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:  

C++

// C++ implementation of the approach
#include <iostream>
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

<script>

// Javascript implementation of the approach

// Function to return the minimized sum
function 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
let n = 3, k = 56;

document.write(minimum_sum(n, k));

// This code is contributed by rishavmahato348

</script>

Output:

1

Time Complexity: O(1)

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