Google – Minimum Word Set with Frequency Sum Greater than Target


Google – Minimum Word Set with Frequency Sum Greater than Target
给一篇文章,里面有很多单词,整篇文章可以全部装入内存,再给一个target number, 要求返回一个最小的set,使得set里那些单词的出现频率总和大于等于target number
[Solution]
#1 sort
#2 quick select
这道题quick select的思路的确很巧。但是仔细想一下这道题除了sort之外一时半会儿也想不到其他方法。而凡是能用sort做的题目,都多根经尝试一下quick select。
思路就是每次partition出来一个index, 计算[L, index]这个区间的sum和target的大小关系,来判断递归左边还是右边。区间sum在partition的过程中就可以算出来,所以其实就是一个quick select过程,没有其他额外开销。
[Note]
注意在getResult和getSum之间取index的不同,getSum取的区间应该是[L, index], 而getResult必须要从0开始,取[0, index]。
因为再递归过程中,如果当前sum < target, 递归右边的时候target会变成target – sum,就像kth smallest number那种做法一样,所以在getSum和target比较的时候,就是在当前[L, R]的期间里。
但是到了getResult, 情况就不一样了,必须得从0开始来对应最开始Input的target number.
[Code]
下面的code是quick select的算法,但是偷懒了getSum没有在partition的过程中算,这样的话时间复杂度可能和sort差不多。
// Quick Select, O(m), where m is the number of distinct words
class Solution2 {
  public Set<String> getWords(List<String> file, int target) {
    Set<String> result = new HashSet<>();
    if (file == null || file.isEmpty()) {
      return result;
    }
     
    Map<String, Integer> cntMap = new HashMap<>();
    for (String word : file) {
      cntMap.put(word, cntMap.getOrDefault(word, 0) + 1);
    }
     
    List<Map.Entry<String, Integer>> freq = new ArrayList<>(cntMap.entrySet());
     
    quickSelect(freq, 0, freq.size() - 1, target);
    return result;
  }
   
  private void quickSelect(List<Map.Entry<String, Integer>> freq, int l, int r, int target, Set<String> result) {
    if (l > r) {
      return;
    }
    if (l == r) {
      getResult(freq, 0, l, result);
      return
    }
     
    int index = partition(freq, l, r);
     
    int sum = getSum(freq, l, index);
    if (sum == target) {
      getResult(freq, 0, index, result);
    }
    else if (sum < target) {
      quickSelect(freq, index + 1, r, target - sum, result);
    }
    else {
      quickSelect(freq, l, index, sum, result);
    }
  }
   
  private int partition(List<Map.Entry<String, Integer>> freq, int l, int r) {
    int pivot = freq.get(l).getValue();
    int left = l;
    int right = r + 1;
    while (true) {
      while (++left < r && freq.get(left).getValue() < pivot);
      while (--right > l && freq.get(right).getValue() > pivot);
      if (left < right) {
        swap(freq, left, right);
      }
      else {
        break;
      }
    }
    swap(freq, l, rigth);
     
    return right;
  }
   
  private int getSum(List<Map.Entry<String, Integer>> freq, int l, int r) {
    int sum = 0;
    for (int i = l; i <= r; i++) {
      sum += freq.get(i).getValue();
    }
    return sum;
  }
   
  private void getResult(List<Map.Entry<String, Integer>> freq, int l, int r, Set<String> result) {
    for (int i = l; i <= r; i++) {
      result.add(freq.get(i).getKey());
    }
  }
}
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