There are N Mice and N holes are placed in a straight line.
Each hole can accomodate only 1 mouse.
A mouse can stay at his position, move one step right from x to x + 1, or move one step left from x to x − 1. Any of these moves consumes 1 minute.
Assign mice to holes so that the time when the last mouse gets inside a hole is minimized.
Input Format
First line contains the integer N. Next line contains N integers, the position of the mice.
Third line contains N integers, the position of the holes.
Constraints
1 <= N <= 105
Output Format
Output one number corresponding to the minimum number of minutes it will take for the last mice to go into the hole.
Sample Input 0
3
4 -4 2
4 0 5
Sample Output 0
4#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
void sort(int *arr, int left, int right) {
int p, i, j, temp;
if (left < right) {
i = left;
j = right;
p = left;
while (i < j) {
while (arr[i] <= arr[p] && i < right) {
i++;
}
while (arr[j] > arr[p]) {
j--;
}
if (i < j) {
temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
temp = arr[p];
arr[p] = arr[j];
arr[j] = temp;
sort(arr, left, j - 1);
sort(arr, j + 1, right);
}
}
int solveCase(int n, int *hole, int *mouse) {
int time = 0;
sort(hole, 0, n - 1);
sort(mouse, 0, n - 1);
for (int j = 0; j < n; j++) {
int g = mouse[j] - hole[j];
if (g < 0) {
g = -g;
}
if (g > time) {
time = g;
}
}
return time;
}
int main() {
int _n;
cin >> _n;
int _hole[_n], _mouse[_n], _time;
for(int j = 0; j < _n; j++) {
cin >> _hole[j];
}
for(int j = 0; j < _n; j++) {
cin >> _mouse[j];
}
cout << solveCase(_n, _hole, _mouse) << endl;
return 0;
}
