Squirrel Simulation in C++

There's a tree, a squirrel, and several nuts. Positions are represented by the cells in a 2D grid. Your goal is to find the minimal distance for the squirrel to collect all the nuts and put them under the tree one by one. The squirrel can only take at most one nut at one time and can move in four directions - up, down, left, and right, to the adjacent cell. The distance is represented by the number of moves.

So, if the input is like Height: 5 Width: 7 Tree position: [2,2] Squirrel: [4,4] Nuts: [[3,0], [2,5]], then the output will be 12,

To solve this, we will follow these steps −

• Define a function calc(), this will take x1, y1, x2, y2,

• return |x1 - x2| + |y1 - y2|

• Define a function minDistance(), this will take height, width, an array tree, an array sq, one 2D array nuts,

• ret := 0

• maxDiff := -inf

• for initialize i := 0, when i < size of nuts, update (increase i by 1), do −

  • dist := calc(tree[0], tree[1], nuts[i, 0], nuts[i, 1])

  • ret := ret + 2 * dist

  • maxDiff := maximum of maxDiff and 2 * dist - (dist + calc(nuts[i, 0], nuts[i, 1], sq[0], sq[1]))

• return ret - maxDiff

Example

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   int calc(int x1, int y1, int x2, int y2){
      return abs(x1 - x2) + abs(y1 - y2);
   }
   int minDistance(int height, int width, vector<int>& tree, vector<int>& sq, vector<vector>& nuts) {
      int ret = 0;
      int maxDiff = INT_MIN;
      for (int i = 0; i < nuts.size(); i++) {
         int dist = calc(tree[0], tree[1], nuts[i][0],
         nuts[i][1]);
         ret += 2 * dist;
         maxDiff = max(maxDiff, 2 * dist - (dist + calc(nuts[i][0], nuts[i][1], sq[0], sq[1])));
      }
      return ret - maxDiff;
   }
};
main(){
   Solution ob;
   vector<int> v = {2,2}, v1 = {4,4};
   vector<vector<int>> v2 = {{3,0}, {2,5}};
   cout << (ob.minDistance(5,7,v, v1, v2));
}

Input

5, 7, {2,2},{4,4}, {{3,0}, {2,5}}

Output

12