Split array into K subarrays with minimum sum of absolute difference between adjacent elements

0 0 6 minutes read

Given an array, arr[] of size N and an integer K, the task is to split the array into K subarrays minimizing the sum of absolute difference between adjacent elements of each subarray.

Examples:

Input: arr[] = {1, 3, -2, 5, -1}, K = 2
Output: 13
Explanation: Split the array into following 2 subarrays: {1, 3, -2} and {5, -1}.

Input: arr[] = {2, 14, 26, 10, 5, 12}, K = 3
Output: 24
Explanation: Splitting array into following 3 subarrays: {2, 14}, {26}, {10, 5, 12}.

Approach: The given problem can be solved based on the following observations:

The idea is to slice the array arr[] at i^th index which gives maximum absolute difference of adjacent elements. Subtract it from the result. Slicing at K – 1 places will give K subarrays with minimum sum of absolute difference of adjacent elements.

Follow the steps below to solve the problem:

Initialize an array, say new_Arr[], and an integer variable, say ans, to store total absolute difference sum.
Traverse the array .
- Store absolute difference of adjacent elements, say arr[i+1] and arr[i] in new_Arr[] array.
- Increment ans by absolute difference of arr[i] and arr[i + 1]
Sort the new_Arr[] array in descending order.
Traverse the array from i = 0 to i = K-1
- Decrement ans by new_Arr[i].
Finally, print ans.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to split an array into K subarrays
// with minimum sum of absolute difference
// of adjacent elements in each of K subarrays
void absoluteDifference(int arr[], int N, int K)
{
 
    // Stores the absolute differences
    // of adjacent elements
    int new_Arr[N - 1];
 
    // Stores the answer
    int ans = 0;
 
    // Stores absolute differences of
    // adjacent elements in new_Arr
    for (int i = 0; i < N - 1; i++) {
        new_Arr[i] = abs(arr[i] - arr[i + 1]);
 
        // Stores the sum of all absolute
        // differences of adjacent elements
        ans += new_Arr[i];
    }
 
    // Sorting the new_Arr
    // in decreasing order
    sort(new_Arr, new_Arr + N - 1,
         greater<int>());
 
    for (int i = 0; i < K - 1; i++) {
 
        // Removing K - 1 elements
        // with maximum sum
        ans -= new_Arr[i];
    }
 
    // Prints the answer
    cout << ans << endl;
}
 
// Driver code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 3, -2, 5, -1 };
 
    // Given K
    int K = 2;
 
    // Size of array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    absoluteDifference(arr, N, K);
    return 0;
}

Java

//  java program for the above approach
import java.util.*;
import java.util.Arrays;
class GFG
{
     
 public static void reverse(int[] array) 
 { 
 
   // Length of the array 
   int n = array.length; 
 
   // Swapping the first half elements with last half 
   // elements 
   for (int i = 0; i < n / 2; i++)
   { 
 
     // Storing the first half elements temporarily 
     int temp = array[i]; 
 
     // Assigning the first half to the last half 
     array[i] = array[n - i - 1]; 
 
     // Assigning the last half to the first half 
     array[n - i - 1] = temp; 
   } 
 } 
 
  // Function to split an array into K subarrays
  // with minimum sum of absolute difference
  // of adjacent elements in each of K subarrays
  public static void absoluteDifference(int arr[],
                                        int N, int K)
  {
 
    // Stores the absolute differences
    // of adjacent elements
    int new_Arr[] = new int[N - 1];
 
    // Stores the answer
    int ans = 0;
 
    // Stores absolute differences of
    // adjacent elements in new_Arr
    for (int i = 0; i < N - 1; i++)
    {
      new_Arr[i] = Math.abs(arr[i] - arr[i + 1]);
 
      // Stores the sum of all absolute
      // differences of adjacent elements
      ans += new_Arr[i];
    }
 
    // Sorting the new_Arr
    // in decreasing order
    Arrays.sort(new_Arr); 
    reverse(new_Arr);
 
    for (int i = 0; i < K - 1; i++)
    {
 
      // Removing K - 1 elements
      // with maximum sum
      ans -= new_Arr[i];
    }
 
    // Prints the answer
    System.out.println(ans);
  }
 
  // Driver code
  public static void main (String[] args)
  {
 
    // Given array arr[]
    int arr[] = { 1, 3, -2, 5, -1 };
 
    // Given K
    int K = 2;
 
    // Size of array
    int N = arr.length;
 
    // Function Call
    absoluteDifference(arr, N, K);
   }
}
 
// This code is contributed by AnkThon

Python3

# Python3 program for the above approach
 
# Function to split an array into K subarrays
# with minimum sum of absolute difference
# of adjacent elements in each of K subarrays
def absoluteDifference(arr, N, K):
     
    # Stores the absolute differences
    # of adjacent elements
    new_Arr = [0 for i in range(N - 1)]
 
    # Stores the answer
    ans = 0
 
    # Stores absolute differences of
    # adjacent elements in new_Arr
    for i in range(N - 1):
        new_Arr[i] = abs(arr[i] - arr[i + 1])
 
        # Stores the sum of all absolute
        # differences of adjacent elements
        ans += new_Arr[i]
 
    # Sorting the new_Arr
    # in decreasing order
    new_Arr = sorted(new_Arr)[::-1]
 
    for i in range(K - 1):
 
        # Removing K - 1 elements
        # with maximum sum
        ans -= new_Arr[i]
 
    # Prints the answer
    print (ans)
 
# Driver code
if __name__ == '__main__':
 
    # Given array arr[]
    arr = [ 1, 3, -2, 5, -1 ]
 
    # Given K
    K = 2
 
    # Size of array
    N = len(arr)
 
    # Function Call
    absoluteDifference(arr, N, K)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
  
public static void reverse(int[] array) 
{ 
     
    // Length of the array 
    int n = array.Length; 
     
    // Swapping the first half elements with 
    // last half elements 
    for(int i = 0; i < n / 2; i++)
    { 
     
        // Storing the first half elements 
        // temporarily 
        int temp = array[i]; 
         
        // Assigning the first half to 
        // the last half 
        array[i] = array[n - i - 1]; 
         
        // Assigning the last half to 
        // the first half 
        array[n - i - 1] = temp; 
    } 
} 
     
// Function to split an array into K subarrays
// with minimum sum of absolute difference
// of adjacent elements in each of K subarrays
public static void absoluteDifference(int[] arr,
                                      int N, int K)
{
     
    // Stores the absolute differences
    // of adjacent elements
    int[] new_Arr = new int[N - 1];
     
    // Stores the answer
    int ans = 0;
     
    // Stores absolute differences of
    // adjacent elements in new_Arr
    for(int i = 0; i < N - 1; i++)
    {
        new_Arr[i] = Math.Abs(arr[i] - 
                              arr[i + 1]);
         
        // Stores the sum of all absolute
        // differences of adjacent elements
        ans += new_Arr[i];
    }
     
    // Sorting the new_Arr
    // in decreasing order
    Array.Sort(new_Arr); 
    reverse(new_Arr);
     
    for(int i = 0; i < K - 1; i++)
    {
         
        // Removing K - 1 elements
        // with maximum sum
        ans -= new_Arr[i];
    }
     
    // Prints the answer
    Console.WriteLine(ans);
}
 
// Driver Code
public static void Main ()
{
     
    // Given array arr[]
    int[] arr = { 1, 3, -2, 5, -1 };
     
    // Given K
    int K = 2;
     
    // Size of array
    int N = arr.Length;
     
    // Function Call
    absoluteDifference(arr, N, K);
}
}
 
// This code is contributed by susmitakundugoaldanga

Javascript

<script>
 
// Javascript program for the above approach
function reverse(array) 
{ 
 
    // Length of the array 
    let n = array.length; 
     
    // Swapping the first half elements
    // with last half elements 
    for(let i = 0; i < n / 2; i++)
    { 
     
        // Storing the first half 
        // elements temporarily 
        let temp = array[i]; 
         
        // Assigning the first half
        // to the last half 
        array[i] = array[n - i - 1]; 
         
        // Assigning the last half 
        // to the first half 
        array[n - i - 1] = temp; 
    } 
} 
 
// Function to split an array into K subarrays
// with minimum sum of absolute difference
// of adjacent elements in each of K subarrays
function absoluteDifference(arr, N, K)
{
 
    // Stores the absolute differences
    // of adjacent elements
    let new_Arr = new Array(N - 1).fill(0);
     
    // Stores the answer
    let ans = 0;
     
    // Stores absolute differences of
    // adjacent elements in new_Arr
    for(let i = 0; i < N - 1; i++)
    {
        new_Arr[i] = Math.abs(arr[i] - arr[i + 1]);
         
        // Stores the sum of all absolute
        // differences of adjacent elements
        ans += new_Arr[i];
    }
     
    // Sorting the new_Arr
    // in decreasing order
    new_Arr.sort(); 
    reverse(new_Arr);
     
    for(let i = 0; i < K - 1; i++)
    {
     
        // Removing K - 1 elements
        // with maximum sum
        ans -= new_Arr[i];
    }
     
    // Print the answer
    document.write(ans);
}
 
// Driver Code
 
// Given array arr[]
let arr = [ 1, 3, -2, 5, -1 ];
 
// Given K
let K = 2;
 
// Size of array
let N = arr.length;
 
// Function Call
absoluteDifference(arr, N, K);
 
// This code is contributed by splevel62
 
</script>

Output:

Time Complexity: O(N * Log(N))
Auxiliary Space: O(N)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!