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62. Unique Paths 👍

Approach 1: 2D DP

  • Time: $O(mn)$
  • Space: $O(mn)$
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class Solution {
 public:
  int uniquePaths(int m, int n) {
    // dp[i][j] := the number of unique paths from (0, 0) to (i, j)
    vector<vector<int>> dp(m, vector<int>(n, 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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class Solution {
  public int uniquePaths(int m, int n) {
    // dp[i][j] := the number of unique paths from (0, 0) to (i, j)
    int[][] dp = new int[m][n];
    Arrays.stream(dp).forEach(A -> Arrays.fill(A, 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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class Solution:
  def uniquePaths(self, m: int, n: int) -> int:
    # dp[i][j] := the number of unique paths from (0, 0) to (i, j)
    dp = [[1] * n for _ in range(m)]

    for i in range(1, m):
      for j in range(1, n):
        dp[i][j] = dp[i - 1][j] + dp[i][j - 1]

    return dp[-1][-1]

Approach 2: 1D DP

  • Time: $O(mn)$
  • Space: $O(n)$
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class Solution {
 public:
  int uniquePaths(int m, int n) {
    vector<int> dp(n, 1);

    for (int i = 1; i < m; ++i)
      for (int j = 1; j < n; ++j)
        dp[j] += dp[j - 1];

    return dp[n - 1];
  }
};

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class Solution {
  public int uniquePaths(int m, int n) {
    int[] dp = new int[n];
    Arrays.fill(dp, 1);

    for (int i = 1; i < m; ++i)
      for (int j = 1; j < n; ++j)
        dp[j] += dp[j - 1];

    return dp[n - 1];
  }
}

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class Solution:
  def uniquePaths(self, m: int, n: int) -> int:
    dp = [1] * n

    for _ in range(1, m):
      for j in range(1, n):
        dp[j] += dp[j - 1]

    return dp[n - 1]