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Unique Paths.cpp
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Unique Paths.cpp
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Method 1: Recurtion
// TLE
int dfs(int i, int j, int m, int n)
{
if(i == m-1 && j == n-1)return 1;
if(i >= m || j >= n)return 0;
return dfs(i+1, j, m, n) + dfs(i, j+1, m, n);
}
int uniquePaths(int m, int n) {
return dfs(0,0,m,n);
}
Method 2: DP recursive
vector<vector<int>> dp;
int dfs(int i, int j, int m, int n)
{
if(i == m-1 && j == n-1)return 1;
if(i >= m || j >= n)return 0;
if(dp[i][j] != -1)return dp[i][j];
return dp[i][j] = (dfs(i+1, j, m, n) + dfs(i, j+1, m, n));
}
int uniquePaths(int m, int n) {
// set -1 to dp vector
dp.resize(m,vector<int> (n,-1));
return dfs(0,0,m,n);
}
Method 3: DP iterative
int uniquePaths(int m, int n)
{
int dp[m][n];
for(int i=0;i<n;i++)
{
dp[m-1][i]=1;
}
for(int i=0;i<m;i++)
{
dp[i][n-1]=1;
}
for(int i=n-2;i>=0;i--)
{
for(int j=m-2;j>=0;j--)
{
dp[j][i]=dp[j+1][i]+dp[j][i+1];
}
}
return dp[0][0];
}
Method 4: Combanitation.
int uniquePaths(int m, int n)
{
double ans = 1;
// total we take n + m - 2 turn. from start to end we reached.
// So in n + m -2 , we take select m -1 position with right.
// calculate (n+m-2)C(m-1).
// ex. m = 3, n = 3.
// we calculate 4C2.
// i start from 3 to 4 and m from 1 to 2.
for(double i = n, j = 1; j <= m-1; j++,i++)
{
ans = (ans * i)/j;
}
return ans;
}