[LC] Minimum Path Sum

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom Right which minimizes the sum of any numbers along its path.

Note: You can only move either down or right at any point in time.

The answer at the beginning is this, it will time out.

Class Solution {public:    int minpathsum (vector<vector<int>>& grid) {        int m = grid. Size ();        int n = grid[0].size ();                int dp[m][n];        Dp[0][0] = grid[0][0];        for (int i = 1; i < m; ++i)             dp[i][0] = dp[i-1][0] + grid[i][0];         for (int i = 1; i < n; ++i)            dp[0][i] = dp[0][i-1] + grid[0][i];         for (int i = 1;i<m;++m) for            (int j = 1; j<n;++n)                 dp[i][j] = min (dp[i-1][j],dp[i][j-1]) + grid[i][j];                return dp[m-1][n-1];}    ;

You see, this requires m*n space, in fact, we do not need to maintain the entire m*n, just to ensure that the outermost row and a column is good.

Class Solution {public:    int minpathsum (vector<vector<int>>& grid) {        int m = grid. Size ();        int n = grid[0].size ();                int dp[m],hr[n];        Dp[0] = hr[0] = grid[0][0];        for (int i = 1; i < m; ++i)             dp[i] = dp[i-1] + grid[i][0];         for (int i = 1; i < n; ++i) {            hr[i] = hr[i-1] + grid[0][i];         }        Update horizontal line every time for next use        int tmp;        for (int i = 1;i<m;++i) {            hr[0] = dp[i];            for (int j=1;j<n;++j) {                Hr[j] = min (hr[j-1],hr[j]) + grid[i][j];              }        }        return hr[n-1];}    ;

Does space have an effect on time?

[LC] Minimum Path Sum