you can use bit masking here.

for any number i from 1 — 2^10

ans[i] = maximum number in the original array which consist of digits same as the set bits in number i,

ex — i = 7, ans[i] = maximum number in the original array which consists of digits 0, 1, 2 as 2^0 + 2^1 + 2^2 = 7

now after calculating maximums you can find maximum ans by considering two numbers i,j in range 1 — 2^10

if((i&j)==0) final_ans = max(final_ans,ans[i]+ans[j])

def max_sum_without_common_digit(nums): max_num = {} max_sum = -1

for num in nums:
    max_val = int(''.join(sorted(str(num), reverse=True)))
    max_num[max_val] = max(max_num.get(max_val, 0), num)

for num in nums:
    for key in max_num.keys():
        if not any(digit in str(num) for digit in str(key)):
            max_sum = max(max_sum, num + max_num[key])

return max_sum

nums = [15, 0, 105] print(max_sum_without_common_digit(nums))

int solution(vector<int>& nums) {
    unordered_map<int, int> max_num;
    int max_sum = -1;
    for (int num : nums) {
        string num_str = to_string(num);
        sort(num_str.rbegin(), num_str.rend());
        int max_val = stoi(num_str);
        max_num[max_val] = max(max_num[max_val], num); 
    }
    for (int num : nums) {
        string num_str = to_string(num);
        for (auto& it : max_num) {
            string key_str = to_string(it.first);
            if (none_of(key_str.begin(), key_str.end(), [&num_str](char c) { return num_str.find(c) != string::npos; })) { 
                max_sum = max(max_sum, num + it.second);
            }
        }
    }

    return max_sum;
}

Working well on TC's