Given a matrix of integers where every element represents 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 following 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 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.
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) { // creat 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 paht 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 rech // 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 int 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; } |
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Java
// Java Code for Maximum weight path ending at // any element of last row in a matrix import 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. |
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Python3
# Python3 program to find the path # having the maximum weight in matrix MAX = 1000 # Function which return the # maximum weight path sum def maxCost(mat, N): # creat 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 Program 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 = 5print('Maximum Path Sum :', maxCost(mat, N)) # This code is contributed by Soumen Ghosh. |
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C#
// C# Code for Maximum weight path // ending at any element of last // row in a matrix using 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. |
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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) { // creat 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 rech // 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 ?> |
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Output:
Maximum Path Sum : 255
Time complexity : O(N*N)
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