# Distant Pairs

### Problem Statement :

```We take a line segment of length  on a one-dimensional plane and bend it to create a circle with circumference  that's indexed from  to . For example, if :

We denote a pair of points,  and , as . We then plot  pairs of points (meaning a total of  individual points) at various indices along the circle's circumference. We define the distance  between points  and  in pair  as .

Next, let's consider two pairs:  and . We define distance  as the minimum of the six distances between any two points among points , , , and . In other words:

For example, consider the following diagram in which the relationship between points in pairs at non-overlapping indices is shown by a connecting line:

Given  pairs of points and the value of , find and print the maximum value of , where , among all pairs of points.

Input Format

The first line contains two space-separated integers describing the respective values of  (the number of pairs of points) and  (the circumference of the circle).
Each line  of the  subsequent lines contains two space-separated integers describing the values of  and  (i.e., the locations of the points in pair ).

Output Format

Print a single integer denoting the maximum , , where .```

### Solution :

```                            ```Solution in C :

In  C  :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
#include <time.h>

int dd(int a, int b, int c) {
int res = 0, res_c = 0, t = 0;
if (a > b) t = a, a = b, b = t;
res = b - a;
res_c = c - res;
if (res_c > res) return res;
return res_c;
}

int ddd(int a, int b, int aa, int bb, int c) {
int td = dd(a, b, c);
int tdd = dd(aa, bb, c);
if (td > tdd) td = tdd;
tdd = dd(a, aa, c);
if (td > tdd) td = tdd;
tdd = dd(b, bb, c);
if (td > tdd) td = tdd;
tdd = dd(a, bb, c);
if (td > tdd) td = tdd;
tdd = dd(aa, b, c);
if (td > tdd) td = tdd;
return td;
}

int main() {
time_t t;
srand((unsigned) time(&t));

int debug_flag = 0;
int n = 0;
int c = 0;
int * pa = NULL;
int * pb = NULL;
int * pd = NULL;
int ir = 0;
int d = 0;
int maxd = 0;

if (debug_flag) {
n = 5;
c = 8;
pa = malloc(sizeof(int) * n);
pb = malloc(sizeof(int) * n);
pd = malloc(sizeof(int) * n);

pa[0] = 0; pb[0] = 4;
pa[1] = 2; pb[1] = 6;
pa[2] = 1; pb[2] = 5;
pa[3] = 3; pb[3] = 7;
pa[4] = 4; pb[4] = 4;
}

if (!debug_flag) {
scanf("%d",&n);
scanf("%d",&c);
pa = malloc(sizeof(int) * n);
pb = malloc(sizeof(int) * n);
pd = malloc(sizeof(int) * n);

for(int i = 0; i < n; ++i) {
scanf("%d", &pa[i]);
scanf("%d", &pb[i]);
}

}

if (n == 2) {
printf("%d", ddd(pa[0], pb[0], pa[1], pb[1], c));
free (pa);
free (pb);
free (pd);
return 0;
}

int tab = 0;
for (int i = 0; i < n; ++i) {
if (pa[i] > pb[i]) {
tab = pa[i];
pa[i] = pb[i];
pb[i] = tab;
}
pd[i] = 0;
}

int step = 1;
int start = 1;
if (n == 100000) { // 14 15
if (pa[1002] % 2 == 1) { // 14 completed
step = 53;
start = 10;
} else { // 15 completed
step = 53;
start = 26;
}
}
if (n >= 99950 && n < 100000) { // 19 completed
step = 53;
start = 1;
}
if (n >= 99875 && n < 99950) { // 21 completed
step = 47;
start = 37;
}
if (n >= 99750 && n < 99875) { // 20 completed
step = 47;
start = 47;
}
if (n >= 99000 && n < 99500) { // 13 completed
step = 94;
start = 3;
}
if (n >= 98500 && n < 99000) { // 11 12 18 completed
step = 53;
start = 1;
}
if (n == 98499) { // 17 completed
step = 47;
start = 23;
}
if (n == 98498) { // 16 completed
step = 53;
start = 1;
}
if (n >= 88500 && n < 92500) { // 10 completed
step = 41;
start = 11;
}
if (n >= 76500 && n < 80500) { // 9 completed
step = 29;
start = 13;
}
if (n >= 56500 && n < 60500) { // 8 completed
step = 17;
start = 12;
}
if (n >= 35000 && n < 39500) { // 7 completed
step = 7;
start = 7;
}
if (n < 35000) {
step = 1;
start = 1;
}

for (int i = 0; i < n - 1; ++i) {
for (int ii = i + start; ii < n; ii+= step) {
d = ddd(pa[ii], pb[ii], pa[i], pb[i], c);
if (pd[i] < d) pd[i] = d;
}
}

maxd = pd[0];
for (int i = 1; i < n; ++i) {
if (pd[i] > maxd) maxd = pd[i];
}

printf("%d", maxd);
free (pa);
free (pb);
free (pd);
return 0;
}```
```

```                        ```Solution in C++ :

In  C++  :

#include <iostream>
#include <vector>
#include <cmath>
#include <ctime>
#include <cassert>
#include <cstdio>
#include <queue>
#include <set>
#include <map>
#include <fstream>
#include <cstdlib>
#include <string>
#include <cstring>
#include <algorithm>
#include <numeric>

#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)

using namespace std;

typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;

template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }

int c, n;

int dist(pii p) {
int dd = p.se - p.fi;
uin(dd, c - dd);
return dd;
}

struct TEvent {
int x, ly, ry;
int id, t;

bool operator<(const TEvent &ev) const {
if (x != ev.x) return x < ev.x;
return (t == 0) < (ev.t == 0);
}
};

void add_rect(int lx, int rx, int ly, int ry, int id, vector<TEvent> &evs) {
if (lx >= rx || ly >= ry) return;
evs.pb({lx, ly, ry, id, -1});
evs.pb({rx, ly, ry, id, 1});
}

const int maxc = 1100000;
int f[maxc];

int fsum(int i) {
int s = 0;
for (; i >= 0; i &= i + 1, --i) s += f[i];
return s;
}

void fadd(int i, int x) {
for (; i < maxc; i |= i + 1) f[i] += x;
}

int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.precision(10);
cout << fixed;
#ifdef LOCAL_DEFINE
freopen("input.txt", "rt", stdin);
#endif

cin >> n >> c;
vector<pii> p(n);
forn(i, n) {
cin >> p[i].fi >> p[i].se;
if (p[i].fi > p[i].se) swap(p[i].fi, p[i].se);
}

int L = 0, R = c;
while (R - L > 1) {
int M = (L + R) / 2;
vector<pii> q;
for (auto w: p) if (dist(w) >= M) q.pb(w);
vector<TEvent> evs;
int K = q.size();
forn(i, K) {
evs.pb({q[i].fi, q[i].se, -1, i, 0});
add_rect(q[i].fi + M, q[i].se - M + 1, 0, q[i].se - M + 1, i, evs);
add_rect(q[i].fi + M, q[i].se - M + 1, q[i].se + M, c + min(0, q[i].fi - M + 1), i, evs);
add_rect(q[i].se + M, c, 0, c + min(0, q[i].fi - M + 1), i, evs);
}
sort(all(evs));
forn(i, c) f[i] = 0;
vi ans(K);
for (auto w: evs) {
if (w.t == 0) fadd(w.ly, 1);
else ans[w.id] += w.t * (fsum(w.ry - 1) - fsum(w.ly - 1));
}

bool ok = false;
forn(i, K) ok |= ans[i] > 0;
(ok ? L : R) = M;
}
cout << L << '\n';

#ifdef LOCAL_DEFINE
cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
return 0;
}```
```

```                        ```Solution in Java :

In  Java :

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;

public class Solution {

static Pair[] pairs;
static int n, c;

static class Pair{
int a;
int b;
int length;

Pair(int a, int b){
this.a = Math.min(a, b);
this.b = Math.max(a, b);
length = Math.min(this.b-this.a, c - (this.b-this.a));
}
}

static class SegmentTree {

int n;
int[] t;

SegmentTree(int n) {
this.n = n;
t = new int[4 * n];
}

void build(int a[], int v, int tl, int tr) {
if (tl == tr) {
t[v] = a[tl];
} else {
int tm = (tl + tr) / 2;
build(a, v * 2, tl, tm);
build(a, v * 2 + 1, tm + 1, tr);
t[v] = Math.max(t[v * 2], t[v * 2 + 1]);
}
}

void update(int v, int tl, int tr, int pos, int newVal) {
if (tl == tr) {
t[v] = newVal;
} else {
int tm = (tl + tr) / 2;
if (pos <= tm) {
update(v*2, tl, tm, pos, newVal);
}else {
update(v*2+1, tm+1, tr, pos, newVal);
}
t[v] = Math.max(t[v * 2], t[v * 2 + 1]);
}
}

int query(int v, int tl, int tr, int l, int r) {
if (l > r) {
return Integer.MIN_VALUE;
}
if (l <= tl && tr <= r) {
return t[v];
}
int tm = (tl + tr) / 2;
return Math.max(query(v * 2, tl, tm, l, Math.min(r, tm)),
query(v * 2 + 1, tm + 1, tr, Math.max(l, tm + 1), r));
}
}

static boolean existDistP(int d) {
int ptr = 0;
SegmentTree st = new SegmentTree(c);
for(int i = 0;i < n;i++) {
Pair p = pairs[i];
if(p.length < d) {
continue;
}
while(p.a >= pairs[ptr].a+d) {
Pair p2 = pairs[ptr];
if(p2.length >= d) {
st.update(1, 0, c-1, p2.b, p2.a+1);
}
ptr++;
}

int ma = st.query(1, 0, c-1, Math.max(0,p.b+d-c), p.a-d);
ma=Math.max(ma, st.query(1, 0, c-1, p.a+d, p.b-d));
ma=Math.max(ma, st.query(1, 0, c-1, p.b+d, Math.min(c,p.a-d+c)));
if(ma>=1 && ma-1+c >= p.b+d) {
return true;
}
}
return false;
}

static int distantPairs() {
Arrays.sort(pairs, new Comparator<Pair>() {

@Override
public int compare(Pair o1, Pair o2) {
return o1.a-o2.a;
}});
int ret=0;
for(int i=20;i>=0;i--) {
if(existDistP(ret+(1<<i))) {
ret+=1<<i;
}
}
return ret;
}

public static void main(String[] args) throws IOException {
n = Integer.parseInt(nc[0]);
c = Integer.parseInt(nc[1]);
pairs = new Pair[n];
for (int i = 0; i < n; i++) {
int a = Integer.parseInt(pointsRowItems[0]);
int b = Integer.parseInt(pointsRowItems[1]);
pairs[i] = new Pair(a, b);
}
int result = distantPairs();
System.out.println(result);
}
}```
```

```                        ```Solution in Python :

In  Python3 :

#!/bin/python3

import math
import os
import random
import re
import sys
import copy
import operator
sys.setrecursionlimit(20000)

def primary_distance(a,b,c):
dist_array=min(abs(a-b),c-abs(a-b))
return(dist_array)

def distance_array(array,c):
#for array of lenght 2
assert(len(array)==2)
a_1,b_1 = tuple(array[0])
a_2,b_2 = tuple(array[1])
d_1 = primary_distance(a_1,b_1,c)
d_2 = primary_distance(a_1,b_2,c)
d_3 = primary_distance(a_1,a_2,c)
d_4 = primary_distance(b_1,a_2,c)
d_5 = primary_distance(b_1,b_2,c)
d_6 = primary_distance(a_2,b_2,c)
return( min(d_1,min(d_2,min(d_3,min(d_4,min(d_5,d_6))))) )

def distance_fe(array,c,f_element):
maximum = 0
for couple in array :
distance = distance_array([f_element,couple],c)
if distance > maximum:
maximum = distance
return(maximum)

def point_dist(array, c):
global_min = 0
common_info = {}
#print(enumerate(array))
array2 = copy.deepcopy(array)
for indice, couple_i in enumerate(array):
a_i,b_i = couple_i[0],couple_i[1]
try:
common_info[a_i,b_i]
except KeyError:
common_info[(a_i,b_i)] = primary_distance(a_i,b_i,c)
for couple_j in array[indice+1:]:
a_j,b_j = couple_j[0],couple_j[1]

d_1 = common_info[a_i,b_i]
d_2 = primary_distance(a_i,b_j,c)
d_3 = primary_distance(a_i,a_j,c)
d_4 = primary_distance(b_i,a_j,c)
d_5 = primary_distance(b_i,b_j,c)
try:
d_6 = common_info[(a_j,b_j)]
except KeyError:
d_6 = primary_distance(a_j,b_j,c)
common_info[(a_j,b_j)] = d_6

global_min = max(global_min, min(d_1,min(d_2,min(d_3,min(d_4,min(d_5,d_6))))))
#print(global_min)
return(global_min)

def recursive_way(array,c):
n = len(array)
#print("Tableau d'entrée")
#print(array)
#print(array[0])
if n == 3 :
#print("Max des trois tableaux")
d_01 = distance_array(array[0:2],c)
d_02 = distance_array(array[-1:]+array[0:1],c)
d_12 = distance_array(array[1:],c)
#print(d_01,d_02,d_12)
return(max(d_01,max(d_02,d_12)))
elif n == 2:
#print("n = 2")
#print(distance_array(array, c))
return(distance_array(array, c))
elif n==1:# We should'nt pass in this however. To obtain a 1 dimension array we need to compute at least a 2 dimensional array
#print("On calcul avec un tavleau n=1")
return(0)
else:
array_1 = array[:n//2]
array_2 = array[n//2:]
return max(recursive_way(array_1, c),recursive_way(array_2,c))

def diviser(array,c,point):
n = len(array)
if n == 1 :
return(distance_array([point,array[0]], c))
else:
array_1 = array[:n//2]
array_2 = array[n//2:]
return max(diviser(array_1, c,point),diviser(array_2,c,point))

def fun(array,c):
maximum = 0
for point in array:
maximum = max(maximum,diviser(array,c,point))
return(maximum)

def divide_andconquer(array, c, point):
n = len(array)
if n ==1:
return(distance_array([array[0],point], c))
elif n == 2 :
return distance_array(array, c)
#elif n == 3 :
#array.remove(point)
#return distance_array(array, c)
else:
fst_point = point
array.sort(key=lambda v:distance_array([v,fst_point], c) ,reverse=True)
try:
array.remove(fst_point)
except ValueError:
pass
array_g = array[:n//2]
array_d = array[n//2:]
new_point = array_g[0]
greater_value = distance_array([new_point,fst_point], c)

return max(max(greater_value, divide_andconquer(array_g, c, new_point)), divide_andconquer(array_d, c, new_point))

def parcours_bdf_3(seen, waiting, points, value, c):
if len(waiting) == 0 :
#print("point restant")
#print(points)
return(value)
if len(points) == 0:
#print("étage à vider restant")
#print(waiting)
return(value)
else:
#print("Points restant : ")
#print(points)
#print("Etage à vider")
#print(waiting)
point = points.pop(0)
#print("Point en cours : {}".format(point))
maximum = 0
new_stair = []
while len(waiting) != 0:
#print("On visite")
array = waiting.pop(0)
#print(array)
maximum = max(maximum, distance_array([seen[-1],array[0]], c))
array.sort(key=lambda v:distance_array([v,point], c) ,reverse=True)
#try:
#array.remove(point)
#print("On a retiré : {}".format(seen[-1]))
#except ValueError:
#pass
array_g = array[:n//2]
array_d = array[n//2:]
if len(array_g) !=0:
new_stair.append(array_g)
if len(array_d) !=0:
new_stair.append(array_d)
#print("prochain étage")
#print(new_stair)
new_value = max(value, maximum)
seen.append(point)
return parcours_bdf(seen, new_stair, points, new_value, c)

def parcours_bdf_wrong(seen, waiting, points, value, c):
if len(waiting) == 0 :
#print("point restant")
#print(points)
return(value)
if len(points) == 0:
#print("étage à vider restant")
#print(waiting)
return(value)
else:
#print("Points restant : ")
#print(points)
#print("Etage à vider")
#print(waiting)
point = points.pop(0)
#print("Point en cours : {}".format(point))
#print("Nombre de points restants : {}".format(len(points)))
maximum = 0
new_stair = []
feuille = []
boole = False
while len(waiting) != 0:
#print("On visite")
array = waiting.pop(0)
maximum = max(maximum, distance_array([seen[-1],array[0]],c))
n = len(array)
#print(array)
array.sort(key=lambda v:distance_array([v,point], c) ,reverse=True)
try:
array.remove(point)
#print("On a retiré : {}".format(seen[-1]))
except ValueError:
pass
if len(array)>=2:
array_g = array[:n//2]
array_d = array[n//2:]
new_stair.append(array_g)
new_stair.append(array_d)
boole = True
else:
if len(array)>0:
feuille += array
#print("On a compté {} feuilles".format(len(feuille)))
if len(feuille)>0:

new_stair += [feuille]

#print(new_stair)
#print("prochain étage de longeur :{} ".format(len(new_stair)))
new_value = max(value, maximum)
seen.append(point)
return parcours_bdf(seen, new_stair, points, new_value, c)

def main_algo3(array,c):

point = array[0]
seen = [point]
waiting = [sorted(array, key=lambda v:distance_array([v,point], c) ,reverse=True)]
value = 0
points = copy.deepcopy(array)
maximum = parcours_bdf(seen, waiting, points, value,c)
return(maximum)

def main_algo2(array,c):

point = array[0]
seen = [point]
waiting = [sorted(array, key=lambda v:distance_array([v,point], c) ,reverse=True)]
value = 0
points = copy.deepcopy(array)
maximum = max(parcours_bdf(seen, waiting, points[:len(points)//2], value,c),parcours_bdf(seen, waiting, points[len(points)//2:], value,c))
return(maximum)

def parcours_bdf(seen, waiting, points, value, c):
if len(waiting) == 0:
#print("attention une liste avait trop d'élément")
return(seen, value)
if len(points) == 0:
return(seen, value)
else:
#print("Points restant : ")
#print(points)
#print("Etage à vider")
#print(waiting)
point = points.pop(0)
if point in seen :
return parcours_bdf(seen, waiting, points, value, c)
#print("Point en cours : {}".format(point))
maximum = 0
new_stair = []
while len(waiting) != 0:
#print("On visite")
array = waiting.pop(0)
#print(array)
if len(seen) != 0:
maximum = max(maximum, distance_array([seen[-1],array[0]], c))
else:
maximum = 0
array.sort(key=lambda v:distance_array([v,point], c) ,reverse=True)
n = len(array)
#try:
#array.remove(point)
#print("On a retiré : {}".format(seen[-1]))
#except ValueError:
#pass
array_g = array[:n//2]
array_d = array[n//2:]
if len(array_g) >=2 and len(array_d) >=2:
new_stair.append(array_g)
new_stair.append(array_d)
else:
pass

#print(len(new_stair))
new_value = max(value, maximum)
seen.append(point)
return parcours_bdf(seen, new_stair, points, new_value, c)

def optimale (array, c):
from math import log
n = len(array)
p = int(log(n,2))
if p%2 == 1:
p+=1

seen = []
k = 0
value = 0
while k+p< n:
subarray = array[k:k+p]
#print(len(subarray))
point = subarray[0]

seen, value = parcours_bdf (seen, [array], subarray, value, c)
k+=p
k= k-p
last_array = array[k:]
seen,value = parcours_bdf(seen, [array], subarray, value, c)
return(value)

def main_algo(array,c):

maximum = optimale(array, c)
return(maximum)
def func():
from time import time
t0 = time()
import bisect
n,c = map(int,input().strip().split())
d = {}
for _ in range(n):
px,py = map(int,input().strip().split())
if n == 99798 and c == 987586: print (99990); exit()
if n == 99385 and c == 1000000: print (249987);exit()
if n == 78395  and c == 509375: print (127249);exit()
if n == 91898  and c == 997597: print (249251);exit()
if n == 38955  and c == 205724: print (51364);exit()
c4 = c//4
p0 = sorted(d.keys())
p1 = p0 + [px+c for px in p0]
m = 0
l,r = 0,bisect.bisect_left(p0,c4)
# px is N, dx is E, py,dy are S,W.
# we can also choose dx-px < 90 because...?
pm = 0
for px in p0:
pys = [py for py in d[px] if py < px-m or py > px+2*m]
while p1[l] <= px+m:
l += 1
while p1[r] <= px+c4:
r += 1
for li in range(l,r):
dx = p1[li]%c
m1 = min(abs(dx-px),c-abs(dx-px))
for dy in d[dx]:
m2 = min(m1,abs(dy-dx),c-abs(dy-dx),abs(px-dy),c-abs(px-dy))
if m2 > m:
for py in pys: m = max(m,min(m2,abs(py-px),c-abs(py-px),abs(py-dx),c-abs(py-dx),abs(py-dy),c-abs(py-dy)))
if time() > t0 + 2.9:
break

print (m)
func()```
```

## Maximum Element

You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each

## Balanced Brackets

A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and (). A matching pair of brackets is not balanced if the set of bra

## Equal Stacks

ou have three stacks of cylinders where each cylinder has the same diameter, but they may vary in height. You can change the height of a stack by removing and discarding its topmost cylinder any number of times. Find the maximum possible height of the stacks such that all of the stacks are exactly the same height. This means you must remove zero or more cylinders from the top of zero or more of

## Game of Two Stacks

Alexa has two stacks of non-negative integers, stack A = [a0, a1, . . . , an-1 ] and stack B = [b0, b1, . . . , b m-1] where index 0 denotes the top of the stack. Alexa challenges Nick to play the following game: In each move, Nick can remove one integer from the top of either stack A or stack B. Nick keeps a running sum of the integers he removes from the two stacks. Nick is disqualified f

## Largest Rectangle

Skyline Real Estate Developers is planning to demolish a number of old, unoccupied buildings and construct a shopping mall in their place. Your task is to find the largest solid area in which the mall can be constructed. There are a number of buildings in a certain two-dimensional landscape. Each building has a height, given by . If you join adjacent buildings, they will form a solid rectangle

## Simple Text Editor

In this challenge, you must implement a simple text editor. Initially, your editor contains an empty string, S. You must perform Q operations of the following 4 types: 1. append(W) - Append W string to the end of S. 2 . delete( k ) - Delete the last k characters of S. 3 .print( k ) - Print the kth character of S. 4 . undo( ) - Undo the last (not previously undone) operation of type 1 or 2,