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LESSON 134 OF 671 · DICTIONARIES - CHALLENGES

Count a Vote in Java: getOrDefault and put

Practice Read the full explanation

You get votes, a HashMap from names to how many votes each person has. You also get the name on a new vote.

Count the new vote and return votes.

The person called name gets one more vote. A name with no votes yet isn't in votes. It starts from 0, so after this vote it has 1. The order of the entries doesn't matter.

 

Example 1:

Input: votes = {"Ana": 3, "Ben": 5}, name = "Ana"
Output: {"Ana": 4, "Ben": 5}
Explanation: Ana had 3 votes. Now she has 3 + 1 = 4.

Example 2:

Input: votes = {"Ana": 3, "Ben": 5}, name = "Cara"
Output: {"Ana": 3, "Ben": 5, "Cara": 1}
Explanation: Cara had no votes yet. She starts from 0 and now has 1.

Example 3:

Input: votes = {}, name = "Max"
Output: {"Max": 1}
Explanation: This is the very first vote. Max now has 1.

 

Constraints:

  • votes can be empty.
  • Every count in votes is at least 1.
  • Names are single words made of English letters.
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    JAVA · DICTIONARIES - CHALLENGES

    Count A Vote in Java, explained

    Problem overview

    The method countVote receives votes, a HashMap from names to how many votes each person has, and the name on a new vote. It gives that person one more vote and returns the map. A person with no votes yet is not in the map; they start from 0, so after this vote they have 1. The map may be empty, and the order of entries does not matter.

    With Ana at 3 and Ben at 5, a vote for Ana makes it 4. A vote for Cara, who is missing, adds Cara with 1. In an empty map, the first vote for Max gives Max 1.

    Approach

    Both cases, a known name and a new one, follow the same rule: new count = old count + 1, where the old count is 0 for a missing name. If you read the old count with get, a missing name gives null, and unboxing that into an int throws a NullPointerException. getOrDefault(name, 0) returns the stored count, or 0 when the name is missing, and it does not add the key by itself. Then one put stores the new count: it adds a new key, or replaces the value of an existing one. No if is needed.

    Step by step

    1. Read the count so far: int count = votes.getOrDefault(name, 0);.
    2. Store the increased count: votes.put(name, count + 1).
    3. Return votes. Only the entry for name changed.

    The solution in Java

    class Solution {
    public HashMap<String, Integer> countVote(HashMap<String, Integer> votes, String name) {
        int count = votes.getOrDefault(name, 0);
        votes.put(name, count + 1);
        return votes;
    }
}

    Details worth knowing:

    • The map holds Integer objects, and Java converts them to and from int automatically (boxing and unboxing). That conversion is why a null read is a problem and a default is the fix.
    • The method changes the map it receives and returns that same map.
    • count + 1 is computed before put runs, so the same line works whether the key is new or old.

    Edge cases and common mistakes

    • Using get for the first read. int count = votes.get(name); works for Ana but throws NullPointerException for Cara, who has no entry.
    • Storing the old count. put(name, count) without the + 1 never changes anything.
    • Forgetting to put. Computing count + 1 and not storing it leaves the map as it was.
    • Empty map. getOrDefault returns 0, so the first vote creates the entry with 1.

    One pattern for two jobs

    Read with a default, add one, store: this three-step shape appears whenever you count things in a map, such as letters in a word or votes in a poll. The default of 0 is what makes the first occurrence behave like every later one. The map returned is the same object that was passed in: after the call, votes already holds the new count.

    Testing

    Results, printed as key: value lines sorted by key, as the lesson's tests print them:

    • Ana 3, Ben 5, vote for Ana gives Ana: 4, Ben: 5.
    • Ana 3, Ben 5, vote for Cara gives Ana: 3, Ben: 5, Cara: 1.
    • An empty map, vote for Max gives Max: 1.
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