# Maximum number of candies that can be bought

Given an array arr[] of size n where arr[i] is the amount of candies of type i. You have unlimited amount of money. The task is to buy as many candies as possible satisfying the following conditions:
If you buy x(i) candies of type i (clearly, 0 ≤ x(i) ≤ arr[i]), then for all j (1 ≤ j ≤ i) at least one of the following must hold:

1. x(j) < x(i) (you bought less candies of type j than of type i)
2. x(j) = 0 (you bought 0 candies of type j)

Examples:

Input:arr[] = {1, 2, 1, 3, 6}
Output: 10
x[] = {0, 0, 1, 3, 6} where x[i] is the number of candies bought of type i

Input: arr[] = {3, 2, 5, 4, 10}
Output: 20

Input: arr[] = {1, 1, 1, 1}
Output: 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: We can use greedy approach and start from the end of the array. If we have taken x candies of the type i + 1 then we can only take min(arr[i], x – 1) candies of type i. If this value is negative, we cannot buy candies of the current type.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the maximum candies ` `// that can be bought ` `int` `maxCandies(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Buy all the candies of the last type ` `    ``int` `prevBought = arr[n - 1]; ` `    ``int` `candies = prevBought; ` ` `  `    ``// 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 = min(prevBought - 1, arr[i]); ` ` `  `        ``if` `(x >= 0) { ` ` `  `            ``// Add candies of current type ` `            ``// that can be bought ` `            ``candies += x; ` ` `  `            ``// Update the previous bought amount ` `            ``prevBought = x; ` `        ``} ` `    ``} ` ` `  `    ``return` `candies; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 1, 3, 6 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << maxCandies(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` `     `  `// Function to return the maximum candies ` `// that can be bought ` `static` `int` `maxCandies(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Buy all the candies of the last type ` `    ``int` `prevBought = arr[n - ``1``]; ` `    ``int` `candies = prevBought; ` ` `  `    ``// 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 = Math.min(prevBought - ``1``, arr[i]); ` ` `  `        ``if` `(x >= ``0``)  ` `        ``{ ` ` `  `            ``// Add candies of current type ` `            ``// that can be bought ` `            ``candies += x; ` ` `  `            ``// Update the previous bought amount ` `            ``prevBought = x; ` `        ``} ` `    ``} ` ` `  `    ``return` `candies; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``1``, ``2``, ``1``, ``3``, ``6` `}; ` `    ``int` `n = arr.length; ` `    ``System.out.println(maxCandies(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech. `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the maximum candies  ` `# that can be bought  ` `def` `maxCandies(arr, n) : ` `     `  `    ``# Buy all the candies of the last type  ` `    ``prevBought ``=` `arr[n ``-` `1``]; ` `    ``candies ``=` `prevBought;  ` `     `  `    ``# Starting from second last  ` `    ``for` `i ``in` `range``(n ``-` `2``, ``-``1``, ``-``1``) : ` `         `  `        ``# Amount of candies of the current ` `        ``# type that can be bought  ` `        ``x ``=` `min``(prevBought ``-` `1``, arr[i]);  ` `        ``if` `(x >``=` `0``) : ` `             `  `            ``# Add candies of current type  ` `            ``# that can be bought ` `            ``candies ``+``=` `x;  ` `             `  `            ``# Update the previous bought amount  ` `            ``prevBought ``=` `x;  ` `             `  `    ``return` `candies;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` `     `  `    ``arr ``=` `[ ``1``, ``2``, ``1``, ``3``, ``6` `]; ` `    ``n ``=` `len``(arr) ` `    ``print``(maxCandies(arr, n));  ` ` `  `# This code is contributed by Ryuga `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the maximum candies ` `// that can be bought ` `static` `int` `maxCandies(``int``[] arr, ``int` `n) ` `{ ` ` `  `    ``// Buy all the candies of the last type ` `    ``int` `prevBought = arr[n - 1]; ` `    ``int` `candies = prevBought; ` ` `  `    ``// 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 = Math.Min(prevBought - 1, arr[i]); ` ` `  `        ``if` `(x >= 0)  ` `        ``{ ` ` `  `            ``// Add candies of current type ` `            ``// that can be bought ` `            ``candies += x; ` ` `  `            ``// Update the previous bought amount ` `            ``prevBought = x; ` `        ``} ` `    ``} ` ` `  `    ``return` `candies; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int``[] arr= { 1, 2, 1, 3, 6 }; ` `    ``int` `n = arr.Length; ` `    ``Console.WriteLine(maxCandies(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech. `

## PHP

 `= 0; ``\$i``--)  ` `    ``{ ` ` `  `        ``// Amount of candies of the current ` `        ``// type that can be bought ` `        ``\$x` `= min(``\$prevBought` `- 1, ``\$arr``[``\$i``]); ` ` `  `        ``if` `(``\$x` `>= 0)  ` `        ``{ ` ` `  `            ``// Add candies of current type ` `            ``// that can be bought ` `            ``\$candies` `+= ``\$x``; ` ` `  `            ``// Update the previous bought amount ` `            ``\$prevBought` `= ``\$x``; ` `        ``} ` `    ``} ` ` `  `    ``return` `\$candies``; ` `} ` ` `  `// Driver code ` `\$arr` `= ``array``(1, 2, 1, 3, 6 ); ` `\$n` `= sizeof(``\$arr``); ` `echo``(maxCandies(``\$arr``, ``\$n``)); ` ` `  `// This code is contributed by Code_Mech. ` `?> `

Output:

```10
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : AnkitRai01, Code_Mech