Score an Integer by Digit Frequencies

# Score an Integer by Digit Frequencies You are given a positive integer `n` written in decimal. For each distinct digit `d` that appears in `n`, let `freq(d)` be its number of occurrences. Define the score as the sum of `d * freq(d)` over those distinct digits. Return the score of `n`. ### Function `int digitFrequencyScore(int n)` ### Example 1 **Input:** `n = 700707` **Output:** `21` **Explanation:** Digit `7` appears three times, contributing `7 * 3 = 21`. Digit `0` also appears three times and contributes `0`. Their total is `21`. ### Example 2 **Input:** `n = 1000000000` **Output:** `1` **Explanation:** Digit `1` appears once and digit `0` appears nine times. The score is `1 * 1 + 0 * 9 = 1`. ### Constraints - `1 <= n <= 1000000000` ### Clarifications - Frequencies are taken from the ordinary base-10 representation of the supplied integer. - A repeated digit contributes its value once for every occurrence, through its frequency. - A zero digit contributes `0`, regardless of how many times it appears. ### What a Strong Answer Covers - Applying the frequency weight to every distinct digit correctly. - Handling repeated digits, internal zeros and the ten-digit upper endpoint. - Returning the score without changing the supplied integer outside the function.