Finding the lower bound of vector of pairs based on first value of pair

Question

I have a vector of pair "v" which is sorted according to the first value of the pair. Now I want to find out the lower bound of vector "v" according to the first value of the pair -- while finding the lower bound I want to ignore the second.

int pl; cin>>pl;
vector<pair<int,int>> v;
for (int i = 0; i < n; i++){
    int p,c; cin>>p>>c;
    v.push_back({p,c});
}

sort(v.begin(),v.end());

    
auto it = lower_bound(v.begin(),v.end(),pl);

cout<<(*it)<<endl;

I want to find the lower bound of pl in the vector pair,according to the 1st value of pairs.I'm a newbie so please elaborate

Too bad your question is tagged C++17, because C++20 ranges projection would also be great for this. — JHBonarius, Jan 31 '22 at 09:08
Aside: `sort(v.begin(),v.end());` sorts by the first then the second of the pairs, but as you have no duplicate firsts, it isn't obvious. — Caleth, Feb 01 '22 at 10:06

JHBonarius · Answer 1 · 2022-02-01T09:29:27.723

I know your question is marked C++17, but for completeness: the C++20 solution could be to use the "projection" feature that the new ranges offer. E.g.:

#include <algorithm>
#include <iostream>
#include <vector>

int main() {
    auto v{std::vector<std::pair<int, int>>{
        {3, 1},
        {1, 2},
        {4, 5},
    }};

    std::ranges::sort(v);

    auto const pl{3};

    auto const it{std::ranges::lower_bound(v, pl, {},
        &decltype(v)::value_type::first)}; // last param = projection

    std::cout << (*it).second << '\n';
}

prints: 1

score 1 · Answer 2 · answered Jan 31 '22 at 08:48

1

Since you don't care about the second element of your pair, you can find a value whose lower bound is (pl, 0):

#include <vector>
#include <algorithm>
#include <iostream>

int main()
{
    std::vector<std::pair<int,int>> v;
    for (int i = 0; i < 11; i++)
    {
        v.push_back({ (17*i) % 59, (i*i) % 13});
    }

    std::sort(v.begin(),v.end());
    auto const it = std::lower_bound(v.begin(), v.end(), std::make_pair(11, 0)); // pl=11 in this example
    
    std::cout << "(" << it->first << ", " << it->second << ")\n"; // (17, 1)
}

Live demo

This works because std::pair::operator< is defined such that

(0,0) < (0,1) < (1,0) < (1,1)

answered Jan 31 '22 at 08:48

YSC

38,212
9
96
149

what about (1,-1)? – n. m. could be an AI Feb 01 '22 at 08:02
@n.1.8e9-where's-my-sharem. I've linked to the documentation of that operator. Go and read the fine manual. – YSC Feb 01 '22 at 20:26
I know what that operator does. I am asking what will happen if a pair with a negative second element is in the array. Will your method work? [Let's find out](https://godbolt.org/z/hdzb1W8q7). – n. m. could be an AI Feb 01 '22 at 22:06

score 1 · Answer 3 · answered Feb 01 '22 at 10:03

You can pass a custom comparator to std::lower_bound, so long as it is consistent with your sort order.

#include <algorithm>
#include <iostream>
#include <vector>
#include <utility>

bool first_less(std::pair<int, int> lhs, int rhs) {
    return lhs.first < rhs;
}

int main() {
    std::vector<std::pair<int, int>> v{
        {3, 1},
        {1, 2},
        {4, 5},
    };
    auto pl = 3;

    std::sort(v.begin(), v.end());

    auto it = std::lower_bound(v.begin(), v.end(), pl, first_less);

    std::cout << it->second << '\n';
}

See it live

score 0 · Answer 4 · answered Feb 01 '22 at 13:22

JHBonarius' answer applies to C++17 too, as long as you are ok with using Range-v3:

#include <iostream>
#include <iterator>
#include <vector>
#include <utility>
#include <range/v3/algorithm/lower_bound.hpp>

int main() {
    std::vector<std::pair<int,int>> v{{0,3}, {0,2}, {1,1}, {1,2}};

    auto constexpr first = &decltype(v)::value_type::first;

    auto itFor0 = ranges::lower_bound(v, 0, std::less<>{}, first);
    auto itFor1 = ranges::lower_bound(v, 1, std::less<>{}, first);

    std::cout << std::distance(std::begin(v), itFor0) << std::endl; // 0
    std::cout << std::distance(std::begin(v), itFor1) << std::endl; // 2
}

Finding the lower bound of vector of pairs based on first value of pair

4 Answers4