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# Sequence with sum K and minimum sum of absolute differences between consecutive elements

• Last Updated : 12 Jul, 2022

Given two integers N and K, the task is to find a sequence of integers of length N such that the sum of all the elements of the sequence is K and the sum of absolute differences between all consecutive elements is minimum. Print this minimized sum.

Examples:

Input: N = 3, K = 56
Output:
The sequence is {19, 19, 18} and the sum of absolute
differences of all the consecutive elements is
|19 – 19| + |19 – 18| = 0 + 1 = 1 which is the minimum possible.

Input: N = 12, K = 48
Output:
The sequence is {4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4}.

Approach: There can be two cases:

1. When K % N = 0 then the answer will be 0 as K can be evenly divided into N parts i.e. every element of the sequence will be equal.
2. When K % N != 0 then the answer will be 1 because the sequence can be arranged in a way:
• (K – (K % N)) is divisible by N so it can be divided evenly among all the N parts.
• And the rest (K % N) value can be divided in such a way to minimize the consecutive absolute difference i.e. add 1 to the first or the last (K % N) elements and the sequence will be of the type {x, x, x, x, y, y, y, y, y} yielding the minimum sum as 1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the minimized sum``int` `minimum_sum(``int` `n, ``int` `k)``{` `    ``// If k is divisible by n``    ``// then the answer will be 0``    ``if` `(k % n == 0)``        ``return` `0;` `    ``// Else the answer will be 1``    ``return` `1;``}` `// Driver code``int` `main()``{``    ``int` `n = 3, k = 56;` `    ``cout << minimum_sum(n, k);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.io.*;` `class` `GFG``{``    ` `// Function to return the minimized sum``static` `int` `minimum_sum(``int` `n, ``int` `k)``{` `    ``// If k is divisible by n``    ``// then the answer will be 0``    ``if` `(k % n == ``0``)``        ``return` `0``;` `    ``// Else the answer will be 1``    ``return` `1``;``}` `// Driver code``public` `static` `void` `main (String[] args)``{` `    ``int` `n = ``3``, k = ``56``;``    ``System.out.println (minimum_sum(n, k));``}``}` `// This code is contributed By @ajit_23`

## Python3

 `# Python3 implementation of the approach` `# Function to return the minimized sum``def` `minimum_sum(n, k):` `    ``# If k is divisible by n``    ``# then the answer will be 0``    ``if` `(k ``%` `n ``=``=` `0``):``        ``return` `0``;` `    ``# Else the answer will be 1``    ``return` `1` `# Driver code` `n ``=` `3``k ``=` `56` `print``(minimum_sum(n, k))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``        ` `// Function to return the minimized sum``static` `int` `minimum_sum(``int` `n, ``int` `k)``{` `    ``// If k is divisible by n``    ``// then the answer will be 0``    ``if` `(k % n == 0)``        ``return` `0;` `    ``// Else the answer will be 1``    ``return` `1;``}` `// Driver code``static` `public` `void` `Main ()``{``    ``int` `n = 3, k = 56;``    ``Console.Write(minimum_sum(n, k));``}``}` `// This code is contributed By Tushil`

## Javascript

 ``

Output:

`1`

Time Complexity: O(1)

Auxiliary Space: O(1), since no extra space has been taken.

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