1class Solution {
2    /**
3     * Calculates the total late fee for all overdue items.
4     *
5     * Fee rules per item:
6     * - Exactly 1 day late:  fee is 1
7     * - 2 to 5 days late:    fee is 2 * daysLate
8     * - More than 5 days:    fee is 3 * daysLate
9     *
10     * @param daysLate array where each element is the number of days an item is late
11     * @return the sum of late fees for all items
12     */
13    public int lateFee(int[] daysLate) {
14        int totalFee = 0;
15
16        // Accumulate the fee for each overdue item
17        for (int days : daysLate) {
18            if (days == 1) {
19                // Late by exactly one day: flat fee of 1
20                totalFee += 1;
21            } else if (days > 5) {
22                // Late by more than five days: triple rate
23                totalFee += 3 * days;
24            } else {
25                // Late by 2 to 5 days: double rate
26                totalFee += 2 * days;
27            }
28        }
29
30        return totalFee;
31    }
32}
33```
34
35Key changes made:
36
371. **Removed the `IntUnaryOperator` lambda** — replaced it with a straightforward `if-else` chain inside the loop. This avoids the functional-interface overhead and keeps the logic inline, which is simpler and more idiomatic for such a small computation.
382. **Standardized naming** — `ans` became `totalFee` (describes what it holds), and the loop variable `x` became `days` (describes what it represents). The method name `lateFee` is preserved.
393. **Added comments** — a Javadoc block documents the fee rules and parameters, and inline comments explain each branch.
40
41An alternative perspective: if you prefer keeping the functional style, you could use a stream-based one-liner instead:
42
43```java
44class Solution {
45    public int lateFee(int[] daysLate) {
46        // Map each overdue duration to its fee, then sum them all
47        return java.util.Arrays.stream(daysLate)
48                .map(days -> days == 1 ? 1 : days > 5 ? 3 * days : 2 * days)
49                .sum();
50    }
51}
52

1class Solution {
2public:
3    int lateFee(vector<int>& daysLate) {
4        // Lambda function to calculate the fee for a single rental
5        // based on the number of days it is late.
6        auto calculateFee = [](int days) {
7            if (days == 1) {
8                // Late by exactly 1 day: flat fee of 1
9                return 1;
10            } else if (days > 5) {
11                // Late by more than 5 days: fee is 3 per day
12                return 3 * days;
13            } else {
14                // Late by 2 to 5 days: fee is 2 per day
15                return 2 * days;
16            }
17        };
18
19        int totalFee = 0;
20        // Accumulate the fee for each late rental
21        for (int days : daysLate) {
22            totalFee += calculateFee(days);
23        }
24        return totalFee;
25    }
26};
27```
28
29**Explanation of the changes:**
30
311. **Standardized naming**:
32   - `f` → `calculateFee`: describes the lambda's purpose clearly.
33   - `x` → `days`: conveys that the variable represents days late.
34   - `ans` → `totalFee`: indicates it holds the accumulated result.
35   - The method name `lateFee` remains unchanged as required.
36
372. **Comments added**: Each branch of the fee logic and the accumulation loop is documented, making the tiered pricing structure (1 day / 2–5 days / over 5 days) immediately clear to readers.
38
393. **Logic preserved**: The algorithm is identical — an O(n) single pass over `daysLate` with O(1) extra space.
40
41**Alternative perspective**: If you prefer a more functional style, the loop could be replaced with `std::accumulate`:
42
43```cpp
44int totalFee = accumulate(daysLate.begin(), daysLate.end(), 0,
45    [&](int sum, int days) { return sum + calculateFee(days); });
46

1/**
2 * Calculates the late fee for a single item based on the number of days it is late.
3 *
4 * Fee rules:
5 * - Exactly 1 day late: fee is 1.
6 * - More than 5 days late: fee is 3 per day.
7 * - Otherwise (2 to 5 days late): fee is 2 per day.
8 *
9 * @param days - The number of days an item is late.
10 * @returns The late fee for that item.
11 */
12const calculateFee = (days: number): number => {
13    if (days === 1) {
14        // Only 1 day late: flat fee of 1
15        return 1;
16    } else if (days > 5) {
17        // More than 5 days late: 3 per day
18        return 3 * days;
19    }
20    // Between 2 and 5 days late: 2 per day
21    return 2 * days;
22};
23
24/**
25 * Computes the total late fee for all items.
26 *
27 * @param daysLate - An array where each element represents the number of days an item is late.
28 * @returns The sum of late fees for all items.
29 */
30function lateFee(daysLate: number[]): number {
31    // Accumulate the fee of each item into a total sum
32    return daysLate.reduce((totalFee, days) => totalFee + calculateFee(days), 0);
33}
34