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Problem solution · C++

Find the Index of Permutation

Find the Index of Permutation: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
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

Direct simulation

For Find the Index of Permutation, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 53 lines of C++ from the credited upstream file 3109.cpp.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 Index of Permutation · C++C++
Use this to learn the idea, then write your own version.
class FenwickTree { public:  FenwickTree(int n) : sums(n + 1) {}   void add(int i, int delta) {    while (i < sums.size()) {      sums[i] += delta;      i += lowbit(i);    }  }   int get(int i) const {    int sum = 0;    while (i > 0) {      sum += sums[i];      i -= lowbit(i);    }    return sum;  }  private:  vector<int> sums;   static inline int lowbit(int i) {    return i & -i;  }}; class Solution { public:  int getPermutationIndex(vector<int>& perm) {    constexpr int kMod = 1'000'000'007;    const int n = perm.size();    int ans = 0;    FenwickTree tree(n);    vector<int> fact(n + 1, 1);  // fact[i] := i!     for (int i = 2; i <= n; ++i)      fact[i] = (fact[i - 1] * static_cast<long>(i)) % kMod;     for (int i = 0; i < n; ++i) {      const int num = perm[i];      // the number of unused numbers less than `num`      const int unusedNums = num - 1 - tree.get(num - 1);      const int suffixLength = fact[n - 1 - i];      ans = (ans + unusedNums * static_cast<long>(suffixLength)) % kMod;      tree.add(num, 1);    }     return ans;  }};

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