We are given an array of candies[] of length stored in ‘size’. Each element candies[i] has a number for candies of type i.The goal is to buy as many candies as possible for any amount of money. The conditions are as given −
If you purchase X[i] of type i (0<= X[i] <= candies[i] ), then for all j ( 1<=j<=i ) at least on of the following conditions must be true −
X(j) < X(i) ( candies purchased of type j less than of type i )
X(j)=0, no candies of type j purchased
Let’s understand with examples.
Input − Arr[] = { 1,3,5,2,6,7 }.
Output − Maximum Candies that can be bought − 16
Explanation − Candies bought of type i { 0,3,5,2,6,0 }
Input − Arr[] = { 5,7,7,3,4 }.
Output − Maximum Candies that can be bought − 10
Explanation − Candies bought of type i { 0,0,7,3,0 }
Approach used in the below program is as follows
The integer array candies[] is used to store the number of candies of type i.
Variable ‘size’ stores the length of array candies.
Function maxCandies(int arr[], int n) is used to return the total number of candies that can be bought.
First suppose we bought the last type of candies. bought=arr[n-1]
Starting from the second last element, for(i=n-2;i>=0;i--)
Variable x stores the amount of candies of current type that can be bought. x=arr[i] or bought-1 whichever is less.
If x is non-zeo then add this to total.
If the total is more than the previous bought value then bought=x.
Return the bought result.
Example
#include <stdio.h>
int maxCandies(int arr[], int n){
int bought = arr[n - 1];
int total = bought;
// Starting from second last
for (int i = n - 2; i >= 0; i--) {
// Amount of candies of the current
// type that can be bought
int x = arr[i]<bought-1?arr[i]:bought-1;
if (x >= 0) {
total += x;
bought = x;
}
}
return total;
}
int main(){
int candies[] = { 1,2,4,3,7 };
int size = 5;
printf("Total Candies that can be bought: %d", maxCandies(candies, size));
return 0;
}Output
If we run the above code it will generate the following output −
Total Candies that can be bought: 13