2522-partition-string-into-substrings-with-values-at-most-k.cpp

class Solution {
public:
    int minimumPartition(string s, int k) {
        long long n = 0, res = 1;
        for (int i = 0; i < s.size(); ++i) {
            if (n * 10 + (s[i] - '0') > k) {
                ++res;
                n = 0;
            }
            n = n * 10 + (s[i] - '0');
            if (n > k)
                return -1;
        }
        return res;
    }
};