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Home Code Solutions Hackerrank Algorithms

Recording Episodes – HackerRank Solution

Recording Episodes - HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

bhautik bhalala by bhautik bhalala
May 26, 2022
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Table of Contents

  • Recording Episodes – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • Solutions of Algorithms Data Structures Hard HackerRank:
    • Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts
  • C++ Recording Episodes HackerRank Solution
  • Java Recording Episodes HackerRank Solution
  • Python 3 Recording Episodes HackerRank Solution
  • Python 2 Recording Episodes HackerRank Solution
  • C Recording Episodes HackerRank Solution
    • Leave a comment below
      • Related posts:

Recording Episodes – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Solutions of Algorithms Data Structures Hard HackerRank:

Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts

C++ Recording Episodes HackerRank Solution


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#include <bits/stdc++.h>
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define REP(I, N) for (int I = 0; I < (N); ++I)
#define REPP(I, A, B) for (int I = (A); I < (B); ++I)
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define RS(X) scanf("%s", (X))
#define CASET int ___T, case_n = 1; scanf("%d ", &___T); while (___T-- > 0)
#define MP make_pair
#define PB push_back
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define PII pair<int,int>
#define VI vector<int>
#define VPII vector<pair<int,int> >
#define PLL pair<long long,long long>
#define VPLL vector<pair<long long,long long> >
#define F first
#define S second
typedef long long LL;
using namespace std;
const int MOD = 1e9+7;
const int SIZE = 1e6+10;
struct SCC{
    int n,used[SIZE],order[SIZE],gg[SIZE];
    vector<int>e[SIZE],ae[SIZE],ge[SIZE],emp;
    int id,gn;
    void init(int _n){
        n=_n;
        memset(used,0,sizeof(int)*n);
        REP(i,n){
            e[i]=ae[i]=ge[i]=emp;
        }
    }
    void add_edge(int x,int y){
        e[x].PB(y^1);
        ae[y^1].PB(x);
        e[y].PB(x^1);
        ae[x^1].PB(y);
    }
    void dfs1(int x){
        if(used[x]==1)return;
        used[x]=1;
        REP(i,SZ(e[x])){
            int y=e[x][i];
            dfs1(y);
        }
        order[--id]=x;
    }
    void dfs2(int x){
        if(used[x]==2)return;
        gg[x]=gn;
        used[x]=2;
        REP(i,SZ(ae[x])){
            int y=ae[x][i];
            if(used[y]!=2)dfs2(y);
        }
    }
    bool good(){
        gn=0;
        id=n;
        REP(i,n)
            dfs1(i);
        REP(i,n){
            if(used[order[i]]!=2){
                dfs2(order[i]);
                gn++;
            }
        }
        REP(i,n){
            if(gg[i]==gg[i^1])return 0;
            i++;
        }
        return 1;
    }
}scc;
int input[100][2][2];
bool XX(int x1,int y1,int x2,int y2){
    return !((y1<x2)||(y2<x1));
}
int main(){
    CASET{
        DRI(n);
        REP(i,n)REP(j,4)RI(input[i][j>>1][j&1]);
        int rr=1;
        int an1=1,an2=0;
        REP(i,n){
            if(i+an1>=n)break;
            while(rr<n){
                scc.init((rr-i)*2+2);
                REPP(k2,i,rr+1)
                    REPP(k1,i,k2){
                        REP(j1,2)REP(j2,2){
                            if(XX(input[k1][j1][0],input[k1][j1][1],input[k2][j2][0],input[k2][j2][1])){
                                scc.add_edge((k1-i)*2+j1,(k2-i)*2+j2);
                            }
                        }
                    }
                if(!scc.good())break;
                else rr++;
            }
            if(rr-i>an1){an1=rr-i;an2=i;}
        }
        an2++;
        printf("%d %d\n",an2,an2+an1-1);
    }
    return 0;
}

Java Recording Episodes HackerRank Solution


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import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class F {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		for(int T = ni();T > 0;T--){
			int n = ni();
			int[][] rs = new int[n][];
			for(int i = 0;i < n;i++){
				rs[i] = na(4);
			}
			int[] from = new int[20005];
			int[] to = new int[20005];
			// !(a and b)
			int maxlen = 0;
			int argi = -1;
			int argj = -1;
			outer:
			for(int i = 0;i < n;i++){
				int p = 0;
				for(int j = i;j < n;j++){
					for(int k = i;k < j;k++){
						for(int u = 0;u < 2;u++){
							for(int v = 0;v < 2;v++){
								if(Math.max(rs[k][u*2], rs[j][v*2]) <=
										Math.min(rs[k][u*2+1], rs[j][v*2+1])){
									from[p] = k*2+u; to[p] = j*2+v^1; p++;
									from[p] = j*2+v; to[p] = k*2+u^1; p++;
								}
							}
						}
					}
					if(j-i+1 <= maxlen)continue;
//					tr(Arrays.copyOf(from, p));
//					tr(Arrays.copyOf(to, p));
					int[][] g = packD(n*2, from, to, p);
					int[] clus = decomposeToSCC(g);
					for(int k = i;k <= j;k++){
						if(clus[2*k] == clus[2*k+1]){
							continue outer;
						}
					}
					if(j-i+1 > maxlen){
						maxlen = j-i+1;
						argi = i;
						argj = j;
					}
				}
			}
			out.println(argi+1 + " " + (argj+1));
		}
	}
	
	public static int[] decomposeToSCC(int[][] g)
	{
		int n = g.length;
		int[] stack = new int[n+1];
		int[] ind = new int[n+1];
		int[] ord = new int[n];
		Arrays.fill(ord, -1);
		int[] low = new int[n];
		Arrays.fill(low, -1);
		int sp = 0;
		int id = 0; // preorder
		int[] clus = new int[n];
		int cid = 0;
		int[] cstack = new int[n+1];
		int csp = 0;
		boolean[] incstack = new boolean[n];
		for(int i = 0;i < n;i++){
			if(ord[i] == -1){
				ind[sp] = 0;
				cstack[csp++] = i;
				stack[sp++] = i;
				incstack[i] = true;
				while(sp > 0){
					int cur = stack[sp-1];
					if(ind[sp-1] == 0){
						ord[cur] = low[cur] = id++;
					}
					if(ind[sp-1] < g[cur].length){
						int nex = g[cur][ind[sp-1]];
						if(ord[nex] == -1){
							ind[sp-1]++;
							ind[sp] = 0;
							incstack[nex] = true;
							cstack[csp++] = nex;
							stack[sp++] = nex;
						}else{
							// shortcut
//							U.tr(cur, nex, incstack[nex], low[nex], stack);
							if(incstack[nex])low[cur] = Math.min(low[cur], low[nex]);
							ind[sp-1]++;
						}
					}else{
						if(ord[cur] == low[cur]){
							while(csp > 0){
								incstack[cstack[csp-1]] = false;
								clus[cstack[--csp]] = cid;
								if(cstack[csp] == cur)break;
							}
							cid++;
						}
						if(--sp >= 1)low[stack[sp-1]] = Math.min(low[stack[sp-1]], low[stack[sp]]);
					}
				}
			}
		}
		return clus;
	}

	
	public static int[][] packD(int n, int[] from, int[] to, int sup)
	{
		int[][] g = new int[n][];
		int[] p = new int[n];
		for(int i = 0;i < sup;i++)p[from[i]]++;
		for(int i = 0;i < n;i++)g[i] = new int[p[i]];
		for(int i = 0;i < sup;i++){
			g[from[i]][--p[from[i]]] = to[i];
		}
		return g;
	}


	
	void run() throws Exception
	{
		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		long s = System.currentTimeMillis();
		solve();
		out.flush();
		if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
	}
	
	public static void main(String[] args) throws Exception { new F().run(); }
	
	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;
	
	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	
	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	
	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }
	
	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	
	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private static void tr(Object... o) { System.out.println(Arrays.deepToString(o)); }
}

 



Python 3 Recording Episodes HackerRank Solution


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#True if two intervals interesects
def has_intersection(I1,I2):
    if I1[0] >= I2[0] and I1[0] <= I2[1]: return True
    if I1[1] >= I2[0] and I1[1] <= I2[1]: return True
    if I2[0] >= I1[0] and I2[0] <= I1[1]: return True
    if I2[1] >= I1[0] and I2[1] <= I1[1]: return True
    return False

#Given the SAT Graph, return the longest interval with a dual-complete graph 
def max_range(G,GT):
    size = len(G)/4
    maxRange = [0,1]
    maxSize = 1

    a = 0
    b = 2

    while b <= size:
        SCC = []
        SCC = get_strong_connected_components(G,GT, a*4, b*4)
        newSize = b-a
        if is_satisfiable(SCC):
            if newSize > maxSize:
                maxRange = [a,b]
                maxSize = newSize
            b+=1
        else:
            if newSize == maxSize + 1:
                b+=1
            a+=1

    return maxRange

#Get the components only considering vertices between a (inclusive) and b (exclusive) indices
def get_strong_connected_components(G, GT, a, b):
    SCC = []
    numVertices = b-a
    compQueue = []
    Visited = [0 for c1 in range(numVertices)]
    for c1 in range(a,b):
        build_queue(G, a, b, compQueue, Visited, c1)
    while len(compQueue) > 0:
        i = compQueue.pop(-1)
        C = []
        build_component(GT, a, b, Visited, C, i)
        if len(C)>0: SCC.append(C)
    return SCC

def build_queue(G,a,b,compQueue, Visited, i):
    if Visited[i-a] == 1: return
    Visited[i-a] += 1

    for j in G[i]:
        if j>=a and j<b:
            build_queue(G, a, b, compQueue, Visited, j)
    compQueue.append(i)
    return

def build_component(GT, a, b, Visited, C, i):
    if Visited[i-a] == 2: return
    Visited[i-a] += 1
    for j in GT[i]:
        if j>=a and j<b:
            build_component(GT, a, b, Visited, C, j)
    C.append(i)
    return

    #returns False if for some a, a and not a are in the same component
def is_satisfiable(SCC):
    for C in SCC:
        C.sort()
        for c1 in range(len(C)-1):
            if C[c1+1]-C[c1] == 1 and C[c1]%2==0: return False
    return True

def build_digraph(S):
    size = len(S)*4
    G = []
    GT = []
    for c1 in range(size):
        G.append([])
        GT.append([])

    for c1 in range(len(S)):
        i = c1*4
        G[i+1].append(i+2)
        GT[i+2].append(i+1)
        G[i+3].append(i)
        GT[i].append(i+3)

    for c1 in range(len(S)):
        for c2 in range(2):
            i = c2*2
            tr1 = [S[c1][i],S[c1][i+1]]
            for c3 in range(c1, len(S)):
                for c4 in range(2):
                    if c1==c3 and c2==c4: continue
                    j = c4*2
                    tr2 = [S[c3][j],S[c3][j+1]]
                    if has_intersection(tr1,tr2):
                        k = c1*4
                        l = c3*4
                        G[k+i].append(l+j+1)
                        GT[l+j+1].append(k+i)
                        if c1!=c3:
                            G[l+j].append(k+i+1)
                            GT[k+i+1].append(l+j)

    return [G,GT]

    #main
q = int(input())
for c1 in range(q):
    n = int(input())
    S = []
    for c2 in range(n):
        s = input().split(' ')
        for c3 in range(len(s)):
            s[c3] = int(s[c3])
        S.append(s)
    G,GT = build_digraph(S)
    maxRange = max_range(G,GT)
    print(f'{maxRange[0]+1} {maxRange[1]}')
    



Python 2 Recording Episodes HackerRank Solution


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import sys
import copy

T = input()

def overlaps(t1, t2):
    if t1[0] > t2[0]:
        t1,t2 = t2,t1
    if t1[1] >= t2[0]:
        return True
    return False

for _ in range(T):
    times = []
    epi = input()
    for _ in range(epi):
        s1,e1,s2,e2 = map(int, raw_input().strip().split(' '))
        times.append([(s1, e1), (s2, e2)])
    #print 'times:'
    #print times

    cur_seq = 0
    max_seq = float('-inf')
    current_t = []
    max_s_e = []
    for i in range(len(times)):
        #print cur_seq
        if len(current_t) == 0:
            current_t = [[times[i][0]], [times[i][1]]]
            cur_s_e = i
            cur_seq = 1
        else:
            for j in range(len(current_t)):
                overlap1 = False
                overlap2 = False
                for k in range(len(current_t[j])):
                    if overlaps(current_t[j][k], times[i][0]) is True:
                        # create new slot if both are able to get in
                        overlap1 = True
                    if overlaps(current_t[j][k], times[i][1]) is True:
                        overlap2 = True
                    if overlap1 is True and overlap2 is True:
                        break
                if overlap1 is False and overlap2 is True:
                    current_t[j].append(times[i][0])
                    cur_seq += 1
                    #print '1: '
                    #print current_t
                elif overlap1 is True and overlap2 is False:
                    current_t[j].append(times[i][1])
                    cur_seq += 1
                    #print '2: '
                    #print current_t
                elif overlap1 is False and overlap2 is False:
                    current_t.append(copy.copy(current_t[j]))
                    current_t[j].append(times[i][0])
                    current_t[len(current_t) - 1].append(times[i][1])
                    cur_seq += 1
                    #print '3: '
                    #print current_t
                # both overlap, cannot continue
                #else:
                #    current_t = [[times[i][0]], [times[i][1]]]
                #    cur_s_e = i
                #    cur_seq = 1
                #    print '4: '
                #    print current_t
            if overlap1 is True and overlap2 is True:
                current_t = []
            else:
                current_t = [x for x in current_t if len(x) == cur_seq]
            # cannot continue, start new
            if len(current_t) == 0:
                if cur_seq > max_seq:
                    max_seq = cur_seq
                    max_s_e = [cur_s_e, i-1]
                current_t = [[(times[i][0][0], times[i][0][1])], [(times[i][1][0], times[i][1][1])]]
        if cur_seq > max_seq:
            max_seq = cur_seq
            max_s_e = [cur_s_e, i-1]
    #print current_t
    #print max_seq
    #print cur_s_e
    print max_s_e[0] + 1, max_s_e[1] + 2



C Recording Episodes HackerRank Solution


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#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct _lnode{
  int x;
  int w;
  struct _lnode *next;
} lnode;
void clean_table();
void insert_edge(int x,int y,int w);
int check();
void dfs1(int x);
void dfs2(int x);
int checkx(int x);
int l,r,n,st,c,a[4][100],mark[400],component[400],s[400];
lnode *table[400]={0},*rtable[400]={0};

int main(){
  int q,max,maxi,i,j;
  scanf("%d",&q);
  while(q--){
    scanf("%d",&n);
    for(i=0;i<n;i++)
      scanf("%d%d%d%d",&a[0][i],&a[1][i],&a[2][i],&a[3][i]);
    for(i=0;i<n;i++){
      insert_edge(2*n+i,n+i,1);
      insert_edge(3*n+i,i,1);
      for(j=0;j<n;j++){
        if(i==j)
          continue;
        if(!(a[0][i]>a[1][j] || a[0][j]>a[1][i]))
          insert_edge(i,2*n+j,1);
        if(!(a[2][i]>a[1][j] || a[0][j]>a[3][i]))
          insert_edge(n+i,2*n+j,1);
        if(!(a[0][i]>a[3][j] || a[2][j]>a[1][i]))
          insert_edge(i,3*n+j,1);
        if(!(a[2][i]>a[3][j] || a[2][j]>a[3][i]))
          insert_edge(n+i,3*n+j,1);
      }
    }
    max=1;
    maxi=0;
    for(i=0;i<n;i++)
      for(j=max+1;i+j<=n;j++){
        l=i;
        r=i+j-1;
        if(check()){
          max=j;
          maxi=i;
        }
        else
          break;
      }
    printf("%d %d\n",maxi+1,maxi+max);
    clean_table();
  }
  return 0;
}
void clean_table(){
  int i;
  lnode *p,*pp;
  for(i=0;i<400;i++)
    if(table[i]){
      p=table[i];
      while(p){
        pp=p->next;
        free(p);
        p=pp;
      }
      table[i]=NULL;
    }
  for(i=0;i<400;i++)
    if(rtable[i]){
      p=rtable[i];
      while(p){
        pp=p->next;
        free(p);
        p=pp;
      }
      rtable[i]=NULL;
    }
  return;
}
void insert_edge(int x,int y,int w){
  lnode *t=malloc(sizeof(lnode));
  t->x=y;
  t->w=w;
  t->next=table[x];
  table[x]=t;
  t=malloc(sizeof(lnode));
  t->x=x;
  t->w=w;
  t->next=rtable[y];
  rtable[y]=t;
  return;
}
int check(){
  int i;
  st=c=0;
  memset(mark,0,sizeof(mark));
  memset(component,0,sizeof(component));
  for(i=0;i<4*n;i++)
    if(!mark[i] && checkx(i))
      dfs1(i);
  memset(mark,0,sizeof(mark));
  while(st){
    if(!mark[s[st-1]]){
      c++;
      dfs2(s[st-1]);
    }
    st--;
  }
  for(i=0;i<2*n;i++)
    if(component[i]==component[i+2*n] && component[i] && checkx(i) && checkx(i+2*n))
      return 0;
  return 1;
}
void dfs1(int x){
  lnode *p;
  mark[x]=1;
  for(p=table[x];p;p=p->next)
    if(!mark[p->x] && checkx(p->x))
      dfs1(p->x);
  s[st++]=x;
  return;
}
void dfs2(int x){
  lnode *p;
  mark[x]=1;
  for(p=rtable[x];p;p=p->next)
    if(!mark[p->x] && checkx(p->x))
      dfs2(p->x);
  component[x]=c;
  return;
}
int checkx(int x){
  return (x>=l && x<=r || x>=l+n && x<=r+n || x>=l+2*n && x<=r+2*n || x>=l+3*n && x<=r+3*n);
}

 

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