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3754. Concatenate Non-Zero Digits and Multiply by Sum I

Description

You are given an integer n.

Form a new integer x by concatenating all the non-zero digits of n in their original order. If there are no non-zero digits, x = 0.

Let sum be the sum of digits in x.

Return an integer representing the value of x * sum.

 

Example 1:

Input: n = 10203004

Output: 12340

Explanation:

  • The non-zero digits are 1, 2, 3, and 4. Thus, x = 1234.
  • The sum of digits is sum = 1 + 2 + 3 + 4 = 10.
  • Therefore, the answer is x * sum = 1234 * 10 = 12340.

Example 2:

Input: n = 1000

Output: 1

Explanation:

  • The non-zero digit is 1, so x = 1 and sum = 1.
  • Therefore, the answer is x * sum = 1 * 1 = 1.

 

Constraints:

  • 0 <= n <= 109

Solutions

Solution 1: Simulation

Thinking

\(n\) has few digits, so we may follow the definition directly. Peeling digits from the low end, each nonzero digit updates both the concatenated integer \(x\) and the digit sum \(s\); the answer is \(x\cdot s\).

We can simulate the required operation by processing the number digit by digit. While processing each digit, we concatenate non-zero digits to form a new integer \(x\) and calculate the digit sum \(s\). Finally, we return \(x \times s\).

The time complexity is \(O(\log n)\) and the space complexity is \(O(1)\).

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class Solution:
    def sumAndMultiply(self, n: int) -> int:
        p = 1
        x = s = 0
        while n:
            n, v = divmod(n, 10)
            if v:
                s += v
                x += p * v
                p *= 10
        return x * s

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class Solution {
    public long sumAndMultiply(int n) {
        int p = 1;
        int x = 0, s = 0;
        for (; n > 0; n /= 10) {
            int v = n % 10;
            if (v != 0) {
                s += v;
                x += p * v;
                p *= 10;
            }
        }
        return 1L * x * s;
    }
}

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class Solution {
public:
    long long sumAndMultiply(int n) {
        int p = 1;
        int x = 0, s = 0;
        for (; n > 0; n /= 10) {
            int v = n % 10;
            if (v != 0) {
                s += v;
                x += p * v;
                p *= 10;
            }
        }
        return 1LL * x * s;
    }
};

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func sumAndMultiply(n int) int64 {
    p := 1
    x := 0
    s := 0
    for n > 0 {
        v := n % 10
        if v != 0 {
            s += v
            x += p * v
            p *= 10
        }
        n /= 10
    }
    return int64(x) * int64(s)
}

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function sumAndMultiply(n: number): number {
    let p = 1;
    let x = 0;
    let s = 0;

    while (n > 0) {
        const v = n % 10;
        if (v !== 0) {
            s += v;
            x += p * v;
            p *= 10;
        }
        n = Math.floor(n / 10);
    }

    return x * s;
}

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impl Solution {
    pub fn sum_and_multiply(mut n: i32) -> i64 {
        let mut p = 1;
        let mut x = 0;
        let mut s = 0;

        while n > 0 {
            let v = n % 10;
            if v != 0 {
                s += v;
                x += p * v;
                p *= 10;
            }
            n /= 10;
        }

        x as i64 * s as i64
    }
}

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public class Solution {
    public long SumAndMultiply(int n) {
        int p = 1;
        int x = 0, s = 0;

        while (n > 0) {
            int v = n % 10;
            if (v != 0) {
                s += v;
                x += p * v;
                p *= 10;
            }
            n /= 10;
        }

        return 1L * x * s;
    }
}

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/**
 * @param {number} n
 * @return {number}
 */
var sumAndMultiply = function (n) {
    let p = 1;
    let x = 0;
    let s = 0;

    while (n > 0) {
        const v = n % 10;
        if (v !== 0) {
            s += v;
            x += p * v;
            p *= 10;
        }
        n = Math.floor(n / 10);
    }

    return x * s;
};

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