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LESSON 133 OF 667 · DICTIONARIES - CHALLENGES

Count a Vote with an unordered_map in C++

Practice Read the full explanation

You get votes, an unordered_map 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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    C++ · DICTIONARIES - CHALLENGES

    Count A Vote in C++, explained

    Problem overview

    The method countVote gets votes, an unordered_map<string, int> from names to vote counts, and the name on a new vote. The person called name gets one more vote, and the method returns votes. A name that has no votes yet is not in the map, so it starts from 0 and ends with 1. The order of entries does not matter.

    With {"Ana": 3, "Ben": 5} and "Ana", Ana goes to 4. With "Cara", a new key appears with 1. With an empty map and "Max", the result is {"Max": 1}.

    Approach

    This is the "default, then check" pattern. Start with a count of 0, then read the real count only if the name is a key. Whatever the count is, store count + 1 under the name. For a new name that adds the key, and for an existing one it replaces the old number with the new one.

    Step by step

    1. Create int count = 0;. It will hold the votes the person has so far.
    2. Write an if whose condition is votes.count(name) > 0. Inside it, set count to votes[name]. The check keeps the map unchanged for a missing name.
    3. After the if, store count + 1 under the key: votes[name] = count + 1;.
    4. Return votes. Only the entry of name changed.

    The solution in C++

    class Solution {
public:
    unordered_map<string, int> countVote(unordered_map<string, int>& votes, string name) {
        int count = 0;
        if (votes.count(name) > 0) {
            count = votes[name];
        }
        votes[name] = count + 1;
        return votes;
    }
};

    Running the method on the statement's examples, with entries printed as name: count lines sorted by name, gives:

    Ana: 4
Ben: 5

    For "Cara" the map becomes Ana: 3, Ben: 5, Cara: 1. For the empty map and "Max" it becomes Max: 1.

    Another way

    For an unordered_map<string, int>, a missing key read with [] is created with 0, so votes[name]++ does the whole job in one line. Applied twice to the key "x" of an empty map, it gives x: 2. The lesson's version spells out each step, which is clearer while you are learning how maps behave. Both give the same results.

    Trying more inputs

    Start with an empty map and count three votes for "Ana" and one for "Bo". The first call creates Ana with 1, the next two raise her to 3, and Bo is created with 1. The final map is Ana: 3 and Bo: 1. Each call is independent: it needs only the current map and the name.

    The same method also works if the map already holds many names, since only one key is looked up and changed each time.

    Complexity

    Time is O(1) on average and space is O(1) extra, because each step is one hash-map lookup or store.

    Edge cases and common mistakes

    • Forgetting the first vote. Storing the new name with 0 instead of 1 loses the vote. The expression is count + 1, and count is 0 for a new name.
    • Putting the store inside the if. Then new names are never added. The store must come after the if.
    • Returning a different value. The method returns the whole map, not the new count.
    • Empty map. It needs no special case, because count(name) is 0 and the store creates the entry.
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