Return the minimum, maximum, mean, median, and mode of the sample respectively, as an array of floating point numbers.  The mode is guaranteed to be unique.

(Recall that the median of a sample is:

• The middle element, if the elements of the sample were sorted and the number of elements is odd;
• The average of the middle two elements, if the elements of the sample were sorted and the number of elements is even.)

Example 1:

Input: count = [0,1,3,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: [1.00000,3.00000,2.37500,2.50000,3.00000]

Example 2:

Input: count = [0,4,3,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: [1.00000,4.00000,2.18182,2.00000,1.00000]

Constraints:

1. count.length == 256
2. 1 <= sum(count) <= 10^9
3. The mode of the sample that count represents is unique.
4. Answers within 10^-5 of the true value will be accepted as correct.

Solution: TreeMap

Time complexity: O(1)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  vector<double> sampleStats(vector<int>& count) {
    int counts = 0;
    int minVal = 255;
    int maxVal = 0;
    int mode = -1;
    int modeCount = 0;
    long sum = 0;
    map<int, int> m;
    for (int i = 0; i <= 255; ++i) {      
      if (!count[i]) continue;
      sum += static_cast<long>(i) * count[i];
      minVal = min(minVal, i);
      maxVal = max(maxVal, i);
      if (count[i] > modeCount) {
        mode = i;
        modeCount = count[i];
      }
      m[counts += count[i]] = i;
    }
    auto it1 = m.lower_bound(counts / 2);
    double median = it1->second;    
    auto it2 = next(it1);
    if (counts % 2 == 0 && it2 != m.end() && it1->first == counts / 2) {  
      median = (median + it2->second) / 2;
    }
    return {minVal, maxVal, static_cast<double>(sum) / counts, median, mode};
  }
};

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