Unique Paths LeetCode Solution

Problem – Unique Paths LeetCode Solution

There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The test cases are generated so that the answer will be less than or equal to 2 * 109.

Example 1:

Input: m = 3, n = 7
Output: 28

Example 2:

Input: m = 3, n = 2
Output: 3
Explanation: From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down

Constraints:

  • 1 <= m, n <= 100

Unique Paths LeetCode Solution in Java

public class Solution {
    public int uniquePaths(int m, int n) {
        if(m == 1 || n == 1)
            return 1;
        m--;
        n--;
        if(m < n) {              // Swap, so that m is the bigger number
            m = m + n;
            n = m - n;
            m = m - n;
        }
        long res = 1;
        int j = 1;
        for(int i = m+1; i <= m+n; i++, j++){       // Instead of taking factorial, keep on multiply & divide
            res *= i;
            res /= j;
        }
            
        return (int)res;
    }
}

Unique Paths LeetCode Solution in C++

class Solution {
public:
    int uniquePaths(int m, int n) {
        int N = n+m-2; // total steps = n-1 + m-1
        int r = min(n,m) - 1; // will iterate on the minimum for efficiency = (total) C (min(right, down))
        
        double res = 1;
        
		// compute nCr
        for(int i=1; i<=r; ++i, N--){
            
            res = res * (N) / i;
        }
        
        return (int)res;
    }
};

Unique Paths LeetCode Solution in Python

class Solution:
    # @return an integer
    def uniquePaths(self, m, n):
        aux = [[1 for x in range(n)] for x in range(m)]
        for i in range(1, m):
            for j in range(1, n):
                aux[i][j] = aux[i][j-1]+aux[i-1][j]
        return aux[-1][-1]
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