Find out in linear time whether there is a pair in sorted vector that adds up to certain value - c++

Question

Given an std::vector of distinct elements sorted in ascending order, I want to develop an algorithm that determines whether there are two elements in the collection whose sum is a certain value, sum.
I've tried two different approaches with their respective trade-offs:
I can scan the whole vector and, for each element in the vector, apply binary search (std::lower_bound) on the vector for searching an element corresponding to the difference between sum and the current element. This is an O(n log n) time solution that requires no additional space.
I can traverse the whole vector and populate an std::unordered_set. Then, I scan the vector and, for each element, I look up in the std::unordered_set for the difference between sum and the current element. Since searching on a hash table runs in constant time on average, this solution runs in linear time. However, this solution requires additional linear space because of the std::unordered_set data structure.
Nevertheless, I'm looking for a solution that runs in linear time and requires no additional linear space. Any ideas? It seems that I'm forced to trade speed for space.

Answer 1

As the std::vector is already sorted and you can calculate the sum of a pair on the fly, you can achieve a linear time solution in the size of the vector with O(1) space.
The following is an STL-like implementation that requires no additional space and runs in linear time:
template<typename BidirIt, typename T>
bool has_pair_sum(BidirIt first, BidirIt last, T sum) {
if (first == last)
return false; // empty range
for (--last; first != last;) {
if ((*first + *last) == sum)
return true; // pair found
if ((*first + *last) > sum)
--last; // decrease pair sum
else // (*first + *last) < sum (trichotomy)
++first; // increase pair sum
}
return false;
}
The idea is to traverse the vector from both ends – front and back – in opposite directions at the same time and calculate the sum of the pair of elements while doing so.
At the very beginning, the pair consists of the elements with the lowest and the highest values, respectively. If the resulting sum is lower than sum, then advance first – the iterator pointing at the left end. Otherwise, move last – the iterator pointing at the right end – backward. This way, the resulting sum progressively approaches to sum. If both iterators end up pointing at the same element and no pair whose sum is equal to sum has been found, then there is no such a pair.
auto main() -> int {
std::vector<int> vec{1, 3, 4, 7, 11, 13, 17};
std::cout << has_pair_sum(vec.begin(), vec.end(), 2) << ' ';
std::cout << has_pair_sum(vec.begin(), vec.end(), 7) << ' ';
std::cout << has_pair_sum(vec.begin(), vec.end(), 19) << ' ';
std::cout << has_pair_sum(vec.begin(), vec.end(), 30) << '\n';
}
The output is:
0 1 0 1
Thanks to the generic nature of the function template has_pair_sum() and since it just requires bidirectional iterators, this solution works with std::list as well:
std::list<int> lst{1, 3, 4, 7, 11, 13, 17};
has_pair_sum(lst.begin(), lst.end(), 2);

Answer 2

I had the same idea as the one in the answer of 眠りネロク, but with a little bit more comprehensible implementation.
bool has_pair_sum(std::vector<int> v, int sum){
if(v.empty())
return false;
std::vector<int>::iterator p1 = v.begin();
std::vector<int>::iterator p2 = v.end(); // points to the End(Null-terminator), after the last element
p2--; // Now it points to the last element.
while(p1 != p2){
if(*p1 + *p2 == sum)
return true;
else if(*p1 + *p2 < sum){
p1++;
}else{
p2--;
}
}
return false;
}

Answer 3

well, since we are already given sorted array, we can do it with two pointer approach, we first keep a left pointer at start of the array and a right pointer at end of array, then in each iteration we check if sum of value of left pointer index and value of right pointer index is equal or not , if yes, return from here, otherwise we have to decide how to reduce the boundary, that is either increase left pointer or decrease right pointer, so we compare the temporary sum with given sum and if this temporary sum is greater than the given sum then we decide to reduce the right pointer, if we increase left pointer the temporary sum will remain same or only increase but never lesser, so we decide to reduce the right pointer so that temporary sum decrease and we reach near our given sum, similary if temporary sum is less than given sum, so no meaning of reducing the right pointer as temporary sum will either remain sum or decrease more but never increase so we increase our left pointer so our temporary sum increase and we reach near given sum, and we do the same process again and again unless we get the equal sum or left pointer index value becomes greater than right right pointer index or vice versa
below is the code for demonstration, let me know if something is not clear
bool pairSumExists(vector<int> &a, int &sum){
if(a.empty())
return false;
int len = a.size();
int left_pointer = 0 , right_pointer = len - 1;
while(left_pointer < right_pointer){
if(a[left_pointer] + a[right_pointer] == sum){
return true;
}
if(a[left_pointer] + a[right_pointer] > sum){
--right_pointer;
}
else
if(a[left_pointer] + a[right_poitner] < sum){
++left_pointer;
}
}
return false;
}