CPMagical CP Library

Problem solution · C++

Power Update After K Th Largest Insertion I

Power Update After K Th Largest Insertion I: a C++ solution using segment tree or range structure. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Segment tree or range structure
Source
Kamyu LeetCode Solutions
Length
137 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

Segment tree or range structure

For Power Update After K Th Largest Insertion I, the implementation stores interval information in a range-query data structure so updates and queries avoid rescanning the full input.

  1. Choose the aggregate stored for each interval or prefix.
  2. Build or initialize the structure from the input.
  3. Apply updates and combine the affected nodes to answer each query.

Code notes

  • 137 lines of C++ from the credited upstream file power-update-after-k-th-largest-insertion-i.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 11 loop blocks detected.

Complexity

Count the build once, then multiply the logarithmic update or query path by the number of operations.

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 codePower Update After K Th Largest Insertion I · C++C++
Use this to learn the idea, then write your own version.
// Time:  O((n + q) * log(n * q) + q * logr)// Space: O(n + q) #include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>using namespace __gnu_pbds; // ordered set, fast exponentiationclass Solution {public:    vector<int> powerUpdate(vector<int>& nums, int p, vector<vector<int>>& queries) {        static const uint32_t MOD = 1e9 + 7;         const auto& powmod = [&](uint32_t a, uint32_t b, uint32_t mod) {            a %= mod;            uint32_t result = 1;            while (b) {                if (b & 1) {                    result = (static_cast<uint64_t>(result) * a) % mod;                }                a = (static_cast<uint64_t>(a) * a) % mod;                b >>= 1;            }            return result;        };         using ordered_set = tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update>;        ordered_set os;        int i = 0;        for (; i < size(nums); ++i) {            os.insert({nums[i], i});        }        vector<int> result;        result.reserve(size(queries));        for (const auto& q : queries) {            os.insert({q[0], i++});            p = powmod(p, os.find_by_order(size(os) - q[1])->first, MOD);            result.emplace_back(p);        }        return result;    }}; // Time:  O((n + q) * log(n * q) + q * logr)// Space: O(n + q)// sort, coordinate compression, fenwick tree, fast exponentiationclass Solution2 {public:    vector<int> powerUpdate(vector<int>& nums, int p, vector<vector<int>>& queries) {        static const uint32_t MOD = 1e9 + 7;         const auto& powmod = [&](uint32_t a, uint32_t b, uint32_t mod) {            a %= mod;            uint32_t result = 1;            while (b) {                if (b & 1) {                    result = (static_cast<uint64_t>(result) * a) % mod;                }                a = (static_cast<uint64_t>(a) * a) % mod;                b >>= 1;            }            return result;        };         vector<int> sorted_vals(nums);        for (const auto& q : queries) {            sorted_vals.emplace_back(q[0]);        }        ranges::sort(sorted_vals);        sorted_vals.erase(unique(begin(sorted_vals), end(sorted_vals)), end(sorted_vals));        unordered_map<int, int> val_to_idx;        for (int i = 0; i < size(sorted_vals); ++i) {            val_to_idx[sorted_vals[i]] = i;        }        BIT bit(size(val_to_idx));        for (const auto& x : nums) {            bit.add(val_to_idx[x], +1);        }        vector<int> result;        result.reserve(size(queries));        int total = size(nums);        for (const auto& q : queries) {            bit.add(val_to_idx[q[0]], +1);            const auto& i = bit.kth_element(++total - q[1] + 1);            p = powmod(p, sorted_vals[i], MOD);            result.emplace_back(p);        }        return result;    } private:    class BIT {    public:        BIT(int n) : bit_(n + 1) {  // 0-indexed        }                void add(int i, int val) {            ++i;            for (; i < size(bit_); i += lower_bit(i)) {                bit_[i] += val;            }        }         int query(int i) const {            ++i;            int total = 0;            for (; i > 0; i -= lower_bit(i)) {                total += bit_[i];            }            return total;        }                int kth_element(int k) const {            int total = 0;            int pos = 0;            for (int i = floor_log2_x(size(bit_) - 1); i >= 0; --i) {                if (pos + (1 << i) < size(bit_) && !(total + bit_[pos + (1 << i)] >= k)) {                    total += bit_[pos + (1 << i)];                    pos += (1 << i);                }            }            return (pos + 1) - 1;        }        private:        int lower_bit(int i) const {            return i & -i;        }                int floor_log2_x(int x) const {            return bit_width(static_cast<uint32_t>(x)) - 1;        };                vector<int> bit_;    };};

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗