62. Unique Paths - DevCodeEx

Explanation:

To solve this problem, we can use dynamic programming. We will create a 2D array to store the number of unique paths to reach each cell in the grid. The number of unique paths to reach a cell is the sum of the number of paths to reach the cell above it and the cell to the left of it.

Initialize a 2D array dp of size m x n where dp[i][j] represents the number of unique paths to reach cell (i, j).
Initialize the first row and first column of dp with 1 since there is only one way to reach any cell in the first row or first column.
Iterate through the grid starting from (1, 1) and update dp[i][j] as dp[i-1][j] + dp[i][j-1].
Return dp[m-1][n-1] which represents the number of unique paths to reach the bottom-right corner.

Time complexity: O(m x n) where m is the number of rows and n is the number of columns.
Space complexity: O(m x n) for the 2D array dp.

:

View Leetcode Problem

class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        
        for (int i = 0; i < m; i++) {
            dp[i][0] = 1;
        }
        
        for (int j = 0; j < n; j++) {
            dp[0][j] = 1;
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        
        return dp[m-1][n-1];
    }
}

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