# Total number of decreasing paths in a matrix

• Difficulty Level : Hard
• Last Updated : 23 Jul, 2021

Given a matrix of size N X N of integers. The task is to find the number of decreasing path in the matrix. You are allowed to start from any cell and from the cell (i, j), you are allowed to move to (i + 1, j), (i – 1, j), (i, j + 1) and (i, j – 1) cell.
Examples:

```Input : m[][] = { { 1, 2 },
{ 1, 3 } }
Output : 8
Explanation : Decreasing paths are { 1 }, { 1 }, { 2 }, { 3 },
{ 2, 1 }, { 3, 1 }, { 3, 2 }, { 3, 2, 1 }

Input : m[][] = { { 1, 2, 3 },
{ 1, 3, 4 },
{ 1, 5, 6 } }
Output : 41```

The idea to solve this problem is to use Dynamic Programming. Declare a dp[][] array, where dp[i][j] stores the number of decreasing path that can be formed from cell (i, j). So, we will define a recursive function to evaluate the number of decreasing paths with parameters, say i, j, the row number and column number of the current cell. Make every possible move from the cell(i,j) and keep a count of the total number of paths. First, we will check in the function that the number of decreasing paths for input position (i, j) is already calculated or not. If yes, return the value dp[i][j] else find the number of decreasing sequence is allowed four directions and return the value. Meanwhile, we will also store the number of decreasing for intermediate cells. Since DP[i][j] stores the number of decreasing paths for every cell, so the summation of all the cells of DP[][] will answer to count of decreasing paths in the complete matrix.
Below is the implementation of the above approach:

## C++

 `// CPP program to count number``// of decreasing path in a matrix``#include ``using` `namespace` `std;``#define MAX 100` `// Function that returns the number of``// decreasing paths from a cell(i, j)``int` `CountDecreasingPathsCell(``int` `mat[MAX][MAX], ``int` `dp[MAX][MAX],``                                              ``int` `n, ``int` `x, ``int` `y)``{``    ``// checking if already calculated``    ``if` `(dp[x][y] != -1)``        ``return` `dp[x][y];` `    ``// all possible paths``    ``int` `delta[4][2] = { { 0, 1 }, { 1, 0 }, { -1, 0 }, { 0, -1 } };``    ``int` `newx, newy;` `    ``// counts the total number of paths``    ``int` `ans = 1;` `    ``// In all four allowed direction.``    ``for` `(``int` `i = 0; i < 4; i++) {` `        ``// new co-ordinates``        ``newx = x + delta[i][0];``        ``newy = y + delta[i][1];` `        ``// Checking if not going out of matrix and next``        ``// cell value is less than current cell value.``        ``if` `(newx >= 0 && newx < n && newy >= 0``            ``&& newy < n && mat[newx][newy] < mat[x][y]) {``            ``ans += CountDecreasingPathsCell(mat, dp, n, newx, newy);``        ``}``    ``}``    ``// function that returns the answer``    ``return` `dp[x][y] = ans;``}` `// Function that counts the total``// decreasing path in the matrix``int` `countDecreasingPathsMatrix(``int` `n,``                               ``int` `mat[MAX][MAX])``{``    ``int` `dp[MAX][MAX];` `    ``// Initialising dp[][] to -1.``    ``for` `(``int` `i = 0; i < n; i++)``        ``for` `(``int` `j = 0; j < n; j++)``            ``dp[i][j] = -1;` `    ``int` `sum = 0;` `    ``// Calculating number of decreasing path from each cell.``    ``for` `(``int` `i = 0; i < n; i++)``        ``for` `(``int` `j = 0; j < n; j++)``            ``sum += CountDecreasingPathsCell(mat, dp, n, i, j);` `    ``return` `sum;``}` `// Driver Code``int` `main()``{``    ``int` `n = 2;` `    ``int` `mat[MAX][MAX] = { { 1, 2 }, { 1, 3 } };``    ``// function call that returns the``    ``// count of decreasing paths in a matrix``    ``cout << countDecreasingPathsMatrix(n, mat)``         ``<< endl;``    ``return` `0;``}`

Java

``````
// Java program to count number
// of decreasing path in a matrix
import java.util.*;
import java.lang.*;
import java.io.*;

class GFG
{
public static Scanner scn =
new Scanner(System.in);

// Function that returns the number of
// decreasing paths from a cell(i, j)
public static int CountDecreasingPathsCell(int mat[][], int dp[][],
int n, int x, int y)
{
if (dp[x][y] != -1)
return dp[x][y];

// all possible paths
int delta[][] = { { 0, 1 }, { 1, 0 },
{ -1, 0}, { 0, -1}};
int newx, newy;

// counts the total
// number of paths
int ans = 1;

// In all four allowed direction.
for (int i = 0; i < 4; i++)
{

// new co-ordinates
newx = x + delta[i][0];
newy = y + delta[i][1];

// Checking if not going out
// of matrix and next cell
// value is less than current
// cell value.
if (newx >= 0 && newx < n && newy >= 0 &&
newy < n && mat[newx][newy] < mat[x][y])
{
ans += CountDecreasingPathsCell(mat, dp, n,
newx, newy);
}
}

// function that
return dp[x][y] = ans;
}

// Function that counts the total
// decreasing path in the matrix
public static int countDecreasingPathsMatrix(int n,
int mat[][])
{
int dp[][] = new int[n][n];

// Initialising dp[][] to -1.
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
dp[i][j] = -1;

int sum = 0;

// Calculating number of
// decreasing path from each cell.
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
sum += CountDecreasingPathsCell(mat, dp,
n, i, j);

return sum;
}

// Driver Code
public static void main(String[] args)
{
int n = 2;

int mat[][]= {{1, 2},
{1, 3}};

// function call that returns the
// count of decreasing paths in a matrix
System.out.println(countDecreasingPathsMatrix(n, mat));

}
}

// This code is contributed by khyati grover

``````

## Python3

 `# Python3 program to count number``# of decreasing path in a matrix``MAX` `=` `100` `# Function that returns the number of``# decreasing paths from a cell(i, j)``def` `CountDecreasingPathsCell(mat, dp, n, x, y):``    ` `    ``# checking if already calculated``    ``if` `(dp[x][y] !``=` `-``1``):``        ``return` `dp[x][y]``        ` `    ``# all possible paths``    ``delta ``=` `[[``0``, ``1``], [``1``, ``0``],``             ``[``-``1``, ``0``], [``0``, ``-``1``]]``    ``newx, newy ``=` `0``, ``0``    ` `    ``# counts the total number of paths``    ``ans ``=` `1``    ` `    ``# In all four allowed direction.``    ``for` `i ``in` `range``(``4``):``        ` `        ``# new co-ordinates``        ``newx ``=` `x ``+` `delta[i][``0``]``        ``newy ``=` `y ``+` `delta[i][``1``]``        ` `        ``# Checking if not going out of matrix and next``        ``# cell value is less than current cell value.``        ``if` `(newx >``=` `0` `and` `newx < n ``and` `newy >``=` `0` `and``            ``newy < n ``and` `mat[newx][newy] < mat[x][y]):``            ``ans ``+``=` `CountDecreasingPathsCell(mat, dp, n,``                                            ``newx, newy)``                                            ` `    ``# function that returns the answer``    ``dp[x][y] ``=` `ans``    ``return` `dp[x][y]` `# Function that counts the total``# decreasing path in the matrix``def` `countDecreasingPathsMatrix(n,mat):``    ``dp ``=` `[]``    ` `    ``# Initialising dp[][] to -1.``    ``for` `i ``in` `range``(n):``        ``l ``=` `[]``        ``for` `j ``in` `range``(n):``            ``l.append(``-``1``)``        ``dp.append(l)``    ``sum` `=` `0``    ` `    ``# Calculating number of decreasing``    ``# path from each cell.``    ``for` `i ``in` `range``(n):``        ``for` `j ``in` `range``(n):``            ``sum` `+``=` `CountDecreasingPathsCell(mat, dp,``                                            ``n, i, j)``    ``return` `sum``    ` `# Driver Code``n ``=` `2``mat ``=` `[[``1``, ``2``], [``1``, ``3``]]` `# function call that returns the``# count of decreasing paths in a matrix``print``(countDecreasingPathsMatrix(n, mat))` `# This code is contributed by SHUBHAMSINGH10`

## C#

 `// C# program to count number``// of decreasing path in a matrix``using` `System;` `class` `GFG``{``    ` `// Function that returns``// the number of decreasing``// paths from a cell(i, j)``public` `static` `int` `CountDecreasingPathsCell(``int``[,] mat, ``int``[,] dp,``                                           ``int` `n, ``int` `x, ``int` `y)``{``    ``// checking if already``    ``// calculated``    ``if` `(dp[x, y] != -1)``        ``return` `dp[x, y];` `    ``// all possible paths``    ``int``[,] delta = {{0, 1}, {1, 0},``                    ``{-1, 0},{0, -1}};``    ``int` `newx, newy;` `    ``// counts the total``    ``// number of paths``    ``int` `ans = 1;` `    ``// In all four``    ``// allowed direction.``    ``for` `(``int` `i = 0; i < 4; i++)``    ``{` `        ``// new co-ordinates``        ``newx = x + delta[i,0];``        ``newy = y + delta[i,1];` `        ``// Checking if not going out``        ``// of matrix and next cell``        ``// value is less than current``        ``// cell value.``        ``if` `(newx >= 0 && newx < n &&``            ``newy >= 0 && newy < n &&``            ``mat[newx,newy] < mat[x,y])``        ``{``            ``ans += CountDecreasingPathsCell(mat, dp, n,``                                            ``newx, newy);``        ``}``    ``}``    ` `    ``// function that``    ``// returns the answer``    ``return` `dp[x,y] = ans;``}` `// Function that counts the total``// decreasing path in the matrix``public` `static` `int` `countDecreasingPathsMatrix(``int` `n,``                                        ``int``[,] mat)``{``    ``int``[,] dp = ``new` `int``[n, n];` `    ``// Initialising dp[][] to -1.``    ``for` `(``int` `i = 0; i < n; i++)``        ``for` `(``int` `j = 0; j < n; j++)``            ``dp[i, j] = -1;` `    ``int` `sum = 0;` `    ``// Calculating number of``    ``// decreasing path from each cell.``    ``for` `(``int` `i = 0; i < n; i++)``        ``for` `(``int` `j = 0; j < n; j++)``            ``sum += CountDecreasingPathsCell(mat, dp,``                                            ``n, i, j);` `    ``return` `sum;``}` `// Driver code``static` `public` `void` `Main ()``{``    ``int` `n = 2;``    ` `    ``int``[,] mat= {{1, 2},``                ``{1, 3}};``    ` `    ``// function call that returns the``    ``// count of decreasing paths in a matrix``    ``Console.WriteLine(countDecreasingPathsMatrix(n, mat));``}``}` `// This code is contributed by vij.`

## Javascript

 ``

Output:

`8`

Time Complexity : O(N2
Auxiliary Space : O(N2)

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