I have a vector of pairs
. Suppose it is like this:
vector<pair<int,int>> vec = { {1,12}, {1,5}, {1,6}, {1,9}, {3,9}, {3,11}, {3,13}, {3,4}, {5,9}, {5,91}, {13,8}, {16,8}, {20,8}, {20,81} };
The pairs
are sorted by first element.
Given a pair
, I need to find the index of the last pair
in the vector whose first element is less than or equal to first element of the given pair. If for that last pair
, other pairs lie to its left with the same value of the first element, I need the first of all those pairs:
<4,10> => 4 (vec[4] is <3,9>, the elements with the largest first value less than or equal to 4 are those with first element as 3, and there are 4 pairs with a 3 in the first element, at indices 4-7, so return the first of those pairs)
<0,10> => -1, since no element exists to its right.
<1,6> => 0 (vec[0] is <1,12>. There is no pair whose first element is less than 1, and there are 4 pairs, including <1,6> whose first element is 1. So we need the first of these 4 pairs.)
<23,81> => 12 (vec[12] is <20,8>)
Condition: I need to use only standard algorithms like upper_bound
, binary_search
and lower_bound
. I tried this, but it is failing badly:
vector<pair<int,int>> vec = { {1,12}, {1,5}, {1,6},{1,9}, {3,9}, {3,11}, {3,13}, {3,4}, {5,9}, {5,91}, {13,8}, {16,8}, {20,8}, {20,81} };
auto i = std::lower_bound(vec.begin(), vec.end(), make_pair<int,int>(4,10),
[](const pair<int,int>& f1, const pair<int,int>& f2) { return f1.first < f2.first; });
cout << i-vec.begin();