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 ith 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 approachÂ
#include <bits/stdc++.h>using namespace std;Â
// Function to split an array into K subarrays// with minimum sum of absolute difference// of adjacent elements in each of K subarraysvoid absoluteDifference(int arr[], int N, int K){Â
    // Stores the absolute differences    // of adjacent elements    int new_Arr[N - 1];Â
    // Stores the answer    int ans = 0;Â
    // 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]);Â
        // Stores the sum of all absolute        // differences of adjacent elements        ans += new_Arr[i];    }Â
    // Sorting the new_Arr    // in decreasing order    sort(new_Arr, new_Arr + N - 1,         greater<int>());Â
    for (int i = 0; i < K - 1; i++) {Â
        // Removing K - 1 elements        // with maximum sum        ans -= new_Arr[i];    }Â
    // Prints the answer    cout << ans << endl;}Â
// Driver codeint main(){Â Â Â Â // Given array arr[]Â Â Â Â int arr[] = { 1, 3, -2, 5, -1 };Â
    // Given K    int K = 2;Â
    // Size of array    int N = sizeof(arr) / sizeof(arr[0]);Â
    // Function Call    absoluteDifference(arr, N, K);    return 0;} |
Java
//Â java program for the above approachimport java.util.*;import java.util.Arrays;class GFG{Â Â Â Â Â Â public static void reverse(int[] array) Â { Â
   // Length of the array    int n = array.length; Â
   // Swapping the first half elements with last half    // elements    for (int i = 0; i < n / 2; i++)   { Â
     // Storing the first half elements temporarily      int temp = array[i]; Â
     // Assigning the first half to the last half      array[i] = array[n - i - 1]; Â
     // Assigning the last half to the first half      array[n - i - 1] = temp;    }  } Â
  // 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)  {Â
    // Stores the absolute differences    // of adjacent elements    int new_Arr[] = new int[N - 1];Â
    // Stores the answer    int ans = 0;Â
    // 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]);Â
      // Stores the sum of all absolute      // differences of adjacent elements      ans += new_Arr[i];    }Â
    // Sorting the new_Arr    // in decreasing order    Arrays.sort(new_Arr);     reverse(new_Arr);Â
    for (int i = 0; i < K - 1; i++)    {Â
      // Removing K - 1 elements      // with maximum sum      ans -= new_Arr[i];    }Â
    // Prints the answer    System.out.println(ans);  }Â
  // Driver code  public static void main (String[] args)  {Â
    // Given array arr[]    int arr[] = { 1, 3, -2, 5, -1 };Â
    // Given K    int K = 2;Â
    // Size of array    int N = arr.length;Â
    // Function Call    absoluteDifference(arr, N, K);   }}Â
// This code is contributed by AnkThon |
Python3
# Python3 program for the above approachÂ
# Function to split an array into K subarrays# with minimum sum of absolute difference# of adjacent elements in each of K subarraysdef absoluteDifference(arr, N, K):         # Stores the absolute differences    # of adjacent elements    new_Arr = [0 for i in range(N - 1)]Â
    # Stores the answer    ans = 0Â
    # Stores absolute differences of    # adjacent elements in new_Arr    for i in range(N - 1):        new_Arr[i] = abs(arr[i] - arr[i + 1])Â
        # Stores the sum of all absolute        # differences of adjacent elements        ans += new_Arr[i]Â
    # Sorting the new_Arr    # in decreasing order    new_Arr = sorted(new_Arr)[::-1]Â
    for i in range(K - 1):Â
        # Removing K - 1 elements        # with maximum sum        ans -= new_Arr[i]Â
    # Prints the answer    print (ans)Â
# Driver codeif __name__ == '__main__':Â
    # Given array arr[]    arr = [ 1, 3, -2, 5, -1 ]Â
    # Given K    K = 2Â
    # Size of array    N = len(arr)Â
    # Function Call    absoluteDifference(arr, N, K)Â
# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;Â
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 subarrayspublic 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);}Â
// Driver Codepublic 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);}}Â
// This code is contributed by susmitakundugoaldanga |
Javascript
<script>Â
// Javascript program for the above approachfunction reverse(array) { Â
    // 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;     } } Â
// Function to split an array into K subarrays// with minimum sum of absolute difference// of adjacent elements in each of K subarraysfunction absoluteDifference(arr, N, K){Â
    // 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);}Â
// Driver CodeÂ
// Given array arr[]let arr = [ 1, 3, -2, 5, -1 ];Â
// Given Klet K = 2;Â
// Size of arraylet N = arr.length;Â
// Function CallabsoluteDifference(arr, N, K);Â
// This code is contributed by splevel62Â
</script> |
Output:Â
13
Â
Time Complexity: O(N * Log(N))
Auxiliary Space: O(N)
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