# Daily Coding Challenge #108

This is a series of Daily Coding Challenge. Each day I show a few solutions written in C++. The questions are from coding practice/contest sites such as HackerRank, LeetCode, Codeforces, Atcoder and etc.

/*
Sorting and Searching - Sliding Median
*/

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> Ordered_set;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<string, string> pss;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vii;
typedef vector<ll> vl;
typedef vector<vl> vvl;

double EPS=1e-9;
int INF=1000000005;
long long INFF=1000000000000000005ll;
double PI=acos(-1);
int dirx[8]={ -1, 0, 0, 1, -1, -1, 1, 1 };
int diry[8]={ 0, 1, -1, 0, -1, 1, -1, 1 };
const ll MOD = 1000000007;

#define DEBUG fprintf(stderr, "====TESTING====\n")
#define VALUE(x) cerr << "The value of " << #x << " is " << x << endl
#define OUT(x) cout << x << endl
#define OUTH(x) cout << x << " "
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define READ(x) for(auto &(z):x) cin >> z;
#define FOR(a, b, c) for (int(a)=(b); (a) < (c); ++(a))
#define FORN(a, b, c) for (int(a)=(b); (a) <= (c); ++(a))
#define FORD(a, b, c) for (int(a)=(b); (a) >= (c); --(a))
#define FORSQ(a, b, c) for (int(a)=(b); (a) * (a) <= (c); ++(a))
#define FORC(a, b, c) for (char(a)=(b); (a) <= (c); ++(a))
#define EACH(a, b) for (auto&(a) : (b))
#define REP(i, n) FOR(i, 0, n)
#define REPN(i, n) FORN(i, 1, n)
#define MAX(a, b) a=max(a, b)
#define MIN(a, b) a=min(a, b)
#define SQR(x) ((ll)(x) * (x))
#define RESET(a, b) memset(a, b, sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ALL(v) v.begin(), v.end()
#define ALLA(arr, sz) arr, arr + sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(ALL(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr, sz) sort(ALLA(arr, sz))
#define REVERSEA(arr, sz) reverse(ALLA(arr, sz))
#define PERMUTE next_permutation
#define TC(t) while (t--)
#define FAST_INP  ios_base::sync_with_stdio(false);cin.tie(NULL)

void solve() {

}

int main()
{
FAST_INP;
// #ifndef ONLINE_JUDGE
// freopen("input.txt","r", stdin);
// freopen("output.txt","w", stdout);
// #endif

//    int tc; cin >> tc;
//    TC(tc) solve();
int n, k;
cin >> n >> k;
vi x(n);
Ordered_set s;
// ordered set + sliding window
REP(i,k) s.insert(x[i]); // insert first k elements to ordered_set
OUTH(*s.find_by_order((k+1)/2-1));
REP(i,n-k){
s.erase(s.find_by_order(s.order_of_key(x[i]))); // O(logN)
s.insert(x[i+k]);
OUTH(*s.find_by_order((k+1)/2-1));
}
OUT("");
return 0;
}



/*
Sorting and Searching - Sliding Cost
*/

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> Ordered_set;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<string, string> pss;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vii;
typedef vector<ll> vl;
typedef vector<vl> vvl;

double EPS=1e-9;
int INF=1000000005;
long long INFF=1000000000000000005ll;
double PI=acos(-1);
int dirx[8]={ -1, 0, 0, 1, -1, -1, 1, 1 };
int diry[8]={ 0, 1, -1, 0, -1, 1, -1, 1 };
const ll MOD = 1000000007;

#define DEBUG fprintf(stderr, "====TESTING====\n")
#define VALUE(x) cerr << "The value of " << #x << " is " << x << endl
#define OUT(x) cout << x << endl
#define OUTH(x) cout << x << " "
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define READ(x) for(auto &(z):x) cin >> z;
#define FOR(a, b, c) for (int(a)=(b); (a) < (c); ++(a))
#define FORN(a, b, c) for (int(a)=(b); (a) <= (c); ++(a))
#define FORD(a, b, c) for (int(a)=(b); (a) >= (c); --(a))
#define FORSQ(a, b, c) for (int(a)=(b); (a) * (a) <= (c); ++(a))
#define FORC(a, b, c) for (char(a)=(b); (a) <= (c); ++(a))
#define EACH(a, b) for (auto&(a) : (b))
#define REP(i, n) FOR(i, 0, n)
#define REPN(i, n) FORN(i, 1, n)
#define MAX(a, b) a=max(a, b)
#define MIN(a, b) a=min(a, b)
#define SQR(x) ((ll)(x) * (x))
#define RESET(a, b) memset(a, b, sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ALL(v) v.begin(), v.end()
#define ALLA(arr, sz) arr, arr + sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(ALL(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr, sz) sort(ALLA(arr, sz))
#define REVERSEA(arr, sz) reverse(ALLA(arr, sz))
#define PERMUTE next_permutation
#define TC(t) while (t--)
#define FAST_INP  ios_base::sync_with_stdio(false);cin.tie(NULL)

void solve() {

}

int main()
{
FAST_INP;
// #ifndef ONLINE_JUDGE
// freopen("input.txt","r", stdin);
// freopen("output.txt","w", stdout);
// #endif

//    int tc; cin >> tc;
//    TC(tc) solve();
int n, k;
cin >> n >> k;
vi x(n);
Ordered_set s;
// ordered set + sliding window
REP(i,k) s.insert(x[i]); // insert first k elements to ordered_set
ll m = *s.find_by_order((k+1)/2-1);
ll d = 0;
REP(i,k) d+=abs(m-x[i]);
OUTH(d);
REP(i,n-k){
s.erase(s.find_by_order(s.order_of_key(x[i]))); // O(logN)
s.insert(x[i+k]);
ll cur_m = *s.find_by_order((k+1)/2-1);
d+=abs(cur_m-x[i+k])-abs(m-x[i]);
if((k&1)^1) d-=(cur_m-m);
m=cur_m;
OUTH(d);
}
OUT("");
return 0;
}



/*
LeetCode - Sliding Window Median

Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value.

Examples:
[2,3,4] , the median is 3

[2,3], the median is (2 + 3) / 2 = 2.5

Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Your job is to output the median array for each window in the original array.

For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

Window position                Median
---------------               -----
[1  3  -1] -3  5  3  6  7       1
1 [3  -1  -3] 5  3  6  7       -1
1  3 [-1  -3  5] 3  6  7       -1
1  3  -1 [-3  5  3] 6  7       3
1  3  -1  -3 [5  3  6] 7       5
1  3  -1  -3  5 [3  6  7]      6
Therefore, return the median sliding window as [1,-1,-1,3,5,6].

Note:
You may assume k is always valid, ie: k is always smaller than input array's size for non-empty array.
Answers within 10^-5 of the actual value will be accepted as correct.
*/

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> Ordered_set;

class Solution {
public:
vector<double> medianSlidingWindow(vector<int>& nums, int k) {
Ordered_set s;
vector<double> ans;
int n = (int)nums.size();
for(int i=0;i<k;i++) s.insert(nums[i]);
if(k&1){
// if k is odd, just take the middle one
ans.push_back((double)*s.find_by_order(k/2));
for(int i=0;i<n-k;i++){
s.erase(s.find_by_order(s.order_of_key(nums[i])));
s.insert(nums[i+k]);
ans.push_back((double)*s.find_by_order(k/2));
}
} else {
// if k is even, find the left element and the right element, sum them up and divide by 2
// example: [1,2,3,4] -> (2+3)/2 -> [2.50000]
ans.push_back(((double)*s.find_by_order((k+1)/2-1)+(double)*s.find_by_order(k/2))/2);
for(int i=0;i<n-k;i++){
s.erase(s.find_by_order(s.order_of_key(nums[i])));
s.insert(nums[i+k]);
ans.push_back(((double)*s.find_by_order((k+1)/2-1)+(double)*s.find_by_order(k/2))/2);
}
}
return ans;
}
};


There are other programming solutions in the following repositories below. Star and watch for timely updates!

## wingkwong / competitive-programming

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### Wing-Kam

Consultant by day. Developer by night. AWS certified. Exploring #CloudNative currently.