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:
1 <= m, n <= 100
Intuition
Approach 1: Comninatorics
Approach 2:
Recursively find which paths reach 0,0 and return 1 if yes
Links
Video Links
Approach 1:
Combinatorics
C++
class Solution {
public:
int uniquePaths(int m, int n) {
if(m == 1 || n == 1)
return 1;
m--;
n--;
if(m < n) { // Swap, so that m is the bigger number
swap(m,n);
}
long res = 1;
int j = 1;
for(int i = m+1; i <= m+n; i++, j++){
// Instead of taking factorial, keep on multiply & divide
res *= i;
res /= j;
}
return res;
}
};
Approach 2:
Memoization
C++
class Solution {
public:
int path(int i,int j, vector<vector<int>> &dp){
if(i==0 and j==0)
return 1;
if(i<0 or j<0)
return 0;
if(dp[i][j] != -1)
return dp[i][j];
int l = path(i-1,j,dp);
int r = path(i,j-1,dp);
return dp[i][j] = l+r;
}
int uniquePaths(int m, int n) {
vector<vector<int>> dp(m,vector<int>(n,-1));
dp[0][0] = 1;
return path(m-1,n-1,dp);
}
};
Approach 3:
Tabulation
C++
class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int>> dp(m,vector<int>(n,-1));
dp[0][0] = 1;
for(int i=0; i<m; i++){
for(int j = 0; j < n; j++) {
if(i == 0 and j == 0) continue;
int l=0,r=0;
if(i>0)
l = dp[i-1][j];
if(j>0)
r = dp[i][j-1];
dp[i][j] = l+r;
}
}
return dp[m-1][n-1];
}
};
Approach 4:
Space Optimization
C++
class Solution {
public:
int uniquePaths(int m, int n) {
// vector<vector<int>> dp(m,vector<int>(n,-1));
// dp[0][0] = 1;
vector<int> dp(n,1);
dp[0] = 1;
for(int i=1; i<m; i++){
vector<int> temp(n);
for(int j = 0; j < n; j++) {
if(i == 0 and j==0) continue;
int up=0,left=0;
if(i>0)
up = dp[j];
if(j>0)
left = temp[j-1];
temp[j] = up+left;
}
dp = temp;
}
return dp[n-1];
}
};
