Medium
Design a map that allows you to do the following:
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
50
calls will be made to insert
and sum
.class MapSum {
internal class Node {
var `val`: Int = 0
var child: Array<Node?> = arrayOfNulls(26)
}
private val root: Node = Node()
fun insert(key: String, `val`: Int) {
var curr: Node? = root
for (c in key.toCharArray()) {
if (curr!!.child[c.code - 'a'.code] == null) {
curr.child[c.code - 'a'.code] = Node()
}
curr = curr.child[c.code - 'a'.code]
}
curr!!.`val` = `val`
}
private fun sumHelper(root: Node?): Int {
var o = 0
for (i in 0..25) {
if (root!!.child[i] != null) {
o += root.child[i]!!.`val` + sumHelper(root.child[i])
}
}
return o
}
fun sum(prefix: String): Int {
var curr: Node? = root
for (c in prefix.toCharArray()) {
if (curr!!.child[c.code - 'a'.code] == null) {
return 0
}
curr = curr.child[c.code - 'a'.code]
}
val sum = curr!!.`val`
return sum + sumHelper(curr)
}
}
/*
* Your MapSum object will be instantiated and called as such:
* var obj = MapSum()
* obj.insert(key,`val`)
* var param_2 = obj.sum(prefix)
*/