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

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: 1
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: 0
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:

## CPP

 `// 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 `

Output:

```1
```

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Improved By : mohit kumar 29, jit_t

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