# Candies

### Problem Statement :

```Alice is a kindergarten teacher. She wants to give some candies to the children in her class.  All the children sit in a line and each of them has a rating score according to his or her performance in the class.  Alice wants to give at least 1 candy to each child. If two children sit next to each other, then the one with the higher rating must get more candies. Alice wants to minimize the total number of candies she must buy.

Example

arr =  [ 4, 6, 4, 5 , 6 , 2 ]

She gives the students candy in the following minimal amounts: [ 1, 2, 1, 2,  3, 1 ] . She must buy a minimum of 10 candies.

Function Description

Complete the candies function in the editor below.

candies has the following parameter(s):

int n: the number of children in the class
int arr[n]: the ratings of each student
Returns

int: the minimum number of candies Alice must buy

Input Format

The first line contains an integer, n , the size of arr.
Each of the next n lines contains an integer arr[ i ] indicating the rating of the student at position i.

Constraints

1  <=   n  <=  10^5
1  <=  arr[ i ]  <=  10^5

Sample Input 0

3
1
2
2

Sample Output 0

4```

### Solution :

```                            ```Solution in C :

In C  :

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

#define MAXN 100000

struct stu{
int id;
int rating;
};

int comp( const void*p1, const void*p2){
return ((struct stu*)p1)->rating - ((struct stu*)p2)->rating;
}

struct stu student[MAXN];
int candies[MAXN];
int ratings[MAXN];

int main(int argc, char *argv[])
{
int N;
int i,j;
int candy;
int id;
long long sumcandies = 0;

scanf("%d",&N);
for(i = 0; i < N; i++){
scanf("%d",&student[i].rating);
student[i].id = i;
ratings[i] = student[i].rating;
}

qsort(student,N,sizeof(struct stu),comp);

for(i = 0; i < N; i++){
id = student[i].id;
candy = 1;

if(id > 0){
if(candies[id-1] != 0 && ratings[id] > ratings[id-1]){
candy = candies[id-1] + 1;
}
}

if(id < N-1){
if(candies[id+1] != 0 && ratings[id] > ratings[id+1] &&
candy <= candies[id+1]){
candy = candies[id+1] + 1;
}
}

sumcandies += candy;
candies[id] = candy;
}

printf("%lld\n",sumcandies);
return 0;
}```
```

```                        ```Solution in C++ :

In  C  ++  :

#include <cstdio>
#include <string>
using namespace std;

typedef long long ll;

int a[100005],b[100005];

inline int maxi(int x,int y) {
return (x>y)?x:y;
}

int main() {
int i,j,n;
ll ans;

scanf("%d",&n);
for (i=1;i<=n;++i) {
scanf("%d",&a[i]);
}
a[0]=a[1];
a[n+1]=a[n];
for (i=1;i<=n;++i) {
if ((a[i]<=a[i+1]) && (a[i]<=a[i-1])) {
b[i]=1;
for (j=i-1;j && (a[j]>a[j+1]);--j) {
b[j]=b[j+1]+1;
}
for (;i<n && (a[i+1]>a[i]);++i) {
b[i+1]=b[i]+1;
}
}
}
ans=0;
for (i=1;i<=n;++i) {
if ((a[i]>a[i-1]) && (a[i]>a[i+1])) {
b[i]=maxi(b[i-1],b[i+1])+1;
}
ans+=b[i];
}
printf("%Ld\n",ans);
return 0;

}```
```

```                        ```Solution in Java :

In  Java  :

import java.util.Scanner;
public class Solution{

public static void main(String[] args){

Scanner in = new Scanner(System.in);

int N, K;
N = in.nextInt();
//K = in.nextInt();

int C[] = new int[N];
for(int i=0; i<N; i++){
C[i] = in.nextInt();
}

int res[] = new int[N];
for(int i=0 ; i < N ;i++) res[i] =1 ;

int d = N ;
while(d <= N ){

boolean changed = false;
//forward
for(int i=1 ; i < N ;i++){
if(C[i-1] < C[i] && res[i-1] >= res[i] ){
res[i] = res[i-1] +1 ;
changed = true;
}
}
//backward
for(int i=N-1 ; i > 0 ;i--){
if(C[i-1] > C[i] && res[i-1] <= res[i] ){
res[i-1] = res[i] +1 ;
changed = true;
}
}

if(!changed)break;
}

int sum = 0;
for(int j : res ) sum += j;

System.out.println(sum);
}
}```
```

```                        ```Solution in Python :

In  Python3 :

N = int(input())
kids = []
for n in range(N):
kids.append(int(input()))

def candy(ratings):
count = 1
candies = [1]
for index in range(1, len(ratings)):
if ratings[index] > ratings[index - 1]:
count += 1
else:
count = 1
candies.append(count)

return candies

ll = candy(kids)
kids.reverse()
lr = candy(kids)
print(sum(map(max, zip(ll, reversed(lr)))))```
```

## Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

## Left Rotation

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d

## Sparse Arrays

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## Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

## Print the Elements of a Linked List

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