LeetCode – Unique Paths (Java)

 

A robot is located at the top-left corner of a m x n grid. It can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid.

How many possible unique paths are there?

Java Solution 1 - DFS

A depth-first search solution is pretty straight-forward. However, the time of this solution is too expensive, and it didn't pass the online judge.

public int uniquePaths(int m, int n) {
    return dfs(0,0,m,n);
}
 
public int dfs(int i, int j, int m, int n){
    if(i==m-1 && j==n-1){
        return 1;
    }
 
    if(i<m-1 && j<n-1){
        return dfs(i+1,j,m,n) + dfs(i,j+1,m,n);
    }
 
    if(i<m-1){
        return dfs(i+1,j,m,n);
    }
 
    if(j<n-1){
        return dfs(i,j+1,m,n);
    }
 
    return 0;
}

Java Solution 2 - Dynamic Programming

public int uniquePaths(int m, int n) {
    if(m==0 || n==0) return 0;
    if(m==1 || n==1) return 1;
 
    int[][] dp = new int[m][n];
 
    //left column
    for(int i=0; i<m; i++){
        dp[i][0] = 1;
    }
 
    //top row
    for(int j=0; j<n; j++){
        dp[0][j] = 1;
    }
 
    //fill up the dp table
    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];
}

Related posts:

  1. LeetCode – Unique Paths II (Java)
  2. LeetCode – Binary Tree Paths (Java)
  3. LeetCode – Unique Binary Search Trees (Java)
  4. LeetCode – Unique Binary Search Trees II (Java)
Category >> Algorithms >> Interview  
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  1. Leo on 2015-8-18

    There’s an O(1) solution. Remember math.

    The robot has to make (m+n) moves to get from the top left to the bottom right corner. It must make exactly m moves across and n moves down. Thus, to find the number of possible traversals, consider how many ways the robot can order its m right moves and n down moves. Pick m of the (m+n) to be the right moves, and the rest must be the n down. This is (m+n) choose m = (m+n)! / (m!n!).

  2. sansor on 2015-9-16

    This code is running perfectly..

    public int numberOfPaths(int m, int n) {
    if (m == 1 || n == 1) {
    return 1;
    }
    int temp = numberOfPaths(m – 1, n) + numberOfPaths(m, n – 1);
    return temp;

    }

  3. sherif on 2015-10-20

    to compute factorial won’t be a o(1), this will take o(m+n) but still thumbs up for the maths solution!

  4. Ugur Basak on 2015-10-28

    Hi Sansor, as you’ve said this code will work but only for small numbers. Because it is re-calculating same values again and again. Dynamic Programming solution will be much faster.

  5. Vasyl Grygoryev on 2015-12-23

    Good point! But there are (m-1)+(n-1) moves. So we need decrease m and n by one and then can use the formula m+n)! / (m!n!)

    My algorithm with O(min(m,n)) time complexity and O(1) extra space:

    public int uniquePaths(int m, int n) {
    m--;
    n--;

    if (m > n) {
    int t = m;
    m = n;
    n = t;
    }

    int divider = 2;
    int result = 1;
    for (int i = n + m; i > n; i--) {
    result *= i;
    while (divider <= m && result % divider == 0)
    result /= divider++;
    }

    return result;
    }

  6. Prasad Bhagwat on 2016-9-5

    Does not work!! Take 2×2, Solution is 2, your solution gives 1.

  7. Vasyl Grygoryev on 2016-9-6

    It works and it gives 2 in case 2×2.
    Lets check formula:
    0) m = 2; n = 2
    1) Decrease n and m by one => m = 1; n =1
    2) use formula (m+n)! / (m!n!) = (1+1)!/(1!1!)= 2

    Lets check the code:
    > m–; // => m=1
    > n–; // => n=1
    if-condition doesn’t work in this case:
    > if (m > n) { // in our case m = n = 1
    Set variables:
    > int divider = 2;
    > int result = 1;
    Count combinations (loop):
    > for (int i = n + m; i > n; i–) { // in our case i = 2; 2 > 1; i–
    > result *= i; // => result =2;
    While loop doesn’t work in our case. Condition divider <= m (2 while (divider for (int i = n + m; i > n; i–) { // in our case i = 1; 1 > 1; i–
    1 > 1 – is false, so for-loop is finished
    > return result; // result = 2
    So it gives 2.

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