Question

62. Unique Paths

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Description

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

Solutions

Solution 1: Dynamic Programming

We define $f[i][j]$ to represent the number of paths from the top left corner to $(i, j)$, initially $f[0][0] = 1$, and the answer is $f[m - 1][n - 1]$.

Consider $f[i][j]$:

• If $i > 0$, then $f[i][j]$ can be reached by taking one step from $f[i - 1][j]$, so $f[i][j] = f[i][j] + f[i - 1][j]$;
• If $j > 0$, then $f[i][j]$ can be reached by taking one step from $f[i][j - 1]$, so $f[i][j] = f[i][j] + f[i][j - 1]$.

Therefore, we have the following state transition equation:

$$ f[i][j] = \begin{cases} 1 & i = 0, j = 0 \ f[i - 1][j] + f[i][j - 1] & \text{otherwise} \end{cases} $$

The final answer is $f[m - 1][n - 1]$.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns of the grid, respectively.

We notice that $f[i][j]$ is only related to $f[i - 1][j]$ and $f[i][j - 1]$, so we can optimize the first dimension space and only keep the second dimension space, resulting in a time complexity of $O(m \times n)$ and a space complexity of $O(n)$.

class Solution:
  def uniquePaths(self, m: int, n: int) -> int:
    f = [[0] * n for _ in range(m)]
    f[0][0] = 1
    for i in range(m):
      for j in range(n):
        if i:
          f[i][j] += f[i - 1][j]
        if j:
          f[i][j] += f[i][j - 1]
    return f[-1][-1]

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

class Solution {
public:
  int uniquePaths(int m, int n) {
    vector<vector<int>> f(m, vector<int>(n));
    f[0][0] = 1;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (i) {
          f[i][j] += f[i - 1][j];
        }
        if (j) {
          f[i][j] += f[i][j - 1];
        }
      }
    }
    return f[m - 1][n - 1];
  }
};

func uniquePaths(m int, n int) int {
  f := make([][]int, m)
  for i := range f {
    f[i] = make([]int, n)
  }
  f[0][0] = 1
  for i := 0; i < m; i++ {
    for j := 0; j < n; j++ {
      if i > 0 {
        f[i][j] += f[i-1][j]
      }
      if j > 0 {
        f[i][j] += f[i][j-1]
      }
    }
  }
  return f[m-1][n-1]
}

function uniquePaths(m: number, n: number): number {
  const f: number[][] = Array(m)
    .fill(0)
    .map(() => Array(n).fill(0));
  f[0][0] = 1;
  for (let i = 0; i < m; ++i) {
    for (let j = 0; j < n; ++j) {
      if (i > 0) {
        f[i][j] += f[i - 1][j];
      }
      if (j > 0) {
        f[i][j] += f[i][j - 1];
      }
    }
  }
  return f[m - 1][n - 1];
}

impl Solution {
  pub fn unique_paths(m: i32, n: i32) -> i32 {
    let (m, n) = (m as usize, n as usize);
    let mut f = vec![1; n];
    for i in 1..m {
      for j in 1..n {
        f[j] += f[j - 1];
      }
    }
    f[n - 1]
  }
}

/**
 * @param {number} m
 * @param {number} n
 * @return {number}
 */
var uniquePaths = function (m, n) {
  const f = Array(m)
    .fill(0)
    .map(() => Array(n).fill(0));
  f[0][0] = 1;
  for (let i = 0; i < m; ++i) {
    for (let j = 0; j < n; ++j) {
      if (i > 0) {
        f[i][j] += f[i - 1][j];
      }
      if (j > 0) {
        f[i][j] += f[i][j - 1];
      }
    }
  }
  return f[m - 1][n - 1];
};

Solution 2

class Solution:
  def uniquePaths(self, m: int, n: int) -> int:
    f = [[1] * n for _ in range(m)]
    for i in range(1, m):
      for j in range(1, n):
        f[i][j] = f[i - 1][j] + f[i][j - 1]
    return f[-1][-1]

class Solution {
  public int uniquePaths(int m, int n) {
    var f = new int[m][n];
    for (var g : f) {
      Arrays.fill(g, 1);
    }
    for (int i = 1; i < m; ++i) {
      for (int j = 1; j < n; j++) {
        f[i][j] = f[i - 1][j] + f[i][j - 1];
      }
    }
    return f[m - 1][n - 1];
  }
}

class Solution {
public:
  int uniquePaths(int m, int n) {
    vector<vector<int>> f(m, vector<int>(n, 1));
    for (int i = 1; i < m; ++i) {
      for (int j = 1; j < n; ++j) {
        f[i][j] = f[i - 1][j] + f[i][j - 1];
      }
    }
    return f[m - 1][n - 1];
  }
};

func uniquePaths(m int, n int) int {
  f := make([][]int, m)
  for i := range f {
    f[i] = make([]int, n)
    for j := range f[i] {
      f[i][j] = 1
    }
  }
  for i := 1; i < m; i++ {
    for j := 1; j < n; j++ {
      f[i][j] = f[i-1][j] + f[i][j-1]
    }
  }
  return f[m-1][n-1]
}

function uniquePaths(m: number, n: number): number {
  const f: number[][] = Array(m)
    .fill(0)
    .map(() => Array(n).fill(1));
  for (let i = 1; i < m; ++i) {
    for (let j = 1; j < n; ++j) {
      f[i][j] = f[i - 1][j] + f[i][j - 1];
    }
  }
  return f[m - 1][n - 1];
}

/**
 * @param {number} m
 * @param {number} n
 * @return {number}
 */
var uniquePaths = function (m, n) {
  const f = Array(m)
    .fill(0)
    .map(() => Array(n).fill(1));
  for (let i = 1; i < m; ++i) {
    for (let j = 1; j < n; ++j) {
      f[i][j] = f[i - 1][j] + f[i][j - 1];
    }
  }
  return f[m - 1][n - 1];
};

Solution 3

class Solution:
  def uniquePaths(self, m: int, n: int) -> int:
    f = [1] * n
    for _ in range(1, m):
      for j in range(1, n):
        f[j] += f[j - 1]
    return f[-1]

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

class Solution {
public:
  int uniquePaths(int m, int n) {
    vector<int> f(n, 1);
    for (int i = 1; i < m; ++i) {
      for (int j = 1; j < n; ++j) {
        f[j] += f[j - 1];
      }
    }
    return f[n - 1];
  }
};

func uniquePaths(m int, n int) int {
  f := make([]int, n+1)
  for i := range f {
    f[i] = 1
  }
  for i := 1; i < m; i++ {
    for j := 1; j < n; j++ {
      f[j] += f[j-1]
    }
  }
  return f[n-1]
}

function uniquePaths(m: number, n: number): number {
  const f: number[] = Array(n).fill(1);
  for (let i = 1; i < m; ++i) {
    for (let j = 1; j < n; ++j) {
      f[j] += f[j - 1];
    }
  }
  return f[n - 1];
}

/**
 * @param {number} m
 * @param {number} n
 * @return {number}
 */
var uniquePaths = function (m, n) {
  const f = Array(n).fill(1);
  for (let i = 1; i < m; ++i) {
    for (let j = 1; j < n; ++j) {
      f[j] += f[j - 1];
    }
  }
  return f[n - 1];
};