# Find the maximum possible value of last element of the Array

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 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 = 5Output: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 3Input:arr[] = {1, 100}, M = 2Output: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:**

3

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