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Problem solution · C++

Find the Sum of Subsequence Powers

Find the Sum of Subsequence Powers: a C++ solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sorting and greedy selection
Source
walkccc LeetCode Solutions
Length
53 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

Sorting and greedy selection

For Find the Sum of Subsequence Powers, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 53 lines of C++ from the credited upstream file 3098.cpp.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

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

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind the Sum of Subsequence Powers · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int sumOfPowers(vector<int>& nums, int k) {    const int n = nums.size();    ranges::sort(nums);    vector<vector<vector<vector<int>>>> mem(        n + 1, vector<vector<vector<int>>>(                   n + 1, vector<vector<int>>(n + 1, vector<int>(k + 1, -1))));    return sumOfPowers(nums, 0, k, -1, -1, -1, mem);  }  private:  static constexpr int kMod = 1'000'000'007;   // Returns the sum of powers of all subsequences of nums[i..n) which  // have length equal to k, where `lastPickedIndex` is the index of the last  // picked number and nums[secondIndex] - nums[firstIndex] is the minimum power  // so far.  int sumOfPowers(const vector<int>& nums, int i, int k, int lastPickedIndex,                  int firstIndex, int secondIndex,                  vector<vector<vector<vector<int>>>>& mem) {    if (k == 0)      return nums[secondIndex] - nums[firstIndex];    if (i == nums.size())      return 0;    const int a = hash(lastPickedIndex);    const int b = hash(firstIndex);    const int c = hash(secondIndex);    if (mem[a][b][c][k] != -1)      return mem[a][b][c][k];    int newFirstIndex = firstIndex;    int newSecondIndex = secondIndex;    if (firstIndex == -1) {      newFirstIndex = i;    } else if (secondIndex == -1) {      newSecondIndex = i;    } else if (nums[i] - nums[lastPickedIndex] <               nums[secondIndex] - nums[firstIndex]) {      newFirstIndex = lastPickedIndex;      newSecondIndex = i;    }    const int pick =        sumOfPowers(nums, i + 1, k - 1, i, newFirstIndex, newSecondIndex, mem);    const int skip = sumOfPowers(nums, i + 1, k, lastPickedIndex, firstIndex,                                 secondIndex, mem);    return mem[a][b][c][k] = (pick + skip) % kMod;  }   constexpr int hash(int x) {    return x + 1;  }};

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