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Merge pull request #25 from colorbox/373
373. Find K Pairs with Smallest Sums
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‎373/step1.cpp

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/*
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1時間以上
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Time: O( log(max(nums1,nums2)) * n log(m) )
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Space: O( k )
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n = nums1.size()
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m = nums2.size()
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nums1,nums2のすべてのペアを検証すると時間が足りないので、それを補う方法を考える。
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ペアの数がk個以下となるような値Lを二分探索で探す。
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合計値がL以下のペアをpriority_queueにいれて最小k個のペアを探す。
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・二分探索での桁溢れ
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・同値のペアが大量に作成される
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などの問題を都度修正したら過アンリ時間がかかってしまったがかろうじてパスはしたものの
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色々と筋悪な回答なのでstep2以降で修正する。
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*/
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class Solution {
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public:
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vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
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long long int left = nums1.front() + nums2.front();
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long long int right = nums1.back() + nums2.back() + 1;
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long long int mid;
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while (left + 1 < right) {
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mid = left + (right - left) / 2;
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int count = count_less_pairs(nums1, nums2, mid, k);
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if (count >= k) {
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right = mid;
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} else {
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left = mid;
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}
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}
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priority_queue<SumPair> sum_pairs;
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for (int num1 : nums1) {
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for (int num2 : nums2) {
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if (num1 + num2 > right) {
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break;
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}
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sum_pairs.push({num1 + num2, {num1, num2}});
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if (sum_pairs.size() > k) {
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sum_pairs.pop();
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if (num1 + num2 == sum_pairs.top().sum) {
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break;
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}
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}
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}
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}
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vector<vector<int>> answer;
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while (!sum_pairs.empty()) {
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answer.push_back(sum_pairs.top().pair);
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sum_pairs.pop();
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}
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return answer;
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}
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private:
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struct SumPair {
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int sum;
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vector<int> pair;
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bool operator< (const SumPair& other) const {
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return sum < other.sum;
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}
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};
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int count_less_pairs(vector<int> nums1, vector<int>nums2, int num, int limit) {
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int count_sum = 0;
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for (int num1 : nums1) {
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auto it = upper_bound(nums2.begin(), nums2.end(), num - num1);
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if (it == nums2.end()) {
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count_sum += nums2.size();
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if (count_sum > limit) {
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return limit;
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}
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continue;
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}
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int count = it - nums2.begin();
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count_sum += count;
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}
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return count_sum;
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}
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};

‎373/step1_fix.cpp

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/*
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step1 after review
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*/
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class Solution {
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public:
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vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
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int64_t left = nums1.front() + nums2.front();
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int64_t right = nums1.back() + nums2.back() + 1;
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while (left < right) {
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int64_t mid = left + (right - left) / 2;
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int count = count_less_pairs(nums1, nums2, mid, k);
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if (count >= k) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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int max_sum = left;
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priority_queue<SumPair> sum_pairs;
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for (int num1 : nums1) {
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for (int num2 : nums2) {
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if (num1 + num2 > max_sum) {
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break;
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}
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sum_pairs.push({num1 + num2, {num1, num2}});
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if (sum_pairs.size() > k) {
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sum_pairs.pop();
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if (num1 + num2 == sum_pairs.top().sum) {
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break;
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}
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}
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}
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}
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vector<vector<int>> answer;
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while (!sum_pairs.empty()) {
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answer.push_back(sum_pairs.top().pair);
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sum_pairs.pop();
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}
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return answer;
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}
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private:
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struct SumPair {
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int sum;
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vector<int> pair;
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bool operator< (const SumPair& other) const {
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return sum < other.sum;
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}
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};
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int count_less_pairs(vector<int> nums1, vector<int>nums2, int num, int limit) {
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int num_pairs = 0;
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for (int num1 : nums1) {
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auto it = upper_bound(nums2.begin(), nums2.end(), num - num1);
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num_pairs += distance(nums2.begin(), it);
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if (num_pairs > limit) {
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return limit;
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}
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}
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return num_pairs;
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}
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};

‎373/step2.cpp

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/*
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他の方の回答を参考にやり直したもの
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nums1,nums2のインデックスのペアをダイクストラ法っぽく処理していく
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Time: O(k logk)
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Space: O(k)
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*/
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class Solution {
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public:
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vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
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priority_queue<SumIndex, vector<SumIndex>, greater<SumIndex>> sum_index;
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set<vector<int>> visited;
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vector<vector<int>> answer;
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sum_index.push({nums1[0] + nums2[0], {0, 0}});
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while (answer.size() < k) {
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int i = sum_index.top().index[0];
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int j = sum_index.top().index[1];
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sum_index.pop();
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answer.push_back({nums1[i], nums2[j]});
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if (i + 1 < nums1.size() && !visited.contains({i + 1, j})) {
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sum_index.push({nums1[i + 1] + nums2[j], {i + 1, j}});
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visited.insert({i + 1, j});
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}
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if (j + 1 < nums2.size() && !visited.contains({i, j + 1})) {
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sum_index.push({nums1[i] + nums2[j + 1], {i, j + 1}});
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visited.insert({i, j + 1});
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}
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}
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return answer;
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}
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private:
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struct SumIndex {
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int sum;
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vector<int> index;
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bool operator> (const SumIndex& other) const {
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return sum > other.sum;
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}
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};
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};

‎373/step2_2.cpp

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/*
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step2 after review
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*/
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class Solution {
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public:
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vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
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priority_queue<SumIndex> sum_index;
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set<vector<int>> visited;
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vector<vector<int>> answer;
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sum_index.push({nums1[0] + nums2[0], {0, 0}});
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while (answer.size() < k) {
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int i = sum_index.top().index[0];
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int j = sum_index.top().index[1];
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sum_index.pop();
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answer.push_back({nums1[i], nums2[j]});
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if (i + 1 < nums1.size() && !visited.contains({i + 1, j})) {
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sum_index.push({nums1[i + 1] + nums2[j], {i + 1, j}});
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visited.insert({i + 1, j});
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}
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if (j + 1 < nums2.size() && !visited.contains({i, j + 1})) {
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sum_index.push({nums1[i] + nums2[j + 1], {i, j + 1}});
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visited.insert({i, j + 1});
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}
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}
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return answer;
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}
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private:
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struct SumIndex {
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int sum;
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vector<int> index;
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bool operator< (const SumIndex& other) const {
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return sum > other.sum;
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}
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};
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};

‎373/step3_1.cpp

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/*
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struct不使用パターン
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*/
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class Solution {
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public:
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vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
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priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> sum_index;
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set<vector<int>> visited;
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vector<vector<int>> answer;
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sum_index.push({0, 0, 0});
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while (answer.size() < k && sum_index.size() > 0) {
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int i = sum_index.top()[1];
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int j = sum_index.top()[2];
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sum_index.pop();
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answer.push_back({nums1[i], nums2[j]});
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if (i + 1 < nums1.size() && !visited.contains({i + 1, j})) {
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sum_index.push({nums1[i + 1] + nums2[j], i + 1, j});
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visited.insert({i + 1, j});
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}
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if (j + 1 < nums2.size() && !visited.contains({i, j + 1})) {
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sum_index.push({nums1[i] + nums2[j + 1], i, j + 1});
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visited.insert({i, j + 1});
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}
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}
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return answer;
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}
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};

‎373/step3_2.cpp

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/*
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structを使用したパターン
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priority_queueやstructの宣言で若干冗長に感じるが、こちらのほうが読みやすい
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*/
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class Solution {
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public:
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vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
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priority_queue<SumIndex, vector<SumIndex>, greater<SumIndex>> sum_index;
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set<vector<int>> visited;
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vector<vector<int>> minimum_k_pairs;
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sum_index.push({nums1[0] + nums2[0], 0, 0});
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while (minimum_k_pairs.size() < k && sum_index.size() > 0) {
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int i = sum_index.top().num1_index;
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int j = sum_index.top().num2_index;
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sum_index.pop();
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minimum_k_pairs.push_back({nums1[i], nums2[j]});
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if (i + 1 < nums1.size() && !visited.contains({i + 1, j})) {
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sum_index.push({nums1[i + 1] + nums2[j], i+1, j});
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visited.insert({i + 1, j});
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}
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if (j + 1 < nums2.size() && !visited.contains({i, j + 1})) {
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sum_index.push({nums1[i] + nums2[j + 1], i, j + 1});
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visited.insert({i, j + 1});
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}
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}
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return minimum_k_pairs;
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}
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private:
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struct SumIndex {
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int sum;
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int num1_index;
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int num2_index;
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bool operator> (const SumIndex& other) const {
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return sum > other.sum;
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}
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};
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};

‎373/step3_3.cpp

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/*
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{i + 1, j}と{i, j + 1}をpriority_queueにいれる時に、{i + 1, j - 1}と{i - 1, j + 1}がvisitedに含まれているかを確認して効率化するパターン
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条件判定部分が無駄に複雑なので関数化したいが、いい感じにできないのでそのままにしている
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*/
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class Solution {
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public:
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vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
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priority_queue<SumIndex, vector<SumIndex>, greater<SumIndex>> sum_index;
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vector<vector<int>> minimum_k_pair;
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set<vector<int>> visited;
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sum_index.push({nums1[0] + nums2[0], 0, 0});
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while (minimum_k_pair.size() < k && sum_index.size() > 0) {
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int i = sum_index.top().num1_index;
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int j = sum_index.top().num2_index;
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sum_index.pop();
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minimum_k_pair.push_back({nums1[i], nums2[j]});
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if (i + 1 < nums1.size() && !visited.contains({i + 1, j}) && (j == 0 || visited.contains({i + 1, j - 1}))) {
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sum_index.push({nums1[i + 1] + nums2[j], i + 1, j});
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visited.insert({i + 1, j});
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}
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if (j + 1 < nums2.size() && !visited.contains({i, j + 1}) && (i == 0 || visited.contains({i - 1, j + 1}))) {
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sum_index.push({nums1[i] + nums2[j + 1], i, j + 1});
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visited.insert({i, j + 1});
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}
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}
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return minimum_k_pair;
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}
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private:
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struct SumIndex {
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int sum;
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int num1_index;
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int num2_index;
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bool operator> (const SumIndex& other) const {
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return sum > other.sum;
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}
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};
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};

‎373/step3_3_fix.cpp

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/*
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{i + 1, j}と{i, j + 1}をpriority_queueにいれる時に、{i + 1, j - 1}と{i - 1, j + 1}がvisitedに含まれているかを確認して効率化するパターン
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条件判定部分が無駄に複雑なので関数化したいが、いい感じにできないのでそのままにしている
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*/
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class Solution {
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public:
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vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
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priority_queue<SumIndex, vector<SumIndex>, greater<SumIndex>> sum_index;
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vector<vector<int>> minimum_k_pair;
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set<vector<int>> visited;
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sum_index.push({nums1[0] + nums2[0], 0, 0});
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while (minimum_k_pair.size() < k && sum_index.size() > 0) {
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int i = sum_index.top().num1_index;
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int j = sum_index.top().num2_index;
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sum_index.pop();
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minimum_k_pair.push_back({nums1[i], nums2[j]});
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if (i + 1 < nums1.size() && !visited.contains({i + 1, j}) && (j == 0 || visited.contains({i + 1, j - 1}))) {
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sum_index.push({nums1[i + 1] + nums2[j], i + 1, j});
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visited.insert({i + 1, j});
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}
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if (j + 1 < nums2.size() && !visited.contains({i, j + 1}) && (i == 0 || visited.contains({i - 1, j + 1}))) {
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sum_index.push({nums1[i] + nums2[j + 1], i, j + 1});
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visited.insert({i, j + 1});
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}
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}
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return minimum_k_pair;
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}
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private:
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struct SumIndex {
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int sum;
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int num1_index;
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int num2_index;
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bool operator> (const SumIndex& other) const {
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return sum > other.sum;
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}
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};
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};

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