Finding longest increasing path in a matrix

Question

This is the online judge, https://leetcode.com/problems/longest-increasing-path-in-a-matrix/

why I cannot get the result using DFS?

As you kown from each cell, either move to four directions: left, right, up or down.

Store the length of the longest increasing path.

/*

for each elem, neighbours dfs

*/
class Solution {
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int row = matrix.size();
        int col = matrix[0].size();
        int x[] = {0,1,0,-1};// l-r       -1,1    
        int y[] = {1,0,-1,0};// up-down   +1,-1
        int maxlen = 0;

        for(int i = 0; i < row; i++){
            for(int j = 0; j< col; j++){
                // each node in the matrix[i][j], neighbours
                int len = 0;
                dfs(maxlen, len, i, j, x, y, matrix);
            }
        }
        return maxlen;
    }

private:
    bool isIn(int x, int y, int row, int col){
        if(x>=0&&x<=col && y>=0&&y<=row) return true;
        else return false;
    }

    void dfs(int& maxlen, int len, int i, int j,int* x, int* y, vector<vector<int>> matrix){
        int row = matrix.size();
        int col = matrix[0].size();

        for(int k = 0; k < 4; k++){
            int i_t = i+x[k];//the current position
            int j_t = j+y[k];
            if(isIn(i_t,j_t,row,col)&& (matrix[i_t][j_t]>matrix[i][j]) ){ // if inside the matrix, within the boundary&& the value of (i_t,j_t)> 
                len+=1;
                maxlen = max(len,maxlen);
                dfs(maxlen, len, i_t, j_t, x, y, matrix);
            }
        }
    }
};

Answer 1

There are multiple problems with this code.

1. if(x>=0&&x<=col && y>=0&&y<=row) should be changed to if(x>=0&&x<col && y>=0&&y<row)

2. You are adding all the path that originate from one element together which result in a wrong answer. This part of code

   len+=1;
   maxlen = max(len,maxlen);
   dfs(maxlen, len, i_t, j_t, x, y, matrix);
   

   should be changed to:

   //len+=1;
   maxlen = max(len+1,maxlen);
   dfs(maxlen, len+1, i_t, j_t, x, y, matrix);
   

   so that you don't add all the paths in different directions together.

   3. You are solving a lot of overlapping problems. Once you call dfs(r,c) you can save its result, and use that value for future references (dynamic programming).

This is how I would have implemented it:

#include <vector>
#include <iostream>
#include <map>
using namespace std;

map< pair<int,int>, int > dp;
pair<int,int> moves[] = {{0,1},{0,-1},{1,0},{-1,0}};
vector<vector<int> > matrix = { {3,4,5},
                                {3,2,6},
                                {2,2,1}};

int dfs(int r, int c, int n_rows, int n_cols){
    pair<int,int> p = make_pair(r,c);
    if ( dp.count(p) ){
        return dp[p];
    }
    int mx = 0;
    for ( int i=0; i<4; ++i ){
        int next_r = r+moves[i].first;
        int next_c = c+moves[i].second;
        if ( 0<=next_r && next_r < n_rows && 0<=next_c && next_c < n_cols ){
            if ( matrix[next_r][next_c] > matrix[r][c] )
                mx = max(mx, dfs(next_r, next_c, n_rows, n_cols));
        }
    }
    mx++;
    dp[p] = mx;
    return mx;
}

int main(){
    int rows = matrix.size();
    int cols = matrix[0].size();
    int result = 0;
    for ( int i=0; i<rows; ++i ){
        for ( int j=0; j<cols; ++j ){
            result = max(result, dfs(i,j,rows,cols));
        }
    }
    cout << result << endl;                                    
}