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:

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.

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; 
    int a[10000];
    for(int score_i = 0; score_i < n; score_i++){
    int i,min,max,c1,c2;
    printf("%d %d\n",c1,c2);
    // your code goes here
    return 0;

C++  :

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;
    ll mx;
    ll mn=mx;
    ll a1=0,a2=0;
    ll x;
    cout<<a1<<" "<<a2<<endl;

Java  :

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(;
        int n = in.nextInt();
        int[] score = new int[n];
        for(int score_i=0; score_i < n; score_i++){
            score[score_i] = in.nextInt();
        // your code goes here
        int most = score[0];
        int least = score[0];
        int mr = 0;
        int lr = 0;
        for( int i = 1; i < n; i++ ){
            if( score[i] > most ){
                most = score[i];
            if( score[i] < least ){
                least = score[i];
        System.out.print(mr + " " + lr);

python3  :

a = list(map(int, input().split()))
m = M = a[0]
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)

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