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3270. Find the Key of the Numbers

SourceBiweekly Contest 138 Q1DifficultyEasyRating1205

Description

You are given three positive integers num1, num2, and num3.

The key of num1, num2, and num3 is defined as a four-digit number such that:

  • Initially, if any number has less than four digits, it is padded with leading zeros.
  • The ith digit (1 <= i <= 4) of the key is generated by taking the smallest digit among the ith digits of num1, num2, and num3.

Return the key of the three numbers without leading zeros (if any).

 

Example 1:

Input: num1 = 1, num2 = 10, num3 = 1000

Output: 0

Explanation:

On padding, num1 becomes "0001", num2 becomes "0010", and num3 remains "1000".

  • The 1st digit of the key is min(0, 0, 1).
  • The 2nd digit of the key is min(0, 0, 0).
  • The 3rd digit of the key is min(0, 1, 0).
  • The 4th digit of the key is min(1, 0, 0).

Hence, the key is "0000", i.e. 0.

Example 2:

Input: num1 = 987, num2 = 879, num3 = 798

Output: 777

Example 3:

Input: num1 = 1, num2 = 2, num3 = 3

Output: 1

 

Constraints:

  • 1 <= num1, num2, num3 <= 9999

Solutions

Solution 1: Simulation

Thinking

Digit \(d\) of the key is the minimum of the three numbers' \(d\)-th digits, missing high digits treated as zero. There are four digits, so take minima and add with place weights.

For \(k=1,10,100,1000\) add \(\min_i (x_i//k)\% 10\) times \(k\). No string conversion is required.

We can directly simulate this process by defining a variable \(\textit{ans}\) to store the answer and a variable \(\textit{k}\) to represent the current digit place, where \(\textit{k} = 1\) represents the units place, \(\textit{k} = 10\) represents the tens place, and so on.

Starting from the units place, for each digit place, we calculate the current digit of \(\textit{num1}\), \(\textit{num2}\), and \(\textit{num3}\), take the minimum of the three, and then add this minimum value multiplied by \(\textit{k}\) to the answer. Then, multiply \(\textit{k}\) by 10 and continue to the next digit place.

Finally, return the answer.

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

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class Solution:
    def generateKey(self, num1: int, num2: int, num3: int) -> int:
        ans, k = 0, 1
        for _ in range(4):
            x = min(num1 // k % 10, num2 // k % 10, num3 // k % 10)
            ans += x * k
            k *= 10
        return ans

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class Solution {
    public int generateKey(int num1, int num2, int num3) {
        int ans = 0, k = 1;
        for (int i = 0; i < 4; ++i) {
            int x = Math.min(Math.min(num1 / k % 10, num2 / k % 10), num3 / k % 10);
            ans += x * k;
            k *= 10;
        }
        return ans;
    }
}

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class Solution {
public:
    int generateKey(int num1, int num2, int num3) {
        int ans = 0, k = 1;
        for (int i = 0; i < 4; ++i) {
            int x = min({num1 / k % 10, num2 / k % 10, num3 / k % 10});
            ans += x * k;
            k *= 10;
        }
        return ans;
    }
};

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func generateKey(num1 int, num2 int, num3 int) (ans int) {
    k := 1
    for i := 0; i < 4; i++ {
        x := min(min(num1/k%10, num2/k%10), num3/k%10)
        ans += x * k
        k *= 10
    }
    return
}

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function generateKey(num1: number, num2: number, num3: number): number {
    let [ans, k] = [0, 1];
    for (let i = 0; i < 4; ++i) {
        const x = Math.min(((num1 / k) | 0) % 10, ((num2 / k) | 0) % 10, ((num3 / k) | 0) % 10);
        ans += x * k;
        k *= 10;
    }
    return ans;
}

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