum of weight matrix for (int i=1; i<N; i++) for (int j=1; j<i+1&&j<N; j++) dp[i][j] = mat[i][j] + max(dp[i-1][j-1], dp[i-1][j]); // find the max weight path sum to reach // the last row int result = 0; for (int i=0; i<N; i++) if (result < dp[N-1][i]) result = dp[N-1][i]; // return maximum weight path sum return result;}// Driver programint main(){ int mat[MAX][MAX] = { { 4, 1 ,5 ,6 , 1 }, { 2 ,9 ,2 ,11 ,10 }, { 15,1 ,3 ,15, 2 }, { 16, 92, 41,4,3}, { 8, 142, 6, 4, 8 } }; int N = 5; cout << "Maximum Path Sum : " << maxCost(mat, N)<<endl; return 0;} |
C
// C program to find the path having the// maximum weight in matrix#include <stdio.h>#include <string.h>#define MAX 1000int max(int a, int b){ int max = a; if(max<b) max = b; return max;}/* Function which return the maximum weight path sum */int maxCost(int mat[][MAX], int N){ // create 2D matrix to store the sum of the path int dp[N][N]; memset(dp, 0, sizeof(dp)); dp[0][0] = mat[0][0]; // Initialize first column of total weight // array (dp[i to N][0]) for (int i=1; i<N; i++) dp[i][0] = mat[i][0] + dp[i-1][0]; // Calculate rest path sum of weight matrix for (int i=1; i<N; i++) for (int j=1; j<i+1&&j<N; j++) dp[i][j] = mat[i][j] + max(dp[i-1][j-1], dp[i-1][j]); // find the max weight path sum to reach // the last row int result = 0; for (int i=0; i<N; i++) if (result < dp[N-1][i]) result = dp[N-1][i]; // return maximum weight path sum return result;}// Driver programint main(){ int mat[MAX][MAX] = { { 4, 1 ,5 ,6 , 1 }, { 2 ,9 ,2 ,11 ,10 }, { 15,1 ,3 ,15, 2 }, { 16, 92, 41,4,3}, { 8, 142, 6, 4, 8 } }; int N = 5; printf("Maximum Path Sum : %d\n",maxCost(mat, N)); return 0;}// This code is contributed by kothavvsaakash. |
Java
// Java Code for Maximum weight path ending at // any element of last row in a matriximport java.util.*;class GFG { /* Function which return the maximum weight path sum */ public static int maxCost(int mat[][], int N) { // create 2D matrix to store the sum of // the path int dp[][]=new int[N][N]; dp[0][0] = mat[0][0]; // Initialize first column of total // weight array (dp[i to N][0]) for (int i = 1; i < N; i++) dp[i][0] = mat[i][0] + dp[i-1][0]; // Calculate rest path sum of weight matrix for (int i = 1; i < N; i++) for (int j = 1; j < i + 1 && j < N; j++) dp[i][j] = mat[i][j] + Math.max(dp[i-1][j-1], dp[i-1][j]); // find the max weight path sum to reach // the last row int result = 0; for (int i = 0; i < N; i++) if (result < dp[N-1][i]) result = dp[N-1][i]; // return maximum weight path sum return result; } /* Driver program to test above function */ public static void main(String[] args) { int mat[][] = { { 4, 1 ,5 ,6 , 1 }, { 2 ,9 ,2 ,11 ,10 }, { 15,1 ,3 ,15, 2 }, { 16, 92, 41,4,3}, { 8, 142, 6, 4, 8 } }; int N = 5; System.out.println("Maximum Path Sum : "+ maxCost(mat, N)); }}// This code is contributed by Arnav Kr. Mandal. |
Python3
# Python3 program to find the path # having the maximum weight in matrixMAX = 1000# Function which return the # maximum weight path sumdef maxCost(mat, N): # create 2D matrix to store the sum of the path dp = [[0 for i in range(N)] for j in range(N)] dp[0][0] = mat[0][0] # Initialize first column of total weight # array (dp[i to N][0]) for i in range(1, N): dp[i][0] = mat[i][0] + dp[i - 1][0] # Calculate rest path sum of weight matrix for i in range(1, N): for j in range(1, min(i + 1, N)): dp[i][j] = mat[i][j] + \ max(dp[i - 1][j - 1], dp[i - 1][j]) # find the max weight path sum to reach # the last row result = 0 for i in range(N): if (result < dp[N - 1][i]): result = dp[N - 1][i] # return maximum weight path sum return result# Driver Programmat = [ [4, 1 ,5 ,6 , 1], [2 ,9 ,2 ,11 ,10], [15,1 ,3 ,15, 2], [16, 92, 41,4,3], [8, 142, 6, 4, 8]]N = 5print('Maximum Path Sum :', maxCost(mat, N))# This code is contributed by Soumen Ghosh. |
C#
// C# Code for Maximum weight path// ending at any element of last// row in a matrixusing System;class GFG { /* Function which return the maximum weight path sum */ public static int maxCost(int [,] mat, int N) { // create 2D matrix to store the // sum of the path int [,] dp = new int[N,N]; dp[0,0] = mat[0,0]; // Initialize first column of total // weight array (dp[i to N][0]) for (int i = 1; i < N; i++) dp[i,0] = mat[i,0] + dp[i-1,0]; // Calculate rest path sum of weight matrix for (int i = 1; i < N; i++) for (int j = 1; j < i + 1 && j < N; j++) dp[i,j] = mat[i,j] + Math.Max(dp[i-1,j-1], dp[i-1,j]); // find the max weight path sum to reach // the last row int result = 0; for (int i = 0; i < N; i++) if (result < dp[N-1,i]) result = dp[N-1,i]; // return maximum weight path sum return result; } /* Driver program to test above function */ public static void Main() { int [,] mat = { { 4, 1 ,5 ,6 , 1 }, { 2 ,9 ,2 ,11 ,10 }, { 15,1 ,3 ,15, 2 }, { 16, 92, 41,4,3}, { 8, 142, 6, 4, 8 } }; int N = 5; Console.Write("Maximum Path Sum : " + maxCost(mat, N)); }}// This code is contributed by KRV. |
PHP
<?php// PHP program to find the path having the // maximum weight in matrix /* Function which return the maximum weight path sum */function maxCost($mat, $N) { // create 2D matrix to store the sum of the path $dp = array(array()) ; //memset(dp, 0, sizeof(dp)); $dp[0][0] = $mat[0][0]; // Initialize first column of total weight // array (dp[i to N][0]) for ($i=1; $i<$N; $i++) $dp[$i][0] = $mat[$i][0] + $dp[$i-1][0]; // Calculate rest path sum of weight matrix for ($i=1; $i<$N; $i++) { for ($j=1; $j<$i+1 && $j<$N; $j++) $dp[$i][$j] = $mat[$i][$j] + max($dp[$i-1][$j-1], $dp[$i-1][$j]); } // find the max weight path sum to reach // the last row $result = 0; for ($i=0; $i<$N; $i++) if ($result < $dp[$N-1][$i]) $result = $dp[$N-1][$i]; // return maximum weight path sum return $result; } // Driver program $mat = array( array( 4, 1 ,5 ,6 , 1 ), array( 2 ,9 ,2 ,11 ,10 ), array( 15,1 ,3 ,15, 2 ), array( 16, 92, 41,4,3), array( 8, 142, 6, 4, 8 ) ); $N = 5; echo "Maximum Path Sum : ", maxCost($mat, $N) ; // This code is contributed by Ryuga?> |
Javascript
<script> // JavaScript Code for Maximum weight path ending at // any element of last row in a matrix /* Function which return the maximum weight path sum */ function maxCost(mat, N) { // create 2D matrix to store the sum of // the path let dp = new Array(N); for (let i = 0; i < N; i++) { dp[i] = new Array(N); for (let j = 0; j < N; j++) { dp[i][j] = 0; } } dp[0][0] = mat[0][0]; // Initialize first column of total // weight array (dp[i to N][0]) for (let i = 1; i < N; i++) dp[i][0] = mat[i][0] + dp[i-1][0]; // Calculate rest path sum of weight matrix for (let i = 1; i < N; i++) for (let j = 1; j < i + 1 && j < N; j++) dp[i][j] = mat[i][j] + Math.max(dp[i-1][j-1], dp[i-1][j]); // find the max weight path sum to reach // the last row let result = 0; for (let i = 0; i < N; i++) if (result < dp[N-1][i]) result = dp[N-1][i]; // return maximum weight path sum return result; } let mat = [ [ 4, 1 ,5 ,6 , 1 ], [ 2 ,9 ,2 ,11 ,10 ], [ 15,1 ,3 ,15, 2 ], [ 16, 92, 41,4,3], [ 8, 142, 6, 4, 8 ] ]; let N = 5; document.write("Maximum Path Sum : "+ maxCost(mat, N));</script> |
Output
Maximum Path Sum : 255
Time Complexity: O(N*N)
Auxiliary Space: O(N*N)
Method – 2 O(1) extra space solution
In the above solution, we use an extra dp array to store ans for a particular dp state but here we use input mat as dp so we don’t need that extra N*N space so this solution works in O(1) space complexity.
C++
// C++ program to find the path having the maximum weight in matrix#include<bits/stdc++.h>using namespace std;const int MAX = 1000;// Function which return the maximum weight path sumint maxCost(int mat[][MAX], int N) { // Initialize first column of total weight array (dp[i to N][0]) for (int i = 1; i < N; i++) mat[i][0] = mat[i][0] + mat[i - 1][0]; // Calculate rest path sum of weight matrix for (int i = 1; i < N; i++) { for (int j = 1; j < i + 1 && j < N; j++) { mat[i][j] = mat[i][j] + max(mat[i - 1][j - 1], mat[i - 1][j]); } } // find the max weight path sum to reach the last row int result = 0; for (int i = 0; i < N; i++) { if (result < mat[N - 1][i]) { result = mat[N - 1][i]; } } // return maximum weight path sum return result;}// Driver programint main(){ int mat[MAX][MAX] = { { 4, 1 , 5 , 6 , 1 }, { 2 , 9 , 2 , 11 , 10 }, { 15, 1 , 3 , 15, 2 }, { 16, 92, 41, 4, 3}, { 8, 142, 6, 4, 8 } }; int N = 5; cout << "Maximum Path Sum : " << maxCost(mat, N) << endl; return 0;} |
Java
// Java program to find the path having the maximum weight in matriximport java.io.*;class GFG { public static int MAX = 1000; // Function which return the maximum weight path sum public static int maxCost(int mat[][], int N) { // Initialize first column of total weight array (dp[i to N][0]) for (int i = 1; i < N; i++) mat[i][0] = mat[i][0] + mat[i - 1][0]; // Calculate rest path sum of weight matrix for (int i = 1; i < N; i++) { for (int j = 1; j < i + 1 && j < N; j++) { mat[i][j] = mat[i][j] + Math.max(mat[i - 1][j - 1], mat[i - 1][j]); } } // find the max weight path sum to reach the last row int result = 0; for (int i = 0; i < N; i++) { if (result < mat[N - 1][i]) { result = mat[N - 1][i]; } } // return maximum weight path sum return result; } public static void main(String[] args) { int[][] mat = { { 4, 1 , 5 , 6 , 1 }, { 2 , 9 , 2 , 11 , 10 }, { 15, 1 , 3 , 15, 2 }, { 16, 92, 41, 4, 3}, { 8, 142, 6, 4, 8 } }; int N = 5; System.out.println("Maximum Path Sum : " + maxCost(mat, N)); }}// This code is contributed by ajaymakvana. |
C#
// C# program to find the path having the maximum weight in// matrixusing System;class GFG { // Function which return the maximum weight path sum static int maxCost(int[, ] mat, int N) { // Initialize first column of total weight array // (dp[i to N][0]) for (int i = 1; i < N; i++) mat[i, 0] = mat[i, 0] + mat[i - 1, 0]; // Calculate rest path sum of weight matrix for (int i = 1; i < N; i++) { for (int j = 1; j < i + 1 && j < N; j++) { mat[i, j] = mat[i, j] + Math.Max(mat[i - 1, j - 1], mat[i - 1, j]); } } // find the max weight path sum to reach the last row int result = 0; for (int i = 0; i < N; i++) { if (result < mat[N - 1, i]) { result = mat[N - 1, i]; } } // return maximum weight path sum return result; } static void Main() { int[, ] mat = { { 4, 1, 5, 6, 1 }, { 2, 9, 2, 11, 10 }, { 15, 1, 3, 15, 2 }, { 16, 92, 41, 4, 3 }, { 8, 142, 6, 4, 8 } }; int N = 5; Console.Write("Maximum Path Sum : " + maxCost(mat, N)); }}// This code is contributed by garg28harsh. |
Javascript
// javascript program to find the path having the maximum weight in matrix// Function which return the maximum weight path sumfunction maxCost(mat, N) { // Initialize first column of total weight array (dp[i to N][0]) for (let i = 1; i < N; i++) mat[i][0] = mat[i][0] + mat[i - 1][0]; // Calculate rest path sum of weight matrix for (let i = 1; i < N; i++) { for (let j = 1; j < i + 1 && j < N; j++) { mat[i][j] = mat[i][j] + Math.max(mat[i - 1][j - 1], mat[i - 1][j]); } } // find the max weight path sum to reach the last row let result = 0; for (let i = 0; i < N; i++) { if (result < mat[N - 1][i]) { result = mat[N - 1][i]; } } // return maximum weight path sum return result;}// Driver program let mat = [[ 4, 1 , 5 , 6 , 1 ], [ 2 , 9 , 2 , 11 , 10 ], [ 15, 1 , 3 , 15, 2], [ 16, 92, 41, 4, 3], [ 8, 142, 6, 4, 8 ]]; let N = 5; console.log("Maximum Path Sum : " + maxCost(mat, N)) ; // This code is contributed by garg28harsh. |
Python3
# Python program to find the path having the maximum weight in matrix# Function which return the maximum weight path sumdef maxCost(mat, N): # Initialize first column of total weight array (dp[i to N][0]) for i in range(1, N): mat[i][0] = mat[i][0] + mat[i-1][0] # Calculate rest path sum of weight matrix for i in range(1, N): for j in range(1, i+1): mat[i][j] = mat[i][j] + max(mat[i-1][j-1], mat[i - 1][j]) # find the max weight path sum to reach the last row result = 0 for i in range(N): if result < mat[N-1][i]: result = mat[N-1][i] # return maximum weight path sum return result# Driver programif __name__ == "__main__": mat = [[4, 1, 5, 6, 1], [2, 9, 2, 11, 10], [15, 1, 3, 15, 2], [16, 92, 41, 4, 3], [8, 142, 6, 4, 8] ] N = 5 print("Maximum Path Sum: ", maxCost(mat, N)) |
Output
Maximum Path Sum : 255
Time Complexity : O(N*N)
Auxiliary Space: O(1)
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