LeetCode--1240--hard--TilingRectangleWithFewestSquares

package com.app.main.LeetCode;

import java.util.Collection;
import java.util.Collections;

/**
 *
 * 1240
 *
 * hard
 *
 * https://leetcode.com/problems/tiling-a-rectangle-with-the-fewest-squares/
 *
 * Given a rectangle of size n x m, find the minimum number of integer-sided squares that tile the rectangle.
 *
 *
 *
 * Example 1:
 *
 *
 *
 * Input: n = 2, m = 3
 * Output: 3
 * Explanation: 3 squares are necessary to cover the rectangle.
 * 2 (squares of 1x1)
 * 1 (square of 2x2)
 * Example 2:
 *
 *
 *
 * Input: n = 5, m = 8
 * Output: 5
 * Example 3:
 *
 *
 *
 * Input: n = 11, m = 13
 * Output: 6
 *
 *
 * Constraints:
 *
 * 1 <= n <= 13
 * 1 <= m <= 13
 *
 *
 * Created with IDEA
 * author:Dingsheng Huang
 * Date:2020/1/6
 * Time:下午8:11
 */
public class TilingRectangleWithFewestSquares {

    private int result = Integer.MAX_VALUE;

    private int hMax;


    public int tilingRectangle(int n, int m) {
        if (n > m) {
            return tilingRectangle(m, n);
        }
        int[] hArray = new int[n];
        hMax = m;
        dfs(hArray, 0);
        return result;
    }

    private void dfs(int[] hAarray, int currRes) {
        // pruning
        if (currRes > result) {
            return;
        }
        // search the h array , find the bottom
        int bottom = 0;
        for (int i = 1; i < hAarray.length; i++) {
            if (hAarray[i] < hAarray[bottom]) {
                bottom = i;
            }
        }
        if (hAarray[bottom] == hMax) {
            result = currRes;
            return;
        }
        // find the right border
        int temp = bottom;
        while (temp <= hAarray.length - 1 && (temp -  bottom + 1) <= hMax - hAarray[bottom] && hAarray[temp] == hAarray[bottom]) {
            temp++;
        }
        temp--;
        for (int i = 1; i <= temp - bottom + 1; i++) {
            // update h array
            for (int j = bottom; j < bottom + i; j++) {
                hAarray[j] += i;
            }
            currRes++;
            dfs(hAarray, currRes);
            // backtrack
            for (int j = bottom; j < bottom + i; j++) {
                hAarray[j] -= i;
            }
            currRes--;
        }

    }

}