The purpose is to find min `d(ak,q)+d(ak,t)+2*sum{d(ai,t),i!=k}`

, which is equal to `2*sum{d(ai,t),all i}+d(ak,q)-d(ak,t)`

, so we only need to find `min{d(ak,q)-d(ak,t)}`

.

`d(ai,t)`

is the distance of nuts[i] to tree, `d(ai,q)`

is the distance of nuts[i] to squirrel.

```
public int minDistance(int height, int width, int[] tree, int[] squirrel, int[][] nuts) {
int[] d=new int[nuts.length];//distance of nuts to tree
int minDiff=Integer.MAX_VALUE;
int result=0;
for (int i = 0; i < nuts.length; i++) {
d[i]=Math.abs(nuts[i][0]-tree[0])+Math.abs(nuts[i][1]-tree[1]);
int disSqui=Math.abs(nuts[i][0]-squirrel[0])+Math.abs(nuts[i][1]-squirrel[1]);//distance of nut to squirrel
int diff=disSqui-d[i];
if(diff<minDiff){
minDiff=diff;
}
result+=(2*d[i]);
}
result+=minDiff;
return result;
}
```