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 <bits/stdc++.h>
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)
    {
        // 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
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

<script>
 
// Javascript program to count number
// of decreasing path in a matrix
var MAX = 100
 
// Function that returns the number of
// decreasing paths from a cell(i, j)
function CountDecreasingPathsCell(mat, dp, n, x, y)
{
    // checking if already calculated
    if (dp[x][y] != -1)
        return dp[x][y];
 
    // all possible paths
    var delta = [ [ 0, 1 ], [ 1, 0 ], [ -1, 0 ],
    [ 0, -1 ] ];
    var newx, newy;
 
    // counts the total number of paths
    var ans = 1;
 
    // In all four allowed direction.
    for (var 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
    dp[x][y] = ans;
    return ans;
}
 
// Function that counts the total
// decreasing path in the matrix
function countDecreasingPathsMatrix(n, mat)
{
    var dp = Array.from(Array(MAX),
    ()=> Array(MAX));
 
    // Initialising dp[][] to -1.
    for (var i = 0; i < n; i++)
        for (var j = 0; j < n; j++)
            dp[i][j] = -1;
 
    var sum = 0;
 
    // Calculating number of decreasing
    // path from each cell.
    for (var i = 0; i < n; i++)
        for (var j = 0; j < n; j++)
            sum += CountDecreasingPathsCell
            (mat, dp, n, i, j);
 
    return sum;
}
 
// Driver Code
 
var n = 2;
var mat = [ [ 1, 2 ], [ 1, 3 ] ];
 
// function call that returns the
// count of decreasing paths in a matrix
document.write( countDecreasingPathsMatrix(n, mat));
 
 
</script>

Output:

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

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