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3945. Digit Frequency Score

Description

You are given an integer n.

The score of n is defined as the sum of d * freq(d) over all distinct digits d, where freq(d) denotes the number of times the digit d appears in n.

Return an integer denoting the score of n.

Β 

Example 1:

Input: n = 122

Output: 5

Explanation:

  • The digit 1 appears 1 time, contributing 1 * 1 = 1.
  • The digit 2 appears 2 times, contributing 2 * 2 = 4.
  • Thus, the score of n is 1 + 4 = 5.

Example 2:

Input: n = 101

Output: 2

Explanation:

  • The digit 0 appears 1 time, contributing 0 * 1 = 0.
  • The digit 1 appears 2 times, contributing 1 * 2 = 2.
  • Thus, the score of n is 2.

Β 

Constraints:

  • 1 <= n <= 109

Solutions

Solution 1: Simulation

The problem is equivalent to finding the sum of each digit of a number. We can obtain each digit by repeatedly taking the modulus and dividing by 10, and accumulate the result.

The time complexity is \(O(\log n)\), where \(\log n\) is the number of digits in \(n\). The space complexity is \(O(1)\).

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class Solution:
    def digitFrequencyScore(self, n: int) -> int:
        ans = 0
        while n:
            n, x = divmod(n, 10)
            ans += x
        return ans

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class Solution {
    public int digitFrequencyScore(int n) {
        int ans = 0;
        for (; n > 0; n /= 10) {
            ans += n % 10;
        }
        return ans;
    }
}

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class Solution {
public:
    int digitFrequencyScore(int n) {
        int ans = 0;
        for (; n > 0; n /= 10) {
            ans += n % 10;
        }
        return ans;
    }
};

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func digitFrequencyScore(n int) (ans int) {
    for ; n > 0; n /= 10 {
        ans += n % 10
    }
    return
}

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function digitFrequencyScore(n: number): number {
    let ans = 0;
    for (; n; n = Math.floor(n / 10)) {
        ans += n % 10;
    }
    return ans;
}

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