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Number of Unique Paths

Tags : dynamic-programming, gfg, cpp, easy

Problem Statement - link #

Given a A X B matrix with your initial position at the top-left cell, find the number of possible unique paths to reach the bottom-right cell of the matrix from the initial position.

Note: Possible moves can be either down or right at any point in time, i.e., we can move to matrix[i+1][j] or matrix[i][j+1] from matrix[i][j].

Examples #

Example 1:

Input:
A = 2, B = 2
Output: 2
Explanation: There are only two unique
paths to reach the end of the matrix of
size two from the starting cell of the
matrix.

Example 2:

Input:
A = 3, B = 4
Output: 10
Explanation: There are only 10 unique
paths to reach the end of the matrix of
size two from the starting cell of the
matrix.

Your Task: #

Complete NumberOfPath() function which takes 2 arguments(A and B) and returns the number of unique paths from top-left to the bottom-right cell.

Expected Time Complexity: O(A*B). Expected Auxiliary Space: O(A*B).

Constraints #

Solutions #

// { Driver Code Starts
//Initial template for C++
#include<bits/stdc++.h>
using namespace std;

 // } Driver Code Ends
//User function template in C++

class Solution
{
    public:
    //Function to find total number of unique paths.
    int NumberOfPath(int a, int b)
    {
        int dp[a][b];
        for(int i=0;i<b;i++) dp[0][i] = 1;
        for(int i=0;i<a;i++) dp[i][0] = 1;
        for(int i=1;i<a;i++)
        for(int j=1;j<b;j++)
        dp[i][j] = dp[i][j-1]+dp[i-1][j];
        return dp[a-1][b-1];
        //code here
    }
};


// { Driver Code Starts.
int main()
{
    //taking total testcases
    int t;
    cin>>t;
    while(t--)
    {
        //taking dimensions of the matrix
        int a,b;
        cin>>a>>b;
        Solution ob;
        //calling NumberOfPath() function
        cout << ob.NumberOfPath(a,b) << endl;
    }
}

  // } Driver Code Ends