# Gridland Metro

### Problem Statement :

```The city of Gridland is represented as an n x m matrix where the rows are numbered from 1 to n and the columns are numbered from 1 to m.

Gridland has a network of train tracks that always run in straight horizontal lines along a row. In other words, the start and end points of a train track are  and , where  represents the row number,  represents the starting column, and  represents the ending column of the train track.

The mayor of Gridland is surveying the city to determine the number of locations where lampposts can be placed. A lamppost can be placed in any cell that is not occupied by a train track.

Given a map of Gridland and its  train tracks, find and print the number of cells where the mayor can place lampposts.

Note: A train track may overlap other train tracks within the same row.

In this case, there are five open cells (red) where lampposts can be placed.

Function Description

Complete the gridlandMetro function in the editor below.

gridlandMetro has the following parameter(s):

int n:: the number of rows in Gridland
int m:: the number of columns in Gridland
int k:: the number of tracks
track[k]: each element contains  integers that represent , all 1-indexed
Returns

int: the number of cells where lampposts can be installed
Input Format

The first line contains three space-separated integers  and , the number of rows, columns and tracks to be mapped.

Each of the next  lines contains three space-separated integers,  and , the row number and the track column start and end.```

### Solution :

```                            ```Solution in C :

In    C++  :

#include <bits/stdc++.h>
using namespace std;

int main() {
long long h, w, K;
cin >> h >> w >> K;

long long ans = h * w;

map<int, vector<pair<int, int>>> mp;

for (int i = 0; i < K; i++) {
int y, l, r;
scanf("%d %d %d", &y, &l, &r);
mp[y].emplace_back(l, r);
}

for (auto &kv : mp) {
vector<pair<int, int>> evs;

for (auto p : kv.second) {
evs.emplace_back(p.first, 0);
evs.emplace_back(p.second, 1);
}

sort(evs.begin(), evs.end());

int cnt = 0;
int l = 0;

for (auto e : evs) {
if (e.second == 0) {
if (cnt == 0) l = e.first;
cnt++;
} else {
cnt--;
if (cnt == 0) ans -= e.first - l + 1;
}
}
}

cout << ans << endl;
}

In   Java :

import java.util.*;
import java.awt.*;

public class Solution {
ArrayList<Point> locs;

public Solution(){
locs = new ArrayList<>();
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
long rows = in.nextInt(), cols = in.nextInt();
int tracks = in.nextInt();
ArrayList<Integer> check = new ArrayList<>();
HashMap<Integer, Solution> found = new HashMap<>();

for(int i = 0; i<tracks; i++){
int curRow = in.nextInt(), start = in.nextInt(), end = in.nextInt();
if(!found.containsKey(curRow)){
Solution sol = new Solution();
found.put(curRow,sol);
}else{
Solution sol = found.get(curRow);
}
}

//Computing the total # of tracks - those of train tracks
long total = rows * cols;
for(int r : check){
//  System.out.println(r);
Solution sol = found.get(r);
ArrayList<Point> myPoints = sol.locs;
Collections.sort(myPoints, new myComparator());
Point first = myPoints.get(0);
total -= (first.y - first.x+1);
int lastEnd = first.y+1;
for(int i = 1; i<myPoints.size(); i++){
Point curPoint = myPoints.get(i);
if(curPoint.y< lastEnd) continue;
int begin = Math.max(curPoint.x, lastEnd);
total -=(curPoint.y - begin + 1);
lastEnd = curPoint.y+1;
}
}

System.out.println(total);
}
}
class myComparator implements Comparator<Point>{
public int compare(Point p1, Point p2){
return p1.x - p2.x;
}
}

In   C  :

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

struct points
{
long int r;
long int c1;
long int c2;
};

typedef struct points points;

int comp(const void* a,const void* b)
{
points* a1=(points*)a;
points* b1=(points*)b;
if((a1->r)!=(b1->r))
return (a1->r)-(b1->r);
else
{
if((a1->c1)!=(b1->c1))
return (a1->c1)-(b1->c1);
else
return (a1->c2)-(b1->c2);
}
}

long int min(long int a,long int b)
{
if(a<b)
return a;
return b;
}

long int max(long int a,long int b)
{
if(a>b)
return a;
return b;
}

int main() {

long int n,m,r,c1,c2,temp,r_s,r_e;
int k,i,j;
scanf("%ld %ld %d",&n,&m,&k);
points p[k];
for(i=0;i<k;i++)
scanf("%ld %ld %ld",&p[i].r,&p[i].c1,&p[i].c2);
qsort(p,k,sizeof(p),comp);
int visited[k];
memset(visited,0,sizeof(visited));
long long ans=n*m;
for(i=0;i<k;i++)
{
long int sub=0;
if(visited[i]==0){
for(j=i+1;j<k;j++)
{
if(p[i].r==p[j].r)
{
if(p[i].c2<p[j].c1)
{
sub+=p[i].c2-p[i].c1+1;
p[i].c1=p[j].c1;
p[i].c2=p[j].c2;
}
else
{
r_s=min(p[i].c1,p[j].c1);
r_e=max(p[i].c2,p[j].c2);
p[i].c1=r_s;
p[i].c2=r_e;
}
visited[j]=1;
}
else
break;
}
ans-=sub;
ans-=(p[i].c2-p[i].c1+1);
}
}
printf("%lld",ans);
return 0;
}

In   Python3  :

from collections import defaultdict

n, m, k = map(int, input().split())

tracks = defaultdict(list)
for _ in range(k):
row, c1, c2 = map(int, input().split())
tracks[row].append((c1, -1))
tracks[row].append((c2 + 1, 1))

ans = 0
for row in tracks:
tracks[row].sort()

prev = tracks[row]
depth = 1
for event in tracks[row][1:]:
if depth > 0:
ans += (event - prev)

depth -= event
prev = event

print(n * m - ans)```
```

## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

## Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

## Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

## Binary Search Tree : Lowest Common Ancestor

You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b

## Swap Nodes [Algo]

A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from