## 62. Unique Paths ###### tags: `Dynamic Programming` `Medium` >[62. Unique Paths](https://leetcode.com/problems/unique-paths/description/) <br> :::info :::spoiler *Optimal Space & Time Complexity* <br> ``` - Time complexity: O() - Space complexity: O() ``` ::: <br> ## Thoughts & Solutions ### YC :::spoiler Code ```javascript= ``` ::: <hr/> ### Sol :::spoiler Code ```javascript= ``` ::: <hr/> ### 東 :::spoiler Code ```javascript= // Time O(m*n) | Space O(m*n) var uniquePaths = function(m, n) { const dp = Array.from({length: m}, () => Array(n).fill(0)); for(let i = 0; i < m; i++) { dp[i][0] = 1; } for(let j = 0; j < n; j++) { dp[0][j] = 1; } for(let i = 1; i < m; i++) { for(let j = 1; j < n; j++) { dp[i][j] = dp[i-1][j] + dp[i][j-1]; } } return dp[m-1][n-1]; }; ``` ::: <hr/> ### Jessie :::spoiler Code ```javascript= ``` ::: <hr/> ### Howard :::spoiler Code ```python= ``` ::: <hr/> ### Haoyu :::spoiler Code ```javascript= var uniquePaths = function(m, n) { const dp = new Array(m).fill(new Array(n).fill(1)); for (let i = 0; i < m; i += 1) { for (let j = 0; j < n; j += 1) { if (i === 0 && j === 0) continue; else if (i === 0) dp[i][j] = dp[i][j - 1]; else if (j === 0) dp[i][j] = dp[i - 1][j]; else dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; } } return dp[m - 1][n - 1]; }; ``` ::: <hr/> <br> ## Live Coding :::spoiler (name) ``` // write your code here ``` :::
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