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package class_2022_06_1_week;
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// 字符串的 波动 定义为子字符串中出现次数 最多 的字符次数与出现次数 最少 的字符次数之差。
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// 给你一个字符串 s ，它只包含小写英文字母。请你返回 s 里所有 子字符串的 最大波动 值。
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// 子字符串 是一个字符串的一段连续字符序列。
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// 注意：必须同时有，最多字符和最少字符的字符串才是有效的
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// 测试链接 : https://leetcode.cn/problems/substring-with-largest-variance/
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public class Code04_SubstringWithLargestVariance {
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public static int largestVariance1(String s) {
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if (s == null || s.length() == 0) {
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return 0;
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}
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int n = s.length();
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// a b a c b b a
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// 0 1 0 2 1 1 0
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int[] arr = new int[n];
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for (int i = 0; i < n; i++) {
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arr[i] = s.charAt(i) - 'a';
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}
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int ans = 0;
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// 26 * 26 * n O(N)
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for (int more = 0; more < 26; more++) {
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for (int less = 0; less < 26; less++) {
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if (more != less) {
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int continuousA = 0;
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boolean appearB = false;
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int max = 0;
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// 从左到右遍历，
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for (int i = 0; i < n; i++) {
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if (arr[i] != more && arr[i] != less) {
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continue;
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}
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if (arr[i] == more) { // 当前字符是more
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continuousA++;
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if (appearB) {
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max++;
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}
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} else { // 当前字符是B
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max = Math.max(max, continuousA) - 1;
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continuousA = 0;
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appearB = true;
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}
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ans = Math.max(ans, max);
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}
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}
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}
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}
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return ans;
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}
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public static int largestVariance2(String s) {
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if (s == null || s.length() == 0) {
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return 0;
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}
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int n = s.length();
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// a b a c b b a
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// 0 1 0 2 1 1 0
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int[] arr = new int[n];
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for (int i = 0; i < n; i++) {
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arr[i] = s.charAt(i) - 'a';
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}
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// dp[a][b] = more a less b max
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// dp[b][a] = more b less a max
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int[][] dp = new int[26][26];
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// continuous[a][b] more a less b 连续出现a的次数
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// continuous[b][a] more b less a 连续出现b的次数
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int[][] continuous = new int[26][26];
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// appear[a][b] more a less b b有没有出现过
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// appear[b][a] more b less a a有没有出现过
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boolean[][] appear = new boolean[26][26];
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int ans = 0;
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// 26 * N
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for (int i : arr) {
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for (int j = 0; j < 26; j++) {
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if (j != i) {
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// i,j
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// more i less j 三个变量 连续出现i，j有没有出现过，i-j max
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// more j less i 三个变量 连续出现j，i有没有出现过，j-i max
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++continuous[i][j];
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if (appear[i][j]) {
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++dp[i][j];
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}
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if (!appear[j][i]) {
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appear[j][i] = true;
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dp[j][i] = continuous[j][i] - 1;
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} else {
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dp[j][i] = Math.max(dp[j][i], continuous[j][i]) - 1;
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}
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continuous[j][i] = 0;
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ans = Math.max(ans, Math.max(dp[j][i], dp[i][j]));
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}
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}
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}
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return ans;
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}
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}
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