Given a N X N matrix Mat[N][N] of positive integers. There are only three possible moves from a cell (i, j)
- (i+1, j)
- (i+1, j-1)
- (i+1, j+1)
Starting from any column in row 0, return the largest sum of any of the paths up to row N-1.
Examples:
Input : mat[4][4] = { {4, 2, 3, 4}, {2, 9, 1, 10}, {15, 1, 3, 0}, {16, 92, 41, 44} }; Output :120 path : 4 + 9 + 15 + 92 = 120
Asked in: Amazon interview
The above problem can be recursively defined.
Let initial position be MaximumPathSum(N-1, j), where j varies from 0 to N-1. We return maximum value between all path that we start traversing (N-1, j) [ where j varies from 0 to N-1]
i = N-1, j = 0 to N -1 int MaximumPath(Mat[][N], I, j) // IF we reached to first row of // matrix then return value of that // element IF ( i == 0 && j = 0 ) return Mat[i][j] // out of matrix bound IF( i = N || j < 0 ) return 0; // call all rest position that we reached // from current position and find maximum // between them and add current value in // that path return max(MaximumPath(Mat, i-1, j), MaximumPath(Mat, i-1, j-1), MaximumPath(Mat, i-1, j+1))) + Mat[i][j];
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 Maximum path sum // start any column in row '0' and ends // up to any column in row 'n-1' #include<bits/stdc++.h> using namespace std; #define N 4 // function find maximum sum path int MaximumPath( int Mat[][N]) { int result = 0 ; // creat 2D matrix to store the sum // of the path int dp[N][N+2]; // initialize all dp matrix as '0' memset (dp, 0, sizeof (dp)); // copy all element of first column into // 'dp' first column for ( int i = 0 ; i < N ; i++) dp[0][i+1] = Mat[0][i]; for ( int i = 1 ; i < N ; i++) for ( int j = 1 ; j <= N ; j++) dp[i][j] = max(dp[i-1][j-1], max(dp[i-1][j], dp[i-1][j+1])) + Mat[i][j-1] ; // Find maximum path sum that end ups // at any column of last row 'N-1' for ( int i=0; i<=N; i++) result = max(result, dp[N-1][i]); // return maximum sum path return result ; } // driver program to test above function int main() { int Mat[4][4] = { { 4, 2 , 3 , 4 }, { 2 , 9 , 1 , 10}, { 15, 1 , 3 , 0 }, { 16 ,92, 41, 44 } }; cout << MaximumPath ( Mat ) <<endl ; return 0; } |
Java
// Java program to find Maximum path sum // start any column in row '0' and ends // up to any column in row 'n-1' import java.util.*; class GFG { static int N = 4 ; // function find maximum sum path static int MaximumPath( int Mat[][]) { int result = 0 ; // creat 2D matrix to store the sum // of the path int dp[][] = new int [N][N + 2 ]; // initialize all dp matrix as '0' for ( int [] rows : dp) Arrays.fill(rows, 0 ); // copy all element of first column into // 'dp' first column for ( int i = 0 ; i < N; i++) dp[ 0 ][i + 1 ] = Mat[ 0 ][i]; for ( int i = 1 ; i < N; i++) for ( int j = 1 ; j <= N; j++) dp[i][j] = Math.max(dp[i - 1 ][j - 1 ], Math.max(dp[i - 1 ][j], dp[i - 1 ][j + 1 ])) + Mat[i][j - 1 ]; // Find maximum path sum that end ups // at any column of last row 'N-1' for ( int i = 0 ; i <= N; i++) result = Math.max(result, dp[N - 1 ][i]); // return maximum sum path return result; } // driver code public static void main(String arg[]) { int Mat[][] = { { 4 , 2 , 3 , 4 }, { 2 , 9 , 1 , 10 }, { 15 , 1 , 3 , 0 }, { 16 , 92 , 41 , 44 } }; System.out.println(MaximumPath(Mat)); } } // This code is contributed by Anant Agarwal. |
Python3
# Python3 program to find # Maximum path sum # start any column in # row '0' and ends # up to any column in row 'n-1' N = 4 # function find maximum sum path def MaximumPath(Mat): result = 0 # create 2D matrix to store the sum # of the path # initialize all dp matrix as '0' dp = [[ 0 for i in range (N + 2 )] for j in range (N)] # copy all element of first column into # dp first column for i in range (N): for j in range ( 1 , N + 1 ): dp[i][j] = max (dp[i - 1 ][j - 1 ], max (dp[i - 1 ][j], dp[i - 1 ][j + 1 ])) + \ Mat[i][j - 1 ] # Find maximum path sum that end ups # at any column of last row 'N-1' for i in range (N + 1 ): result = max (result, dp[N - 1 ][i]) # return maximum sum path return result # driver program to test above function Mat = [[ 4 , 2 , 3 , 4 ], [ 2 , 9 , 1 , 10 ], [ 15 , 1 , 3 , 0 ], [ 16 , 92 , 41 , 44 ]] print (MaximumPath(Mat)) # This code is contributed by Soumen Ghosh. |
C#
// C# program to find Maximum path sum // start any column in row '0' and ends // up to any column in row 'n-1' using System; class GFG { static int N = 4; // function find maximum sum path static int MaximumPath( int [,] Mat) { int result = 0; // creat 2D matrix to store the sum // of the path int [,]dp = new int [N,N + 2]; // initialize all dp matrix as '0' //for (int[] rows : dp) // Arrays.fill(rows, 0); // copy all element of first column into // 'dp' first column for ( int i = 0; i < N; i++) dp[0,i + 1] = Mat[0,i]; for ( int i = 1; i < N; i++) for ( int j = 1; j <= N; j++) dp[i,j] = Math.Max(dp[i - 1,j - 1], Math.Max(dp[i - 1,j], dp[i - 1,j + 1])) + Mat[i,j - 1]; // Find maximum path sum that end ups // at any column of last row 'N-1' for ( int i = 0; i <= N; i++) result = Math.Max(result, dp[N - 1,i]); // return maximum sum path return result; } // driver code public static void Main() { int [,]Mat = { { 4, 2, 3, 4 }, { 2, 9, 1, 10 }, { 15, 1, 3, 0 }, { 16, 92, 41, 44 } }; Console.WriteLine(MaximumPath(Mat)); } } // This code is contributed by Ryuga. |
PHP
<?php // PHP program to find Maximum path sum // start any column in row '0' and ends // up to any column in row 'n-1' $N = 4; // function find maximum sum path function MaximumPath(& $Mat ) { global $N ; $result = 0; // creat 2D matrix to store the sum // of the path $dp = array_fill (0, $N , array_fill (0, $N + 2, NULL)); // copy all element of first column // into 'dp' first column for ( $i = 0 ; $i < $N ; $i ++) $dp [0][ $i + 1] = $Mat [0][ $i ]; for ( $i = 1 ; $i < $N ; $i ++) for ( $j = 1 ; $j <= $N ; $j ++) $dp [ $i ][ $j ] = max( $dp [ $i - 1][ $j - 1], max( $dp [ $i - 1][ $j ], $dp [ $i - 1][ $j + 1])) + $Mat [ $i ][ $j - 1] ; // Find maximum path sum that end ups // at any column of last row 'N-1' for ( $i = 0; $i <= $N ; $i ++) $result = max( $result , $dp [ $N - 1][ $i ]); // return maximum sum path return $result ; } // Driver Code $Mat = array ( array (4, 2 , 3 , 4), array (2 , 9 , 1 , 10), array (15, 1 , 3 , 0), array (16 ,92, 41, 44)); echo MaximumPath ( $Mat ) . "\n" ; // This code is contributed by ita_c ?> |
Output:
120
Time complexity : O(N2)
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