**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

## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

View Solution →## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func

View Solution →## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

View Solution →## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

View Solution →## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

View Solution →## 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 →