Paint House II

/*
Logic:
Logic is same as paint house 1 but we need minimum from previous row but we can get the first and second minimum
because if the minimum cost is of different color then cost to color is current_cost+previous min
        else current_cost+second_previous_min;
*/
#include <iostream>
#include <bits/stdc++.h>
using namespace std ;
int main()
{
  int n ;
  cin >> n ;
  int k ;
  cin >> k ;
  vector<vector<int>>  arr(n, vector<int>(k, 0));
  for (int i = 0 ; i < arr.size() ; i++)
  {
    for (int j = 0 ; j < arr[0].size() ; j++)
    {
      cin >> arr[i][j] ;
    }
  }

  // min values ;
  int min = INT_MAX;
  int smin = INT_MAX;
  //get min and second min from first column
  for (int j = 0; j < arr[0].size(); j++) {
    if (arr[0][j] <= min) {
      smin = min;
      min = arr[0][j];
    } else if (arr[0][j] <= smin) {
      smin = arr[0][j];
    }
  }
  // get answer
  for (int i = 1; i < arr.size(); i++) {
    int cmin = INT_MAX;
    int csmin = INT_MAX;

    for (int j = 0; j < arr[i].size(); j++) {
      if (arr[i - 1][j] != min) {
        arr[i][j] += min;
      } else {
        arr[i][j] += smin;
      }

      if (arr[i][j] <= cmin) {
        csmin = cmin;
        cmin = arr[i][j];
      } else if (arr[i][j] <= csmin) {
        csmin = arr[i][j];
      }
    }

    min = cmin;
    smin = csmin;
  }

  c