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Form Smallest Number From Two Digit Arrays — Easy Problem & Solution

nums1 and nums2 each hold distinct digits from 1 to 9. Return the smallest positive integer that contains at least one digit from nums1 and at least one…

Problem statement

nums1 and nums2 each hold distinct digits from 1 to 9.

Return the smallest positive integer that contains at least one digit from nums1 and at least one digit from nums2.

Example 1

Input: nums1 = [4,1,3], nums2 = [5,7]
Output: 15
Explanation: No shared digit, so pair the two smallest: `15` beats `51`.

Example 2

Input: nums1 = [3,5,2,6], nums2 = [3,1,7]
Output: 3
Explanation: `3` is in both arrays, so one digit suffices.

Example 3

Input: nums1 = [9], nums2 = [2]
Output: 29

Constraints

  • 1 <= nums1.length, nums2.length <= 9
  • 1 <= nums1[i], nums2[i] <= 9
  • All digits in each array are unique.

How to solve Form Smallest Number From Two Digit Arrays

A shared digit gives a one-digit answer, and one digit always beats two — so take the smallest common digit if there is one. Otherwise pick the smallest digit from each array and return the smaller of the two arrangements.

Approach

  1. Find the smallest digit present in both arrays; return it if one exists.
  2. Otherwise take a = min(nums1) and b = min(nums2).
  3. Return min(10a + b, 10b + a).

Why it works

Fewer digits always wins for positive integers, which is why the shared-digit case is checked first and needs no comparison against any two-digit candidate. In the two-digit case, using anything but the minimum of each array can only make one of the two positions larger, so the smallest digits are forced and only their order is a real choice.

Complexity

  • Time — O(n · m), or O(n + m) with a set
  • Space — O(1)

Pitfalls

  • A shared digit beats every two-digit answer, however small the digits.
  • Both orders must be tried: min(nums1) is not always the leading digit.
  • The digits are 1–9, so there is no leading-zero case to worry about.

Reference solution

Python

from typing import List

def minNumber(nums1: List[int], nums2: List[int]) -> int:
    shared = set(nums1) & set(nums2)
    if shared:
        return min(shared)
    a, b = min(nums1), min(nums2)
    return min(a * 10 + b, b * 10 + a)

JavaScript

var minNumber = function(nums1, nums2) {
    var shared = 10, i, j;
    for (i = 0; i < nums1.length; i++) {
        for (j = 0; j < nums2.length; j++) {
            if (nums1[i] === nums2[j] && nums1[i] < shared) shared = nums1[i];
        }
    }
    if (shared < 10) return shared;
    var a = 10, b = 10;
    for (i = 0; i < nums1.length; i++) if (nums1[i] < a) a = nums1[i];
    for (j = 0; j < nums2.length; j++) if (nums2[j] < b) b = nums2[j];
    var first = a * 10 + b, second = b * 10 + a;
    return first < second ? first : second;
};

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