# Codingame: Temperatures

## The Goal

In this exercise, you have to analyze records of temperature to find the closest to zero.

## Rules

**If two numbers are equally close to zero, positive integer has to be considered closest to zero**(for instance, if the temperatures are -5 and 5, then display 5).

## Game Input

**Input**

Line 1: `N`, the number of temperatures to analyze

Line 2: A string with the `N` temperatures expressed as integers ranging from -273 to 5526

**Output**

**Constraints**

`N`< 10000

## Solution

When we start reading the temperatures \(t_i\) with the first iteration, we don't have a current minimum \(m\), so that the first element will always update the minimum. From now on the minimum will only be set to the current element \(t_i\) if \(|t_i| < |m|\). When \(|t_i| = |m|\) we need to set the minimum to the absolute value \(|t_i|\). That means we can implement the check like this:

#include <iostream> #include <sstream> using namespace std; int main() { int N; cin >> N; cin.ignore(); string TEMPS; getline(cin, TEMPS); istringstream buf(TEMPS); int t, m = 0; for (int i = 0; i < N; ++i) { buf >> t; if (i == 0 || abs(t) < abs(m)) { m = t; } else if (abs(t) == abs(m)) { m = abs(t); } } cout << m << endl; return 0; }

This algorithm works fine, but we call the abs function quite often. We can remove the else-if branch by dissecting all cases for \(|t_i| = |m|\):

t > 0 m > 0: keep m m < 0: take t m = 0: keep m t < 0 m > 0: keep m m < 0: keep m m = 0: keep m t = 0 m > 0: keep m m < 0: keep m m = 0: keep m

That means we can remove the else-if branch and add `t == -m && t > 0` or'ed to the first. We can do the same for \(|t| < |m|\):

t > 0 m > 0 t < m m < 0 t < -m m = 0 false t < 0 m > 0 -t < m m < 0 -t < -m m = 0 false t = 0 m > 0 t < m m < 0 t < -m m = 0 false => t > 0, m > 0, t < m t > 0, m < 0, t < -m t < 0, m > 0,-t < m t < 0, m < 0,-t < -m t = 0, m > 0, t < m t = 0, m < 0, t < -m => t > 0, m > t t > 0, m < -t t < 0, m > -t t < 0, m < t t = 0, m > t t = 0, m < t => t >= 0, m > t t < =0, m < t t > 0, m < -t t < 0, m > -t t > 0, t = -m # from first equation => t >= 0, m > t t <= 0, m < t t > 0, m <= -t t < 0, m > -t

Stating the improved algorithm again:

#include <iostream> #include <sstream> using namespace std; int main() { int N; cin >> N; cin.ignore(); string TEMPS; if (N == 0) { cout << 0 << endl; return 0; } getline(cin, TEMPS); istringstream buf(TEMPS); int t, m; buf >> m; for (int i = 1; i < N; ++i) { buf >> t; if (t >= 0 && m > t || t <= 0 && m < t || t > 0 && m <= -t || t < 0 && m > -t) { m = t; } } cout << m << endl; return 0; }

Well, the original code was much more readable and is thus of more practical value, but it was nice to see how far we can go with the optimization.

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