测试用例不是吃素的

public String modifyString(String s) {
    if (s.equals("?")) return "a";

    char[] arr = s.toCharArray();
    for (int i = 0; i < arr.length; i++) {
        if (arr[i] == '?') {
            if (i == 0) arr[i] = get(arr[i + 1]);
            else if (i == arr.length - 1) arr[i] = get(arr[i - 1]);
            else arr[i] = get(arr[i - 1], arr[i + 1]);
        }
    }
    return new String(arr);
}

private char get(char l, char r) {
    if (r == '?') return get(l);
    char res = get(r);
    while (res == r || res == l)
        res = get(res);
    return res;
}

private char get(char nb) {
    if (nb == '?') return 'a';
    if (nb == 'z') return 'a';
    return (char) (nb + 1);
}

public int numTriplets(int[] nums1, int[] nums2) {
    Arrays.sort(nums1);
    Arrays.sort(nums2);

    Map<Integer, List<Integer>> map1 = getMap(nums1), map2 = getMap(nums2);

    return cnt(nums1, nums2, map1, map2) + cnt(nums2, nums1, map2, map1);
}

private int cnt(int[] nums1, int[] nums2, Map<Integer, List<Integer>> map1, Map<Integer, List<Integer>> map2) {
    int cnt = 0;
    int last = 0;
    
    // 枚举 i
    for (int i = 0; i < nums1.length; i++) {
        int v = nums1[i];
        if (i > 0 && v == nums1[i - 1]) { // nums1 同值，快速计算
            cnt += last;
            continue;
        }

        long v2 = (long) v * (long) v;
        int curr = 0; // 当前 i 的结果
        
        // 枚举 j
        for (int j = 0; j < nums2.length - 1; j++) {
        
            // 二分查找 k：从 j + 1 开始，第一个满足的
            int l = j + 1, r = nums2.length - 1;
            while (l < r) {
                int mid = l + (r - l) / 2;
                if ((long) nums2[mid] * (long) nums2[j] < v2) l = mid + 1;
                else r = mid;
            }
            if ((long) nums2[l] * (long) nums2[j] == v2) {
                // 还要看 l 后面有几个与 nums2[l] 相同
                List<Integer> list = map2.get(nums2[l]);
                curr += list.get(list.size() - 1) - l + 1;
            }
        }
        
        cnt += curr; // 累计
        last = curr; // 记录缓存
    }
    // System.out.println();
    return cnt;
}

// (value, indexList)
private Map<Integer, List<Integer>> getMap(int[] arr) {
    Map<Integer, List<Integer>> map = new HashMap<>();
    for (int i = 0; i < arr.length; i++) {
        if (!map.containsKey(arr[i]))
            map.put(arr[i], new ArrayList<>());
        map.get(arr[i]).add(i);
    }
    return map;
}

public int minCost(String s, int[] cost) {
    char[] arr = s.toCharArray();
    int n = arr.length;
    if (n == 1) return 0;

    int res = 0;
    char last = '0';
    int lastIndex = -1;
    for (int i = 0; i < n; i++) {
        if (arr[i] == last) {
            if (cost[i] < cost[lastIndex]) {
                // 删 arr[i]
                res += cost[i];
                // lastIndex = lastIndex 无变化
            } else {
                // 删 arr[lastIndex]
                res += cost[lastIndex];
                lastIndex = i; // 变化
            }
        } else lastIndex = i; // 更新
        last = arr[i]; // 总是更新
    }
    return res;
}

class UnionFind {
    private int count;
    private int[] parent;

    public UnionFind(int n) {
        count = n;
        parent = new int[n + 1];
        for (int i = 1; i <= n; i++)
            parent[i] = i;
    }

    // false: 已可通，无需新增连接
    public boolean union(int p, int q) {
        int rootP = find(p);
        int rootQ = find(q);
        if (rootP == rootQ) return false;

        parent[rootP] = rootQ;
        count--;
        return true;
    }

    public int find(int p) {
        int root = p;
        while (root != parent[root])
            root = parent[root];
        while (p != parent[p]) {
            int x = p;
            p = parent[p];
            parent[x] = root;
        }
        return root;
    }
}

public int maxNumEdgesToRemove(int n, int[][] edges) {
    if (edges.length == 1) return 0;

    // 各点的入度情况
    int[][] ins = new int[n + 1][3 + 1];
    for (int[] e : edges) {
        int type = e[0], a = e[1], b = e[2];
        ins[a][type]++;
        ins[b][type]++;
    }

    // 检查无法完全遍历
    for (int i = 1; i <= n; i++) {
        int[] in = ins[i];
        if (in[3] == 0 && (in[1] == 0 || in[2] == 0))
            return -1;
    }

    // 可完全遍历
    int res = 0;
    // 并查集
    UnionFind ufa = new UnionFind(n), ufb = new UnionFind(n);
    // 贪心：先选共用 e[0] desc
    Arrays.sort(edges, (e1, e2) -> e2[0] - e1[0]);
    for (int[] e : edges) {
        int type = e[0], a = e[1], b = e[2];
        if (type == 1) {
            if (!ufa.union(a, b)) res++;
        } else if (type == 2) {
            if (!ufb.union(a, b)) res++;
        } else {
            boolean ua = ufa.union(a, b), ub = ufb.union(a, b);
            if (!ua && !ub) res++; // Alice, Bob 都不再需要这条边
        }
    }
    return res;
}