**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)))))
```

## View More Similar Problems

## 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 <

View Solution →## 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

View Solution →## 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

View Solution →## 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

View Solution →## Kitty's Calculations on a Tree

Kitty has a tree, T , consisting of n nodes where each node is uniquely labeled from 1 to n . Her friend Alex gave her q sets, where each set contains k distinct nodes. Kitty needs to calculate the following expression on each set: where: { u ,v } denotes an unordered pair of nodes belonging to the set. dist(u , v) denotes the number of edges on the unique (shortest) path between nodes a

View Solution →## Is This a Binary Search Tree?

For the purposes of this challenge, we define a binary tree to be a binary search tree with the following ordering requirements: The data value of every node in a node's left subtree is less than the data value of that node. The data value of every node in a node's right subtree is greater than the data value of that node. Given the root node of a binary tree, can you determine if it's also a

View Solution →