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:
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 $O(n)$.
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| 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]
|
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| 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];
}
}
|
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| 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];
}
};
|
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| 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]
}
|
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| 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];
}
|
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| 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]
}
}
|
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| /**
* @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#
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| 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]
|
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| 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];
}
}
|
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| 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];
}
};
|
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| 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]
}
|
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| 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];
}
|
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| /**
* @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#
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| 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]
|
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| 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];
}
}
|
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| 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];
}
};
|
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| 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]
}
|
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| 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];
}
|
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| /**
* @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];
};
|
