Find minimum value in vector - c++

Question

Below I have attached code for a project that is intended to find the lowest value in a user-inputed vector, return -1 if the vector is empty, and 0 if the vector only has one index. I have run into an issue with the condition in which a vector is empty as the unit test continues to fail the returns_negative_one_for_empty_vector test.
main.cc
#include <iostream>
#include <vector>
#include "minimum.h"
int main() {
int size;
std::cout << "How many elements? ";
std::cin >> size;
std::vector<double> numbers(size);
for (int i = 0; i < size; i++) {
double value;
std::cout << "Element " << i << ": ";
std::cin >> value;
numbers.at(i) = value;
}
double index;
index = IndexOfMinimumElement(numbers);
std::cout << "The minimum value in your vector is at index" << index << std::endl;
}
minimum.cc
#include "minimum.h"
#include <vector>
int IndexOfMinimumElement(std::vector<double> input) {
int i, min_index;
double min_ = input.at(0);
for (int i = 0; i < input.size(); i++) {
if (input.at(i) < min_) {
min_index = i;
return min_index;
}
else if (input.size() == 0) {
return -1;
}
else if(input.size() == 1) {
return 0;
}
}
};
minimum.h
#include <vector>
int IndexOfMinimumElement(std::vector<double> input);

Answer 1

find the lowest value in a user-inputed vector, return -1 if the
vector is empty, and 0 if the vector only has one index.
Instead of writing raw for loops, this can be accomplished much more easily by using the STL algorithm functions.
There are other issues, one being that the vector should be passed by const reference, not by value. Passing the vector by-value incurs an unnecessary copy.
#include <algorithm>
#include <vector>
#include <iostream>
int IndexOfMinimumElement(const std::vector<double>& input)
{
if (input.empty())
return -1;
auto ptrMinElement = std::min_element(input.begin(), input.end());
return std::distance(input.begin(), ptrMinElement);
}
int main()
{
std::cout << IndexOfMinimumElement({ 1.2, 3.4, 0.8, 7.8 }) << std::endl;
std::cout << IndexOfMinimumElement({}) << std::endl; // empty
std::cout << IndexOfMinimumElement({3}) << std::endl; // only 1 element
return 0;
}
Output:
2
-1
0
The relevant functions are std::min_element and std::distance. The std::min_element returns an iterator (similar to a pointer) to the minimum element in the range.
The code is written with a clear understanding of what each function does -- it is practically self-documenting. To get the minimum element, you call std::min_element. To get the distance from the first to the found minimum element, you call std::distance with an iterator to the starting position and an iterator to the ending position.
The bottom line is this: the STL algorithm functions rarely, if ever, fail when given the proper input parameters. Writing raw for loops will always have a much greater chance of failure, as you have witnessed. Thus the goal is to minimize having to write such for loops.

Answer 2

In IndexOfMinimumElement you return on the very first iteration, as all branches of your if/else lead to a return.
If your vector contained {14, 2, 10, 1} the index it would return would be 1, because 2 is less than 14.
Instead, you want to have a couple of conditional checks at the top of your function that return based on the length of the vector.
If the function call gets past those, it should iterate over the values in the vector, checking if they are less than the running minimum value, and update the minimum index accordingly.
int IndexOfMinimumElement(std::vector<double> input) {
if (input.size() == 0) return -1;
if (input.size() == 1) return 0;
int i = 0;
double min = input[0];
int min_idx = 0;
for (auto &v : input) {
if (v < min) {
min = v;
min_idx = i;
}
++i;
}
return min_idx;
}
A minimal test:
int main() {
std::vector<double> foo { 1.2, 3.4, 0.8, 7.8 };
std::cout << IndexOfMinimumElement(foo) << std::endl;
return 0;
}
Prints, as expected:
2