Question

Suppose you have a sorted range (x to y) of values in an array.

x = 3; y = 11;  array == 3, 4, 5, 6, 7, 8, 9, 10, 11 

But it is possible that some values are duplicated and some are missing, so you might have:

array == 4, 5, 5, 5, 7, 8, 9, 10, 10 

What’s the best way in your language to find all duplicates and missing values so you get:

resultMissingValuesArray == 3, 6, 11 resultDuplicatesArray == 5, 5, 10 

Here’s some C++ code to get you started:

#include <vector> #include <iostream> #include <algorithm>  using namespace std;  const int kLastNumber = 50000; // last number expected in array const int kFirstNumber = 3; // first number expected in array  int main() {     vector<int> myVector;      // fill up vector, skip values at the beginning and end to check edge cases     for(int x = kFirstNumber + 5; x < kLastNumber - 5; x++)     {            if(x % 12 != 0 &&  x % 13 != 0 && x % 17 != 0)             myVector.push_back(x);  // skip some values          else if(x % 9 == 0)         {             myVector.push_back(x);  // add duplicates             myVector.push_back(x);           }          else if(x % 16 == 0)         {             myVector.push_back(x);  // add multiple duplicates             myVector.push_back(x);               myVector.push_back(x);               myVector.push_back(x);           }     }      // put the results in here     vector<int> missingValues;     vector<int> duplicates;      //  YOUR CODE GOES HERE               // validate missingValues for false positives     for(int x = 0; x < (int) missingValues.size(); ++x)     {         if(binary_search(myVector.begin(), myVector.end(), missingValues.at(x)))             cout << 'Oh noes! You missed an unmissed value. Something went horribly, horribly wrong.';     }      // validate duplicates (I think... errr)     vector<int>::iterator vecItr = myVector.begin();     vector<int>::iterator dupItr = duplicates.begin();      while(dupItr < duplicates.end())     {         vecItr = adjacent_find(vecItr, myVector.end());               if(*vecItr != *dupItr)             cout << 'Oh noes! Something went horribly, horribly wrong.';          // oh god         while(++dupItr != duplicates.end() && *(--dupItr) == *(++dupItr) && *vecItr == *(++vecItr));                      ++vecItr;     }      return 0; } 

I didn’t test the validation parts much, so there may be be something wrong with them (especially with the duplicates one).

I will post my own solution as an answer.

Answer 1

Since you’ve marked it language-agnostic, here’s the algorithm I’d use.

# Get numbers and sort them in ascending order.  input x,y; input number[1..n]; sort number[1..n];  # Set dups and missing to empty sets.  dups = []; missing = [];  # Get edge cases.  if number[1] > x:     foreach i x .. number[1] - 1:         missing.add(i) if number[n] < y:     foreach i number[n] + 1 .. y:         missing.add(i)  # Process all numbers starting at second one.  foreach i 2 .. n:     # If number same as last and not already in dups set, add it.      if number[i] == number[i-1] and not dups.contains(number[i]):         if number[i] >= x and number[i] <= y:             dups.add(number[i])      # If number not last number plus one, add all between the two     #   to missing set.      if number[i] != number[i-1] + 1:         foreach j number[i-1] + 1 .. number[i] - 1:             if j >= x and j <= y:                 missing.add(j)