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Problem solution · C++

Minimum Cost to Make Array Equalindromic

Minimum Cost to Make Array Equalindromic: 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
33 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 Minimum Cost to Make Array Equalindromic, 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

  • 33 lines of C++ from the credited upstream file 2967.cpp.
  • The implementation visibly relies on sequence storage.
  • 1 loop block 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 codeMinimum Cost to Make Array Equalindromic · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  long long minimumCost(vector<int>& nums) {    ranges::sort(nums);    const int median = nums[nums.size() / 2];    const int nextPalindrome = getPalindrome(median, /*delta=*/1);    const int prevPalindrome = getPalindrome(median, /*delta=*/-1);    return min(cost(nums, nextPalindrome), cost(nums, prevPalindrome));  }  private:  // Returns the cost to change all the numbers to `palindrome`.  long cost(const vector<int>& nums, int palindrome) {    return accumulate(nums.begin(), nums.end(), 0L,                      [palindrome](long acc, int num) {      return acc + abs(palindrome - num);    });  }   // Returns the palindrome `p`, where p = num + a * delta and a > 0.  int getPalindrome(int num, int delta) {    while (!isPalindrome(num))      num += delta;    return num;  }   bool isPalindrome(int num) {    const string original = to_string(num);    const string reversed = {original.rbegin(), original.rend()};    return original == reversed;  }};

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