# Assign Mice to Holes

There are N Mice and N holes are placed in a straight line. Each hole can accommodate 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.

Examples:

Input : positions of mice are: 4 -4 2 positions of holes are: 4 0 5 Output : 4 Assign mouse at position x = 4 to hole at position x = 4 : Time taken is 0 minutes Assign mouse at position x=-4 to hole at position x = 0 : Time taken is 4 minutes Assign mouse at position x=2 to hole at position x = 5 : Time taken is 3 minutes After 4 minutes all of the mice are in the holes. Since, there is no combination possible where the last mouse's time is less than 4, answer = 4. Input : positions of mice are: -10, -79, -79, 67, 93, -85, -28, -94 positions of holes are: -2, 9, 69, 25, -31, 23, 50, 78 Output : 102

This problem can be solved using greedy strategy. We can put every mouse to its nearest hole to minimize the time. This can be done by sorting the positions of mice and holes. This allows us to put the ith mice to the corresponding hole in the holes list. We can then find the maximum difference between the mice and corresponding hole position.

In example 2, on sorting both the lists, we find that the mouse at position -79 is the last to travel to hole 23 taking time 102.

sort mice positions (in any order) sort hole positions Loop i = 1 to N: update ans according to the value of |mice(i) - hole(i)|. It should be maximum of all differences.

**Proof of correctness:**

Let i1 < i2 be the positions of two mice and let j1 < j2 be the positions of two holes.
It suffices to show via case analysis that

max(|i1-j1|, |i2-j2|) <= max(|i1-j2|, |i2-j1|), where '|a - b|' represent absolute value of (a - b)

Since it follows by induction that every assignment can be transformed by a series of swaps into the sorted assignment, where none of these swaps increases the span.

`// Java program to find the minimum time to place ` `// all mice in all holes. ` `import` `java.util.* ; ` ` ` `public` `class` `GFG ` `{ ` ` ` `// Returns minimum time required to place mice ` ` ` `// in holes. ` ` ` `public` `int` `assignHole(ArrayList<Integer> mice, ` ` ` `ArrayList<Integer> holes) ` ` ` `{ ` ` ` `if` `(mice.size() != holes.size()) ` ` ` `return` `-` `1` `; ` ` ` ` ` `/* Sort the lists */` ` ` `Collections.sort(mice); ` ` ` `Collections.sort(holes); ` ` ` ` ` `int` `size = mice.size(); ` ` ` ` ` `/* finding max difference between ith mice and hole */` ` ` `int` `max = ` `0` `; ` ` ` `for` `(` `int` `i=` `0` `; i<size; i++) ` ` ` `if` `(max < Math.abs(mice.get(i)-holes.get(i))) ` ` ` `max = Math.abs(mice.get(i)-holes.get(i)); ` ` ` ` ` `return` `Math.abs(max); ` ` ` `} ` ` ` ` ` `/* Driver Function to test other functions */` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `GFG gfg = ` `new` `GFG(); ` ` ` `ArrayList<Integer> mice = ` `new` `ArrayList<Integer>(); ` ` ` `mice.add(` `4` `); ` ` ` `mice.add(-` `4` `); ` ` ` `mice.add(` `2` `); ` ` ` `ArrayList<Integer> holes= ` `new` `ArrayList<Integer>(); ` ` ` `holes.add(` `4` `); ` ` ` `holes.add(` `0` `); ` ` ` `holes.add(` `5` `); ` ` ` `System.out.println(` `"The last mouse gets into "` `+ ` ` ` `"the hole in time: "` `+gfg.assignHole(mice, holes)); ` ` ` `} ` `} ` |

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Output:

The last mouse gets into the hole in time: 4

This article is contributed by **Saloni Baweja**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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