Maximum number of candies that can be bought

Given an array arr[] of size n where arr[i] is the number of candies of type i. You have an 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 

Approach: We can use a greedy approach and start from the end of the array. If we have taken x candies of 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 <bits/stdc++.h>
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[0]);
    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

<?php
// PHP implementation of the approach

// Function to return the maximum candies
// that can be bought
function maxCandies($arr, $n)
{

    // Buy all the candies of the last type
    $prevBought = $arr[$n - 1];
    $candies = $prevBought;

    // Starting from second last
    for ($i = $n - 2; $i >= 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.
?>

Javascript

<script>

// Javascript implementation of the approach

// Function to return the maximum candies
// that can be bought
function maxCandies(arr, n)
{

    // Buy all the candies of the last type
    let prevBought = arr[n - 1];
    let candies = prevBought;

    // Starting from second last
    for (let i = n - 2; i >= 0; i--)
    {

        // Amount of candies of the current
        // type that can be bought
        let 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

      let arr = [ 1, 2, 1, 3, 6 ];
    let n = arr.length;
    document.write(maxCandies(arr, n));

</script>

Output:

10

Time Complexity: O(n)

Auxiliary Space: O(1)