# Mr. X and His Shots

### Problem Statement :

```A cricket match is going to be held. The field is represented by a 1D plane. A cricketer, Mr. X has N favorite shots. Each shot has a particular range. The range of the ith shot is from Ai to Bi. That means his favorite shot can be anywhere in this range. Each player on the opposite team can field only in a particular range. Player i can field from Ci to Di. You are given the N favorite shots of Mr. X and the range of M players.

Si represents the strength of each player i.e. the number of shots player  can stop.

Game Rules: A player can stop the ith shot if the range overlaps with the player's fielding range.

Input Format

The first line consists of two space separated integers, N and M.
Each of the next N lines contains two space separated integers. The ith line contains  Ai and Bi.
Each of the next M  lines contains two integers. The ith line contains integers Ci and Di.

Output Format

You need to print the sum of the strengths of all the players.

Constraints:

1  <=  N, M  <= 10^5
1  <=  Ai, Bi, Ci , Di  <= 10^8```

### Solution :

```                            ```Solution in C :

In C++ :

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std ;

struct segment
{
bool isleft ;
int point ;
bool isfielder ;
} ;
vector < segment > v ( 400010 ) ;
vector < segment > :: iterator it ;

bool cmpr ( segment & a , segment & b )
{
if ( a.point != b.point )
{
return a.point < b.point ;
}
if ( a.isleft != b.isleft )
{
return a.isleft ;
}
return ! ( a.isfielder ) ;
}

int main ()
{
int n , m ;
cin >> n >> m ;
int counter = 0 ;
for ( int i = 0 ; i < n ; i ++ )
{
int a , b;
cin >> a >> b ;
segment temp ;
temp.isleft = true ;
temp.point = a ;
temp.isfielder = false ;
v [ counter ++ ] = temp ;
temp.isleft = false ;
temp.point = b ;
v [ counter ++ ] = temp ;
}
for ( int i = 0 ; i < m ; i ++ )
{
int a , b;
cin >> a >> b ;
segment temp ;
temp.isleft = true ;
temp.point = a ;
temp.isfielder = true ;
v [ counter ++ ] = temp ;
temp.isleft = false ;
temp.point = b ;
v [ counter ++ ] = temp ;
}
it = v.begin () ;
advance ( it , counter ) ;
sort ( v.begin () , it , cmpr ) ;       // segment intersection problem
int bat = 0 , field = 0 ;
long long int ans = 0 ;
for ( int i = 0 ; i < counter ; i ++ )
{
if ( v [ i ].isleft == true )
{
if ( v [ i ].isfielder )
{
ans += bat ;                // intersects with all the batsman
field ++ ;
}
else
{
ans += field ;
bat ++ ;
}
}
else
{
if ( v [ i ].isfielder )
{
field -- ;
}
else
{
bat -- ;
}
}
}
cout << ans ;
return 0 ;
}

In  Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
public static int binarySearch(long[] arr, long num) {
int head = 0, tail = arr.length - 1;
int mid = (head + tail) / 2;
if (arr[mid] <= num) {
} else {
tail = mid - 1;
}
}
}
public static long strenth(long[] heads, long[] tails, long C, long D) {
result -= (tails.length - Solution.binarySearch(tails, -C));
return result;
}

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);
int N = sc.nextInt(), M = sc.nextInt();
long[] A = new long[N], B = new long[N];
for (int i = 0; i < N; i++) {
A[i] = sc.nextLong(); B[i] = -sc.nextLong();
}
Arrays.sort(A); Arrays.sort(B);
long sum = 0L;
for (int i = 0; i < M; i++) {
sum += Solution.strenth(A, B, sc.nextLong(), sc.nextLong());
}
System.out.println(sum);
}
}

In C :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define _DEBUG_     0
typedef struct sRange {
int s;
int e;
} Range;

int cmpfunc(const void *a, const void *b) {
Range *this = (Range *)a;
Range *that = (Range *)b;
if (this->s > that->s) return 1;
if (this->s < that->s) return -1;
return this->e - that->e;
}

__inline__ int overLaps(Range *a, Range *b) {
if ((a->s <= b->e) && (a->e >= b->s)) return 1;
return 0;
}
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int N = 0, M = 0, A = 0, B = 0, C = 0, D = 0, i = 0, j = 0;
int S = 0;
scanf("%d %d", &N, &M);
Range *shots = (Range *)malloc(N * sizeof(Range));
Range *field = (Range *)malloc(M * sizeof(Range));
for (i = 0; i < N; i++) {
scanf("%d %d", &A, &B);
shots[i].s = A;
shots[i].e = B;
}
qsort(shots, N, sizeof(Range), cmpfunc);
#if 0 //_DEBUG_
for (i = 0; i < N; i++) {
printf("%d %d\n", shots[i].s, shots[i].e);
}
#endif
// get fielders
for (i = 0; i < M; i++) {
scanf("%d %d", &C, &D);
field[i].s = C;
field[i].e = D;
}
qsort(field, M, sizeof(Range), cmpfunc);
// Keep moving fielder and shot together
i = 0;
j = 0;
while (i < M && j < N) {
if (field[i].e < shots[j].s) i++; // increment fielder
else if (field[i].s > shots[j].e) j++; //increment shot
else {
// they are overlapping, check all the shots
int tempJ = j;
while (tempJ < N && shots[tempJ].s <= field[i].e) {
if (overLaps(&shots[tempJ], &field[i])) {
S++;
}
tempJ++;
}
i++;
}
#if _DEBUG_
printf("s %d, f %d, S %d\n", j, i, S);
#endif
}
printf("%d\n", S);
return 0;
}

In Python3 :

import sys
import os

# STEP 1: Take in parameters:
firstLine = input().split()
N = int(firstLine[0])
M = int(firstLine[1])

maxVal = 1000000
tally1 = [0] * maxVal
tally2 = [0] * maxVal

# STEP 2: Take in the ranges of the shots:
for i in range(0,N):
line = input().split()
tally1[int(line[0])] += 1
tally2[int(line[1]) + 1] -= 1

# STEP 3: Tally the strengths of the players:
for i in range(1,maxVal):
tally1[i] += tally1[i-1]
tally2[i] += tally2[i-1]

sumStrengths = 0

for j in range(0,M):
player = input().split()
playerStart = int(player[0])
playerEnd = int(player[1])
sumStrengths += tally1[playerEnd] + tally2[playerStart]

print(sumStrengths)```
```

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