Unique Paths - Array - Medium - LeetCode
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Unique Paths - Array - Medium - LeetCode

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  • 1The robot can only move down or right in an m x n grid to reach the bottom-right corner.
  • 2The number of unique paths can be calculated using dynamic programming with a time complexity of O(m*n).
  • 3The solution involves creating a 2D array to store the number of ways to reach each cell in the grid.

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"The robot can only move down or right in an m x n grid to reach the bottom-right corner."

Unique Paths - Array - Medium - LeetCode

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?

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 Example 3:

Input: m = 7, n = 3 Output: 28 Example 4:

Input: m = 3, n = 3 Output: 6

Constraints:

1 <= m, n <= 100 It's guaranteed that the answer will be less than or equal to 2 * 109.

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

Time Complexity: O(m*n)

Space Complexity: O(m*n)

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sumitc91

Published on 31 October 2020 · 1 min read · 203 words

Part of AskGif Blog · coding

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