Question

64. Minimum Path Sum

中文文档

Description

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right, which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

 

Example 1:

Input: grid = [[1,3,1],[1,5,1],[4,2,1]]
Output: 7
Explanation: Because the path 1 → 3 → 1 → 1 → 1 minimizes the sum.

Example 2:

Input: grid = [[1,2,3],[4,5,6]]
Output: 12

 

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 200
• 0 <= grid[i][j] <= 200

Solutions

Solution 1: Dynamic Programming

We define $f[i][j]$ to represent the minimum path sum from the top left corner to $(i, j)$. Initially, $f[0][0] = grid[0][0]$, and the answer is $f[m - 1][n - 1]$.

Consider $f[i][j]$:

• If $j = 0$, then $f[i][j] = f[i - 1][j] + grid[i][j]$;
• If $i = 0$, then $f[i][j] = f[i][j - 1] + grid[i][j]$;
• If $i > 0$ and $j > 0$, then $f[i][j] = \min(f[i - 1][j], f[i][j - 1]) + grid[i][j]$.

Finally, return $f[m - 1][n - 1]$.

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.

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

class Solution {
  public int minPathSum(int[][] grid) {
    int m = grid.length, n = grid[0].length;
    int[][] f = new int[m][n];
    f[0][0] = grid[0][0];
    for (int i = 1; i < m; ++i) {
      f[i][0] = f[i - 1][0] + grid[i][0];
    }
    for (int j = 1; j < n; ++j) {
      f[0][j] = f[0][j - 1] + grid[0][j];
    }
    for (int i = 1; i < m; ++i) {
      for (int j = 1; j < n; ++j) {
        f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + grid[i][j];
      }
    }
    return f[m - 1][n - 1];
  }
}

class Solution {
public:
  int minPathSum(vector<vector<int>>& grid) {
    int m = grid.size(), n = grid[0].size();
    int f[m][n];
    f[0][0] = grid[0][0];
    for (int i = 1; i < m; ++i) {
      f[i][0] = f[i - 1][0] + grid[i][0];
    }
    for (int j = 1; j < n; ++j) {
      f[0][j] = f[0][j - 1] + grid[0][j];
    }
    for (int i = 1; i < m; ++i) {
      for (int j = 1; j < n; ++j) {
        f[i][j] = min(f[i - 1][j], f[i][j - 1]) + grid[i][j];
      }
    }
    return f[m - 1][n - 1];
  }
};

func minPathSum(grid [][]int) int {
  m, n := len(grid), len(grid[0])
  f := make([][]int, m)
  for i := range f {
    f[i] = make([]int, n)
  }
  f[0][0] = grid[0][0]
  for i := 1; i < m; i++ {
    f[i][0] = f[i-1][0] + grid[i][0]
  }
  for j := 1; j < n; j++ {
    f[0][j] = f[0][j-1] + grid[0][j]
  }
  for i := 1; i < m; i++ {
    for j := 1; j < n; j++ {
      f[i][j] = min(f[i-1][j], f[i][j-1]) + grid[i][j]
    }
  }
  return f[m-1][n-1]
}

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

impl Solution {
  pub fn min_path_sum(mut grid: Vec<Vec<i32>>) -> i32 {
    let m = grid.len();
    let n = grid[0].len();
    for i in 1..m {
      grid[i][0] += grid[i - 1][0];
    }
    for i in 1..n {
      grid[0][i] += grid[0][i - 1];
    }
    for i in 1..m {
      for j in 1..n {
        grid[i][j] += grid[i][j - 1].min(grid[i - 1][j]);
      }
    }
    grid[m - 1][n - 1]
  }
}

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

public class Solution {
  public int MinPathSum(int[][] grid) {
    int m = grid.Length, n = grid[0].Length;
    int[,] f = new int[m, n];
    f[0, 0] = grid[0][0];
    for (int i = 1; i < m; ++i) {
      f[i, 0] = f[i - 1, 0] + grid[i][0];
    }
    for (int j = 1; j < n; ++j) {
      f[0, j] = f[0, j - 1] + grid[0][j];
    }
    for (int i = 1; i < m; ++i) {
      for (int j = 1; j < n; ++j) {
        f[i, j] = Math.Min(f[i - 1, j], f[i, j - 1]) + grid[i][j];
      }
    }
    return f[m - 1, n - 1];
  }
}