Number of Unique Paths

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].

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)

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.

Constraints #

• 1<= A <= 15
• 1<= B <= 15

Solutions #

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