CombinationSumII.java

/* (C) 2024 YourCompanyName */
package backtracking;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
 * Created by gouthamvidyapradhan on 14/03/2017. Given a collection of candidate numbers (C) and a
 * target number (T), find all unique combinations in C where the candidate numbers sums to T.
 *
 * <p>Each number in C may only be used once in the combination.
 *
 * <p>Note: All numbers (including target) will be positive integers. The solution set must not
 * contain duplicate combinations. For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and
 * target 8, A solution set is: [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]
 */
public class CombinationSumII {
  /**
   * Main method
   *
   * @param args
   * @throws Exception
   */
  public static void main(String[] args) throws Exception {
    int[] candidates = {1, 1, 2, 2};
    List<List<Integer>> result = new CombinationSumII().combinationSum2(candidates, 4);
  }

  public List<List<Integer>> combinationSum2(int[] candidates, int target) {
    Arrays.sort(candidates);
    List<List<Integer>> result = new ArrayList<>();
    combination(0, target, candidates, new ArrayList<>(), result);
    return result;
  }

  private void combination(
      int i, int target, int[] candidates, List<Integer> row, List<List<Integer>> result) {
    if (target == 0) {
      result.add(new ArrayList<>(row));
    } else if (target > 0) {
      for (int j = i, l = candidates.length; j < l; j++) {
        if (j > i && candidates[j] == candidates[j - 1]) continue;
        row.add(candidates[j]);
        combination(j + 1, target - candidates[j], candidates, row, result);
        row.remove(row.size() - 1);
      }
    }
  }
}