[high frequency interview questions] difficulty 2.5/5, simple combination of DFS trie template level application questions

Title Description

This is a LeetCode Upper 「677. Key value mapping 」 , The difficulty is 「 secondary 」.

Tag : 「 Dictionary tree 」、「DFS」、「 Hashtable 」

Achieve one MapSum class , Two methods are supported ,insert  and  sum：

• MapSum() initialization MapSum object
• void insert(String key, int val) Insert key-val Key value pair , Strings represent keys key , Integers represent values val . If key key Already exist , Then the original key value pair will be replaced by the new key value pair .
• int sum(string prefix) Returns all with this prefix prefix Key at the beginning key The sum of the values of .

Example ：

 Input ：
["MapSum", "insert", "sum", "insert", "sum"]
[[], ["apple", 3], ["ap"], ["app", 2], ["ap"]]

 Output ：
[null, null, 3, null, 5]

 explain ：
MapSum mapSum = new MapSum();
mapSum.insert("apple", 3);  
mapSum.sum("ap");           // return 3 (apple = 3)
mapSum.insert("app", 2);    
mapSum.sum("ap");           // return 5 (apple + app = 3 + 2 = 5)

Tips ：

• key and prefix It's only made up of lowercase letters
• 1 <= val <= 1000
• Call at most 50 Time insert and sum

Trie + DFS

From the need to achieve 「 Store string （ The mapping relationship ）」 and 「 Retrieve the sum of a string prefix 」 Look at , You can know that this is related to Related topics , I don't know yet Students can look at the front ： Realization Trie ( Prefix tree ) .

Consider how to implement two operations ：

• insert ： In the basic Just expand on the basis of the insertion operation . The only difference from the regular insert operation is , You can't simply record the end position of the word , And store Corresponding How much is the . Specifically, we can use int An array of types To replace the original boolean An array of types ;

• sum ： Enter the parameter first Search the dictionary tree , Use after reaching the tail DFS Search all the following schemes , And accumulate the results .

Code （static See , Avoid every sample new Big array ）：

class MapSum {
    int[][] tr = new int[2510][26];
    int[] hash = new int[2510];
    int idx;
    public void insert(String key, int val) {
        int p = 0;
        for (int i = 0; i < key.length(); i++) {
            int u = key.charAt(i) - 'a';
            if (tr[p][u] == 0) tr[p][u] = ++idx;
            p = tr[p][u];
        }
        hash[p] = val;
    }
    public int sum(String prefix) {
        int p = 0;
        for (int i = 0; i < prefix.length(); i++) {
            int u = prefix.charAt(i) - 'a';
            if (tr[p][u] == 0) return 0;
            p = tr[p][u];
        }
        return dfs(p);
    }
    int dfs(int p) {
        int ans = hash[p];
        for (int u = 0; u < 26; u++) {
            if (tr[p][u] != 0) ans += dfs(tr[p][u]);
        }
        return ans;
    }
}

class MapSum {
    static int[][] tr = new int[2510][26];
    static int[] hash = new int[2510];
    static int idx;
    public MapSum() {
        for (int i = 0; i <= idx; i++) Arrays.fill(tr[i], 0);
        Arrays.fill(hash, 0);
        idx = 0;
    }
    public void insert(String key, int val) {
        int p = 0;
        for (int i = 0; i < key.length(); i++) {
            int u = key.charAt(i) - 'a';
            if (tr[p][u] == 0) tr[p][u] = ++idx;
            p = tr[p][u];
        }
        hash[p] = val;
    }
    public int sum(String prefix) {
        int p = 0;
        for (int i = 0; i < prefix.length(); i++) {
            int u = prefix.charAt(i) - 'a';
            if (tr[p][u] == 0) return 0;
            p = tr[p][u];
        }
        return dfs(p);
    }
    int dfs(int p) {
        int ans = hash[p];
        for (int u = 0; u < 26; u++) {
            if (tr[p][u] != 0) ans += dfs(tr[p][u]);
        }
        return ans;
    }
}

• Time complexity ： Make The maximum length of is , The maximum number of calls is , The character set size is （ This topic Fixed for ）, insert The complexity of the operation is ; from DFS From the perspective of , sum The complexity of the operation is , But in fact , For this problem, there is a clear upper bound on the amount of calculation , The complexity of searching all grids is
• Spatial complexity ：

Trie Record the prefix string sum

To reduce sum The complexity of the operation , We can do it in insert Record at the same time during operation （ Add up ） The sum of each prefix .

Code （static See , Avoid every sample new Big array ）：

class MapSum {
    int N = 2510;
    int[][] tr = new int[N][26];
    int[] hash = new int[N];
    int idx;
    Map<String, Integer> map = new HashMap<>();
    public void insert(String key, int val) {
        int _val = val;
        if (map.containsKey(key)) val -= map.get(key);
        map.put(key, _val);
        for (int i = 0, p = 0; i < key.length(); i++) {
            int u = key.charAt(i) - 'a';
            if (tr[p][u] == 0) tr[p][u] = ++idx;
            p = tr[p][u];
            hash[p] += val;
        }
    }
    public int sum(String prefix) {
        int p = 0;
        for (int i = 0; i < prefix.length(); i++) {
            int u = prefix.charAt(i) - 'a';
            if (tr[p][u] == 0) return 0;
            p = tr[p][u];
        }
        return hash[p];
    }
}

class MapSum {
    static int N = 2510;
    static int[][] tr = new int[N][26];
    static int[] hash = new int[N];
    static int idx;
    static Map<String, Integer> map = new HashMap<>();
    public MapSum() {
        for (int i = 0; i <= idx; i++) Arrays.fill(tr[i], 0);
        Arrays.fill(hash, 0);
        idx = 0;
        map.clear();
    }
    public void insert(String key, int val) {
        int _val = val;
        if (map.containsKey(key)) val -= map.get(key);
        map.put(key, _val);
        for (int i = 0, p = 0; i < key.length(); i++) {
            int u = key.charAt(i) - 'a';
            if (tr[p][u] == 0) tr[p][u] = ++idx;
            p = tr[p][u];
            hash[p] += val;
        }
    }
    public int sum(String prefix) {
        int p = 0;
        for (int i = 0; i < prefix.length(); i++) {
            int u = prefix.charAt(i) - 'a';
            if (tr[p][u] == 0) return 0;
            p = tr[p][u];
        }
        return hash[p];
    }
}

• Time complexity ： Make The maximum length of is , insert The complexity of the operation is ; sum The complexity of the operation is
• Spatial complexity ： Make The maximum length of is , The maximum number of calls is , The character set size is （ This topic Fixed for ）, The complexity is

Last

This is us. 「 Brush through LeetCode」 The first of the series No.677 piece , The series begins with 2021/01/01, As of the start date LeetCode I have in common 1916 questions , Part of it is a locked question , We will finish all the questions without lock first .

In this series , In addition to explaining the idea of solving problems , And give the most concise code possible . If the general solution is involved, there will be corresponding code templates .

In order to facilitate the students to debug and submit code on the computer , I've built a warehouse ：https://github.com/SharingSource/LogicStack-LeetCode .

In the warehouse address , You can see the links to the series 、 The corresponding code of the series 、LeetCode Links to the original problem and other preferred solutions .

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