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Problem solution · Java

Find Array Given Subset Sums

Find Array Given Subset Sums: 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
43 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 Find Array Given Subset Sums, 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

  • 43 lines of Java from the credited upstream file 1982.java.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • 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 codeFind Array Given Subset Sums · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int[] recoverArray(int n, int[] sums) {    Arrays.sort(sums);    return recover(sums).stream().mapToInt(Integer::intValue).toArray();  }   private List<Integer> recover(int[] sums) {    if (sums.length == 1) // sums[0] must be 0.      return new ArrayList<>();     Map<Integer, Long> count = Arrays.stream(sums).boxed().collect(        Collectors.groupingBy(Function.identity(), Collectors.counting()));     // Either num or -num must be in the final array.    //  num + sumsExcludingNum = sumsIncludingNum    // -num + sumsIncludingNum = sumsExcludingNum    final int num = sums[1] - sums[0];    int i = 0; // sumsExcludingNum/sumsIncludingNum's index    int[] sumsExcludingNum = new int[sums.length / 2];    int[] sumsIncludingNum = new int[sums.length / 2];    boolean chooseSumsIncludingNum = false;     for (final int sum : sums) {      if (count.get(sum) == 0)        continue;      count.merge(sum, -1L, Long::sum);      count.merge(sum + num, -1L, Long::sum);      sumsExcludingNum[i] = sum;      sumsIncludingNum[i] = sum + num;      ++i;      if (sum + num == 0)        chooseSumsIncludingNum = true;    }     // Choose `sumsExludingNum` by default since we want to gradually strip    // `num` from each sum in `sums` to have the final array. However, we should    // always choose the group of sums with 0 since it's a must-have.    List<Integer> recovered = recover(chooseSumsIncludingNum ? sumsIncludingNum : sumsExcludingNum);    recovered.add(chooseSumsIncludingNum ? -num : num);    return recovered;  }}

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