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Problem solution · Java

Minimum Cost to Make Array Equalindromic

Minimum Cost to Make Array Equalindromic: a Java 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
28 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

  • 28 lines of Java from the credited upstream file 2967.java.
  • 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 · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public long minimumCost(int[] nums) {    Arrays.sort(nums);    final int median = nums[nums.length / 2];    final int nextPalindrome = getPalindrome(median, 1);    final int prevPalindrome = getPalindrome(median, -1);    return Math.min(cost(nums, nextPalindrome), cost(nums, prevPalindrome));  }   // Returns the cost to change all the numbers to `palindrome`.  private long cost(int[] nums, int palindrome) {    return Arrays.stream(nums).mapToLong(num -> Math.abs(palindrome - num)).sum();  }   // Returns the palindrome `p`, where p = num + a * delta and a > 0.  private int getPalindrome(int num, int delta) {    while (!isPalindrome(num))      num += delta;    return num;  }   private boolean isPalindrome(int num) {    final String original = Integer.toString(num);    final String reversed = new StringBuilder(original).reverse().toString();    return original.equals(reversed);  }}

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