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Problem solution · C++

Threshold Majority Queries

Threshold Majority Queries: a C++ solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Binary search
Source
Kamyu LeetCode Solutions
Length
180 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Binary search

For Threshold Majority Queries, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 180 lines of C++ from the credited upstream file threshold-majority-queries.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 13 loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeThreshold Majority Queries · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(nlogn + qlogq + (n + q) * sqrt(n) + q * n)// Space: O(n + q) // sort, coordinate compression, mo's algorithmclass Solution {public:    vector<int> subarrayMajority(vector<int>& nums, vector<vector<int>>& queries) {        vector<int> sorted_nums(nums);        sort(begin(sorted_nums), end(sorted_nums));        sorted_nums.erase(unique(begin(sorted_nums), end(sorted_nums)), end(sorted_nums));        unordered_map<int, int> num_to_idx;        for (int i = 0; i < size(sorted_nums); ++i) {            num_to_idx[sorted_nums[i]] = i;        }                 // reference: https://cp-algorithms.com/data_structures/sqrt_decomposition.html        const auto& mo_s_algorithm = [&]() {            vector<int> cnt(size(num_to_idx));            vector<int> cnt2(size(nums) + 1);            int max_freq = 0;            const auto& add = [&](int i) {                const auto& idx = num_to_idx[nums[i]];                if (cnt[idx]) {                    --cnt2[cnt[idx]];                }                ++cnt[idx];                ++cnt2[cnt[idx]];                max_freq = max(max_freq, cnt[idx]);            };             const auto& remove = [&](int i) {                const auto& idx = num_to_idx[nums[i]];                --cnt2[cnt[idx]];                if (!cnt2[max_freq]) {                    --max_freq;                }                --cnt[idx];                if (cnt[idx]) {                    ++cnt2[cnt[idx]];                }            };             const auto& get_ans = [&](int t) {                if (max_freq < t) {                    return -1;                }                int i = 0;                for (; i < size(cnt); ++i) {                    if (cnt[i] == max_freq) {                        break;                    }                }                return sorted_nums[i];            };             vector<int> result(size(queries), -1);            const int block_size = sqrt(size(nums)) + 1;            vector<int> idxs(size(queries));            iota(begin(idxs), end(idxs), 0);            sort(begin(idxs), end(idxs), [&](const auto& a, const auto& b) {                const auto& i = queries[a][0] / block_size;                const auto& j = queries[b][0] / block_size;                return i != j ? i < j : (i & 1 ? queries[a][1] < queries[b][1] : queries[a][1] > queries[b][1]);            });            int left = 0, right = -1;            for (const auto& i : idxs) {                const auto& l = queries[i][0];                const auto& r = queries[i][1];                const auto& t = queries[i][2];                while (left > l) {                    left -= 1;                    add(left);                }                while (right < r) {                    ++right;                    add(right);                }                while (left < l) {                    remove(left);                    ++left;                }                while (right > r) {                    remove(right);                    --right;                }                result[i] = get_ans(t);            }            return result;        };         return mo_s_algorithm();    }}; // Time:  O(nlogn + qlogq + (n + q) * sqrt(n) * logn)// Space: O(n + q)// sort, coordinate compression, mo's algorithm, bstclass Solution2 {public:    vector<int> subarrayMajority(vector<int>& nums, vector<vector<int>>& queries) {        vector<int> sorted_nums(nums);        sort(begin(sorted_nums), end(sorted_nums));        sorted_nums.erase(unique(begin(sorted_nums), end(sorted_nums)), end(sorted_nums));        unordered_map<int, int> num_to_idx;        for (int i = 0; i < size(sorted_nums); ++i) {            num_to_idx[sorted_nums[i]] = i;        }             // reference: https://cp-algorithms.com/data_structures/sqrt_decomposition.html        const auto& mo_s_algorithm = [&]() {            vector<int> cnt(size(num_to_idx));            vector<multiset<int>> lookup(size(nums) + 1);            int max_freq = 0;            const auto& add = [&](int i) {                const auto& idx = num_to_idx[nums[i]];                if (cnt[idx]) {                    auto it = lookup[cnt[idx]].find(nums[i]);                    lookup[cnt[idx]].erase(it);                }                ++cnt[idx];                lookup[cnt[idx]].emplace(nums[i]);                max_freq = max(max_freq, cnt[idx]);            };             const auto& remove = [&](int i) {                const auto& idx = num_to_idx[nums[i]];                auto it = lookup[cnt[idx]].find(nums[i]);                lookup[cnt[idx]].erase(it);                if (empty(lookup[max_freq])) {                    --max_freq;                }                --cnt[idx];                if (cnt[idx]) {                    lookup[cnt[idx]].emplace(nums[i]);                }            };             const auto& get_ans = [&](int t) {                return max_freq >= t ? *begin(lookup[max_freq]) : -1;            };             vector<int> result(size(queries), -1);            const int block_size = sqrt(size(nums)) + 1;            vector<int> idxs(size(queries));            iota(begin(idxs), end(idxs), 0);            sort(begin(idxs), end(idxs), [&](const auto& a, const auto& b) {                const auto& i = queries[a][0] / block_size;                const auto& j = queries[b][0] / block_size;                return i != j ? i < j : (i & 1 ? queries[a][1] < queries[b][1] : queries[a][1] > queries[b][1]);            });            int left = 0, right = -1;            for (const auto& i : idxs) {                const auto& l = queries[i][0];                const auto& r = queries[i][1];                const auto& t = queries[i][2];                while (left > l) {                    left -= 1;                    add(left);                }                while (right < r) {                    ++right;                    add(right);                }                while (left < l) {                    remove(left);                    ++left;                }                while (right > r) {                    remove(right);                    --right;                }                result[i] = get_ans(t);            }            return result;        };         return mo_s_algorithm();    }};

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