Find the maximum possible value of last element of the Array

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Given a non-negative array arr of size N and an integer M representing the number of moves such that in one move, the value of any one element in the array decreases by one, and the value of its adjacent element on the right increases by one. The task is to find the maximum possible value of the last element of the array in given M number of moves.
Examples:

Input: arr[] = {2, 3, 0, 1}, M = 5
Output: 3
Move 1: Working on index 1, the element 3 at 1st index reduces to 2 and the element 0 at 2nd index increases to 1. Hence the resultant array after one move = {2, 2, 1, 1}
Move 2: Working on index 2, the element 1 at 2nd index reduces to 0 and the element 1 at 3rd index increases to 2. Hence the resultant array after two moves = {2, 2, 0, 2}
Move 3: Working on index 1, the element 2 at 1st index reduces to 1 and the element 0 at 2nd index increases to 1. Hence the resultant array after three moves {2, 1, 1, 2}
Move 4: Working on index 2, the element 1 at 2nd index reduces to 0 and the element 2 at 3rd index increases to 3. Hence the resultant array after four moves {2, 1, 0, 3}
Move 5: Working on index 1, the element 1 at 1st index reduces to 0 and the element 0 at 2nd index increases to 1. Hence the resultant after five moves {2, 0, 1, 3}
So the maximum value of last element after 5 moves is 3

Input: arr[] = {1, 100}, M = 2
Output: 101

Approach:
The number of moves required to move one value from one element to the last element is calculated by the distance between them. For each element in the array, if the distance between this element and the final element is less than equal to M, then this element can be moved to the last. So in order to move it, increase the last element with the distance and reduce the left number of moves with the distance.
Below is the implementation of the above approach:

CPP

// C++ program to find the maximum possible
// value of last element of the array
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum possible
// value of last element of the array
int maxValue(int arr[], int n, int moves)
{
 
    // Traverse for all element
    for (int i = n - 2; i >= 0; i--) {
        if (arr[i] > 0) {
            // Find the distance
            int distance = n - 1 - i;
 
            // If moves less than distance then
            // we can not move this number to end
            if (moves < distance)
                break;
 
            // How many number we can move to end
            int can_take = moves / distance;
 
            // Take the minimum of both of them
            int take = min(arr[i], can_take);
 
            // Increment in the end
            arr[n - 1] += take;
 
            // Remove taken moves
            moves -= take * distance;
        }
    }
 
    // Return the last element
    return arr[n - 1];
}
 
// Driver code
int main()
{
    int arr[] = { 2, 3, 0, 1 };
    int M = 5;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << maxValue(arr, N, M);
 
    return 0;
}

Java

// Java program to find the maximum possible
// value of last element of the array
import java.util.*;
 
class GFG{
  
// Function to find the maximum possible
// value of last element of the array
static int maxValue(int arr[], int n, int moves)
{
  
    // Traverse for all element
    for (int i = n - 2; i >= 0; i--) {
        if (arr[i] > 0) {
            // Find the distance
            int distance = n - 1 - i;
  
            // If moves less than distance then
            // we can not move this number to end
            if (moves < distance)
                break;
  
            // How many number we can move to end
            int can_take = moves / distance;
  
            // Take the minimum of both of them
            int take = Math.min(arr[i], can_take);
  
            // Increment in the end
            arr[n - 1] += take;
  
            // Remove taken moves
            moves -= take * distance;
        }
    }
  
    // Return the last element
    return arr[n - 1];
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 3, 0, 1 };
    int M = 5;
    int N = arr.length;
  
    // Function call
    System.out.print(maxValue(arr, N, M)); 
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to find the maximum possible
# value of last element of the array
 
# Function to find the maximum possible
# value of last element of the array
def maxValue(arr, n, moves):
 
    # Traverse for all element
    for i in range(n - 2, -1, -1):
        if (arr[i] > 0):
             
            # Find the distance
            distance = n - 1 - i
 
            # If moves less than distance then
            # we can not move this number to end
            if (moves < distance):
                break
 
            # How many number we can move to end
            can_take = moves // distance
 
            # Take the minimum of both of them
            take = min(arr[i], can_take)
 
            # Increment in the end
            arr[n - 1] += take
 
            # Remove taken moves
            moves -= take * distance
 
    # Return the last element
    return arr[n - 1]
 
# Driver code
if __name__ == '__main__':
    arr= [2, 3, 0, 1]
    M = 5
    N = len(arr)
 
    # Function call
    print(maxValue(arr, N, M))
     
# This code is contributed by mohit kumar 29

C#

// C# program to find the maximum possible
// value of last element of the array
using System;
 
class GFG{
   
// Function to find the maximum possible
// value of last element of the array
static int maxValue(int []arr, int n, int moves)
{
   
    // Traverse for all element
    for (int i = n - 2; i >= 0; i--) {
        if (arr[i] > 0) {
            // Find the distance
            int distance = n - 1 - i;
   
            // If moves less than distance then
            // we can not move this number to end
            if (moves < distance)
                break;
   
            // How many number we can move to end
            int can_take = moves / distance;
   
            // Take the minimum of both of them
            int take = Math.Min(arr[i], can_take);
   
            // Increment in the end
            arr[n - 1] += take;
   
            // Remove taken moves
            moves -= take * distance;
        }
    }
   
    // Return the last element
    return arr[n - 1];
}
   
// Driver code
public static void Main(String[] args)
{
    int []arr = { 2, 3, 0, 1 };
    int M = 5;
    int N = arr.Length;
   
    // Function call
    Console.Write(maxValue(arr, N, M)); 
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

<script>
 
// Javascript program to find the maximum possible
// value of last element of the array
 
// Function to find the maximum possible
// value of last element of the array
function maxValue(arr, n, moves)
{
 
    // Traverse for all element
    for (var i = n - 2; i >= 0; i--)
    {
        if (arr[i] > 0)
        {
         
            // Find the distance
            var distance = n - 1 - i;
 
            // If moves less than distance then
            // we can not move this number to end
            if (moves < distance)
                break;
 
            // How many number we can move to end
            var can_take = parseInt(moves / distance);
 
            // Take the minimum of both of them
            var take = Math.min(arr[i], can_take);
 
            // Increment in the end
            arr[n - 1] += take;
 
            // Remove taken moves
            moves -= take * distance;
        }
    }
 
    // Return the last element
    return arr[n - 1];
}
 
// Driver code
var arr = [2, 3, 0, 1];
var M = 5;
var N = arr.length;
 
// Function call
document.write( maxValue(arr, N, M));
 
// This code is contributed by rutvik_56.
</script>

Output:

Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(1), constant extra space is required.

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