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Java-based LeetCode algorithm problem solutions, regularly updated
package g0601_0700.s0677_map_sum_pairs;
// #Medium #String #Hash_Table #Design #Trie #2022_03_22_Time_17_ms_(70.10%)_Space_43.5_MB_(21.07%)
/**
* 677 - Map Sum Pairs\.
*
* Medium
*
* Design a map that allows you to do the following:
*
* * Maps a string key to a given value.
* * Returns the sum of the values that have a key with a prefix equal to a given string.
*
* Implement the `MapSum` class:
*
* * `MapSum()` Initializes the `MapSum` object.
* * `void insert(String key, int val)` Inserts the `key-val` pair into the map. If the `key` already existed, the original `key-value` pair will be overridden to the new one.
* * `int sum(string prefix)` Returns the sum of all the pairs' value whose `key` starts with the `prefix`.
*
* **Example 1:**
*
* **Input**
*
* ["MapSum", "insert", "sum", "insert", "sum"] [ [],
*
* ["apple", 3], ["ap"], ["app", 2], ["ap"]]
*
* **Output:** [null, null, 3, null, 5]
*
* **Explanation:**
*
* 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)
*
* **Constraints:**
*
* * `1 <= key.length, prefix.length <= 50`
* * `key` and `prefix` consist of only lowercase English letters.
* * `1 <= val <= 1000`
* * At most `50` calls will be made to `insert` and `sum`.
**/
public class MapSum {
static class Node {
int val;
Node[] child;
Node() {
child = new Node[26];
val = 0;
}
}
private final Node root;
public MapSum() {
root = new Node();
}
public void insert(String key, int val) {
Node curr = root;
for (char c : key.toCharArray()) {
if (curr.child[c - 'a'] == null) {
curr.child[c - 'a'] = new Node();
}
curr = curr.child[c - 'a'];
}
curr.val = val;
}
private int sumHelper(Node root) {
int o = 0;
for (int i = 0; i < 26; i++) {
if (root.child[i] != null) {
o += root.child[i].val + sumHelper(root.child[i]);
}
}
return o;
}
public int sum(String prefix) {
Node curr = root;
for (char c : prefix.toCharArray()) {
if (curr.child[c - 'a'] == null) {
return 0;
}
curr = curr.child[c - 'a'];
}
int sum = curr.val;
return sum + sumHelper(curr);
}
}
/*
* Your MapSum object will be instantiated and called as such:
* MapSum obj = new MapSum();
* obj.insert(key,val);
* int param_2 = obj.sum(prefix);
*/
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