Root of the Problem

Problem Statement :

Chef has a binary tree. The binary tree consists of 1 or more nodes. Each node has a unique integer id. Each node has up to 2 children, which are identified by their ids, and each node is the child of at most 1 other node. A node X is considered to be an ancestor of node Y if node Y is a child of node X or if there is some node Z for which X is an ancestor of Z and Y is a child of Z. No node is an ancestor of itself. A special node called the root node is an ancestor of all other nodes.

Chef has forgotten which node of his tree is the root, and wants you to help him to figure it out. Unfortunately, Chef's knowledge of the tree is incomplete. He does not remember the ids of the children of each node, but only remembers the sum of the ids of the children of each node.


Input begins with an integer T, the number of test cases. Each test case begins with an integer N, the number of nodes in the tree. N lines follow with 2 integers each: the id of a node, and the sum of the ids of its children. The second number will be 0 if the node has no children.


For each test case, output on a line a space separated list of all possible values for the id of the root node in increasing order. It is guaranteed that at least one such id exists for each test case.


1 ≤ T ≤ 50
1 ≤ N ≤ 30
All node ids are between 1 and 1000, inclusive

Sample Input

4 0
1 5
2 0
3 0
4 0
5 5
6 5

Sample Output


Solution :


                            Solution in C :

#include <stdio.h>

int main(void) {
	int t, n, id_sum, sid_sum, x, y;
	scanf("%d", &t);
	while(t--) {
	    scanf("%d", &n);
	    id_sum = 0, sid_sum = 0;
	    while(n--) {
    	    scanf("%d %d", &x, &y);
    	    id_sum += x;
    	    sid_sum += y;
	    printf("%d\n", id_sum - sid_sum);
	return 0;

                        Solution in C++ :

#include <iostream>
using namespace std;

#define ll long long int

int main() {
	ll t;
	for(ll z=1;z<=t;z++){
		ll n;
		int ans=0;
		for(int j=1;j<=n;j++){
			int x,y;
	return 0;

                        Solution in Java :

/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
	public static void main (String[] args) throws java.lang.Exception
		// your code goes here
		Scanner sc = new Scanner(;
		int t = sc.nextInt();
		while(t-- > 0)
		    int n = sc.nextInt();
		    int node = 0;
		    int idsum= 0;
		    for(int i=0;i<n;i++)
		        node = node + sc.nextInt();
		        idsum = idsum + sc.nextInt();
		    int r = node - idsum;

                        Solution in Python : 
# cook your dish here
for _ in range(t):
    n = int(input())
    root = 0
    for j in range(n):
        ide,c = map(int,input().split(' '))
        root +=ide - c

View More Similar Problems

Tree Coordinates

We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For

View Solution →

Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

View Solution →

Self Balancing Tree

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 →