677. Map Sum Pairs¶

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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.

Solution¶

class MapSum {
    private HashMap<String , Integer> map;
    public MapSum() {
        map = new HashMap<>();
    }
    public void insert(String key, int val) {
        map.put(key , val);
    }
    public int sum(String prefix) {
        int Total_Sum = 0;
        for (Map.Entry<String, Integer> curr : map.entrySet()) {
            String key = curr.getKey();
            if (key.startsWith(prefix)) Total_Sum += curr.getValue();
        }
        return Total_Sum;
    }
}
/**
 * Your MapSum object will be instantiated and called as such:
 * MapSum obj = new MapSum();
 * obj.insert(key,val);
 * int param_2 = obj.sum(prefix);
 */

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here