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Problem solution · C++

Count Number of Balanced Permutations

Count Number of Balanced Permutations: 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
94 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 Count Number of Balanced Permutations, 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

  • 94 lines of C++ from the credited upstream file 3343.cpp.
  • The implementation visibly relies on sequence storage.
  • 6 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 codeCount Number of Balanced Permutations · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int countBalancedPermutations(string num) {    vector<int> nums = getNums(num);    const int sum = accumulate(nums.begin(), nums.end(), 0);    if (sum % 2 == 1)      return 0;     ranges::sort(nums, greater<>());     const int even = (nums.size() + 1) / 2;    const int odd = nums.size() / 2;    const int evenBalance = sum / 2;    vector<vector<vector<long>>> mem(        even + 1,        vector<vector<long>>(odd + 1, vector<long>(evenBalance + 1, -1)));    const long perm = getPerm(nums);    return countBalancedPermutations(nums, even, odd, evenBalance, mem) *           modInverse(perm) % kMod;  }  private:  static constexpr int kMod = 1'000'000'007;   // Returns the number of permutations where there are `even` even indices  // left, `odd` odd indices left, and `evenBalance` is the target sum of the  // remaining numbers to be placed in even indices.  long countBalancedPermutations(const vector<int>& nums, int even, int odd,                                 int evenBalance,                                 vector<vector<vector<long>>>& mem) {    if (evenBalance < 0)      return 0;    if (even == 0)      return evenBalance == 0 ? factorial(odd) : 0;    const int index = nums.size() - (even + odd);    if (odd == 0) {      long remainingSum = 0;      for (int i = index; i < nums.size(); ++i)        remainingSum += nums[i];      return (remainingSum == evenBalance) ? factorial(even) : 0;    }    if (mem[even][odd][evenBalance] != -1)      return mem[even][odd][evenBalance];    const long placeEven =        countBalancedPermutations(nums, even - 1, odd,                                  evenBalance - nums[index], mem) *        even % kMod;    const long placeOdd =        countBalancedPermutations(nums, even, odd - 1, evenBalance, mem) * odd %        kMod;    return mem[even][odd][evenBalance] = (placeEven + placeOdd) % kMod;  }   vector<int> getNums(const string& num) {    vector<int> nums;    for (const char c : num)      nums.push_back(c - '0');    return nums;  }   long getPerm(const vector<int>& nums) {    long res = 1;    vector<int> count(10);    for (const int num : nums)      ++count[num];    for (const int freq : count)      res = res * factorial(freq) % kMod;    return res;  }   long factorial(int n) {    long res = 1;    for (int i = 2; i <= n; ++i)      res = res * i % kMod;    return res;  }   long modInverse(long a) {    long m = kMod;    long y = 0;    long x = 1;    while (a > 1) {      const long q = a / m;      long t = m;      m = a % m;      a = t;      t = y;      y = x - q * y;      x = t;    }    return x < 0 ? x + kMod : x;  }};

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