Split array into K subarrays with minimum sum of absolute difference between adjacent elements
Last Updated :
06 Apr, 2023
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 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 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 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# 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
<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>
Time Complexity: O(N * Log(N))
Auxiliary Space: O(N)
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