Balanced Forest
Problem Statement :
Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to allow creation of a balanced forest. If it's not possible to create a balanced forest, return -1. For example, you are given node values c = [15,12,8,14,13] and edges = [ [1,2], [1, 3], [1, 4]. [4, 5] ]. It is the following tree: The blue node is root, the first number in a node is node number and the second is its value. Cuts can be made between nodes 1 and 3 and nodes 1 and 4 to have three trees with sums 27, 27 and 8. Adding a new node w of c[w] = 19 to the third tree completes the solution. Function Description Complete the balancedForest function in the editor below. It must return an integer representing the minimum value of that can be added to allow creation of a balanced forest, or -1 if it is not possible. balancedForest has the following parameter(s): c: an array of integers, the data values for each node edges: an array of 2 element arrays, the node pairs per edge
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <stdlib.h>
#define HASH_SIZE 123455
typedef struct _lnode{
int x;
int w;
struct _lnode *next;
} lnode;
typedef struct _node{
long long x;
int c;
long long ans;
struct _node *next;
} node;
void clean_table();
void insert_edge(int x,int y,int w);
void dfs0(int x,int y);
void dfs1(int x,int y);
void insert(long long x,long long ans);
void removee(long long x,long long ans);
void search(long long x);
void freehash();
int a[50000];
long long sub[50000],min,sum;
lnode *table[50000]={0};
node *hash[HASH_SIZE]={0};
int main(){
int T,n,x,y,i;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(i=sum=0;i<n;i++){
scanf("%d",a+i);
sum+=a[i];
}
for(i=0;i<n-1;i++){
scanf("%d%d",&x,&y);
insert_edge(x-1,y-1,1);
}
dfs0(0,-1);
min=-1;
dfs1(0,-1);
printf("%lld\n",min);
clean_table();
freehash();
}
return 0;
}
void clean_table(){
int i;
lnode *p,*pp;
for(i=0;i<50000;i++)
if(table[i]){
p=table[i];
while(p){
pp=p->next;
free(p);
p=pp;
}
table[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=table[y];
table[y]=t;
return;
}
void dfs0(int x,int y){
lnode *p;
sub[x]=a[x];
for(p=table[x];p;p=p->next)
if(p->x!=y){
dfs0(p->x,x);
sub[x]+=sub[p->x];
}
return;
}
void dfs1(int x,int y){
lnode *p;
long long down,up;
search(sub[x]);
down=sub[x];
up=sum-sub[x];
if(down==up && min==-1)
min=down;
if(down%2==0 && down/2*3>=sum)
insert(down/2,down/2*3-sum);
if(up<down && up*3>=sum){
insert(up,up*3-sum);
insert(down-up,up*3-sum);
}
for(p=table[x];p;p=p->next)
if(p->x!=y)
dfs1(p->x,x);
if(down%2==0 && down/2*3>=sum)
removee(down/2,down/2*3-sum);
if(up<down && up*3>=sum){
removee(up,up*3-sum);
removee(down-up,up*3-sum);
}
if(up%2==0 && up/2*3>=sum)
insert(up/2,up/2*3-sum);
if(down<up && down*3>=sum){
insert(down,down*3-sum);
insert(up-down,down*3-sum);
}
return;
}
void insert(long long x,long long ans){
int bucket=x%HASH_SIZE;
node *t=hash[bucket];
while(t){
if(t->x==x && t->ans==ans){
t->c++;
return;
}
t=t->next;
}
t=(node*)malloc(sizeof(node));
t->x=x;
t->ans=ans;
t->c=1;
t->next=hash[bucket];
hash[bucket]=t;
return;
}
void removee(long long x,long long ans){
int bucket=x%HASH_SIZE;
node *t=hash[bucket],*p=NULL;
while(t){
if(t->x==x && t->ans==ans){
t->c--;
if(!t->c){
if(!p){
hash[bucket]=t->next;
free(t);
}
else{
p->next=t->next;
free(t);
}
}
return;
}
p=t;
t=t->next;
}
return;
}
void search(long long x){
int bucket=x%HASH_SIZE;
node *t=hash[bucket];
while(t){
if(t->x==x)
if(min==-1 || t->ans<min)
min=t->ans;
t=t->next;
}
return;
}
void freehash(){
int i;
node *t,*p;
for(i=0;i<HASH_SIZE;i++){
t=hash[i];
while(t){
p=t->next;
free(t);
t=p;
}
hash[i]=NULL;
}
return;
}
Solution in C++ :
In C ++ :
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <ctime>
#include <deque>
#include <unordered_set>
using namespace std;
int q;
map <long long, int> Map1, Map2;
long long ctot;
int c[110000];
vector <int> ve[110000];
long long ans;
int n;
long long sum[1100000];
void dfs1(int x, int f) {
sum[x] = c[x];
for (int i = 0; i < (int) ve[x].size(); i++)
if (ve[x][i] != f) {
dfs1(ve[x][i], x);
sum[x] += sum[ve[x][i]];
}
Map1[sum[x]] += 1;
}
void test(long long x) {
long long y = ctot - 2 * x;
if (y > 0 && y <= x)
ans = min(ans, x - y);
}
void dfs2(int x, int f) {
if (Map2[2 * sum[x]])
test(sum[x]);
if (Map2[ctot - sum[x]])
test(sum[x]);
if ((ctot - sum[x]) % 2 == 0 && Map2[ctot - (ctot - sum[x]) / 2])
test((ctot - sum[x]) / 2);
Map2[sum[x]] += 1;
if (Map1[sum[x]] > Map2[sum[x]])
test(sum[x]);
if (ctot - 2 * sum[x] >= sum[x] && Map1[ctot - 2 * sum[x]] > Map2[ctot - 2 * sum[x]])
test(sum[x]);
if ((ctot - sum[x]) % 2 == 0 && (ctot - sum[x]) / 2 >= sum[x] && Map1[(ctot - sum[x]) / 2] > Map2[(ctot - sum[x]) / 2])
test((ctot - sum[x]) / 2);
if (sum[x] * 2 == ctot)
ans = min(ans, sum[x]);
for (int i = 0; i < (int) ve[x].size(); i++)
if (ve[x][i] != f) {
dfs2(ve[x][i], x);
}
Map2[sum[x]] -= 1;
}
int main() {
scanf("%d", &q);
while (q--) {
Map1.clear();
Map2.clear();
ans = 1e18;
scanf("%d", &n);
ctot = 0;
for (int i = 1; i <= n; i++) {
scanf("%d", &c[i]);
ctot += c[i];
}
for (int i = 1; i <= n; i++)
ve[i].clear();
for (int i = 1; i < n; i++) {
int x, y;
scanf("%d%d", &x, &y);
ve[x].push_back(y);
ve[y].push_back(x);
}
dfs1(1, 0);
dfs2(1, 0);
if (ans == 1e18)
printf("-1\n");
else
printf("%lld\n", ans);
}
}
Solution in Java :
In Java :
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.BitSet;
import java.util.Comparator;
import java.util.InputMismatchException;
public class E {
InputStream is;
PrintWriter out;
String INPUT = "";
void solve()
{
for(int T = ni();T > 0;T--){
int n = ni();
int[] a = na(n);
int[] from = new int[n - 1];
int[] to = new int[n - 1];
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], ord = pars[1], dep = pars[2];
int[][] rs = makeRights(g, par, 0);
int[] iord = rs[1], right = rs[2];
int[][] spar = logstepParents(par);
long[] des = new long[n];
for(int i = n-1;i >= 0;i--){
int cur = ord[i];
des[cur] += a[cur];
if(i > 0)des[par[cur]] += des[cur];
}
long W = des[0];
long ret = Long.MAX_VALUE;
// t=u
{
long[] dd = Arrays.copyOf(des, n);
Arrays.sort(dd);
for(int i = 0;i < n-1;i++){
if(dd[i] == dd[i+1] && W-2*dd[i] <= dd[i] && W-2*dd[i] >= 0){
ret = Math.min(ret, dd[i]-(W-2*dd[i]));
}
}
}
// s=t
{
long[][] poss = new long[n][];
for(int i = 0;i < n;i++){
poss[i] = new long[]{des[i], iord[i]};
}
Arrays.sort(poss, new Comparator<long[]>() {
public int compare(long[] a, long[] b) {
if(a[0] != b[0])return Long.compare(a[0], b[0]);
return Long.compare(a[1], b[1]);
}
});
long[] posv = new long[n];
for(int i = 0;i < n;i++)posv[i] = poss[i][0];
for(int i = 0;i < n;i++){
long t = des[i];
long u = W-2L*t;
if(u >= 0 && t >= u){
int lb = lowerBound(posv, u);
int ub = lowerBound(posv, u+1);
if(lb < ub){
if((int)poss[lb][1] < iord[i]){
ret = Math.min(ret, t-u);
}
if((int)poss[ub-1][1] > right[iord[i]]){
ret = Math.min(ret, t-u);
}
}
}
}
}
// ireko
// t=u
{
for(int i = 0;i < n;i++){
long u = des[i];
long s = W-u*2;
if(s >= 0 && s <= u){
int cur = i;
for(int h = spar.length-1;h >= 0;h--){
int anc = spar[h][cur];
if(anc == -1)continue;
if(des[anc] <= 2*u){
cur = anc;
}
}
if(des[cur] == 2*u){
ret = Math.min(ret, u-s);
}
}
}
}
// s=u
{
for(int i = 0;i < n;i++){
long u = des[i];
long t = W-u*2;
if(t >= 0 && t <= u){
int cur = i;
for(int h = spar.length-1;h >= 0;h--){
int anc = spar[h][cur];
if(anc == -1)continue;
if(des[anc] <= t+u){
cur = anc;
}
}
if(des[cur] == t+u){
ret = Math.min(ret, u-t);
}
}
}
}
// s=t
{
for(int i = 0;i < n;i++){
long u = des[i];
if((W-u)%2 != 0)continue;
long t = (W-u)/2;
if(t >= 0 && t >= u){
int cur = i;
for(int h = spar.length-1;h >= 0;h--){
int anc = spar[h][cur];
if(anc == -1)continue;
if(des[anc] <= t+u){
cur = anc;
}
}
if(des[cur] == t+u){
ret = Math.min(ret, t-u);
}
}
}
}
if(ret == Long.MAX_VALUE){
out.println(-1);
}else{
out.println(ret);
}
}
}
public static int lowerBound(long[] a, long v)
{
int low = -1, high = a.length;
while(high-low > 1){
int h = high+low>>>1;
if(a[h] >= v){
high = h;
}else{
low = h;
}
}
return high;
}
public static int[] sortByPreorder(int[][] g, int root){
int n = g.length;
int[] stack = new int[n];
int[] ord = new int[n];
BitSet ved = new BitSet();
stack[0] = root;
int p = 1;
int r = 0;
ved.set(root);
while(p > 0){
int cur = stack[p-1];
ord[r++] = cur;
p--;
for(int e : g[cur]){
if(!ved.get(e)){
stack[p++] = e;
ved.set(e);
}
}
}
return ord;
}
public static int[][] makeRights(int[][] g, int[] par, int root)
{
int n = g.length;
int[] ord = sortByPreorder(g, root);
int[] iord = new int[n];
for(int i = 0;i < n;i++)iord[ord[i]] = i;
int[] right = new int[n];
for(int i = n-1;i >= 0;i--){
int v = i;
for(int e : g[ord[i]]){
if(e != par[ord[i]]){
v = Math.max(v, right[iord[e]]);
}
}
right[i] = v;
}
return new int[][]{ord, iord, right};
}
public static int lca2(int a, int b, int[][] spar, int[] depth) {
if (depth[a] < depth[b]) {
b = ancestor(b, depth[b] - depth[a], spar);
} else if (depth[a] > depth[b]) {
a = ancestor(a, depth[a] - depth[b], spar);
}
if (a == b)
return a;
int sa = a, sb = b;
for (int low = 0, high = depth[a], t = Integer.highestOneBit(high), k = Integer
.numberOfTrailingZeros(t); t > 0; t >>>= 1, k--) {
if ((low ^ high) >= t) {
if (spar[k][sa] != spar[k][sb]) {
low |= t;
sa = spar[k][sa];
sb = spar[k][sb];
} else {
high = low | t - 1;
}
}
}
return spar[0][sa];
}
protected static int ancestor(int a, int m, int[][] spar) {
for (int i = 0; m > 0 && a != -1; m >>>= 1, i++) {
if ((m & 1) == 1)
a = spar[i][a];
}
return a;
}
public static int[][] logstepParents(int[] par) {
int n = par.length;
int m = Integer.numberOfTrailingZeros(Integer.highestOneBit(n - 1)) + 1;
int[][] pars = new int[m][n];
pars[0] = par;
for (int j = 1; j < m; j++) {
for (int i = 0; i < n; i++) {
pars[j][i] = pars[j - 1][i] == -1 ? -1 : pars[j - 1][pars[j - 1][i]];
}
}
return pars;
}
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;
}
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 E().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)); }
}
Solution in Python :
In python3 :
from random import randrange
import sys
sys.setrecursionlimit(10**5)
def r():
return list(map(int, input().split()))
def bin_search(arr, pred, lo=0):
hi = len(arr)
while lo < hi:
mid = (lo+hi)//2
if pred(arr[mid]):
hi = mid
else:
lo = mid + 1
return lo if lo < len(arr) else None
class UndirectedGraph(object):
def __init__(self, size):
self.size = size
self.M = 0
self.edges = [set([]) for _ in range(self.size)]
def add_edge(self, u, v):
"""Adds edge to graph if it doesn't already exist."""
if v not in self.edges[u]:
self.edges[u].add(v)
self.edges[v].add(u)
self.M += 1
def neighbors(self, u):
for v in self.edges[u]:
yield v
class WeightedTree(UndirectedGraph):
def __init__(self, size, weights):
super().__init__(size)
self.weights = weights
self.labels = [None]*self.size
self.cum_weights = [None]*self.size
self.inverted = [False]*self.size
self.nextnode = 0
self.num_children = [None]*self.size
self.root = 0
def initialize(self):
self.set_labels(self.root)
self.sorted_nodes = sorted(list(range(self.size)), key = lambda x : self.cum_weights[x])
def set_labels(self, root):
self.dfs(root, set([]))
self.total_weight = self.cum_weights[self.root]
for i, cw in enumerate(self.cum_weights):
if cw > self.total_weight - cw:
self.cum_weights[i] = self.total_weight - cw
self.inverted[i] = True
def dfs(self, u, visited):
visited.add(u)
self.labels[u] = self.nextnode
self.nextnode += 1
self.cum_weights[u] = self.weights[u]
self.num_children[u] = 0
for v in self.neighbors(u):
if v not in visited:
cw, nchildren = self.dfs(v, visited)
self.cum_weights[u] += cw
self.num_children[u] += nchildren + 1
return self.cum_weights[u], self.num_children[u]
def solve(self):
tw = self.total_weight
first_idx = bin_search(self.sorted_nodes, lambda u : self.cum_weights[u] >= (tw+2)//3)
if first_idx is None:
return None
while first_idx < self.size:
first = self.sorted_nodes[first_idx]
first_cw = self.cum_weights[first]
label1 = self.labels[first]
last_child_label = label1 + self.num_children[first]
for target_w in (tw - first_cw * 2, first_cw):
second_idx = bin_search(self.sorted_nodes, lambda u: self.cum_weights[u] >= target_w)
if second_idx is not None:
while (second_idx < self.size
and self.cum_weights[self.sorted_nodes[second_idx]] == target_w):
second = self.sorted_nodes[second_idx]
is_child = label1 < self.labels[second] <= last_child_label
is_child = is_child if not self.inverted[first] else not is_child
if not is_child and self.labels[second] != label1:
return first_cw * 3 - tw
second_idx += 1
first_idx += 1
if 2*first_cw == tw:
return first_cw
return None
def read_input():
n = int(input())
weights = r()
G = WeightedTree(n, weights)
for _ in range(n-1):
u, v = r()
u -= 1
v -= 1
G.add_edge(u, v)
G.initialize()
return G
def main():
q = int(input())
for _ in range(q):
G = read_input()
ans = G.solve()
print(ans if ans is not None else -1)
main()
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An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ
View Solution →Array and simple queries
Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty
View Solution →Median Updates
The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o
View Solution →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
View Solution →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
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