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Kattis Problem 'Mali'
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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
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;
}
}
c++ programming-challenge time-limit-exceeded
edited Apr 2 at 5:41
200_success
131k17157422
asked Apr 2 at 5:21
Griffin WongGriffin Wong
132
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.
$endgroup$
add a comment |
$begingroup$
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;
}
}
c++ programming-challenge time-limit-exceeded
edited Apr 2 at 5:41
200_success
131k17157422
asked Apr 2 at 5:21
Griffin WongGriffin Wong
132
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.
$endgroup$
$begingroup$
(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 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 at 6:27
add a comment |
$begingroup$
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;
}
}
c++ programming-challenge time-limit-exceeded
edited Apr 2 at 5:41
200_success
131k17157422
asked Apr 2 at 5:21
Griffin WongGriffin Wong
132
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
c++ programming-challenge time-limit-exceeded
edited Apr 2 at 5:41
200_success
131k17157422
asked Apr 2 at 5:21
Griffin WongGriffin Wong
132
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.
edited Apr 2 at 5:41
200_success
131k17157422
asked Apr 2 at 5:21
Griffin WongGriffin Wong
132
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.
edited Apr 2 at 5:41
200_success
131k17157422
edited Apr 2 at 5:41
200_success
131k17157422
edited Apr 2 at 5:41
200_success
131k17157422
131k17157422
asked Apr 2 at 5:21
Griffin WongGriffin Wong
132
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.
asked Apr 2 at 5:21
Griffin WongGriffin Wong
132
asked Apr 2 at 5:21
Griffin WongGriffin Wong
132
132
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.
Griffin Wong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
(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 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 at 6:27
add a comment |
$begingroup$
(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 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 at 6:27
$begingroup$
(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 at 6:21
$begingroup$
(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 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 at 6:27
$begingroup$
(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 at 6:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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):
- The
if (c == 1)special case should be eliminated. - Flag variables (
foundaandfoundb) suck. You would be better off writing two separateforloops.
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
asandbstakes 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.
answered Apr 2 at 5:59
200_success200_success
131k17157422
$endgroup$
$begingroup$
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 at 6:52
add a comment |
Your Answer
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$begingroup$
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):
- The
if (c == 1)special case should be eliminated. - Flag variables (
foundaandfoundb) suck. You would be better off writing two separateforloops.
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
asandbstakes 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.
answered Apr 2 at 5:59
200_success200_success
131k17157422
$endgroup$
$begingroup$
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 at 6:52
add a comment |
$begingroup$
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):
- The
if (c == 1)special case should be eliminated. - Flag variables (
foundaandfoundb) suck. You would be better off writing two separateforloops.
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
asandbstakes 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.
answered Apr 2 at 5:59
200_success200_success
131k17157422
$endgroup$
$begingroup$
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 at 6:52
add a comment |
$begingroup$
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):
- The
if (c == 1)special case should be eliminated. - Flag variables (
foundaandfoundb) suck. You would be better off writing two separateforloops.
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
asandbstakes 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.
answered Apr 2 at 5:59
200_success200_success
131k17157422
$endgroup$
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):
- The
if (c == 1)special case should be eliminated. - Flag variables (
foundaandfoundb) suck. You would be better off writing two separateforloops.
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
asandbstakes 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.
answered Apr 2 at 5:59
200_success200_success
131k17157422
answered Apr 2 at 5:59
200_success200_success
131k17157422
answered Apr 2 at 5:59
200_success200_success
131k17157422
answered Apr 2 at 5:59
200_success200_success
131k17157422
131k17157422
$begingroup$
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 at 6:52
add a comment |
$begingroup$
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 at 6:52
$begingroup$
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 at 6:52
$begingroup$
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 at 6:52
add a comment |
Griffin Wong is a new contributor. Be nice, and check out our Code of Conduct.
Griffin Wong is a new contributor. Be nice, and check out our Code of Conduct.
Griffin Wong is a new contributor. Be nice, and check out our Code of Conduct.
Griffin Wong is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
(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 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 at 6:27