4 sum - array quadruplet - facebook United kingdom intern - bravo idea to build two sum hashmap

4 sum - array quadruplet

Leetcode 4 sum 
          For example, given an input array with the elements of [1, 6, 3, 8, 4, 0, 2], 
          given 4 sum value 14, the ordered quadruplet with 14 in an ascending order can be [0, 2, 4, 8].
            
thirdId = 0
allSums is empty
thirdId = 1
allSums is empty
thirdId = 2
allSums contain sum arr[0] + arr[1]
thirdId = 3
allSums contain arr[0] + arr[1], arr[0] + arr[2], arr[1] + arr[2]
thirdId = 4            
       
            given array, find quadruplet to have a sum matching given value 
            
unordered_max <int, pair <int, int> > allSums;
          
i:
          allSums will contain arr[j] + arr[k] for all (j < i && j < k < i )
            
    HashMap.count(x) -> 0 if x is absent
            non-zero otherwise
            
for (int thirdId = 0; thirdId < n; thirdId++) {
  for (int fourthId = thirdId + 1; fourthId < n; fourthId++) {
    int currSum = arr[thirdId] + arr[fouthId];
    if (allSums.count(sum - currSum)) {
      vector <int> ansIds(4);
      ansIds[0] = allSums[sum - currSum].first;  // ? key value - related to index third 
      ansIds[1] = allSums[sum - currSum].second;
      ansIds[2] = thirdId;
      ansIds[3] = fourthId;
      return ansIds;
    }
  }
  for (int firstId = 0; firstId < thirdId; firstId++) {
    allSums[firstId + thirdId] = make_pair(firstId, thirdId);
  }
}           

return NULL;