Maximum sum path in a matrix from top to bottom
Consider a n*n matrix. Suppose each cell in the matrix has a value assigned. We can go from each cell in row i to a diagonally higher cell in row i+1 only [i.e from cell(i, j) to cell(i+1, j-1) and cell(i+1, j+1) only]. Find the path from the top row to the bottom row following the aforementioned condition such that the maximum sum is obtained.
Examples:
Input : mat[][] = { {5, 6, 1, 7},
{-2, 10, 8, -1},
{3, -7, -9, 11},
{12, -4, 2, 6} }
Output : 28
{5, 6, 1, 7},
{-2, 10, 8, -1},
{3, -7, -9, 11},
{12, -4, 2, 6} }
The highlighted numbers from top to bottom
gives the required maximum sum path.
(7 + 8 + 11 + 2) = 28
Algorithm: The idea is to find maximum sum or all paths starting with every cell of first row and finally return maximum of all values in first row. We use Dynamic Programming as results of many subproblems are needed again and again.
C++
// C++ implementation to find the maximum sum// path in a matrix#include <bits/stdc++.h>using namespace std; #define SIZE 10 // function to find the maximum sum// path in a matricint maxSum(int mat[SIZE][SIZE], int n){ // if there is a single element only if (n == 1) return mat[0][0]; // dp[][] matrix to store the results // of each iteration int dp[n][n]; int maxSum = INT_MIN, max; // base case, copying elements of // last row for (int j = 0; j < n; j++) dp[n - 1][j] = mat[n - 1][j]; // building up the dp[][] matrix from // bottom to the top row for (int i = n - 2; i >= 0; i--) { for (int j = 0; j < n; j++) { max = INT_MIN; // finding the maximum diagonal element in the // (i+1)th row if that cell exists if (((j - 1) >= 0) && (max < dp[i + 1][j - 1])) max = dp[i + 1][j - 1]; if (((j + 1) < n) && (max < dp[i + 1][j + 1])) max = dp[i + 1][j + 1]; // adding that 'max' element to the // mat[i][j] element dp[i][j] = mat[i][j] + max; } } // finding the maximum value from the // first row of dp[][] for (int j = 0; j < n; j++) if (maxSum < dp[0][j]) maxSum = dp[0][j]; // required maximum sum return maxSum;} // Driver program to test aboveint main(){ int mat[SIZE][SIZE] = { { 5, 6, 1, 7 }, { -2, 10, 8, -1 }, { 3, -7, -9, 11 }, { 12, -4, 2, 6 } }; int n = 4; cout << "Maximum Sum = " << maxSum(mat, n); return 0;} |
Java
// Java implementation to find the // maximum sum path in a matriximport java.io.*; class MaxSumPath {//int mat[][]; // function to find the maximum // sum path in a matrixstatic int maxSum(int[][] mat, int n){ // if there is a single element only if (n == 1) return mat[0][0]; // dp[][] matrix to store the results // of each iteration int dp[][] = new int[n][n]; int maxSum = Integer.MIN_VALUE, max; // base case, copying elements of // last row for (int j = 0; j < n; j++) dp[n - 1][j] = mat[n - 1][j]; // building up the dp[][] matrix // from bottom to the top row for (int i = n - 2; i >= 0; i--) { for (int j = 0; j < n; j++) { max = Integer.MIN_VALUE; // finding the maximum diagonal // element in the (i+1)th row // if that cell exists if (((j - 1) >= 0) && (max < dp[i + 1][j - 1])) max = dp[i + 1][j - 1]; if (((j + 1) < n) && (max < dp[i + 1][j + 1])) max = dp[i + 1][j + 1]; // adding that 'max' element // to the mat[i][j] element dp[i][j] = mat[i][j] + max; } } // finding the maximum value from // the first row of dp[][] for (int j = 0; j < n; j++) if (maxSum < dp[0][j]) maxSum = dp[0][j]; // required maximum sum return maxSum;} // Driver code public static void main (String[] args) { int mat[][] = { { 5, 6, 1, 7 }, { -2, 10, 8, -1 }, { 3, -7, -9, 11 }, { 12, -4, 2, 6 } }; int n = 4; System.out.println("Maximum Sum = "+ maxSum(mat , n)); }} // This code is contributed by Prerna Saini |
Python3
# Python3 implementation to find # the maximum sum path in a matrix SIZE=10INT_MIN=-10000000 # function to find the maximum sum# path in a matric def maxSum(mat,n): #if there is a single elementif #there is a single element only only if n==1: return mat[0][0] # dp[][] matrix to store the results # of each iteration dp=[[0 for i in range(n)]for i in range(n)] maxSum=INT_MIN # base case, copying elements of # last row for j in range(n): dp[n - 1][j] = mat[n - 1][j] # building up the dp[][] matrix from # bottom to the top row for i in range(n-2,-1,-1): for j in range(n): maxi=INT_MIN # finding the maximum diagonal # element in the # (i+1)th row if that cell exists if ((((j - 1) >= 0) and (maxi < dp[i + 1][j - 1]))): maxi = dp[i + 1][j - 1] if ((((j + 1) < n) and (maxi < dp[i + 1][j + 1]))): maxi = dp[i + 1][j + 1] # adding that 'max' element to the # mat[i][j] element dp[i][j] = mat[i][j] + maxi # finding the maximum value from the # first row of dp[][] for j in range(n): if (maxSum < dp[0][j]): maxSum = dp[0][j] # required maximum sum return maxSum # Driver program to test above if __name__=='__main__': mat=[[5, 6, 1, 7], [-2, 10, 8, -1], [ 3, -7, -9, 11 ], [12, -4, 2, 6 ]] n=4 print("Maximum Sum=",maxSum(mat,n)) #This code is contributed by sahilshelangia |
C#
// C# implementation to find the // maximum sum path in a matrixusing System; class MaxSumPath { //int mat[][]; // function to find the maximum // sum path in a matrix static int maxSum(int[,] mat, int n) { // if there is a single element only if (n == 1) return mat[0, 0]; // dp[][] matrix to store the results // of each iteration int [,]dp = new int[n, n]; int maxSum = int.MinValue, max; // base case, copying elements of // last row for (int j = 0; j < n; j++) dp[n - 1, j] = mat[n - 1, j]; // building up the dp[][] matrix // from bottom to the top row for (int i = n - 2; i >= 0; i--) { for (int j = 0; j < n; j++) { max = int.MinValue; // finding the maximum diagonal // element in the (i+1)th row // if that cell exists if (((j - 1) >= 0) && (max < dp[i + 1, j - 1])) max = dp[i + 1, j - 1]; if (((j + 1) < n) && (max < dp[i + 1, j + 1])) max = dp[i + 1, j + 1]; // adding that 'max' element // to the mat[i][j] element dp[i, j] = mat[i, j] + max; } } // finding the maximum value from // the first row of dp[][] for (int j = 0; j < n; j++) if (maxSum < dp[0, j]) maxSum = dp[0, j]; // required maximum sum return maxSum; } // Driver code public static void Main () { int [,]mat = { { 5, 6, 1, 7 }, { -2, 10, 8, -1 }, { 3, -7, -9, 11 }, { 12, -4, 2, 6 } }; int n = 4; Console.WriteLine("Maximum Sum = "+ maxSum(mat , n)); }} // This code is contributed by vt_m |
PHP
<?php// PHP implementation to find // the maximum sum path in a matrix $SIZE = 10; // function to find the maximum sum// path in a matricfunction maxSum( $mat, $n){ // if there is a single // element only if ($n == 1) return $mat[0][0]; // dp[][] matrix to store the results // of each iteration $dp = array(array()); $maxSum = PHP_INT_MIN; $max; // base case, copying elements of // last row for($j = 0; $j < $n; $j++) $dp[$n - 1][$j] = $mat[$n - 1][$j]; // building up the dp[][] matrix from // bottom to the top row for ( $i = $n - 2; $i >= 0; $i--) { for ( $j = 0; $j < $n; $j++) { $max = PHP_INT_MIN; // finding the maximum // diagonal element in the // (i+1)th row if that cell // exists if ((($j - 1) >= 0) and ($max < $dp[$i + 1][$j - 1])) $max = $dp[$i + 1][$j - 1]; if ((($j + 1) < $n) and ($max < $dp[$i + 1][$j + 1])) $max = $dp[$i + 1][$j + 1]; // adding that 'max' element to the // mat[i][j] element $dp[$i][$j] = $mat[$i][$j] + $max; } } // finding the maximum value from the // first row of dp[][] for ( $j = 0; $j < $n; $j++) if ($maxSum < $dp[0][$j]) $maxSum = $dp[0][$j]; // required maximum sum return $maxSum;} // Driver Code $mat = array(array(5, 6, 1, 7), array(-2, 10, 8, -1), array(3, -7, -9, 11), array(12, -4, 2, 6)); $n = 4; echo "Maximum Sum = " , maxSum($mat, $n); // This code is contributed by anuj_67.?> |
Output:
Maximum Sum = 28
Time Complexity: O(n2).
Auxiliary Space: O(n2).
Exercise: Try to solve it in auxiliary space of O(n).
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