Kattis Problem 'Mali'












1












$begingroup$


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.



Input



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.



Output



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;
c++;
if (c == 1) {
as.push_back(a);
bs.push_back(b);
}
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)
break;
}
}
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;
}
}









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    1












    $begingroup$


    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.



    Input



    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.



    Output



    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;
    c++;
    if (c == 1) {
    as.push_back(a);
    bs.push_back(b);
    }
    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)
    break;
    }
    }
    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;
    }
    }









    share|improve this question









    New contributor




    Griffin Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      1












      1








      1





      $begingroup$


      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.



      Input



      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.



      Output



      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;
      c++;
      if (c == 1) {
      as.push_back(a);
      bs.push_back(b);
      }
      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)
      break;
      }
      }
      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;
      }
      }









      share|improve this question









      New contributor




      Griffin Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      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.



      Input



      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.



      Output



      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;
      c++;
      if (c == 1) {
      as.push_back(a);
      bs.push_back(b);
      }
      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)
      break;
      }
      }
      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;
      }
      }






      c++ programming-challenge time-limit-exceeded






      share|improve this question









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      Griffin Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      Griffin Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 13 mins ago









      200_success

      131k17156422




      131k17156422






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      asked 33 mins ago









      Griffin WongGriffin Wong

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      New contributor




      Griffin Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Griffin Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.






















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