The Indian Engineer

Problem 3270 Find the Key of Numbers

Posted on 2 mins

String

Table of Contents

Problem Statement

Link - Problem 3270

Question

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:

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`

Solution 

class Solution {
public:
    int generateKey(int num1, int num2, int num3) {
        std::ios::sync_with_stdio(false);
        string str1 = addzeros(num1);
        string str2 = addzeros(num2);
        string str3 = addzeros(num3);

        string key = "";
        for (int i = 0; i < 4; ++i) {
            char mindigit = min({str1[i], str2[i], str3[i]});
            key += mindigit;
        }

        return stoi(key);
    }

    string addzeros(int num) {
        string str = to_string(num);
        while (str.length() < 4) {
            str = "0" + str;
        }
        return str;
    }
};

Complexity Analysis 

| Algorithm | Time Complexity | Space Complexity |
| --------- | --------------- | ---------------- |
| Traverse  | O(1)            | O(1)             |

Explanation

1. Intuition

This is a simple problem where we need to just implement what the question asks for. There is no need for any complex algorithm.

- For easy manipulation, we convert the numbers to strings and add leading zeros.
- We then iterate over the strings and find the minimum digit at each position.
- We then return the key as an integer.

2. Implementation