Problem
Given two arrays of unique digits nums1 and nums2, return **the *smallest* number that contains at least one digit from each array**.
Example 1:
Input: nums1 = [4,1,3], nums2 = [5,7]
Output: 15
Explanation: The number 15 contains the digit 1 from nums1 and the digit 5 from nums2. It can be proven that 15 is the smallest number we can have.
Example 2:
Input: nums1 = [3,5,2,6], nums2 = [3,1,7]
Output: 3
Explanation: The number 3 contains the digit 3 which exists in both arrays.
Constraints:
1 <= nums1.length, nums2.length <= 91 <= nums1[i], nums2[i] <= 9- All digits in each array are unique.
Solution (Java)
class Solution {
public int minNumber(int[] nums1, int[] nums2) {
int temp = getCommon(nums1, nums2);
if(temp > 0) return temp;
int a = getMinimum(nums1);
if(isPresent(nums2, a)){
return a;
}
int b = getMinimum(nums2);
if(a < b){
return a*10+b;
}else{
return b*10+a;
}
}
public int getMinimum(int[] nums){
int min = Integer.MAX_VALUE;
int ans;
for(int i = 0; i < nums.length; i++){
min = Math.min(min, nums[i]);
ans = min;
}
return min;
}
public boolean isPresent(int[] nums, int k){
for(int i = 0; i < nums.length; i++){
if(nums[i] == k) return true;
}
return false;
}
public int getCommon(int[] nums1, int[] nums2){
int min = Integer.MAX_VALUE;
List<Integer> list = new ArrayList<Integer>();
int ans=-1;
for(int i = 0; i < nums1.length; i++){
for(int j = 0; j < nums2.length; j++){
if(nums1[i] == nums2[j]){
list.add(nums1[i]);
break;
}
}
}
for(int i=0; i<list.size(); i++){
min = Math.min(min, list.get(i));
ans = min;
}
return ans;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).