How To Solve Leetcode 62. Unique Paths
An example of recursive algorithms and dynamic programming
Problem statement
A robot is located at the top-left corner of a m x n
grid (marked 'Start' in the diagram below).
The robot 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 (marked 'Finish' in the diagram below).
How many possible unique paths are there?
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
Example 3
Input: m = 7, n = 3
Output: 28
Example 4
Input: m = 3, n = 3
Output: 6
Constraints
1 <= m, n <= 100
.It's guaranteed that the answer will be less than or equal to
2*10^9
.
Solution 1: Recursive
At each point, the robot has two ways of moving: right or down. Let P(m,n)
is the wanted result. Then you have a recursive relationship:
P(m,n) = P(m-1, n) + P(m, n-1)
If the grid has only one row or only one column, then there is only one possible path.
P(1, n) = P(m, 1) = 1.
We have a recursive implementation.
Code
#include <iostream>
#include <vector>
using namespace std;
int uniquePaths(int m, int n) {
if (m == 1 || n == 1) {
return 1;
}
return uniquePaths(m - 1, n) + uniquePaths(m, n - 1);
}
int main() {
std::cout << uniquePaths(3,7) << std::endl;
std::cout << uniquePaths(7,3) << std::endl;
std::cout << uniquePaths(3,2) << std::endl;
std::cout << uniquePaths(3,3) << std::endl;
}
Output:
28
28
3
6
Complexity
Runtime:
O(2^m + 2^n)
, wherem*n
is the size of the grid.Extra space:
O(2^m + 2^n)
.
Solution 2: Dynamic programming
The recursive implementation repeats a lot of computations.
For example, uniquePaths(2,2)
was recomputed in both uniquePaths(2,3)
and uniquePaths(3,2)
when you compute uniquePaths(3,3)
.
One way of storing what has been computed is by using dynamic programming.
Code
#include <iostream>
#include <vector>
using namespace std;
int uniquePaths(int m, int n) {
vector<vector<int> > dp(m, vector<int>(n,1));
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];
}
int main() {
std::cout << uniquePaths(3,7) << std::endl;
std::cout << uniquePaths(7,3) << std::endl;
std::cout << uniquePaths(3,2) << std::endl;
std::cout << uniquePaths(3,3) << std::endl;
}
Output:
28
28
3
6
Complexity
Runtime:
O(m*n)
, wherem*n
is the size of the grid.Extra space:
O(m*n)
.
Solution 3: Reduced dynamic programming
You can rephrase the relationship inside the loop like this:
"new value" = "old value" + "previous value";
Then you do not have to store all values of all rows.
Code
#include <iostream>
#include <vector>
using namespace std;
int uniquePaths(int m, int n) {
vector<int> dp(n, 1);
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[j] += dp[j - 1];
}
}
return dp[n - 1];
}
int main() {
std::cout << uniquePaths(3,7) << std::endl;
std::cout << uniquePaths(7,3) << std::endl;
std::cout << uniquePaths(3,2) << std::endl;
std::cout << uniquePaths(3,3) << std::endl;
}
Output:
28
28
3
6
Complexity
Runtime
O(m*n)
.Memory
O(n)
.
Final thought
I am wondering if there is some mathematics behind this problem. Please share your finding if you find a formula for the solution to this problem.
References
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