# My solutions, what's the time complexity?

• ``````class Solution {
public:
vector<int> maxNumber(vector<int>& A, vector<int>& B, int k) {
vector<int> ans;
int a = -1, b = -1;
int m = A.size(), n = B.size();
for (int c = 0; c < k; c++) {
int ea = m-1, eb = n-1;
while (ea > a || eb > b) {
int i = findMaxDigit(A, a+1, ea, B, b+1, eb); //inclusive
if (i < m) {
if (m - (i + 1) + n - (b + 1) >= k - (c+1)) { // found in A
a = i; ans.push_back(A[i]); break;
} else {
ea = i - 1;
}
} else {
i -= m;
if (m - (a + 1) + n - (i + 1) >= k - (c+1)) { // found in B
b = i; ans.push_back(B[i]); break;
} else {
eb = i - 1;
}
}
}
}
return ans;
}
int findMaxDigit(vector<int>& A, int sa, int ea, vector<int>& B, int sb, int eb) // inclusive
{
int max = INT_MIN, pos = -1;
for (int i = sa; i <= ea; i++) {
if (A[i] > max) {
max = A[i];
pos = i;
}
}
for (int i = sb; i <= eb; i++) {
if (B[i] > max) {
max = B[i];
pos = i + A.size();
}
}
return pos;
}
};``````

• this solution is not cover all

example

``````[3,9,5]
[8,9,9]
5``````

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