For each word in a list, determine the minimum number of character replacements needed so that no two adjacent characters are the same. If any two adjacent characters in a string are equal, one of them must be changed.

Example

words = ["add", "boook", "break"]
Output: [1, 1, 0]

• "add": change one 'd' (1 change)
• "boook": change the middle 'o' (1 change)
• "break": no change needed (0)

Constraints

• 1 ≤ n ≤ 100
• 2 ≤ length of words[i] ≤ 10⁵
• characters are lowercase a-z.

解法

扫描字符串，统计每段同字符极大连续段的长度 L；每段贡献 L / 2 次替换(隔一个改一个即可)。累加所有段。复杂度 O(total length)。

def minimal_operations(words):
    ans = []
    for w in words:
        cnt = 0
        i = 0
        n = len(w)
        while i < n:
            j = i
            while j < n and w[j] == w[i]:
                j += 1
            cnt += (j - i) // 2
            i = j
        ans.append(cnt)
    return ans

class Solution {
    static int[] minimalOperations(String[] words) {
        int[] ans = new int[words.length];
        for (int k = 0; k < words.length; k++) {
            String w = words[k];
            int n = w.length(), cnt = 0, i = 0;
            while (i < n) {
                int j = i;
                while (j < n && w.charAt(j) == w.charAt(i)) j++;
                cnt += (j - i) / 2;
                i = j;
            }
            ans[k] = cnt;
        }
        return ans;
    }
}

vector<int> minimalOperations(vector<string>& words) {
 vector<int> ans;
 for (auto& w : words) {
 int n = w.size(), cnt = 0, i = 0;
 while (i < n) {
 int j = i;
 while (j < n && w[j] == w[i]) j++;
 cnt += (j - i) / 2;
 i = j;
 }
 ans.push_back(cnt);
 }
 return ans;
}

A marketing company tracks advertising campaigns and engagements. Generate a report that, for every active campaign, lists the campaign name, the total number of engagements, and the combined sum of views and clicks. Columns: campaign_name | total_engagements | total_views_and_clicks. Sort by campaign_name ascending.

Schema:

campaigns

engagements

解法

campaigns 与 engagements 按 id = campaign_id 内连接，过滤 is_active = 1，按 campaigns.name 分组，聚合 COUNT(*) 和 SUM(views + clicks)，按 name 排序。若需要展示零互动的 campaign，换成 LEFT JOIN 加 COALESCE。

SELECT
 c.name AS campaign_name,
 COUNT(e.campaign_id) AS total_engagements,
 COALESCE(SUM(e.views + e.clicks), 0) AS total_views_and_clicks
FROM campaigns c
LEFT JOIN engagements e ON e.campaign_id = c.id
WHERE c.is_active = 1
GROUP BY c.id, c.name
ORDER BY c.name ASC;

Q1: Big-O notation definition

Big(O) notation:

• (A) identifies the best algorithm to solve a problem.
• (B) determines maximum size of a problem that can be solved on a system in a given time.
• (C) is the lower bound of the growth rate of an algorithm.
• (D) is the upper bound of the growth rate of an algorithm.
• (E) None of the above.

Answer: D. Big-O describes an asymptotic upper bound of the growth rate; Big-Omega is the lower bound, Big-Theta is tight.

Q2: Quicksort partition pivot index

Given array [76, 97, 25, 115, 83, 51, 18, 36], quicksort with the first element (76) as pivot. After one Lomuto/Hoare partition, what is the 1-indexed position of the pivot?

• (A) 1
• (B) 3
• (C) 5
• (D) 4

Answer: C (5). Elements less than 76 are 25, 51, 18, 36 (4 elements). After partition, 76 sits at index 5 (1-indexed) with the 4 smaller values to its left and 97, 115, 83 to its right.

Q3: SQL IN syntax

Pick the expression with the correct syntax.

• (A) SELECT * FROM logs WHERE priority IN ('high', 'medium')
• (B) SELECT * FROM logs WHERE priority VALUES IN ('high', 'medium')
• (C) SELECT * FROM logs WHERE priority IN VALUES ('high', 'medium')
• (D) SELECT * FROM logs WHERE priority IN SET ('high', 'medium')

Answer: A. Standard SQL IN takes a parenthesized value list directly; VALUES / SET keywords are not valid here.