I wrote a solution to the Mali problem on Kattis:

Given inputs a1, a2, a3, …, an and b1, b2, b3, …, bn, determine n pairings (ai, bj) such that each number in the A sequence is used in exactly one pairing, each number in the B sequence is used in exactly one pairing, and the maximum of all sums ai + bj is minimal.


The first line of input contains a single integer N (1 ≤ N ≤ 100000), the number of rounds.

The next N lines contain two integers A and B (1 ≤ A, B ≤ 100), the numbers given in that round.


The output consists of N lines, one for each round. Each line should contain the smallest maximal sum for that round.

I'm fairly sure I can get the right answer 100% of the time, but the code exceeds the time limit of one second. I was wondering if you could help me optimize my code to help it run in time, or if you could explain why my code is inefficient.

#include <iostream>
#include <vector>

int main() {
    int x, c = 0, big = 0;
    std::cin >> x;
    std::vector<int> as, bs;
    for (int i = 0; i < x; i++) {
        bool founda = false, foundb = false;
        int a, b;
        std::cin >> a >> b;
        if (c == 1) {
        else {
            for (int i = 0; i < c; i++) {
                if (as[i] < a || i == c-1) {
                    as.insert(as.begin()+i, a);
                    founda = true;
                if (bs[i] < b || i == c-1) {
                    bs.insert(bs.begin()+i, b);
                    foundb = true;
                if (founda == true && foundb == true) 
        for (int i = 0; i < c; i++) {
            if (as[i] + bs[c-1-i] > big)
                big = as[i] + bs[c-1-i];
        std::cout << big << std::endl;
  • \$\begingroup\$ (Programming challenges with "online judges" are carefully crafted with one good solution in mind that can be coded in, say, one hour. Problem parameters (size) are chosen such that asymptotically inferior approaches don't complete within the time limit. The implementation platform (language) is figured in: you need to find a solution that re-uses information "between sums/pairs".) \$\endgroup\$
    – greybeard
    Apr 2, 2019 at 6:21
  • \$\begingroup\$ (For lack of code comments, I didn't try to understand your approach upfront. Integrating input and both preparatory sorts is thinking out-of-the-box!) \$\endgroup\$
    – greybeard
    Apr 2, 2019 at 6:27

1 Answer 1


You are storing all of the A and B inputs in sorted vectors, basically doing an insertion sort. A few observations (two minor, one major):

  1. The if (c == 1) special case should be eliminated.
  2. Flag variables (founda and foundb) suck. You would be better off writing two separate for loops.
  3. Each time you insert a value, all values beyond the insertion point need to be shifted over to make room for the insertion. That means that constructing as and bs takes O(N2) time altogether — which is unacceptable, since N can be very large.

    However, there are only 100 possible values of A and B. So, do a counting sort instead, building a histogram with 100 buckets. That would be a much more efficient way to represent the same information, using O(N) time and a fixed amount of space.

  • \$\begingroup\$ Usually a stickler for detail, I commend giving neither more nor less than the principle of (the crucial part of) a good solution. \$\endgroup\$
    – greybeard
    Apr 2, 2019 at 6:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.