Problem
Decision trees split data into groups that are as pure as possible. Gini impurity measures how mixed a group's labels are:
Gini = 1 - Σ pₖ²
where pₖ is the fraction of labels equal to class k. A group with a single class has impurity 0; an even mix of two classes has impurity 0.5.
Implement gini_impurity(labels).
Input
labels: The class labels in the group.
Output
Return the Gini impurity as a decimal number.
Constraints
1 ≤ length ≤ 10,000
Each label is 1 to 20 characters
Examples
Input:
labels = ["yes", "yes", "no", "no"]
Expected output:
expected = 0.5Input:
std::vector<std::string> labels = {"yes", "yes", "no", "no"};
Expected output:
double expected = 0.5;Input:
string[] labels = new string[] {"yes", "yes", "no", "no"};
Expected output:
double expected = 0.5;Explanation
Half are yes and half are no: 1 - (0.5² + 0.5²) = 1 - 0.5 = 0.5, the most mixed a two-class group can be.
Input:
labels = ["cat", "cat", "cat"]
Expected output:
expected = 0.0Input:
std::vector<std::string> labels = {"cat", "cat", "cat"};
Expected output:
double expected = 0.0;Input:
string[] labels = new string[] {"cat", "cat", "cat"};
Expected output:
double expected = 0.0;Explanation
Only one class: 1 - 1² = 0. The group is pure.
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
Count how often each label appears, because impurity depends only on the label distribution.
Hint 2
Convert counts to fractions of the whole group, then measure how far the group is from being pure.
Hint 3
A single-label group should give zero, while a balanced mix should give a larger value.
Expected complexity
O(n) time, O(k) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.