• Home
  • Top Posts
  • Code Solutions
  • How to
  • News
  • Trending
  • Anime
  • Health
  • Education
Wednesday, February 8, 2023
  • Login
Zeroplusfour
No Result
View All Result
  • Home
  • Top Posts
  • Code Solutions
  • How to
  • News
  • Trending
  • Anime
  • Health
  • Education
  • Home
  • Top Posts
  • Code Solutions
  • How to
  • News
  • Trending
  • Anime
  • Health
  • Education
No Result
View All Result
Zeroplusfour
No Result
View All Result
Home Code Solutions Hackerrank Algorithms

Tree Splitting – HackerRank Solution

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

bhautik bhalala by bhautik bhalala
May 27, 2022
Reading Time: 1 min read
0
15 Days to learn SQL Hard SQL(Advanced)-Solution

15 Days to learn SQL Hard SQL(Advanced)-Solution alt text

Spread the love

Table of Contents

  • Tree Splitting – 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++ Tree Splitting HackerRank Solution
  • Java Tree Splitting HackerRank Solution
  • C Tree Splitting  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:

Tree Splitting – 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++ Tree Splitting HackerRank Solution


Copy Code Copied Use a different Browser

#include<cstdio>
#include<ctime>
#include<cstdlib>
#include<vector>

using namespace std;

int INF, N, M, nr, V, v[400009], ap[200009], t[200009];
bool scos[200009];
vector < int > muchii[200009];

struct nod
{
    int K, P, nr;
    nod *l, *r, *t;
    nod (int K, int P, int nr, nod *l, nod *r)
    {
        this->nr = nr;
        this->l = l;
        this->r = r;
        this->P = P;
        this->K = K;
        this->t = 0;
    }
}*adresa1[200009], *adresa2[200009], *nil, *R;

int Rand(){return ((rand()%32768)<<15)+(rand()%32768)+1;}

void reup (nod *&n)
{
    if (n->l != nil) n->l->t = n;
    if (n->r != nil) n->r->t = n;
    n->nr = n->l->nr + n->r->nr + 1;
}

void rot_left (nod *&n)
{
    nod *t=n->l;
    n->l = t->r, t->r = n;
    t->t = n->t;
    reup (n);
    reup (t);
    n = t;
}

void rot_right (nod *&n)
{
    nod *t=n->r;
    n->r = t->l, t->l = n;
    t->t = n->t;
    reup (n);
    reup (t);
    n = t;
}

void balance (nod *&n)
{
    if (n->l->P > n->P)
        rot_left (n);
    else
    if (n->r->P > n->P)
        rot_right (n);
}

void Insert (nod *&n, int Poz, int K, int P)
{
    if (n == nil)
    {
        n = new nod (K, P, 1, nil, nil);
        return ;
    }
    if (n->l->nr >= Poz) Insert (n->l, Poz, K, P);
    else Insert (n->r, Poz - n->l->nr - 1, K, P);
    reup (n);
    balance (n);
}

void Erase (nod *&n, int Poz)
{
    if (n->l->nr >= Poz)
        Erase (n->l, Poz);
    else
    if (n->l->nr + 1 < Poz)
        Erase (n->r, Poz - n->l->nr - 1);
    else
    {
        if (n->l == nil && n->r == nil)
        {
            delete n;
            n = nil;
        }
        else
        {
            if (n->l->P > n->r->P)
                rot_left (n);
            else
                rot_right (n);
            Erase (n, Poz);
        }
    }
    if (n != nil) reup (n);
}

void join(nod *&R, nod *&Rl, nod *&Rr)
{
    R=new nod(0, 0, Rl->nr + Rr->nr + 1, Rl, Rr);
    Erase ( R, Rl->nr+1 );
}

void split(nod *&R, nod *&Rl, nod *&Rr, int Poz)
{
    Insert (R, Poz, 0, INF);
    Rl = R->l, Rl->t = 0;
    Rr = R->r, Rr->t = 0;
    delete R, R = nil;
}

bool nu_radacina (nod *&n)
{
    if (n->t != 0 && (n->t->l == n || n->t->r == n)) return 1;
    return 0;
}

void det_root (nod *n, nod *&R)
{
    while (1)
    {
        if ( nu_radacina (n) )
            n = n->t;
        else
            break;
    }
    R = n;
}

int Get_Pos (nod *n)
{
    int ans = n->l->nr + 1;
    while (1)
    {
        if (nu_radacina(n))
        {
            if (n->t->l == n) n = n->t;
            else ans += n->t->l->nr + 1, n = n->t;
        }
        else break;
    }
    return ans;
}

void dfs (int nod)
{
    ap[nod] = 1;
    v[++nr] = nod;
    for (vector < int > :: iterator it = muchii[nod].begin(); it != muchii[nod].end(); it++)
        if (ap[*it] == 0)
        {
            dfs (*it);
            t[*it] = nod;
        }
    v[++nr] = -nod;
}

void build (nod *&n)
{
    if (n == nil) return ;
    if (n->K < 0)
        adresa2[-n->K] = n;
    else
        adresa1[n->K] = n;
    build (n->l);
    build (n->r);
}

void afis (nod *&n)
{
    if (n == nil) return ;
    afis (n->l);
    printf ("%d ", n->K);
    afis (n->r);
}

void Del (int a, int b)
{
    if (t[b] == a)
    {
        int aux = a;
        a = b;
        b = aux;
    }
    /////t[a] == b, deci vad in ce treap este a, si ii dau split pentru a separa subarborele
    nod *R, *R1, *R2, *R3, *R4;
    det_root (adresa1[a], R);
    /////sunt in treapul cu radacina R
    int p1, p2;
    p1 = Get_Pos (adresa1[a]);
    p2 = Get_Pos (adresa2[a]);
    split (R, R4, R3, p2);
    split (R4, R1, R2, p1-1);
    R2->t = 0;
    join (R, R1, R3);
    /*afis (R);
    printf ("\n");
    afis (R2);
    printf ("\n");*/
}

int main()
{
//freopen ("input", "r", stdin);
//freopen ("output", "w", stdout);

srand(time(0));
INF=(1<<30)+7;
scanf ("%d", &N);
for (int i=1; i<N; i++)
{
    int X, Y;
    scanf ("%d %d", &X, &Y);
    muchii[X].push_back(Y);
    muchii[Y].push_back(X);
}
dfs (1);

nil = new nod (0, 0, 0, 0, 0);
R = nil;
for (int i=1; i<=nr; i++)
    Insert (R, i-1, v[i], Rand());
build (R);
int V = 0;
scanf ("%d", &M);
for (int i=1; i<=M; i++)
{
    int x;
    scanf ("%d", &x), x ^= V;
    nod *R;
    det_root (adresa1[x], R);
    printf ("%d\n", R->nr / 2), scos[x] = 1, V = R->nr / 2;
    for (auto it = muchii[x].begin (); it != muchii[x].end (); it ++)
        if (!scos[*it]) Del (x, *it);
}
return 0;
}

Java Tree Splitting HackerRank Solution


Copy Code Copied Use a different Browser

 

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 E2 {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	// 1-5
	// |
	// 4-3
	// |
	// 2
	
	void solve()
	{
		int n = ni();
		int[] from = new int[n - 1];
		int[] to = new int[n - 1];
		Node[] nodes = new Node[n];
		for(int i = 0;i < n;i++){
			nodes[i] = new Node(i);
		}
		for (int i = 0; i < n - 1; i++) {
			from[i] = ni() - 1;
			to[i] = ni() - 1;
		}
		int[][] g = packU(n, from, to);
		int[][] pars = parents3(g, 0);
		int[] par = pars[0];
		for(int i = 0;i < n;i++){
			if(par[i] != -1){
				link(nodes[par[i]], nodes[i]);
			}
		}
		
		int ans = 0;
		boolean[] deled = new boolean[n];
		for(int Q = ni();Q > 0;Q--){
			int x = ni()^ans;
			x--;
//			int x = ni()-1;
			
//			for(int i = 0;i < n;i++)tr(i, nodes[i]);
			expose(nodes[x]);
//			for(int i = 0;i < n;i++)tr(i, nodes[i]);
//			tr();
			ans = nodes[x].size + nodes[x].sumexsize;
//			tr(nodes[x].size, nodes[x].sumexsize);
//			tr("ans", ans);
			out.println(ans);
			for(int e : g[x]){
				if(!deled[e]){
					if(par[x] == e){
						cut(nodes[x]);
					}else{
						cut(nodes[e]);
					}
				}
			}
			deled[x] = true;
		}
	}

	public static int[][] parents3(int[][] g, int root) {
		int n = g.length;
		int[] par = new int[n];
		Arrays.fill(par, -1);

		int[] depth = new int[n];
		depth[0] = 0;

		int[] q = new int[n];
		q[0] = root;
		for (int p = 0, r = 1; p < r; p++) {
			int cur = q[p];
			for (int nex : g[cur]) {
				if (par[cur] != nex) {
					q[r++] = nex;
					par[nex] = cur;
					depth[nex] = depth[cur] + 1;
				}
			}
		}
		return new int[][] { par, q, depth };
	}

	static int[][] packU(int n, int[] from, int[] to) {
		int[][] g = new int[n][];
		int[] p = new int[n];
		for (int f : from)
			p[f]++;
		for (int t : to)
			p[t]++;
		for (int i = 0; i < n; i++)
			g[i] = new int[p[i]];
		for (int i = 0; i < from.length; i++) {
			g[from[i]][--p[from[i]]] = to[i];
			g[to[i]][--p[to[i]]] = from[i];
		}
		return g;
	}
	
		public static class Node
		{
			public int val;
			public Node left, right, parent;
			public Node exparent;
			
			public int size;
			
			public int sumexsize;
			public int exsize;
			
			public Node(int val)
			{
				this.val = val;
				update();
			}
			
			public void update()
			{
				this.sumexsize = this.exsize + exsize(left) + exsize(right);
				this.size = 1 + size(left) + size(right);
			}

			@Override
			public String toString() {
				return "Node [val=" + val + ", size=" + size
						+ ", exsize=" + exsize
						+ ", sumexsize=" + sumexsize
						+ ", exparent=" + (exparent != null) + "]";
			}

			public String toString(String indent) {
				StringBuilder builder = new StringBuilder();
				if(left != null)builder.append(left.toString(indent + "  "));
				builder.append(indent).append(this.toString()).append("\n");
				if(right != null)builder.append(right.toString(indent + "  "));
				return builder.toString();
			}
		}
		
		public static Node expose(Node x)
		{
			if(x == null)return null;
			while(splay(x).exparent != null)promote(x);
			return x;
		}
		
		private static void promote(Node x)
		{
			Node xp = x.exparent;
			splay(xp);
			xp.exsize -= size(x) + exsize(x);
			xp.exsize += size(xp.right) + exsize(xp.right);
			Node xpr = xp.right;
			xp.right = x;
			x.parent = xp;
			x.exparent = null;
			if(xpr != null){
				xpr.exparent = xp;
				xpr.parent = null;
			}
			xp.update();
		}
		
		public static void cut(Node x)
		{
			expose(x);
			Node left = x.left;
			if(left != null)left.parent = null;
			x.left = null;
			x.update();
		}
		
		public static void link(Node par, Node ch)
		{
			expose(par);
			expose(ch);
			Node xpr = par.right;
			par.right = ch;
			ch.parent = par;
			
			if(xpr != null){
				xpr.exparent = par;
				xpr.parent = null;
				par.exsize += size(xpr) + exsize(xpr);
			}
			par.update();
		}
		
		public static Node root(Node x)
		{
			return first(expose(x));
		}
		
		public static boolean inSameLC(Node x, Node y)
		{
			return first(expose(x)).equals(first(expose(y)));
		}
		
		public static boolean inSameSplay(Node x, Node y)
		{
			Node rx = x, ry = y;
			while(rx.parent != null)rx = rx.parent;
			while(ry.parent != null)ry = ry.parent;
			boolean ret = rx.equals(ry);
			splay(x); splay(y);
			return ret;
		}
		
		public static Node lca(Node x, Node y)
		{
			expose(x);
			Node lastPromoted = null;
			while(splay(y).exparent != null){
				lastPromoted = y.exparent;
				promote(y);
			}
			if(inSameSplay(x, y)){
//			if(first(x).equals(first(y))){
				if(index(x) < index(y)){
					return x;
				}else{
					return y;
				}
			}else{
				return lastPromoted;
			}
		}
		
		public static void update(Node x, int v)
		{
			x.val = v;
			x.update();
			splay(x);
		}
		
		private static int size(Node n){ return n == null ? 0 : n.size; }
		private static int exsize(Node n){ return n == null ? 0 : n.sumexsize; }
		
		public static Node get(Node any, int K)
		{
			splay(any);
			if(K < 0 || K >= size(any))throw new IllegalArgumentException();
			Node cur = any;
			while(true){
				if(K == size(cur.left))break;
				if(K < size(cur.left)){
					cur = cur.left;
				}else{
					K -= size(cur.left) + 1;
					cur = cur.right;
				}
			}
			return splay(cur);
		}
		
		public static int index(Node x)
		{
			return size(splay(x).left);
		}
		
		public static Node first(Node any)
		{
			splay(any);
			while(any.left != null)any = any.left;
			return splay(any);
		}
		
		public static Node last(Node any)
		{
			splay(any);
			while(any.right != null)any = any.right;
			return splay(any);
		}
		
		public static Node erase(Node x)
		{
			splay(x);
			if(x.left != null)x.left.parent = null;
			if(x.right != null)x.right.parent = null;
			if(x.left != null){
				Node last = last(x.left);
				last.right = x.right;
				if(x.right != null)x.right.parent = last;
				last.update();
			}
			Node ret = x.left != null ? x.left : x.right;
			x.left = x.right = null;
			x.update();
			return ret;
		}
		
		public static void insert(Node any, int K, Node x)
		{
			splay(any);
			if(K < 0 || K > size(any))throw new IllegalArgumentException();
			if(any == null)return;
			Node cur = any;
			while(true){
				if(K == 0 && cur.left == null){
					cur.left = x;
					x.parent = cur;
					break;
				}
				if(K == size(cur) && cur.right == null){
					cur.right = x;
					x.parent = cur;
					break;
				}
				if(K <= size(cur.left)){
					cur = cur.left;
				}else{
					K -= size(cur.left) + 1;
					cur = cur.right;
				}
			}
			splay(x);
		}
		
		public static Node splay(Node x)
		{
			if(x == null)return null;
			while(x.parent != null){
				Node p = x.parent;
				if(p.parent == null){
					// zig
					if(p.left == x)rotateRight(p);else rotateLeft(p);
				}else{
					Node g = p.parent;
					if(g.left == p){
						if(p.left == x){
							// zig-zig
							rotateRight(g); rotateRight(p);
						}else{
							// zig-zag
							rotateLeft(p); rotateRight(g);
						}
					}else{
						if(p.left == x){
							// zig-zig
							rotateRight(p); rotateLeft(g);
						}else{
							// zig-zag
							rotateLeft(g); rotateLeft(p);
						}
					}
				}
			}
			return x;
		}
		
		//   x      y
		//  y c -> a x   return y
		// a b      b c
		private static Node rotateRight(Node x)
		{
			if(x == null || x.left == null)return x;
			Node y = x.left;
			x.left = y.right;
			y.right = x;
			
			y.exparent = x.exparent;
			x.exparent = null;
			
			Node par = x.parent;
			if(par != null){
				if(par.left == x)par.left = y;
				if(par.right == x)par.right = y;
			}
			if(x.left != null)x.left.parent = x;
			x.parent = y;
			y.parent = par;
			
			x.update();
			y.update();
			
			return y;
		}
		
		//   x      y
		//  a y -> x c   return y
		//   b c  a b
		private static Node rotateLeft(Node x)
		{
			if(x == null || x.right == null)return null;
			Node y = x.right;
			x.right = y.left;
			y.left = x;
			
			y.exparent = x.exparent;
			x.exparent = null;
			
			Node par = x.parent;
			if(par != null){
				if(par.left == x)par.left = y;
				if(par.right == x)par.right = y;
			}
			if(x.right != null)x.right.parent = x;
			x.parent = y;
			y.parent = par;
			
			x.update();
			y.update();
			
			return y;
		}
		
		
		public static Node[] nodes(Node any)
		{
			splay(any);
			int n = any.size;
			Node[] ret = new Node[n];
			dfsNodes(0, any, ret);
			return ret;
		}
		
		private static void dfsNodes(int offset, Node cur, Node[] ret)
		{
			if(cur == null)return;
			ret[offset + size(cur.left)] = cur;
			dfsNodes(offset, cur.left, ret);
			dfsNodes(offset+size(cur.left)+1, cur.right, ret);
		}
	
	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 E2().run(); }
	
	private byte[] inbuf = new byte[1024];
	private 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)); }
}




C Tree Splitting  HackerRank Solution


Copy Code Copied Use a different Browser

#include <stdlib.h>
#include <stdio.h>

struct Set {
    int count;
};

typedef struct Set Set;

struct node{
    int number;
    struct node * parent;
    struct node * next;
    struct node * prev;
    struct node * first_child;
    Set * set;
};

typedef struct node node;

void print_children(node * n){
    node * child = n->first_child;
    while(child){
        printf("%d\n", child->number);
        child = child->next;
    }
}

void add_child(node * n, node * c){
    node * cur = n->first_child;
    n->first_child = c;
    if (cur){
        cur->prev = c;
        c->next = cur;
    }
}

void fill_children(node * root, node ** nodes, node ** result_nodes){
    node * repr = nodes[root->number];
    if(repr == 0){
        return;
    }
    node * child = repr->first_child;
    while(child){
        if (result_nodes[child->number] != 0){
            child = child->next;
            continue;
        }
        node * c = calloc(1, sizeof(node));
        c->number = child->number;
        c->parent = root;
        result_nodes[c->number] = c;
        add_child(root, c);
        fill_children(c, nodes, result_nodes);
        child = child->next;
    }
}

void compute_below(node * root, Set * set) {
    if (set == 0) {
        set = calloc(1, sizeof(set));
    }
    root->set = set;
    set->count++;
    node * child = root -> first_child;
    while(child){
        compute_below(child, set);
        child = child->next;
    }
}

void remove_node(node * item) {
//    subtract_below(item, item->below+1);
    int everyChild = item->parent != 0;
    node * child = item->first_child;
    int childCount = 0;
    int toRemove = 1;
    while (child) {
        childCount++;
        if (everyChild || childCount > 1) {
            compute_below(child, 0);
            toRemove += child->set->count;
        }
        child->parent = 0;
        child = child->next;
    }
    item->set->count -= toRemove;
    node * parent = item->parent;
    if(parent){
        if(parent->first_child == item){
            parent->first_child = item->next;
        }
        if(item->next){
            item->next->prev = item->prev;
        }
        if(item->prev){
            item->prev->next = item->next;
        }
    }
}

int main(int argc, char **argv){
    int n;
    scanf("%d\n", &n);
    int i = 0;
    node ** nodes = calloc(n+1, sizeof(node *));
    for(i = 0; i < n-1; i++){
        int a,b;
        scanf("%d %d\n", &a, &b);
        node * node_a = nodes[a];
        if(node_a == 0) {
            node_a = calloc(1, sizeof(node));
            node_a->number = a;
            nodes[a] = node_a;
        }
        node * x = calloc(1, sizeof(node));
        x->number = b;
        add_child(node_a,x);
        
        node * node_b = nodes[b];
        if(node_b == 0){
            node_b = calloc(1, sizeof(node));
            node_b->number = b;
            nodes[b] = node_b;
        }
        x = calloc(1, sizeof(node));
        x->number = a;
        add_child(node_b, x);
    }
    
    node * root = calloc(1, sizeof(node));
    root->number = 1;
    node ** result_nodes = calloc(n+1, sizeof(node *));
    result_nodes[1] = root;
    fill_children(root, nodes, result_nodes);
    compute_below(result_nodes[1], 0);
    int ans = 0;
    int num_queries;
    scanf("%d\n", &num_queries);
    for(i = 0; i < num_queries; i++){
        int m;
        scanf("%d\n", &m);
        int q = m^ans;
        node * n = result_nodes[q];
        ans = n->set->count;
        printf("%d\n", ans);
        remove_node(n);
    }
    return 0;
}

 

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:

15 Days to learn SQL Hard SQL(Advanced)-SolutionTree: Preorder Traversal – HackerRank Solution 15 Days to learn SQL Hard SQL(Advanced)-SolutionTree: Huffman Decoding – HackerRank Solution 15 Days to learn SQL Hard SQL(Advanced)-SolutionTree Flow – HackerRank Solution 15 Days to learn SQL Hard SQL(Advanced)-SolutionLetter Islands – HackerRank Solution 15 Days to learn SQL Hard SQL(Advanced)-SolutionMini-Max Sum – HackerRank Solution 15 Days to learn SQL Hard SQL(Advanced)-SolutionComputer Game – HackerRank Solution
Tags: Cc++14full solutionGoHackerRank Solutionjavajava 15java 7java 8java8javascriptpypy 3Python 2python 3Tree Splitting
ShareTweetPin
bhautik bhalala

bhautik bhalala

Related Posts

Leetcode All Problems Solutions
Code Solutions

Exclusive Time of Functions – LeetCode Solution

by admin
October 5, 2022
0
41

Exclusive Time of Functions - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions...

Read more
Leetcode All Problems Solutions

Smallest Range Covering Elements from K Lists – LeetCode Solution

October 5, 2022
32
Leetcode All Problems Solutions

Course Schedule III – LeetCode Solution

October 5, 2022
27
Leetcode All Problems Solutions

Maximum Product of Three Numbers – LeetCode Solution

September 11, 2022
53
Leetcode All Problems Solutions

Task Scheduler – LeetCode Solution

September 11, 2022
119
Leetcode All Problems Solutions

Valid Triangle Number – LeetCode Solution

September 11, 2022
28
Next Post
15 Days to learn SQL Hard SQL(Advanced)-Solution

Ticket - HackerRank Solution

15 Days to learn SQL Hard SQL(Advanced)-Solution

DFS Edges - HackerRank Solution

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

You may also like

15 Days to learn SQL Hard SQL(Advanced)-SolutionTree: Preorder Traversal – HackerRank Solution 15 Days to learn SQL Hard SQL(Advanced)-SolutionTree: Huffman Decoding – HackerRank Solution 15 Days to learn SQL Hard SQL(Advanced)-SolutionTree Flow – HackerRank Solution 15 Days to learn SQL Hard SQL(Advanced)-SolutionLetter Islands – HackerRank Solution 15 Days to learn SQL Hard SQL(Advanced)-SolutionMini-Max Sum – HackerRank Solution 15 Days to learn SQL Hard SQL(Advanced)-SolutionComputer Game – HackerRank Solution

Categories

  • Algorithms
  • Anime
  • Biography
  • Business
  • Code Solutions
  • Cosmos
  • Countdowns
  • Culture
  • Economy
  • Education
  • Entertainment
  • Finance
  • Games
  • Hackerrank
  • Health
  • How to
  • Investment
  • LeetCode
  • Lifestyle
  • LINUX SHELL
  • Manga
  • News
  • Opinion
  • Politics
  • Sports
  • SQL
  • Tech
  • Travel
  • Uncategorized
  • Updates
  • World
  • DMCA
  • Home
  • My account
  • Privacy Policy
  • Top Posts

Recent Blogs

Leetcode All Problems Solutions

Exclusive Time of Functions – LeetCode Solution

October 5, 2022
Leetcode All Problems Solutions

Smallest Range Covering Elements from K Lists – LeetCode Solution

October 5, 2022
Leetcode All Problems Solutions
Code Solutions

Majority Element II – LeetCode Solution

September 6, 2022
23
15 Days to learn SQL Hard SQL(Advanced)-Solution
Code Solutions

Draw The Triangle 2 EasySQL (Advanced) – SQL – HackerRank Solution

May 30, 2022
2
Leetcode All Problems Solutions
Code Solutions

Perfect Rectangle – LeetCode Solution

September 8, 2022
26
Biography

Shilpi Raj Net Worth, Age, Height , Achivements and more

September 8, 2022
1

© 2022 ZeroPlusFour - Latest News & Blog.

No Result
View All Result
  • Home
  • Category
    • Business
    • Culture
    • Economy
    • Lifestyle
    • Health
    • Travel
    • Opinion
    • Politics
    • Tech
  • Landing Page
  • Support Forum
  • Contact Us

© 2022 ZeroPlusFour - Latest News & Blog.

Welcome Back!

Login to your account below

Forgotten Password?

Retrieve your password

Please enter your username or email address to reset your password.

Log In
We use cookies on our website to give you the most relevant experience by remembering your preferences and repeat visits. By clicking “Accept All”, you consent to the use of ALL the cookies. However, you may visit "Cookie Settings" to provide a controlled consent.
Cookie SettingsAccept All
Manage consent

Privacy Overview

This website uses cookies to improve your experience while you navigate through the website. Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. We also use third-party cookies that help us analyze and understand how you use this website. These cookies will be stored in your browser only with your consent. You also have the option to opt-out of these cookies. But opting out of some of these cookies may affect your browsing experience.
Necessary
Always Enabled
Necessary cookies are absolutely essential for the website to function properly. These cookies ensure basic functionalities and security features of the website, anonymously.
CookieDurationDescription
cookielawinfo-checkbox-analytics11 monthsThis cookie is set by GDPR Cookie Consent plugin. The cookie is used to store the user consent for the cookies in the category "Analytics".
cookielawinfo-checkbox-functional11 monthsThe cookie is set by GDPR cookie consent to record the user consent for the cookies in the category "Functional".
cookielawinfo-checkbox-necessary11 monthsThis cookie is set by GDPR Cookie Consent plugin. The cookies is used to store the user consent for the cookies in the category "Necessary".
cookielawinfo-checkbox-others11 monthsThis cookie is set by GDPR Cookie Consent plugin. The cookie is used to store the user consent for the cookies in the category "Other.
cookielawinfo-checkbox-performance11 monthsThis cookie is set by GDPR Cookie Consent plugin. The cookie is used to store the user consent for the cookies in the category "Performance".
viewed_cookie_policy11 monthsThe cookie is set by the GDPR Cookie Consent plugin and is used to store whether or not user has consented to the use of cookies. It does not store any personal data.
Functional
Functional cookies help to perform certain functionalities like sharing the content of the website on social media platforms, collect feedbacks, and other third-party features.
Performance
Performance cookies are used to understand and analyze the key performance indexes of the website which helps in delivering a better user experience for the visitors.
Analytics
Analytical cookies are used to understand how visitors interact with the website. These cookies help provide information on metrics the number of visitors, bounce rate, traffic source, etc.
Advertisement
Advertisement cookies are used to provide visitors with relevant ads and marketing campaigns. These cookies track visitors across websites and collect information to provide customized ads.
Others
Other uncategorized cookies are those that are being analyzed and have not been classified into a category as yet.
SAVE & ACCEPT
Are you sure want to unlock this post?
Unlock left : 0
Are you sure want to cancel subscription?