# Breaking the Records

### Problem Statement :

```Maria plays college basketball and wants to go pro. Each season she maintains a record of her play. She tabulates the number of times she breaks her season record for most points and least points in a game. Points scored in the first game establish her record for the season, and she begins counting from there.

For example, assume her scores for the season are represented in the array scores=[12, 24, 10, 24]. Scores are in the same order as the games played. She would tabulate her results as follows:

Count
Game  Score  Minimum  Maximum   Min Max
0      12     12       12       0   0
1      24     12       24       0   1
2      10     10       24       1   1
3      24     10       24       1   1

Given the scores for a season, find and print the number of times Maria breaks her records for most and least points scored during the season.

Function Description

Complete the breakingRecords function in the editor below. It must return an integer array containing the numbers of times she broke her records. Index 0 is for breaking most points records, and index 1 is for breaking least points records.

breakingRecords has the following parameter(s):

scores: an array of integers

Input Format

The first line contains an integer n, the number of games.
The second line contains n space-separated integers describing the respective values of score0, score1, ....., score n-1.

Constraints
1 <= n <= `1000
0 <= scores[i] <= `10^8

Output Format

Print two space-seperated integers describing the respective numbers of times the best (highest) score increased and the worst (lowest) score decreased.```

### Solution :

```                            ```Solution in C :

C  :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
int n;
scanf("%d",&n);
int a;
for(int score_i = 0; score_i < n; score_i++){
scanf("%d",&a[score_i]);
}
int i,min,max,c1,c2;
min=a;
max=a;
c1=0;
c2=0;
for(i=1;i<n;i++)
{
if(a[i]>max)
{
c1++;
max=a[i];
}
if(a[i]<min)
{
c2++;
min=a[i];
}
}
printf("%d %d\n",c1,c2);
return 0;
}

C++  :

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define lim 10000007
#define pb push_back
#define S second
#define pb push_back
#define mp make_pair
#define INF 1e18
#define fr(i,j,k) for(ll i=j;i<=k;i++)
#define frd(i,j,k) for(ll i=j;i>=k;i--)
#define F first
#define sd(n) scanf("%lld",&n)
#define pd(n) printf("%lld\n",n)
#define db double
#define mod 1000000007
#define pii pair<ll,ll>
int main()
{
ll n;
cin>>n;
ll mx;
cin>>mx;
ll mn=mx;
ll a1=0,a2=0;
n--;
ll x;
while(n--)
{
cin>>x;
if(x>mx)
a1++;
if(x<mn)
a2++;
mx=max(mx,x);
mn=min(mn,x);
}
cout<<a1<<" "<<a2<<endl;
}

Java  :

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

public class Solution {

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] score = new int[n];
for(int score_i=0; score_i < n; score_i++){
score[score_i] = in.nextInt();
}
int most = score;
int least = score;
int mr = 0;
int lr = 0;
for( int i = 1; i < n; i++ ){
if( score[i] > most ){
mr++;
most = score[i];
}
if( score[i] < least ){
lr++;
least = score[i];
}
}

System.out.print(mr + " " + lr);
}
}

python3  :

input()
a = list(map(int, input().split()))
m = M = a
x = y = 0
for i in a[1:]:
if i > M:
M = i
x += 1
elif i < m:
m = i
y += 1
print(x, y)```
```

## Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio