CODEWITHSUNDEEP
c++vector
Paula Perdomo
Paula Perdomo

Reputation: 1

How do I find the smallest number vector with repeated numbers?

I have this vector with these numbers {0, 1, 0, 3, 2, 3} and I'm trying to use this approach to have the minimum number and its index:

int min_num = *min_element(nums.begin(), nums.end());
int min_num_idx = min_element(nums.begin(), nums.end()) - nums.begin();

However, this returns the first smallest number it found so the 0 in index 0. Is there a way to make it return the last smallest number instead? (the 0 in index 2 instead)

Upvotes: 0

Views: 405

Answers (2)

Ruks
Ruks

Reputation: 3956

However, this returns the first smallest number it found so the 0 in index 0. Is there a way to make it return the last smallest number instead? (the 0 in index 2 instead)

You can try std::vector<int>::reverse_iterator:

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

int main() {
    std::vector<int> n { 0, 1, 0, 3, 2, 3 };
    std::vector<int>::reverse_iterator found_it = std::min_element(n.rbegin(), n.rend());
    if (found_it != n.rend()) {
        std::cout << "Minimum element: " << *found_it << std::endl;
        std::cout << "Minimum element index: " << std::distance(n.begin(), std::next(found_it).base()) << std::endl;
    }
}

Output:

Minimum element: 0

Minimum element index: 2

Upvotes: 3

robthebloke
robthebloke

Reputation: 9682

int min_num_idx = nums.size() - 1 - min_element(nums.rbegin(), nums.rend()) - nums.rbegin();

Upvotes: 0

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