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Home Code Solutions Hackerrank Algorithms

The Blacklist- HackerRank Solution

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

admin by admin
August 24, 2022
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15 Days to learn SQL Hard SQL(Advanced)-Solution

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Table of Contents

  • The Blacklist – 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++ replace HackerRank Solution
  • Java rep HackerRank Solution
  • Python 3 rep HackerRank Solution
  • Python 2 rep HackerRank Solution
  • C rep HackerRank Solution
    • Warmup Implementation Strings Sorting Search Graph Theory Greedy Dynamic Programming Constructive Algorithms Bit Manipulation Recursion Game Theory NP Complete Debugging
    • Leave a comment below
      • Related posts:

The Blacklist – 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++ replace HackerRank Solution


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#define TRACE
#define DEBUG

#include <algorithm>
#include <bitset>
#include <deque>
#include <cassert>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <utility>
#include <vector>

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pi;
typedef vector<string> vs;

// Basic macros
#define st          first
#define se          second
#define all(x)      (x).begin(), (x).end()
#define ini(a, v)   memset(a, v, sizeof(a))
#define re(i,s,n)  	for(int i=s;i<(n);++i)
#define rep(i,s,n)  for(int i=s;i<=(n);++i)
#define fr(i,n)     re(i,0,n)
#define repv(i,f,t) for(int i = f; i >= t; --i)
#define rev(i,f,t)  repv(i,f - 1,t)
#define frv(i,n)    rev(i,n,0)
#define pu          push_back
#define mp          make_pair
#define sz(x)       (int)(x.size())

const int oo = 1000000009;
const double eps = 1e-9;

#ifdef TRACE
    #define trace1(x)                cerr << #x << ": " << x << endl;
    #define trace2(x, y)             cerr << #x << ": " << x << " | " << #y << ": " << y << endl;
    #define trace3(x, y, z)          cerr << #x << ": " << x << " | " << #y << ": " << y << " | " << #z << ": " << z << endl;
    #define trace4(a, b, c, d)       cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << endl;
    #define trace5(a, b, c, d, e)    cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << endl;
    #define trace6(a, b, c, d, e, f) cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << " | " << #f << ": " << f << endl;

#else

    #define trace1(x)
    #define trace2(x, y)
    #define trace3(x, y, z)
    #define trace4(a, b, c, d)
    #define trace5(a, b, c, d, e)
    #define trace6(a, b, c, d, e, f)

#endif

int n, k;
const int mx = 25;
int a[mx], b[mx], c[mx][mx];
int dp[mx][(1 << 10) + 10];

int solve(int i, int mask) {
    if(i == n) return 0;
    if(mask == (1 << k) - 1) return oo;
    int &ret = dp[i][mask];
    if(ret != -1) return ret;
    ret = oo;

    fr(nxt, k) if((mask & (1 << nxt)) == 0) {
        int cur = 0;
        re(j, i, n) {
            cur += c[nxt][j];
            ret = min(ret, cur + solve(j + 1, mask | (1 << nxt)));
        }
    }

    return ret;
}

int main() {
    scanf("%d %d", &n, &k);
    fr(i, k) {
        fr(j, n) scanf("%d", &c[i][j]);
    }


    ini(dp, -1);
    printf("%d\n", solve(0, 0));
    
	return 0;
}

Java rep HackerRank Solution


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import java.util.Scanner;

public class Solution {

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int N = input.nextInt();
        int K = input.nextInt();
        int[][] price = new int[K][N];
        for (int k=0; k<K; k++) {
            for (int n=0; n<N; n++) {
                price[k][n] = input.nextInt();
            }
        }
        int limit = 1<<K;
        int[][] dp = new int[N+1][limit+1];
        for (int n=0; n<=N; n++) {
            for (int x=0; x<=limit; x++) {
                dp[n][x] = 1000000;
            }
        }
        dp[0][0] = 0;
        for (int n=1; n<=N; n++) {
            for (int k=0; k<K; k++) {
                int mask = 1 << k;
                for (int from=1; from<=n; from++) {
                    for (int x=0; x<limit; x++) {
                        if ((x&mask) == 0) {
                            int newMask = x|mask;
                            int newValue = dp[from-1][x];
                            for (int i=from; i<=n; i++) {
                                newValue += price[k][i-1];
                            }
                            if (dp[n][newMask] > newValue) {
                                dp[n][newMask] = newValue;
                            }
                        }
                    }
                }
            }
        }
        int min = Integer.MAX_VALUE;
        for (int value : dp[N]) {
            min = Math.min(min, value);
        }
        System.out.println(min);
    }
    
}



Python 3 rep HackerRank Solution


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def bits_free(x, K):
    res = []
    for i in range(K):
        if x % 2 == 0:
            res.append(i)
        x //= 2
    return res

N, K = tuple(int(s) for s in input().strip().split())

costs = []
for i in range(K):
    costs1 = [int(s) for s in input().strip().split()]
    costs.append(costs1)
    
#print(costs)

dpmat = [[[1000000 for i in range(2**K)] for j in range(K)] for k in range(N)]

for mer in range(K):
        dpmat[0][mer][1 << mer] = costs[mer][0]
        
#print(dpmat)
        
for gan in range(1,N):
    for lastmer in range(K):
        for mask in range(2**K):
            if dpmat[gan-1][lastmer][mask] >= 1000000:
                continue
            
            dpmat[gan][lastmer][mask] = min(dpmat[gan][lastmer][mask], dpmat[gan-1][lastmer][mask] + costs[lastmer][gan])
            for newmer in bits_free(mask, K):
                newmask = mask | (1 << newmer)
                dpmat[gan][newmer][newmask] = min(dpmat[gan][newmer][newmask], dpmat[gan-1][lastmer][mask] + costs[newmer][gan])
                
        
best = 1000000
for lastmer in range(K):
    for mask in range(2**K):
        if dpmat[N-1][lastmer][mask] < 0:
            continue  
            
        best = min(best, dpmat[N-1][lastmer][mask])
 
#print(dpmat)
print(best)



Python 2 rep HackerRank Solution


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#I want to use a dynamic program to solve this, since there are
#approximately 29!/20! or about 1.5*10^13 ways to arrange the
#mercenaries (with a slight overcounting of cases where not all
#mercenaries are used)
#I will traverse the list of enemies from start to finish. I will
#keep track of states which represent the mercenaries no longer
#available (2^10-1 cases, since the last guy used always remains
#available) x the last guy used, for about 10,000 active cases.
#For each mercenary and each case, consider the result if each
#available mercenary kills the next enemy, and store the one with
#the least cost in that state.
n,k=map(int,raw_input().strip().split())
costs=[]
for j in xrange(k):
  costs.append(map(int,raw_input().strip().split()))
states={}
#initial state, last mercenary used is -1 to indicate none
states[(True,)*k+(-1,)]=0
for i in xrange(n):
  newstates={}
  for state in states:
    for j in xrange(k):
      #if mercenary j is still available
      if state[j]:
        #generate new state if he kills enemy i
        nstate=list(state)
        nstate[-1]=j
        if state[-1]!=-1 and state[-1]!=j:
          nstate[state[-1]]=False
        nstate=tuple(nstate)
        cost=states[state]+costs[j][i]
        if nstate in newstates:
          if cost<newstates[nstate]:
            newstates[nstate]=cost
        else:
          newstates[nstate]=cost
  states=newstates
#find cheapest result
cheapest=200001
for state in states:
  if states[state]<cheapest:
    cheapest=states[state]
print cheapest



C rep HackerRank Solution


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#include <stdio.h>

#define set(c,d) ((1<<d)|(c))
#define isSet(c,d) ((1<<d)&(c))

int n, k;
int a[20][20];
int memo[20][20][2050];

int bestwithout(int i, int j, int without) {

	int best;
	int cp, x;

	if(i == 0) {

//		printf("best[%d][%d]: %d\n", i, j, a[i][j]);
		return a[i][j];
	}

	cp = memo[i][j][without];
	if(cp != -1) {
		
	//	puts("MEMO found!");
		return cp;
	}

	best = bestwithout(i-1, j, without);


	for(x=0;x<k;x++) {

		if(!isSet(without,x)) {

			cp = bestwithout(i-1, x, set(without,x));

			if(cp < best)
				best = cp;
		}
	}

//	printf("best[%d][%d][%d]: %d\n", i, j, without, best);

	best += a[i][j];
	memo[i][j][without] = best;


return best;
}

int main(void) {

int i, j, h, cp, best;

scanf("%d %d", &n, &k);

//printf("%d %d\n", n, k);

for(i=0;i<k;i++)
	for(j=0;j<n;j++)
		scanf("%d", &a[j][i]);

//printf("%d\n", a[n-1][0]);
for(i=0;i<20;i++)
	for(j=0;j<20;j++)
		for(h=0;h<2050;h++)
			memo[i][j][h] = -1;


best = bestwithout(n-1, 0, set(0,0));


for(i=1;i<k;i++) {

	cp = bestwithout(n-1, i, set(0,i));

	if(cp < best)
		best = cp;
}

printf("%d\n", best);


return 0;
}

 

Warmup
Implementation
Strings
Sorting
Search
Graph Theory
Greedy
Dynamic Programming
Constructive Algorithms
Bit Manipulation
Recursion
Game Theory
NP Complete
Debugging

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