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];
Implementation of naive recursion code:
C++
#include <algorithm> // Required for std::max#include <iostream>#include <vector>int MaximumPath(std::vector<std::vector<int> >& Mat, int i, int j){ int N = Mat.size(); // If we reached the first row of the matrix, return the // value of that element if (i == N - 1 && j == N - 1) return Mat[i][j]; // Out of matrix bound if (i >= N || i < 0 || j >= N || j < 0) return 0; // Call all the rest positions that we reached from the // current position and find the maximum between them // and add the current value in that path return std::max( std::max(MaximumPath(Mat, i + 1, j), MaximumPath(Mat, i + 1, j - 1)), MaximumPath(Mat, i + 1, j + 1)) + Mat[i][j];}int main(){ // Example usage std::vector<std::vector<int> > matrix = { { 4, 2, 3, 4 }, { 2, 9, 1, 10 }, { 15, 1, 3, 0 }, { 16, 92, 41, 44 } }; int N = matrix.size(); int result = MaximumPath(matrix, 0, 0); std::cout << "Maximum Path Sum: " << result << std::endl; return 0;} |
Java
import java.util.*;class Main { public static int maximumPath(int[][] mat, int i, int j) { int N = mat.length; if (i == N - 1 && j == N - 1) return mat[i][j]; if (i >= N || i < 0 || j >= N || j < 0) return 0; return Math.max( Math.max(maximumPath(mat, i + 1, j), maximumPath(mat, i + 1, j - 1)), maximumPath(mat, i + 1, j + 1)) + mat[i][j]; } public static void main(String[] args) { int[][] matrix = { {4, 2, 3, 4}, {2, 9, 1, 10}, {15, 1, 3, 0}, {16, 92, 41, 44} }; int N = matrix.length; int result = maximumPath(matrix, 0, 0); System.out.println("Maximum Path Sum: " + result); }} |
Python3
def maximum_path(mat, i, j): n = len(mat) if i == n - 1 and j == n - 1: return mat[i][j] if i >= n or i < 0 or j >= n or j < 0: return 0 return max(maximum_path(mat, i + 1, j), maximum_path(mat, i + 1, j - 1), maximum_path(mat, i + 1, j + 1)) + mat[i][j]# Example usagematrix = [[4, 2, 3, 4], [2, 9, 1, 10], [15, 1, 3, 0], [16, 92, 41, 44]]result = maximum_path(matrix, 0, 0)print("Maximum Path Sum:", result) |
C#
using System;public class GFG{ public static int MaximumPath(int[][] mat, int i, int j) { int N = mat.Length; // If we reached the last row of the matrix, return the // value of that element if (i == N - 1 && j == N - 1) return mat[i][j]; // Out of matrix bound if (i >= N || i < 0 || j >= N || j < 0) return 0; // Call all the rest positions that we reached from the // current position and find the maximum between them // and add the current value in that path return Math.Max( Math.Max(MaximumPath(mat, i + 1, j), MaximumPath(mat, i + 1, j - 1)), MaximumPath(mat, i + 1, j + 1)) + mat[i][j]; } public static void Main() { // Example usage int[][] matrix = { new int[] { 4, 2, 3, 4 }, new int[] { 2, 9, 1, 10 }, new int[] { 15, 1, 3, 0 }, new int[] { 16, 92, 41, 44 } }; int N = matrix.Length; int result = MaximumPath(matrix, 0, 0); Console.WriteLine("Maximum Path Sum: " + result); }} |
Javascript
function MaximumPath(Mat, i, j) { const N = Mat.length; // If we reached the first row of the matrix, return the value of that element if (i === N - 1 && j === N - 1) return Mat[i][j]; // Out of matrix bound if (i >= N || i < 0 || j >= N || j < 0) return 0; // Call all the rest positions that we reached from the // current position and find the maximum between them // and add the current value in that path return Math.max( Math.max(MaximumPath(Mat, i + 1, j), MaximumPath(Mat, i + 1, j - 1)), MaximumPath(Mat, i + 1, j + 1)) + Mat[i][j];}function main() { // Example usage const matrix = [ [4, 2, 3, 4], [2, 9, 1, 10], [15, 1, 3, 0], [16, 92, 41, 44] ]; const N = matrix.length; const result = MaximumPath(matrix, 0, 0); console.log("Maximum Path Sum: " + result);}main(); |
Output
Maximum Path Sum: 120
Time complexity : O(2N)
Auxiliary space: O(N)
Memoization DP:
In memoization DP, we introduced a dp table, which is a 2D vector of integers. The dp table is initialized with -1 to indicate that the results for each cell in the matrix haven’t been calculated yet.
During the recursive calls to the MaximumPath function, we check if the result for the current position (i, j) is already memoized in the dp table. If it is, we return the memoized result instead of recalculating it.
After calculating the maximum path sum for a position (i, j), we store the result in the corresponding cell of the dp table dp[i][j]. This way, we avoid redundant calculations for the same position and improve the efficiency of the algorithm.
Using memoization with a DP table helps reduce the time complexity of the algorithm by avoiding unnecessary recursive calls and reusing previously calculated results. It ensures that each subproblem is solved only once, leading to a more efficient solution for larger matrices.
C++
#include <algorithm> // Required for std::max#include <iostream>#include <vector>int MaximumPath(std::vector<std::vector<int> >& Mat, int i, int j, std::vector<std::vector<int> >& dp){ int N = Mat.size(); // If we reached the first row of the matrix, return the // value of that element if (i == N - 1 && j == N - 1) return Mat[i][j]; // Out of matrix bound if (i >= N || i < 0 || j >= N || j < 0) return 0; if (dp[i][j] != -1) return dp[i][j]; // Call all the rest positions that we reached from the // current position and find the maximum between them // and add the current value in that path return dp[i][j] = std::max( std::max( MaximumPath(Mat, i + 1, j, dp), MaximumPath(Mat, i + 1, j - 1, dp)), MaximumPath(Mat, i + 1, j + 1, dp)) + Mat[i][j];}int main(){ // Example usage std::vector<std::vector<int> > matrix = { { 4, 2, 3, 4 }, { 2, 9, 1, 10 }, { 15, 1, 3, 0 }, { 16, 92, 41, 44 } }; int N = matrix.size(); std::vector<std::vector<int> > dp( N, std::vector<int>(N, -1)); int result = MaximumPath(matrix, 0, 0, dp); std::cout << "Maximum Path Sum: " << result << std::endl; return 0;} |
Java
import java.util.Arrays;import java.util.List;public class MaximumPathSum { public static int maximumPath(List<List<Integer>> matrix, int i, int j, int[][] dp) { int n = matrix.size(); if (i == n - 1 && j == n - 1) return matrix.get(i).get(j); if (i >= n || i < 0 || j >= n || j < 0) return 0; if (dp[i][j] != -1) return dp[i][j]; int maxPath = Math.max(Math.max(maximumPath(matrix, i + 1, j, dp), maximumPath(matrix, i + 1, j - 1, dp)), maximumPath(matrix, i + 1, j + 1, dp)); dp[i][j] = maxPath + matrix.get(i).get(j); return dp[i][j]; } public static void main(String[] args) { List<List<Integer>> matrix = Arrays.asList( Arrays.asList(4, 2, 3, 4), Arrays.asList(2, 9, 1, 10), Arrays.asList(15, 1, 3, 0), Arrays.asList(16, 92, 41, 44) ); int n = matrix.size(); int[][] dp = new int[n][n]; for (int[] row : dp) { Arrays.fill(row, -1); } int result = maximumPath(matrix, 0, 0, dp); System.out.println("Maximum Path Sum: " + result); }} |
Python3
def maximum_path(matrix, i, j, dp): n = len(matrix) if i == n - 1 and j == n - 1: return matrix[i][j] if i >= n or i < 0 or j >= n or j < 0: return 0 if dp[i][j] != -1: return dp[i][j] max_path = max( max(maximum_path(matrix, i + 1, j, dp), maximum_path(matrix, i + 1, j - 1, dp)), maximum_path(matrix, i + 1, j + 1, dp)) dp[i][j] = max_path + matrix[i][j] return dp[i][j]# Example usagematrix = [[4, 2, 3, 4], [2, 9, 1, 10], [15, 1, 3, 0], [16, 92, 41, 44]]n = len(matrix)dp = [[-1] * n for _ in range(n)]result = maximum_path(matrix, 0, 0, dp)print("Maximum Path Sum:", result) |
C#
using System;using System.Collections.Generic;public class GFG{ public static int MaximumPath(List<List<int>> matrix, int i, int j, int[,] dp) { int n = matrix.Count; // If we reached the first row of the matrix, return the // value of that element if (i == n - 1 && j == n - 1) return matrix[i][j]; // Out of matrix bound if (i >= n || i < 0 || j >= n || j < 0) return 0; if (dp[i, j] != -1) return dp[i, j]; // Call all the rest positions that we reached from the // current position and find the maximum between them // and add the current value in that path int maxPath = Math.Max(Math.Max(MaximumPath(matrix, i + 1, j, dp), MaximumPath(matrix, i + 1, j - 1, dp)), MaximumPath(matrix, i + 1, j + 1, dp)); dp[i, j] = maxPath + matrix[i][j]; return dp[i, j]; } public static void Main(string[] args) { // Example usage List<List<int>> matrix = new List<List<int>>() { new List<int>() { 4, 2, 3, 4 }, new List<int>() { 2, 9, 1, 10 }, new List<int>() { 15, 1, 3, 0 }, new List<int>() { 16, 92, 41, 44 } }; int n = matrix.Count; int[,] dp = new int[n, n]; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { dp[i, j] = -1; } } int result = MaximumPath(matrix, 0, 0, dp); Console.WriteLine("Maximum Path Sum: " + result); }} |
Javascript
function maximum_path(matrix, i, j, dp) { const n = matrix.length; // Base case: If we reach the bottom-right corner, return the matrix value. if (i === n - 1 && j === n - 1) { return matrix[i][j]; } // Base case: If we go out of bounds, return 0. if (i >= n || i < 0 || j >= n || j < 0) { return 0; } // If the value has already been calculated, return it from dp array. if (dp[i][j] !== -1) { return dp[i][j]; } // Calculate the maximum path by recursively exploring three directions. const maxPath = Math.max( maximum_path(matrix, i + 1, j, dp), maximum_path(matrix, i + 1, j - 1, dp), maximum_path(matrix, i + 1, j + 1, dp) ); // Store the result in the dp array and return it. dp[i][j] = maxPath + matrix[i][j]; return dp[i][j];}// Example usageconst matrix = [ [4, 2, 3, 4], [2, 9, 1, 10], [15, 1, 3, 0], [16, 92, 41, 44]];const n = matrix.length;const dp = Array.from({ length: n }, () => Array(n).fill(-1));const result = maximum_path(matrix, 0, 0, dp);console.log("Maximum Path Sum:", result); |
Output
Maximum Path Sum: 120
Time complexity : O(N2)
Auxiliary space: O(N2)
Tabulation DP:
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.
Implementation:
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 pathint MaximumPath(int Mat[][N]){ int result = 0 ; // create 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 row into // 'dp' first row 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 functionint 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.*;public class GFG { private static int N = 4; // function find maximum sum path private static int maximumPath(int Mat[][]){ int result = 0; // create 2D matrix to store the sum // of the path int dp[][] = new int[N][N + 2]; // copy all element of first row into // 'dp' first row 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 pathdef 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 row into # dp first row 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 functionMat = [[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; // create 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 row into // 'dp' first row 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. |
Javascript
<script>// Javascript program to find Maximum path sum// start any column in row '0' and ends// up to any column in row 'n-1' let N = 4; // function find maximum sum path function MaximumPath(Mat) { let result = 0; // create 2D matrix to store the sum // of the path let dp = new Array(N); // initialize all dp matrix as '0' for(let i=0;i<N;i++) { dp[i]=new Array(N+2); for(let j=0;j<N+2;j++) { dp[i][j]=0; } } // copy all element of first row into // 'dp' first row for (let i = 0; i < N; i++) dp[0][i + 1] = Mat[0][i]; for (let i = 1; i < N; i++) for (let 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 (let i = 0; i <= N; i++) result = Math.max(result, dp[N - 1][i]); // return maximum sum path return result; } // driver code let Mat = [[4, 2, 3, 4], [2, 9, 1, 10], [15, 1, 3, 0], [16, 92, 41, 44]] document.write(MaximumPath(Mat)) // This code is contributed by rag2127 </script> |
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'$N = 4;// function find maximum sum pathfunction MaximumPath(&$Mat){ global $N; $result = 0; // create 2D matrix to store the sum // of the path $dp = array_fill(0, $N, array_fill(0, $N + 2, NULL)); // copy all element of first row // into 'dp' first row for ($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)
Auxiliary space: O(N2)
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