Find the Key of the Numbers - Problem

Math Easy

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).

Input & Output

Example 1 — Basic Case

$ Input: num1 = 1, num2 = 22, num3 = 333

› Output: 1

💡 Note: Pad to 4 digits: 0001, 0022, 0333. Position-wise minimums: 0,0,0,1 → Key = 0001 → 1

Example 2 — All Same Length

$ Input: num1 = 1234, num2 = 5678, num3 = 9012

› Output: 1012

💡 Note: Already 4 digits. Position-wise minimums: min(1,5,9)=1, min(2,6,0)=0, min(3,7,1)=1, min(4,8,2)=2 → 1012

Example 3 — Leading Zeros Result

$ Input: num1 = 987, num2 = 123, num3 = 456

› Output: 123

💡 Note: Pad to 4 digits: 0987, 0123, 0456. Position-wise minimums: 0,1,2,3 → Key = 0123 → 123

Constraints

1 ≤ num1, num2, num3 ≤ 9999
All three numbers are positive integers

Visualization

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Asked in

G Google 15 a Amazon 12

The key insight is to pad all numbers to 4 digits and compare position by position. Best approach is string conversion for simplicity. Time: O(1), Space: O(1)

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Mathematical Digit Extraction

⏱️ Time: O(1) Space: O(1)

Extract digits mathematically using modulo and division operations, then compare digits at each position to build the result.

String Conversion Approach

⏱️ Time: O(1) Space: O(1)

Convert each number to a string, pad with leading zeros to make 4 digits, then iterate through each position to find the minimum digit.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int min(int a, int b) {
    return a < b ? a : b;
}

// Convert number to 4-digit string with leading zeros
void to4digits(int num, char* result) {
    sprintf(result, "%04d", num);
}

// Find key by taking minimum digit at each position
int findKey(int num1, int num2, int num3) {
    char str1[5], str2[5], str3[5];
    char key[5] = {0};
    
    // Convert all numbers to 4-digit strings
    to4digits(num1, str1);
    to4digits(num2, str2);
    to4digits(num3, str3);
    
    // Find minimum digit at each position
    for (int i = 0; i < 4; i++) {
        char minDigit = str1[i];
        if (str2[i] < minDigit) minDigit = str2[i];
        if (str3[i] < minDigit) minDigit = str3[i];
        key[i] = minDigit;
    }
    key[4] = '\0';
    
    // Convert back to integer (removes leading zeros automatically)
    return atoi(key);
}

int main() {
    int num1, num2, num3;
    
    // Read three integers from stdin
    scanf("%d", &num1);
    scanf("%d", &num2);
    scanf("%d", &num3);
    
    int result = findKey(num1, num2, num3);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

8.5K Views

Medium Frequency

~8 min Avg. Time

350 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

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