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# Maximize element at index K in an array with at most sum M and difference between adjacent elements at most 1

• Difficulty Level : Hard
• Last Updated : 15 Mar, 2021

Given a positive integer N, the task is to construct an array of length N and find the maximize the value at index K such that the sum of all the array elements is at most M and the absolute difference between any two consecutive array elements is at most 1.

Examples:

Input: N = 3, M = 7, K = 1
Output: 3
Explanation:
According to the given constraints, the array with values {2, 3, 2}maximizes the value at index 1. Therefore, the required output is 3.

Input: N = 3, M = 8, K = 1
Output: 3

Approach: The idea is to achieve the maximum value at index K and to decrease the sum of other elements to meet the criteria of the sum of the array to be at most M. Follow the steps below:

• Let the value at index K be X. So the element at K – 1 is X – 1, at K – 2 is X – 2 and so on.
• At index 0 the value is X – K. Similarly, at index K + 1 the value is X – 1 and so on upto X – (N – K – 1) at index N – 1.
• So to achieve the maximum value at index K, the array structure would be X – K, X – (K – 1), …., X – 2, X – 1, X, X – 1, X – 2, ….., X – (N – K – 1).
• So after arranging the equation, it becomes K * X – (1 + 2 + …. + K) + X + (N – K – 1) * X – (1 + 2 + …. + (N – K – 1)) ≤ M.

Follow the steps to solve the above equation:

• Calculate (1 + 2 + …. + K) using K * (K + 1) / 2 and (1 + 2 + ….. + (N – K – 1)) using (N – K – 1) * (N – K) / 2 and store in S1 and S2 respectively.
• This reduces the equation to X * (K + 1 + N – K – 1) ≤ M + S1 + S2.
• Now, the maximum value of X can be obtained by calculating (M + S1 + S2) / N.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to calculate the maximum``// possible value at index K``void` `maxValueAtIndexK(``int` `N, ``int` `K, ``int` `M)``{``    ``// Stores the sum of elements``    ``// in the left and right of index K``    ``int` `S1 = 0, S2 = 0;` `    ``S1 = K * (K + 1) / 2;``    ``S2 = (N - K - 1) * (N - K) / 2;` `    ``// Stores the maximum``    ``// possible value at index K``    ``int` `X = (M + S1 + S2) / N;` `    ``// Print the answer``    ``cout << X;``}` `// Driver Code``int` `main()``{``    ``// Given N, K & M``    ``int` `N = 3, K = 1, M = 7;``    ``maxValueAtIndexK(N, K, M);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;` `class` `GFG{` `// Function to calculate the maximum``// possible value at index K``static` `void` `maxValueAtIndexK(``int` `N, ``int` `K,``                             ``int` `M)``{``    ` `    ``// Stores the sum of elements``    ``// in the left and right of index K``    ``int` `S1 = ``0``, S2 = ``0``;` `    ``S1 = K * (K + ``1``) / ``2``;``    ``S2 = (N - K - ``1``) * (N - K) / ``2``;` `    ``// Stores the maximum``    ``// possible value at index K``    ``int` `X = (M + S1 + S2) / N;` `    ``// Print the answer``    ``System.out.println(X);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ` `    ``// Given N, K & M``    ``int` `N = ``3``, K = ``1``, M = ``7``;``    ` `    ``maxValueAtIndexK(N, K, M);``}``}` `// This code is contributed by Dharanendra L V`

## Python3

 `# Python program for the above approach` `# Function to calculate the maximum``# possible value at index K``def` `maxValueAtIndexK(N, K, M):` `    ``# Stores the sum of elements``    ``# in the left and right of index K``    ``S1 ``=` `0``; S2 ``=` `0``;``    ``S1 ``=` `K ``*` `(K ``+` `1``) ``/``/` `2``;``    ``S2 ``=` `(N ``-` `K ``-` `1``) ``*` `(N ``-` `K) ``/``/` `2``;` `    ``# Stores the maximum``    ``# possible value at index K``    ``X ``=` `(M ``+` `S1 ``+` `S2) ``/``/` `N;` `    ``# Prthe answer``    ``print``(X);` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:` `    ``# Given N, K & M``    ``N ``=` `3``; K ``=` `1``; M ``=` `7``;``    ``maxValueAtIndexK(N, K, M);` `# This code is contributed by 29AjayKumar`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to calculate the maximum``// possible value at index K``static` `void` `maxValueAtIndexK(``int` `N, ``int` `K,``                             ``int` `M)``{``    ` `    ``// Stores the sum of elements``    ``// in the left and right of index K``    ``int` `S1 = 0, S2 = 0;` `    ``S1 = K * (K + 1) / 2;``    ``S2 = (N - K - 1) * (N - K) / 2;` `    ``// Stores the maximum``    ``// possible value at index K``    ``int` `X = (M + S1 + S2) / N;` `    ``// Print the answer``    ``Console.WriteLine(X);``}` `// Driver Code``static` `public` `void` `Main()``{``    ` `    ``// Given N, K & M``    ``int` `N = 3, K = 1, M = 7;``    ` `    ``maxValueAtIndexK(N, K, M);``}``}` `// This code is contributed by Dharanendra L V`

## Javascript

 ``
Output:
`3`

Time Complexity: O(1)
Auxiliary Space: O(1)

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