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Problem solution · Java

Minimum Array Changes to Make Differences Equal

Minimum Array Changes to Make Differences Equal: a Java solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
37 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

Hash-based lookup

For Minimum Array Changes to Make Differences Equal, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 37 lines of Java from the credited upstream file 3224.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 3 loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

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 Array Changes to Make Differences Equal · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int minChanges(int[] nums, int k) {    final int n = nums.length;    final int pairSize = n / 2;    int ans = n;    Map<Integer, Integer> diffCount = new HashMap<>(); // {abs(nums[-1 - i] - nums[i]): freq}    // oneChangeCount[i] := the number of pairs that need only one change to    // achieve a difference of `i`    int[] oneChangeCount = new int[k + 1];     for (int i = 0; i < pairSize; ++i) {      final int a = nums[i];      final int b = nums[n - 1 - i];      diffCount.merge(Math.abs(a - b), 1, Integer::sum);      ++oneChangeCount[Math.max(Math.max(a, b), Math.max(k - a, k - b))];    }     // prefixOneChangeCount[i] := the number of pairs that need only one change    // to achieve a difference >= `i`    // prefixOneChangeCount[i] = sum(oneChangeCount[i..k])    int[] prefixOneChangeCount = oneChangeCount.clone();     for (int i = k - 1; i >= 0; --i)      prefixOneChangeCount[i] += prefixOneChangeCount[i + 1];     for (Map.Entry<Integer, Integer> entry : diffCount.entrySet()) {      final int diff = entry.getKey();      final int freq = entry.getValue();      final int oneChange = prefixOneChangeCount[diff] - freq;      final int twoChanges = (pairSize - prefixOneChangeCount[diff]) * 2;      ans = Math.min(ans, oneChange + twoChanges);    }     return ans;  }}

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