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package class05;
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import java.util.ArrayList;
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import java.util.Comparator;
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import java.util.HashMap;
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import java.util.List;
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public class Code04_DeleteMinCost {
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// 题目：
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// 给定两个字符串s1和s2，问s2最少删除多少字符可以成为s1的子串？
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// 比如 s1 = "abcde"，s2 = "axbc"
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// 返回 1
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// 解法一
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// 求出str2所有的子序列，然后按照长度排序，长度大的排在前面。
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// 然后考察哪个子序列字符串和s1的某个子串相等(KMP)，答案就出来了。
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// 分析：
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// 因为题目原本的样本数据中，有特别说明s2的长度很小。所以这么做也没有太大问题，也几乎不会超时。
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// 但是如果某一次考试给定的s2长度远大于s1，这么做就不合适了。
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public static int minCost1(String s1, String s2) {
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List<String> s2Subs = new ArrayList<>();
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process(s2.toCharArray(), 0, "", s2Subs);
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s2Subs.sort(new LenComp());
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for (String str : s2Subs) {
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if (s1.indexOf(str) != -1) { // indexOf底层和KMP算法代价几乎一样，也可以用KMP代替
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return s2.length() - str.length();
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}
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}
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return s2.length();
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}
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public static void process(char[] str2, int index, String path, List<String> list) {
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if (index == str2.length) {
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list.add(path);
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return;
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}
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process(str2, index + 1, path, list);
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process(str2, index + 1, path + str2[index], list);
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}
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// x字符串只通过删除的方式，变到y字符串
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// 返回至少要删几个字符
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// 如果变不成，返回Integer.Max
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public static int onlyDelete(char[] x, char[] y) {
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if (x.length < y.length) {
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return Integer.MAX_VALUE;
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}
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int N = x.length;
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int M = y.length;
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int[][] dp = new int[N + 1][M + 1];
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for (int i = 0; i <= N; i++) {
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for (int j = 0; j <= M; j++) {
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dp[i][j] = Integer.MAX_VALUE;
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}
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}
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dp[0][0] = 0;
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// dp[i][j]表示前缀长度
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for (int i = 1; i <= N; i++) {
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dp[i][0] = i;
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}
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for (int xlen = 1; xlen <= N; xlen++) {
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for (int ylen = 1; ylen <= Math.min(M, xlen); ylen++) {
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if (dp[xlen - 1][ylen] != Integer.MAX_VALUE) {
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dp[xlen][ylen] = dp[xlen - 1][ylen] + 1;
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}
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if (x[xlen - 1] == y[ylen - 1] && dp[xlen - 1][ylen - 1] != Integer.MAX_VALUE) {
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dp[xlen][ylen] = Math.min(dp[xlen][ylen], dp[xlen - 1][ylen - 1]);
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}
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}
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}
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return dp[N][M];
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}
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public static class LenComp implements Comparator<String> {
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@Override
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public int compare(String o1, String o2) {
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return o2.length() - o1.length();
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}
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}
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// 解法二
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// 生成所有s1的子串
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// 然后考察每个子串和s2的编辑距离(假设编辑距离只有删除动作且删除一个字符的代价为1)
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// 如果s1的长度较小，s2长度较大，这个方法比较合适
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public static int minCost2(String s1, String s2) {
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if (s1.length() == 0 || s2.length() == 0) {
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return s2.length();
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}
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int ans = Integer.MAX_VALUE;
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char[] str2 = s2.toCharArray();
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for (int start = 0; start < s1.length(); start++) {
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for (int end = start + 1; end <= s1.length(); end++) {
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// str1[start....end]
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// substring -> [ 0,1 )
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ans = Math.min(ans, distance(str2, s1.substring(start, end).toCharArray()));
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}
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}
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return ans == Integer.MAX_VALUE ? s2.length() : ans;
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}
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// 求str2到s1sub的编辑距离
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// 假设编辑距离只有删除动作且删除一个字符的代价为1
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public static int distance(char[] str2, char[] s1sub) {
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int row = str2.length;
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int col = s1sub.length;
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int[][] dp = new int[row][col];
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// dp[i][j]的含义：
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// str2[0..i]仅通过删除行为变成s1sub[0..j]的最小代价
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// 可能性一：
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// str2[0..i]变的过程中，不保留最后一个字符(str2[i])，
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// 那么就是通过str2[0..i-1]变成s1sub[0..j]之后，再最后删掉str2[i]即可 -> dp[i][j] = dp[i-1][j] + 1
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// 可能性二：
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// str2[0..i]变的过程中，想保留最后一个字符(str2[i])，然后变成s1sub[0..j]，
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// 这要求str2[i] == s1sub[j]才有这种可能, 然后str2[0..i-1]变成s1sub[0..j-1]即可
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// 也就是str2[i] == s1sub[j] 的条件下，dp[i][j] = dp[i-1][j-1]
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dp[0][0] = str2[0] == s1sub[0] ? 0 : Integer.MAX_VALUE;
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for (int j = 1; j < col; j++) {
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dp[0][j] = Integer.MAX_VALUE;
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}
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for (int i = 1; i < row; i++) {
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dp[i][0] = (dp[i - 1][0] != Integer.MAX_VALUE || str2[i] == s1sub[0]) ? i : Integer.MAX_VALUE;
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}
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for (int i = 1; i < row; i++) {
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for (int j = 1; j < col; j++) {
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dp[i][j] = Integer.MAX_VALUE;
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if (dp[i - 1][j] != Integer.MAX_VALUE) {
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dp[i][j] = dp[i - 1][j] + 1;
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}
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if (str2[i] == s1sub[j] && dp[i - 1][j - 1] != Integer.MAX_VALUE) {
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dp[i][j] = Math.min(dp[i][j], dp[i - 1][j - 1]);
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}
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}
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}
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return dp[row - 1][col - 1];
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}
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// 解法二的优化
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public static int minCost3(String s1, String s2) {
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if (s1.length() == 0 || s2.length() == 0) {
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return s2.length();
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}
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char[] str2 = s2.toCharArray();
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char[] str1 = s1.toCharArray();
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int M = str2.length;
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int N = str1.length;
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int[][] dp = new int[M][N];
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int ans = M;
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for (int start = 0; start < N; start++) { // 开始的列数
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dp[0][start] = str2[0] == str1[start] ? 0 : M;
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for (int row = 1; row < M; row++) {
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dp[row][start] = (str2[row] == str1[start] || dp[row - 1][start] != M) ? row : M;
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}
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ans = Math.min(ans, dp[M - 1][start]);
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// 以上已经把start列，填好
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// 以下要把dp[...][start+1....N-1]的信息填好
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// start...end end - start +2
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for (int end = start + 1; end < N && end - start < M; end++) {
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// 0... first-1 行 不用管
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int first = end - start;
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dp[first][end] = (str2[first] == str1[end] && dp[first - 1][end - 1] == 0) ? 0 : M;
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for (int row = first + 1; row < M; row++) {
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dp[row][end] = M;
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if (dp[row - 1][end] != M) {
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dp[row][end] = dp[row - 1][end] + 1;
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}
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if (dp[row - 1][end - 1] != M && str2[row] == str1[end]) {
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dp[row][end] = Math.min(dp[row][end], dp[row - 1][end - 1]);
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}
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}
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ans = Math.min(ans, dp[M - 1][end]);
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}
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}
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return ans;
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}
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// 来自学生的做法，时间复杂度O(N * M平方)
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// 复杂度和方法三一样，但是思路截然不同
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public static int minCost4(String s1, String s2) {
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char[] str1 = s1.toCharArray();
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char[] str2 = s2.toCharArray();
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HashMap<Character, ArrayList<Integer>> map1 = new HashMap<>();
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for (int i = 0; i < str1.length; i++) {
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ArrayList<Integer> list = map1.getOrDefault(str1[i], new ArrayList<Integer>());
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list.add(i);
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map1.put(str1[i], list);
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}
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int ans = 0;
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// 假设删除后的str2必以i位置开头
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// 那么查找i位置在str1上一共有几个，并对str1上的每个位置开始遍历
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// 再次遍历str2一次，看存在对应str1中i后续连续子串可容纳的最长长度
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for (int i = 0; i < str2.length; i++) {
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if (map1.containsKey(str2[i])) {
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ArrayList<Integer> keyList = map1.get(str2[i]);
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for (int j = 0; j < keyList.size(); j++) {
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int cur1 = keyList.get(j) + 1;
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int cur2 = i + 1;
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int count = 1;
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for (int k = cur2; k < str2.length && cur1 < str1.length; k++) {
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if (str2[k] == str1[cur1]) {
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cur1++;
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count++;
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}
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}
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ans = Math.max(ans, count);
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}
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}
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}
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return s2.length() - ans;
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}
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public static String generateRandomString(int l, int v) {
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int len = (int) (Math.random() * l);
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char[] str = new char[len];
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for (int i = 0; i < len; i++) {
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str[i] = (char) ('a' + (int) (Math.random() * v));
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}
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return String.valueOf(str);
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}
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public static void main(String[] args) {
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char[] x = { 'a', 'b', 'c', 'd' };
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char[] y = { 'a', 'd' };
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System.out.println(onlyDelete(x, y));
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int str1Len = 20;
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int str2Len = 10;
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int v = 5;
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int testTime = 10000;
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boolean pass = true;
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System.out.println("test begin");
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for (int i = 0; i < testTime; i++) {
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String str1 = generateRandomString(str1Len, v);
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String str2 = generateRandomString(str2Len, v);
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int ans1 = minCost1(str1, str2);
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int ans2 = minCost2(str1, str2);
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int ans3 = minCost3(str1, str2);
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int ans4 = minCost4(str1, str2);
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if (ans1 != ans2 || ans3 != ans4 || ans1 != ans3) {
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pass = false;
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System.out.println(str1);
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System.out.println(str2);
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System.out.println(ans1);
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System.out.println(ans2);
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System.out.println(ans3);
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System.out.println(ans4);
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break;
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}
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}
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System.out.println("test pass : " + pass);
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}
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}
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