## Saturday, November 5, 2016

### Hacker Rank - Interquartile Range

Problem:

Solution:

Because the frequency can be quite large, building the solution based on the previous problem in impractical. Instead, we will simply sort the list and reuse the indices we found in the previous problem to solve this.

Code:

#include "stdafx.h"

// https://www.hackerrank.com/challenges/s10-interquartile-range

#include <iostream>
#include <iomanip>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;

int HACKER_RANK_S10_INTERQUARTILE_RANGE()
{
int n;
int count = 0;
cin >> n;
vector<pair<int, int>> data;
data.resize(n);
for (int i = 0; i < n; i++)
{
cin >> data[i].first;
}
for (int i = 0; i < n; i++)
{
cin >> data[i].second;
count += data[i].second;
}
sort(data.begin(), data.end());
if (count % 4 == 3)
{
int t1 = count / 4;
int t2 = count / 2 + 1 + count / 4;
int g1 = 0;
int g2 = 0;
int begin = 0;
for (int i = 0; i < n; i++)
{
int end = begin + data[i].second;
if (begin <= t1 && t1 < end) { g1 = data[i].first; }
if (begin <= t2 && t2 < end) { g2 = data[i].first; break; }
begin = end;
}
}
else if (count % 4 == 2)
{
int t1 = count / 4;
int t2 = count / 2 + count / 4;
int g1 = 0;
int g2 = 0;
int begin = 0;
for (int i = 0; i < n; i++)
{
int end = begin + data[i].second;
if (begin <= t1 && t1 < end) { g1 = data[i].first; }
if (begin <= t2 && t2 < end) { g2 = data[i].first; break; }
begin = end;
}
}
else if (count % 4 == 1)
{
int t1 = count / 4 - 1;
int t2 = count / 4;
int t3 = count / 2 + count / 4;
int t4 = count / 2 + count / 4 + 1;

int g1 = 0;
int g2 = 0;
int g3 = 0;
int g4 = 0;
int begin = 0;
for (int i = 0; i < n; i++)
{
int end = begin + data[i].second;
if (begin <= t1 && t1 < end) { g1 = data[i].first; }
if (begin <= t2 && t2 < end) { g2 = data[i].first; }
if (begin <= t3 && t3 < end) { g3 = data[i].first; }
if (begin <= t4 && t4 < end) { g4 = data[i].first; break; }
begin = end;
}
answer = (g3 + g4 - g2 - g1) / 2.0;
}
else if (count % 4 == 0)
{
int t1 = count / 4 - 1;
int t2 = count / 4;
int t3 = count / 2 + count / 4 - 1;
int t4 = count / 2 + count / 4;

int g1 = 0;
int g2 = 0;
int g3 = 0;
int g4 = 0;
int begin = 0;
for (int i = 0; i < n; i++)
{
int end = begin + data[i].second;
if (begin <= t1 && t1 < end) { g1 = data[i].first; }
if (begin <= t2 && t2 < end) { g2 = data[i].first; }
if (begin <= t3 && t3 < end) { g3 = data[i].first; }
if (begin <= t4 && t4 < end) { g4 = data[i].first; break; }
begin = end;
}
answer = (g3 + g4 - g2 - g1) / 2.0;
}
cout << fixed << setprecision(1) << answer << endl;

return 0;
}