# Count unique paths is a matrix whose product of elements contains odd number of divisors

• Difficulty Level : Medium
• Last Updated : 26 Nov, 2021

Given a matrix mat[][] of dimension NxM, the task is to count the number of unique paths from the top-left cell, i.e. mat[0][0], to the bottom-right cell, i.e. mat[N – 1][M – 1] of the given matrix such that product of elements in that path contains an odd number of divisors. Possible moves from any cell (i, j) are (i, j + 1) or (i + 1, j).

Examples:

Input: mat[][] = {{1, 1}, {3, 1}, {3, 1}}
Output: 2
Explanation: Two possible paths satisfying the condition:

• 1->3->3->1, Product = 9, Number of Divisors of 9 are 1, 3, 9 which is odd.
• 1->1->1->1, Product = 1, Number of Divisors of 1 is 1 only which is odd.

Input: mat[][] = {{4, 1}, {4, 4}}
Output: 2
Explanation: Two possible paths satisfying the condition:

• 4->4->4, Product = 64, Number of Divisors of 9 are 1, 2, 4, 8, 16, 32, 64 which is odd.
• 4->1->4, Product = 16, Number of Divisors of 16 are 1, 2, 4, 8, 16 which is odd.

Naive Approach: The simplest approach is to generate all possible path from the top-left cell to the bottom-right cell for the given matrix and check if the product of all the elements for all such path has an odd number of divisors by finding all the divisors using the approach discussed in this article.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach#include using namespace std; int countPaths = 0; // Store the product of all pathsvector v; // Function to calculate and store// all the paths product in vectorvoid CountUniquePaths(int a[][105], int i,                      int j, int m,                      int n, int ans){    // Base Case    if (i >= m || j >= n)        return;     // If reaches the bottom right    // corner and product is a    // perfect square    if (i == m - 1 && j == n - 1) {         // Find square root        long double sr = sqrt(ans * a[i][j]);         // If square root is an integer        if ((sr - floor(sr)) == 0)            countPaths++;    }     // Move towards down in the matrix    CountUniquePaths(a, i + 1, j, m,                     n, ans * a[i][j]);     // Move towards right in the matrix    CountUniquePaths(a, i, j + 1, m,                     n, ans * a[i][j]);} // Driver Codeint main(){    int M = 3, N = 2;     // Given matrix mat[][]    int mat[M][105] = { { 1, 1 },                        { 3, 1 },                        { 3, 1 } };     // Function Call    CountUniquePaths(mat, 0, 0, M, N, 1);     // Print the result    cout << countPaths;     return 0;}

## Java

 // Java program for the above approachimport java.io.*;import java.util.*;import java.lang.*; class GFG{     static int countPaths = 0; // Function to calculate and store// all the paths product in vectorstatic void CountUniquePaths(int[][] a, int i,                             int j, int m,                             int n, int ans){         // Base Case    if (i >= m || j >= n)        return;         // If reaches the bottom right    // corner and product is a    // perfect square    if (i == m - 1 && j == n - 1)    {                 // Find square root        double sr = Math.sqrt(ans * a[i][j]);             // If square root is an integer        if ((sr - Math.floor(sr)) == 0)            countPaths++;    }         // Move towards down in the matrix     CountUniquePaths(a, i + 1, j, m,                      n, ans * a[i][j]);         // Move towards right in the matrix     CountUniquePaths(a, i, j + 1, m,                      n, ans * a[i][j]);} // Driver Codepublic static void main (String[] args){    int M = 3, N = 2;         // Given matrix mat[][]    int[][] mat = { { 1, 1 },                    { 3, 1 },                    { 3, 1 } };         // Function call    CountUniquePaths(mat, 0, 0, M, N, 1);         System.out.println(countPaths); }} // This code is contributed by sallagondaavinashreddy7

## Python3

 # Python3 program for# the above approachimport math countPaths = 0; # Function to calculate# and store all the paths# product in vectordef CountUniquePaths(a, i, j,                     m, n, ans):     # Base Case    if (i >= m or j >= n):        return;     # If reaches the bottom    # right corner and product    # is a perfect square    global countPaths;         if (i == m - 1 and        j == n - 1):         # Find square root        sr = math.sqrt(ans * a[i][j]);         # If square root is an integer        if ((sr - math.floor(sr)) == 0):            countPaths += 1;        # Move towards down    # in the matrix    CountUniquePaths(a, i + 1, j,                     m, n, ans * a[i][j]);     # Move towards right    # in the matrix    CountUniquePaths(a, i, j + 1,                     m, n, ans * a[i][j]); # Driver Codeif __name__ == '__main__':       M = 3; N = 2;     # Given matrix mat    mat = [[1, 1],           [3, 1],           [3, 1]];     # Function call    CountUniquePaths(mat, 0,                     0, M, N, 1);     print(countPaths); # This code is contributed by Princi Singh

## C#

 // C# program for the// above approachusing System;class GFG{     static int countPaths = 0; // Function to calculate and store// all the paths product in vectorstatic void CountUniquePaths(int[,] a, int i,                             int j, int m,                             int n, int ans){     // Base Case  if (i >= m || j >= n)    return;   // If reaches the bottom right  // corner and product is a  // perfect square  if (i == m - 1 && j == n - 1)  {     // Find square root    double sr = Math.Sqrt(ans *                          a[i, j]);     // If square root is an integer    if ((sr - Math.Floor(sr)) == 0)      countPaths++;  }   // Move towards down in the matrix  CountUniquePaths(a, i + 1, j, m,                   n, ans * a[i, j]);   // Move towards right in the matrix  CountUniquePaths(a, i, j + 1, m,                   n, ans * a[i, j]);} // Driver Codepublic static void Main (String[] args){  int M = 3, N = 2;   // Given matrix mat[][]  int[,] mat = {{1, 1},                {3, 1},                {3, 1}};   // Function call  CountUniquePaths(mat, 0, 0,                   M, N, 1);   Console.Write(countPaths); }} // This code is contributed by Chitranayal

## Javascript



Output:

2

Time Complexity: O((2N)*sqrt(N))
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming. Initialize an auxiliary space dp[][][] that will store the numbers having a product of all the path from top-left to end of each row for each row of the given matrix. Below are the steps:

• Initialize an 3D auxiliary space dp[][][] where dp[i][j] will store the product of all the path from cell (0, 0) to (i, j).
• To compute each state values calculate dp[i][j] by using all the values from dp(i – 1, j) and dp(i, j – 1).
• For the first row of the given matrix mat[][] store the prefix product of the first row at dp[i][0].
• For the first column of the given matrix mat[][] store the prefix product of the first column at dp[0][i].
• Now iterate over (1, 1) to (N, M) using two nested loops i and j and do the following:
• Store the vector top at index dp[i – 1][j] and left at index dp[i][j – 1].
• Store the product of current element mat[i][j] with the elements stored in top[] and left[] in another auxiliary array curr[].
• Update the current dp state as dp[i][j] = curr.
• Now traverse the array stored at dp(N – 1, M – 1) and count all the numbers which is a perfect square.
• Print the final count after the above steps.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach#include using namespace std; // Stores the resultsvector > >    dp(105, vector >(105)); // Count of unique product pathsint countPaths = 0; // Function to check whether number// is perfect square or notbool isPerfectSquare(int n){    long double sr = sqrt(n);     // If square root is an integer    return ((sr - floor(sr)) == 0);} // Function to calculate and store// all the paths product in vectorvoid countUniquePaths(int a[][105],                      int m, int n,                      int ans){    // Store the value a[0][0]    dp[0][0].push_back(a[0][0]);     // Initialize first row of dp    for (int i = 1; i < m; i++) {         // Find prefix product        a[i][0] *= a[i - 1][0];        dp[i][0].push_back(a[i][0]);    }     // Initialize first column of dp    for (int i = 1; i < n; i++) {         // Find the prefix product        a[0][i] *= a[0][i - 1];        dp[0][i].push_back(a[0][i]);    }     // Iterate over range (1, 1) to (N, M)    for (int i = 1; i < m; i++) {        for (int j = 1; j < n; j++) {             // Copy  dp[i-1][j] in top[]            vector top = dp[i - 1][j];             // Copy dp[i][j-1] into left[]            vector left = dp[i][j - 1];             // Compute the values of current            // state and store it in curr[]            vector curr;             // Find the product of a[i][j]            // with elements at top[]            for (int k = 0;                 k < top.size(); k++) {                curr.push_back(top[k] * a[i][j]);            }             // Find the product of a[i][j]            // with elements at left[]            for (int k = 0;                 k < left.size(); k++) {                curr.push_back(left[k] * a[i][j]);            }             // Update the current state            dp[i][j] = curr;        }    }     // Traverse dp[m - 1][n - 1]    for (auto i : dp[m - 1][n - 1]) {         // Check if perfect square        if (isPerfectSquare(i)) {            countPaths++;        }    }} // Driver Codeint main(){    int M = 3, N = 4;     // Given matrix mat[][]    int mat[M][105] = { { 1, 2, 3, 1 },                        { 3, 1, 2, 4 },                        { 2, 3, 1, 1 } };     // Function Call    countUniquePaths(mat, M, N, 1);     // Print the final count    cout << countPaths;     return 0;}

## Java

 // Java program for the above approachimport java.util.*;class GFG{   // Stores the results  static ArrayList>> dp;   // Count of unique product paths  static int countPaths = 0;   // Function to check whether number  // is perfect square or not  static boolean isPerfectSquare(int n)  {    double  sr = Math.sqrt(n);     // If square root is an integer    return ((sr - Math.floor(sr)) == 0);  }   // Function to calculate and store  // all the paths product in vector  static void countUniquePaths(int a[][],                               int m, int n,                               int ans)  {     // Store the value a[0][0]    dp.get(0).get(0).add(a[0][0]);     // Initialize first row of dp    for (int i = 1; i < m; i++)    {       // Find prefix product      a[i][0] *= a[i - 1][0];      dp.get(i).get(0).add(a[i][0]);    }     // Initialize first column of dp    for (int i = 1; i < n; i++)    {       // Find the prefix product      a[0][i] *= a[0][i - 1];      dp.get(0).get(i).add(a[0][i]);    }     // Iterate over range (1, 1) to (N, M)    for (int i = 1; i < m; i++)    {      for (int j = 1; j < n; j++)      {         // Copy  dp[i-1][j] in top[]        ArrayList top =          dp.get(i - 1).get(j);         // Copy dp[i][j-1] into left[]        ArrayList left =          dp.get(i).get(j - 1);         // Compute the values of current        // state and store it in curr[]        ArrayList curr = new ArrayList<>();         // Find the product of a[i][j]        // with elements at top[]        for (int k = 0;             k < top.size(); k++)        {          curr.add(top.get(k) * a[i][j]);        }         // Find the product of a[i][j]        // with elements at left[]        for (int k = 0;             k < left.size(); k++)        {          curr.add(left.get(k) * a[i][j]);        }         // Update the current state        dp.get(i).set(j, curr);      }    }     // Traverse dp[m - 1][n - 1]    for (Integer i : dp.get(m - 1).get(n - 1))    {       // Check if perfect square      if (isPerfectSquare(i))      {        countPaths++;      }    }  }   // Driver code  public static void main (String[] args)  {    int M = 3, N = 4;     // Given matrix mat[][]    int mat[][] = { { 1, 2, 3, 1 },                   { 3, 1, 2, 4 },                   { 2, 3, 1, 1 } };     dp = new ArrayList<>();     for(int i = 0; i < 105; i++)    {      dp.add(new ArrayList<>());      for(int j = 0; j < 105; j++)        dp.get(i).add(new ArrayList());    }     // Function Call    countUniquePaths(mat, M, N, 1);     // Print the final count    System.out.println(countPaths);  }} // This code is contributed by offbeat

## Python3

 # Python3 program for the above approachfrom math import sqrt, floor # Stores the resultsdp = [[[] for j in range(105)]          for k in range(105)] # Count of unique product pathscountPaths = 0 # Function to check whether number# is perfect square or notdef isPerfectSquare(n):         sr = sqrt(n)     # If square root is an integer    return ((sr - floor(sr)) == 0) # Function to calculate and store# all the paths product in vectordef countUniquePaths(a, m, n, ans):         global dp    global countPaths         # Store the value a[0][0]    dp[0][0].append(a[0][0])     # Initialize first row of dp    for i in range(1, m):                 # Find prefix product        a[i][0] *= a[i - 1][0]        dp[i][0].append(a[i][0])     # Initialize first column of dp    for i in range(1, n):                 # Find the prefix product        a[0][i] *= a[0][i - 1]        dp[0][i].append(a[0][i])     # Iterate over range (1, 1) to (N, M)    for i in range(1, m):        for j in range(1, n):                         # Copy  dp[i-1][j] in top[]            top = dp[i - 1][j]             # Copy dp[i][j-1] into left[]            left = dp[i][j - 1]             # Compute the values of current            # state and store it in curr[]            curr = []             # Find the product of a[i][j]            # with elements at top[]            for k in range(len(top)):                curr.append(top[k] * a[i][j])             # Find the product of a[i][j]            # with elements at left[]            for k in range(len(left)):                curr.append(left[k] * a[i][j])             # Update the current state            dp[i][j] = curr     # Traverse dp[m - 1][n - 1]    for i in dp[m - 1][n - 1]:                 # Check if perfect square        if (isPerfectSquare(i)):            countPaths += 1 # Driver Codeif __name__ == '__main__':         M = 3    N = 4     # Given matrix mat[][]    mat = [ [ 1, 2, 3, 1 ],            [ 3, 1, 2, 4 ],            [ 2, 3, 1, 1 ] ]     # Function Call    countUniquePaths(mat, M, N, 1)     # Print the final count    print(countPaths) # This code is contributed by ipg2016107

## C#

 // C# program for the above approachusing System;using System.Collections.Generic;class GFG{     // Stores the results    static List>> dp;     // Count of unique product paths    static int countPaths = 0;     // Function to check whether number    // is perfect square or not    static bool isPerfectSquare(int n)    {        double sr = Math.Sqrt(n);         // If square root is an integer        return ((sr - Math.Floor(sr)) == 0);    }     // Function to calculate and store    // all the paths product in vector    static void countUniquePaths(int[,] a,                                 int m, int n,                                 int ans)    {         // Store the value a[0][0]        dp[0][0].Add(a[0, 0]);         // Initialize first row of dp        for (int i = 1; i < m; i++)        {             // Find prefix product            a[i, 0] *= a[i - 1, 0];            dp[i][0].Add(a[i, 0]);        }         // Initialize first column of dp        for (int i = 1; i < n; i++)        {             // Find the prefix product            a[0, i] *= a[0, i - 1];            dp[0][i].Add(a[0, i]);        }         // Iterate over range (1, 1) to (N, M)        for (int i = 1; i < m; i++)        {            for (int j = 1; j < n; j++)            {                 // Copy  dp[i-1][j] in top[]                List top = dp[i - 1][j];                 // Copy dp[i][j-1] into left[]                List left = dp[i][j - 1];                 // Compute the values of current                // state and store it in curr[]                List curr = new List();                 // Find the product of a[i][j]                // with elements at top[]                for (int k = 0;                     k < top.Count; k++)                {                    curr.Add(top[k] * a[i, j]);                }                 // Find the product of a[i][j]                // with elements at left[]                for (int k = 0;                     k < left.Count; k++)                {                    curr.Add(left[k] * a[i, j]);                }                 // Update the current state                dp[i][j] = curr;            }        }         // Traverse dp[m - 1][n - 1]        foreach (int i in dp[m - 1][n - 1])        {             // Check if perfect square            if (isPerfectSquare(i))            {                countPaths++;            }        }    }     // Driver code    public static void Main()    {        int M = 3, N = 4;         // Given matrix mat[][]        int[,] mat = { { 1, 2, 3, 1 },                   { 3, 1, 2, 4 },                   { 2, 3, 1, 1 } };         dp = new List>>();         for (int i = 0; i < 105; i++)        {            dp.Add(new List>());            for (int j = 0; j < 105; j++)                dp[i].Add(new List());        }         // Function Call        countUniquePaths(mat, M, N, 1);         // Print the final count        Console.Write(countPaths);    }} // This code is contributed by Saurabh Jaiswal

## Javascript



Output:

3

Time Complexity: O(N3)
Auxiliary Space: O(N3)

My Personal Notes arrow_drop_up