package class24;

import java.util.HashMap;
import java.util.Map.Entry;
import java.util.LinkedList;

public class Code04_MinCoinsOnePaper {

	public static int minCoins(int[] arr, int aim) {
		return process(arr, 0, aim);
	}

	public static int process(int[] arr, int index, int rest) {
		if (rest < 0) {
			return Integer.MAX_VALUE;
		}
		if (index == arr.length) {
			return rest == 0 ? 0 : Integer.MAX_VALUE;
		} else {
			int p1 = process(arr, index + 1, rest);
			int p2 = process(arr, index + 1, rest - arr[index]);
			if (p2 != Integer.MAX_VALUE) {
				p2++;
			}
			return Math.min(p1, p2);
		}
	}

	// dp1时间复杂度为：O(arr长度 * aim)
	public static int dp1(int[] arr, int aim) {
		if (aim == 0) {
			return 0;
		}
		int N = arr.length;
		int[][] dp = new int[N + 1][aim + 1];
		dp[N][0] = 0;
		for (int j = 1; j <= aim; j++) {
			dp[N][j] = Integer.MAX_VALUE;
		}
		for (int index = N - 1; index >= 0; index--) {
			for (int rest = 0; rest <= aim; rest++) {
				int p1 = dp[index + 1][rest];
				int p2 = rest - arr[index] >= 0 ? dp[index + 1][rest - arr[index]] : Integer.MAX_VALUE;
				if (p2 != Integer.MAX_VALUE) {
					p2++;
				}
				dp[index][rest] = Math.min(p1, p2);
			}
		}
		return dp[0][aim];
	}

	public static class Info {
		public int[] coins;
		public int[] zhangs;

		public Info(int[] c, int[] z) {
			coins = c;
			zhangs = z;
		}
	}

	public static Info getInfo(int[] arr) {
		HashMap<Integer, Integer> counts = new HashMap<>();
		for (int value : arr) {
			if (!counts.containsKey(value)) {
				counts.put(value, 1);
			} else {
				counts.put(value, counts.get(value) + 1);
			}
		}
		int N = counts.size();
		int[] coins = new int[N];
		int[] zhangs = new int[N];
		int index = 0;
		for (Entry<Integer, Integer> entry : counts.entrySet()) {
			coins[index] = entry.getKey();
			zhangs[index++] = entry.getValue();
		}
		return new Info(coins, zhangs);
	}

	// dp2时间复杂度为：O(arr长度) + O(货币种数 * aim * 每种货币的平均张数)
	public static int dp2(int[] arr, int aim) {
		if (aim == 0) {
			return 0;
		}
		// 得到info时间复杂度O(arr长度)
		Info info = getInfo(arr);
		int[] coins = info.coins;
		int[] zhangs = info.zhangs;
		int N = coins.length;
		int[][] dp = new int[N + 1][aim + 1];
		dp[N][0] = 0;
		for (int j = 1; j <= aim; j++) {
			dp[N][j] = Integer.MAX_VALUE;
		}
		// 这三层for循环，时间复杂度为O(货币种数 * aim * 每种货币的平均张数)
		for (int index = N - 1; index >= 0; index--) {
			for (int rest = 0; rest <= aim; rest++) {
				dp[index][rest] = dp[index + 1][rest];
				for (int zhang = 1; zhang * coins[index] <= aim && zhang <= zhangs[index]; zhang++) {
					if (rest - zhang * coins[index] >= 0
							&& dp[index + 1][rest - zhang * coins[index]] != Integer.MAX_VALUE) {
						dp[index][rest] = Math.min(dp[index][rest], zhang + dp[index + 1][rest - zhang * coins[index]]);
					}
				}
			}
		}
		return dp[0][aim];
	}

	// dp3时间复杂度为：O(arr长度) + O(货币种数 * aim)
	// 优化需要用到窗口内最小值的更新结构
	public static int dp3(int[] arr, int aim) {
		if (aim == 0) {
			return 0;
		}
		// 得到info时间复杂度O(arr长度)
		Info info = getInfo(arr);
		int[] c = info.coins;
		int[] z = info.zhangs;
		int N = c.length;
		int[][] dp = new int[N + 1][aim + 1];
		dp[N][0] = 0;
		for (int j = 1; j <= aim; j++) {
			dp[N][j] = Integer.MAX_VALUE;
		}
		// 虽然是嵌套了很多循环，但是时间复杂度为O(货币种数 * aim)
		// 因为用了窗口内最小值的更新结构
		for (int i = N - 1; i >= 0; i--) {
			for (int mod = 0; mod < Math.min(aim + 1, c[i]); mod++) {
				// 当前面值 X
				// mod mod + x mod + 2*x mod + 3 * x
				LinkedList<Integer> w = new LinkedList<>();
				w.add(mod);
				dp[i][mod] = dp[i + 1][mod];
				for (int r = mod + c[i]; r <= aim; r += c[i]) {
					while (!w.isEmpty() && (dp[i + 1][w.peekLast()] == Integer.MAX_VALUE
							|| dp[i + 1][w.peekLast()] + compensate(w.peekLast(), r, c[i]) >= dp[i + 1][r])) {
						w.pollLast();
					}
					w.addLast(r);
					int overdue = r - c[i] * (z[i] + 1);
					if (w.peekFirst() == overdue) {
						w.pollFirst();
					}
					if (dp[i + 1][w.peekFirst()] == Integer.MAX_VALUE) {
						dp[i][r] = Integer.MAX_VALUE;
					} else {
						dp[i][r] = dp[i + 1][w.peekFirst()] + compensate(w.peekFirst(), r, c[i]);
					}
				}
			}
		}
		return dp[0][aim];
	}

	public static int compensate(int pre, int cur, int coin) {
		return (cur - pre) / coin;
	}

	// 为了测试
	public static int[] randomArray(int N, int maxValue) {
		int[] arr = new int[N];
		for (int i = 0; i < N; i++) {
			arr[i] = (int) (Math.random() * maxValue) + 1;
		}
		return arr;
	}

	// 为了测试
	public static void printArray(int[] arr) {
		for (int i = 0; i < arr.length; i++) {
			System.out.print(arr[i] + " ");
		}
		System.out.println();
	}

	// 为了测试
	public static void main(String[] args) {
		int maxLen = 20;
		int maxValue = 30;
		int testTime = 300000;
		System.out.println("功能测试开始");
		for (int i = 0; i < testTime; i++) {
			int N = (int) (Math.random() * maxLen);
			int[] arr = randomArray(N, maxValue);
			int aim = (int) (Math.random() * maxValue);
			int ans1 = minCoins(arr, aim);
			int ans2 = dp1(arr, aim);
			int ans3 = dp2(arr, aim);
			int ans4 = dp3(arr, aim);
			if (ans1 != ans2 || ans3 != ans4 || ans1 != ans3) {
				System.out.println("Oops!");
				printArray(arr);
				System.out.println(aim);
				System.out.println(ans1);
				System.out.println(ans2);
				System.out.println(ans3);
				System.out.println(ans4);
				break;
			}
		}
		System.out.println("功能测试结束");

		System.out.println("==========");

		int aim = 0;
		int[] arr = null;
		long start;
		long end;
		int ans2;
		int ans3;

		System.out.println("性能测试开始");
		maxLen = 30000;
		maxValue = 20;
		aim = 60000;
		arr = randomArray(maxLen, maxValue);

		start = System.currentTimeMillis();
		ans2 = dp2(arr, aim);
		end = System.currentTimeMillis();
		System.out.println("dp2答案 : " + ans2 + ", dp2运行时间 : " + (end - start) + " ms");

		start = System.currentTimeMillis();
		ans3 = dp3(arr, aim);
		end = System.currentTimeMillis();
		System.out.println("dp3答案 : " + ans3 + ", dp3运行时间 : " + (end - start) + " ms");
		System.out.println("性能测试结束");

		System.out.println("===========");

		System.out.println("货币大量重复出现情况下，");
		System.out.println("大数据量测试dp3开始");
		maxLen = 20000000;
		aim = 10000;
		maxValue = 10000;
		arr = randomArray(maxLen, maxValue);
		start = System.currentTimeMillis();
		ans3 = dp3(arr, aim);
		end = System.currentTimeMillis();
		System.out.println("dp3运行时间 : " + (end - start) + " ms");
		System.out.println("大数据量测试dp3结束");

		System.out.println("===========");

		System.out.println("当货币很少出现重复，dp2比dp3有常数时间优势");
		System.out.println("当货币大量出现重复，dp3时间复杂度明显优于dp2");
		System.out.println("dp3的优化用到了窗口内最小值的更新结构");
	}

}