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# 63. 不同路径 II

力扣题目链接 (opens new window)

一个机器人位于一个 m x n 网格的左上角 (起始点在下图中标记为“Start” )。

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角(在下图中标记为“Finish”)。

现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径?

网格中的障碍物和空位置分别用 1 和 0 来表示。

示例 1:

  • 输入:obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
  • 输出:2 解释:
  • 3x3 网格的正中间有一个障碍物。
  • 从左上角到右下角一共有 2 条不同的路径:
    1. 向右 -> 向右 -> 向下 -> 向下
    2. 向下 -> 向下 -> 向右 -> 向右

示例 2:

  • 输入:obstacleGrid = [[0,1],[0,0]]
  • 输出:1

提示:

  • m == obstacleGrid.length
  • n == obstacleGrid[i].length
  • 1 <= m, n <= 100
  • obstacleGrid[i][j] 为 0 或 1

# 算法公开课

《代码随想录》算法视频公开课 (opens new window)动态规划,这次遇到障碍了| LeetCode:63. 不同路径 II (opens new window),相信结合视频再看本篇题解,更有助于大家对本题的理解

# 思路

这道题相对于62.不同路径 (opens new window) 就是有了障碍。

第一次接触这种题目的同学可能会有点懵,这有障碍了,应该怎么算呢?

62.不同路径 (opens new window)中我们已经详细分析了没有障碍的情况,有障碍的话,其实就是标记对应的dp table(dp数组)保持初始值(0)就可以了。

动规五部曲:

  1. 确定dp数组(dp table)以及下标的含义

dp[i][j] :表示从(0 ,0)出发,到(i, j) 有dp[i][j]条不同的路径。

  1. 确定递推公式

递推公式和62.不同路径一样,dp[i][j] = dp[i - 1][j] + dp[i][j - 1]。

但这里需要注意一点,因为有了障碍,(i, j)如果就是障碍的话应该就保持初始状态(初始状态为0)。

所以代码为:

if (obstacleGrid[i][j] == 0) { // 当(i, j)没有障碍的时候,再推导dp[i][j]
    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
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  1. dp数组如何初始化

62.不同路径 (opens new window)不同路径中我们给出如下的初始化:

vector<vector<int>> dp(m, vector<int>(n, 0)); // 初始值为0
for (int i = 0; i < m; i++) dp[i][0] = 1;
for (int j = 0; j < n; j++) dp[0][j] = 1;
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因为从(0, 0)的位置到(i, 0)的路径只有一条,所以dp[i][0]一定为1,dp[0][j]也同理。

但如果(i, 0) 这条边有了障碍之后,障碍之后(包括障碍)都是走不到的位置了,所以障碍之后的dp[i][0]应该还是初始值0。

如图:

63.不同路径II

下标(0, j)的初始化情况同理。

所以本题初始化代码为:

vector<vector<int>> dp(m, vector<int>(n, 0));
for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) dp[i][0] = 1;
for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) dp[0][j] = 1;
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注意代码里for循环的终止条件,一旦遇到obstacleGrid[i][0] == 1的情况就停止dp[i][0]的赋值1的操作,dp[0][j]同理

  1. 确定遍历顺序

从递归公式dp[i][j] = dp[i - 1][j] + dp[i][j - 1] 中可以看出,一定是从左到右一层一层遍历,这样保证推导dp[i][j]的时候,dp[i - 1][j] 和 dp[i][j - 1]一定是有数值。

代码如下:

for (int i = 1; i < m; i++) {
    for (int j = 1; j < n; j++) {
        if (obstacleGrid[i][j] == 1) continue;
        dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
    }
}
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  1. 举例推导dp数组

拿示例1来举例如题:

63.不同路径II1

对应的dp table 如图:

63.不同路径II2

如果这个图看不懂,建议再理解一下递归公式,然后照着文章中说的遍历顺序,自己推导一下!

动规五部分分析完毕,对应C++代码如下:

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) //如果在起点或终点出现了障碍,直接返回0
            return 0;
        vector<vector<int>> dp(m, vector<int>(n, 0));
        for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) dp[i][0] = 1;
        for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) dp[0][j] = 1;
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 1) continue;
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
};
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  • 时间复杂度:O(n × m),n、m 分别为obstacleGrid 长度和宽度
  • 空间复杂度:O(n × m)

同样我们给出空间优化版本:

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if (obstacleGrid[0][0] == 1)
            return 0;
        vector<int> dp(obstacleGrid[0].size());
        for (int j = 0; j < dp.size(); ++j)
            if (obstacleGrid[0][j] == 1)
                dp[j] = 0;
            else if (j == 0)
                dp[j] = 1;
            else
                dp[j] = dp[j-1];

        for (int i = 1; i < obstacleGrid.size(); ++i)
            for (int j = 0; j < dp.size(); ++j){
                if (obstacleGrid[i][j] == 1)
                    dp[j] = 0;
                else if (j != 0)
                    dp[j] = dp[j] + dp[j-1];
            }
        return dp.back();
    }
};
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  • 时间复杂度:O(n × m),n、m 分别为obstacleGrid 长度和宽度
  • 空间复杂度:O(m)

# 总结

本题是62.不同路径 (opens new window)的障碍版,整体思路大体一致。

但就算是做过62.不同路径,在做本题也会有感觉遇到障碍无从下手。

其实只要考虑到,遇到障碍dp[i][j]保持0就可以了。

也有一些小细节,例如:初始化的部分,很容易忽略了障碍之后应该都是0的情况。

# 其他语言版本

# Java

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];

        //如果在起点或终点出现了障碍,直接返回0
        if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) {
            return 0;
        }

        for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) {
            dp[i][0] = 1;
        }
        for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) {
            dp[0][j] = 1;
        }

        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = (obstacleGrid[i][j] == 0) ? dp[i - 1][j] + dp[i][j - 1] : 0;
            }
        }
        return dp[m - 1][n - 1];
    }
}
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// 空间优化版本
class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[] dp = new int[n];

        for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) {
            dp[j] = 1;
        }

        for (int i = 1; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (obstacleGrid[i][j] == 1) {
                    dp[j] = 0;
                } else if (j != 0) {
                    dp[j] += dp[j - 1];
                }
            }
        }
        return dp[n - 1];
    }
}
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# Python

动态规划(版本一)

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid):
        m = len(obstacleGrid)
        n = len(obstacleGrid[0])
        if obstacleGrid[m - 1][n - 1] == 1 or obstacleGrid[0][0] == 1:
            return 0
        dp = [[0] * n for _ in range(m)]
        for i in range(m):
            if obstacleGrid[i][0] == 0:  # 遇到障碍物时,直接退出循环,后面默认都是0
                dp[i][0] = 1
            else:
                break
        for j in range(n):
            if obstacleGrid[0][j] == 0:
                dp[0][j] = 1
            else:
                break
        for i in range(1, m):
            for j in range(1, n):
                if obstacleGrid[i][j] == 1:
                    continue
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
        return dp[m - 1][n - 1]

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动态规划(版本二)

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid):
        m = len(obstacleGrid)  # 网格的行数
        n = len(obstacleGrid[0])  # 网格的列数
        
        if obstacleGrid[m - 1][n - 1] == 1 or obstacleGrid[0][0] == 1:
            # 如果起点或终点有障碍物,直接返回0
            return 0
        
        dp = [[0] * n for _ in range(m)]  # 创建一个二维列表用于存储路径数
        
        # 设置起点的路径数为1
        dp[0][0] = 1 if obstacleGrid[0][0] == 0 else 0
        
        # 计算第一列的路径数
        for i in range(1, m):
            if obstacleGrid[i][0] == 0:
                dp[i][0] = dp[i - 1][0]
        
        # 计算第一行的路径数
        for j in range(1, n):
            if obstacleGrid[0][j] == 0:
                dp[0][j] = dp[0][j - 1]
        
        # 计算其他位置的路径数
        for i in range(1, m):
            for j in range(1, n):
                if obstacleGrid[i][j] == 1:
                    continue
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
        
        return dp[m - 1][n - 1]  # 返回终点的路径数


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动态规划(版本三)

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid):
        if obstacleGrid[0][0] == 1:
            return 0
        
        dp = [0] * len(obstacleGrid[0])  # 创建一个一维列表用于存储路径数
        
        # 初始化第一行的路径数
        for j in range(len(dp)):
            if obstacleGrid[0][j] == 1:
                dp[j] = 0
            elif j == 0:
                dp[j] = 1
            else:
                dp[j] = dp[j - 1]

        # 计算其他行的路径数
        for i in range(1, len(obstacleGrid)):
            for j in range(len(dp)):
                if obstacleGrid[i][j] == 1:
                    dp[j] = 0
                elif j != 0:
                    dp[j] = dp[j] + dp[j - 1]
        
        return dp[-1]  # 返回最后一个元素,即终点的路径数

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动态规划(版本四)

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid):
        if obstacleGrid[0][0] == 1:
            return 0
        
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        
        dp = [0] * n  # 创建一个一维列表用于存储路径数
        
        # 初始化第一行的路径数
        for j in range(n):
            if obstacleGrid[0][j] == 1:
                break
            dp[j] = 1

        # 计算其他行的路径数
        for i in range(1, m):
            if obstacleGrid[i][0] == 1:
                dp[0] = 0
            for j in range(1, n):
                if obstacleGrid[i][j] == 1:
                    dp[j] = 0
                else:
                    dp[j] += dp[j - 1]
        
        return dp[-1]  # 返回最后一个元素,即终点的路径数

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动态规划(版本五)

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid):
        if obstacleGrid[0][0] == 1:
            return 0
        
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        
        dp = [0] * n  # 创建一个一维列表用于存储路径数
        
        # 初始化第一行的路径数
        for j in range(n):
            if obstacleGrid[0][j] == 1:
                break
            dp[j] = 1

        # 计算其他行的路径数
        for i in range(1, m):
            if obstacleGrid[i][0] == 1:
                dp[0] = 0
            for j in range(1, n):
                if obstacleGrid[i][j] == 1:
                    dp[j] = 0
                    continue
                
                dp[j] += dp[j - 1]
        
        return dp[-1]  # 返回最后一个元素,即终点的路径数


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# Go

func uniquePathsWithObstacles(obstacleGrid [][]int) int {
	m, n := len(obstacleGrid), len(obstacleGrid[0])
	//如果在起点或终点出现了障碍,直接返回0
	if obstacleGrid[m-1][n-1] == 1 || obstacleGrid[0][0] == 1 {
		return 0
	}
	// 定义一个dp数组
	dp := make([][]int, m)
	for i, _ := range dp {
		dp[i] = make([]int, n)
	}
	// 初始化, 如果是障碍物, 后面的就都是0, 不用循环了
	for i := 0; i < m && obstacleGrid[i][0] == 0; i++ {
		dp[i][0] = 1
	}
	for i := 0; i < n && obstacleGrid[0][i] == 0; i++ {
		dp[0][i] = 1
	}
	// dp数组推导过程
	for i := 1; i < m; i++ {
		for j := 1; j < n; j++ {
			// 如果obstacleGrid[i][j]这个点是障碍物, 那么dp[i][j]保持为0
			if obstacleGrid[i][j] != 1 {
				// 否则我们需要计算当前点可以到达的路径数
				dp[i][j] = dp[i-1][j] + dp[i][j-1]
			}
		}
	}
	return dp[m-1][n-1]
}
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# Javascript

var uniquePathsWithObstacles = function(obstacleGrid) {
    const m = obstacleGrid.length
    const n = obstacleGrid[0].length
    const dp = Array(m).fill().map(item => Array(n).fill(0))

    for (let i = 0; i < m && obstacleGrid[i][0] === 0; ++i) {
        dp[i][0] = 1
    }

    for (let i = 0; i < n && obstacleGrid[0][i] === 0; ++i) {
        dp[0][i] = 1
    }

    for (let i = 1; i < m; ++i) {
        for (let j = 1; j < n; ++j) {
            dp[i][j] = obstacleGrid[i][j] === 1 ? 0 : dp[i - 1][j] + dp[i][j - 1]
        }
    }

    return dp[m - 1][n - 1]
};

// 版本二:内存优化,直接以原数组为dp数组
var uniquePathsWithObstacles = function(obstacleGrid) {
    const m = obstacleGrid.length;
    const n = obstacleGrid[0].length;
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            if (obstacleGrid[i][j] === 0) {
                // 不是障碍物
                if (i === 0) {
                    // 取左边的值
                    obstacleGrid[i][j] = obstacleGrid[i][j - 1] ?? 1;
                } else if (j === 0) {
                    // 取上边的值
                    obstacleGrid[i][j] = obstacleGrid[i - 1]?.[j] ?? 1;
                } else {
                    // 取左边和上边的和
                    obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
                }
            } else {
                // 如果是障碍物,则路径为0
                obstacleGrid[i][j] = 0;
            }
        }
    }
    return obstacleGrid[m - 1][n - 1];
};
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# TypeScript

function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
    /**
        dp[i][j]: 到达(i, j)的路径数
        dp[0][*]: 用u表示第一个障碍物下标,则u之前为1,u之后(含u)为0
        dp[*][0]: 同上
        ...
        dp[i][j]: obstacleGrid[i][j] === 1 ? 0 : dp[i-1][j] + dp[i][j-1];
     */
    const m: number = obstacleGrid.length;
    const n: number = obstacleGrid[0].length;
    const dp: number[][] = new Array(m).fill(0).map(_ => new Array(n).fill(0));
    for (let i = 0; i < m && obstacleGrid[i][0] === 0; i++) {
        dp[i][0] = 1;
    }
    for (let i = 0; i < n && obstacleGrid[0][i] === 0; i++) {
        dp[0][i] = 1;
    }
    for (let i = 1; i < m; i++) {
        for (let j = 1; j < n; j++) {
            if (obstacleGrid[i][j] === 1) continue;
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
        }
    }
    return dp[m - 1][n - 1];
};
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# Rust

impl Solution {
    pub fn unique_paths_with_obstacles(obstacle_grid: Vec<Vec<i32>>) -> i32 {
        let m: usize = obstacle_grid.len();
        let n: usize = obstacle_grid[0].len();
        if obstacle_grid[0][0] == 1 || obstacle_grid[m-1][n-1] == 1 {
            return 0;
        }
        let mut dp = vec![vec![0; n]; m];
        for i in 0..m {
            if obstacle_grid[i][0] == 1 {
                break;
            }
            else { dp[i][0] = 1; }
        }
        for j in 0..n {
            if obstacle_grid[0][j] == 1 {
                break;
            }
            else { dp[0][j] = 1; }
        }
        for i in 1..m {
            for j in 1..n {
                if obstacle_grid[i][j] == 1 {
                    continue;
                }
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        dp[m-1][n-1]
    }
}
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空间优化:

impl Solution {
    pub fn unique_paths_with_obstacles(obstacle_grid: Vec<Vec<i32>>) -> i32 {
        let mut dp = vec![0; obstacle_grid[0].len()];
        for (i, &v) in obstacle_grid[0].iter().enumerate() {
            if v == 0 {
                dp[i] = 1;
            } else {
                break;
            }
        }
        for rows in obstacle_grid.iter().skip(1) {
            for j in 0..rows.len() {
                if rows[j] == 1 {
                    dp[j] = 0;
                } else if j != 0 {
                    dp[j] += dp[j - 1];
                }
            }
        }
        dp.pop().unwrap()
    }
}
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# C

//初始化dp数组
int **initDP(int m, int n, int** obstacleGrid) {
    int **dp = (int**)malloc(sizeof(int*) * m);
    int i, j;
    //初始化每一行数组
    for(i = 0; i < m; ++i) {
        dp[i] = (int*)malloc(sizeof(int) * n);
    }

    //先将第一行第一列设为0
    for(i = 0; i < m; ++i) {
        dp[i][0] = 0;
    }
    for(j = 0; j < n; ++j) {
        dp[0][j] = 0;
    }

    //若碰到障碍,之后的都走不了。退出循环
    for(i = 0; i < m; ++i) {
        if(obstacleGrid[i][0]) {
            break;
        }
        dp[i][0] = 1;
    }
    for(j = 0; j < n; ++j) {
        if(obstacleGrid[0][j])
            break;
        dp[0][j] = 1;
    }
    return dp;
}

int uniquePathsWithObstacles(int** obstacleGrid, int obstacleGridSize, int* obstacleGridColSize){
    int m = obstacleGridSize, n = *obstacleGridColSize;
    //初始化dp数组
    int **dp = initDP(m, n, obstacleGrid);

    int i, j;
    for(i = 1; i < m; ++i) {
        for(j = 1; j < n; ++j) {
            //若当前i,j位置有障碍
            if(obstacleGrid[i][j])
                //路线不同
                dp[i][j] = 0;
            else
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
        }
    }
    //返回最后终点的路径个数
    return dp[m-1][n-1];
}
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空间优化版本:

int uniquePathsWithObstacles(int** obstacleGrid, int obstacleGridSize, int* obstacleGridColSize){
    int m = obstacleGridSize;
    int n = obstacleGridColSize[0];
    int *dp = (int*)malloc(sizeof(int) * n);
    int i, j;

    // 初始化dp为第一行起始状态。
    for (j = 0; j < n; ++j) {
        if (obstacleGrid[0][j] == 1)
            dp[j] = 0;
        else if (j == 0)
            dp[j] = 1;
        else
            dp[j] = dp[j - 1];
    }

    for (i = 1; i < m; ++i) {
        for (j = 0; j < n; ++j) {
            if (obstacleGrid[i][j] == 1)
                dp[j] = 0;
            // 若j为0,dp[j]表示最左边一列,无需改动
            // 此处dp[j],dp[j-1]等同于二维dp中的dp[i-1][j]和dp[i][j-1]
            else if (j != 0)
                dp[j] += dp[j - 1];
        }
    }

    return dp[n - 1];
}
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# Scala

object Solution {
  import scala.util.control.Breaks._
  def uniquePathsWithObstacles(obstacleGrid: Array[Array[Int]]): Int = {
    var (m, n) = (obstacleGrid.length, obstacleGrid(0).length)
    var dp = Array.ofDim[Int](m, n)

    // 比如break、continue这些流程控制需要使用breakable
    breakable(
      for (i <- 0 until m) {
        if (obstacleGrid(i)(0) != 1) dp(i)(0) = 1
        else break()
      }
    )
    breakable(
      for (j <- 0 until n) {
        if (obstacleGrid(0)(j) != 1) dp(0)(j) = 1
        else break()
      }
    )

    for (i <- 1 until m; j <- 1 until n; if obstacleGrid(i)(j) != 1) {
      dp(i)(j) = dp(i - 1)(j) + dp(i)(j - 1)
    }

    dp(m - 1)(n - 1)
  }
}
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# C#

public class Solution
{
    public int UniquePathsWithObstacles(int[][] obstacleGrid)
    {
        int m = obstacleGrid.Length;
        int n = obstacleGrid[0].Length;
        int[,] dp = new int[m, n];
        if (obstacleGrid[0][0] == 1 || obstacleGrid[m - 1][n - 1] == 1) return 0;
        for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) dp[i, 0] = 1;
        for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) dp[0, j] = 1;
        for (int i = 1; i < m; i++)
        {
            for (int j = 1; j < n; j++)
            {
                if (obstacleGrid[i][j] == 1) continue;
                dp[i, j] = dp[i - 1, j] + dp[i, j - 1];
            }
        }
        return dp[m - 1, n - 1];
    }
}
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上次更新:: 4/24/2024, 5:16:16 PM
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