LeetCode - Unique Path
Problem description
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:
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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 -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right
Example 2:
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Input: m = 7, n = 3
Output: 28
Constraints:
- 1 <= m, n <= 100
- It’s guaranteed that the answer will be less than or equal to 2 * 10 ^ 9.
Analysis
The first thought is to use backtrack, however, backtrack solution is TLE, and it’s an O(m!n!) algorithm.
Here’s the code:
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public int uniquePaths(int m, int n) {
return backtrack(0, 0, 0, m, n);
}
int backtrack(int res, int i, int j, int m, int n){
if (i == m - 1 && j == n - 1){
return res + 1;
}
if (i != m-1){
res = backtrack(res, i+1, j, m, n);
}
if (j != n-1){
res = backtrack(res, i, j+1, m, n);
}
return res;
}
Another thought is to use DP.
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public int uniquePaths(int m, int n) {
int[][] dp = new int[m][n];
for (int i = 0; i < m; i++){
dp[i][0] = 1;
}
for (int i = 0; i < n; i++){
dp[0][i] = 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];
}
What to improve
- first analysis the time complexity and then do the coding.