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Split a given array into K subarrays minimizing the difference between their maximum and minimum
  • Last Updated : 14 Jul, 2020
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Given a sorted array arr[] of N integers and an integer K, the task is to split the array into K subarrays such that the sum of the difference of maximum and minimum element of each subarray is minimized.

Examples:

Input: arr[] = {1, 3, 3, 7}, K = 4
Output: 0
Explanation:
The given array can be split into 4 subarrays as {1}, {3}, {3}, and {7}.
The difference between minimum and maximum of each subarray is:
1. {1}, difference = 1 – 1 = 0
2. {3}, difference = 3 – 3 = 0
3. {3}, difference = 3 – 3 = 0
4. {7}, difference = 7 – 7 = 0
Therefore, the sum all the difference is 0 which is minimized.

Input: arr[] = {4, 8, 15, 16, 23, 42}, K = 3
Output: 12
Explanation:
The given array can be split into 3 subarrays as {4, 8, 15, 16}, {23}, and {42}.
The difference between minimum and maximum of each subarray is:
1. {4, 8, 15, 16}, difference = 16 – 12 = 0
2. {23}, difference = 23 – 23 = 0
3. {42}, difference = 42 – 42 = 0
Therefore, the sum all the difference is 12 which is minimized.

Approach: To split the given array into K subarrays with the given conditions, the idea is to split at indexes(say i) where the difference between elements arr[i+1] and arr[i] is largest. Below are the steps to implement this approach:



  1. Store the difference between consecutive pairs of elements in the given array arr[] into another array(say temp[]).
  2. Sort the array temp[] in increasing order.
  3. Intialise the total difference(say diff) as the difference of first and last element of the given array arr[].
  4. Add the first K – 1 values from the array temp[] to the above difference.
  5. The value stored in diff is the minimum sum of the difference of maximum and minimum element of K subarray.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the subarray
int find(int a[], int n, int k)
{
    vector<int> v;
  
    // Add the difference to vectors
    for (int i = 1; i < n; ++i) {
        v.push_back(a[i - 1] - a[i]);
    }
  
    // Sort vector to find minimum k
    sort(v.begin(), v.end());
  
    // Initialize result
    int res = a[n - 1] - a[0];
  
    // Adding first k-1 values
    for (int i = 0; i < k - 1; ++i) {
        res += v[i];
    }
  
// Return the minimized sum
    return res;
}
  
// Driver Code
int main()
{
// Given array arr[]
    int arr[] = { 4, 8, 15, 16, 23, 42 };
  
    int N = sizeof(arr) / sizeof(int);
  
// Given K
int K = 3;
  
// Function Call
    cout << find(arr, N, K) << endl;
  
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to find the subarray
static int find(int a[], int n, int k)
{
    Vector<Integer> v = new Vector<Integer>();
  
    // Add the difference to vectors
    for(int i = 1; i < n; ++i)
    {
       v.add(a[i - 1] - a[i]);
    }
  
    // Sort vector to find minimum k
    Collections.sort(v);
  
    // Initialize result
    int res = a[n - 1] - a[0];
  
    // Adding first k-1 values
    for(int i = 0; i < k - 1; ++i)
    {
       res += v.get(i);
    }
      
    // Return the minimized sum
    return res;
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given array arr[]
    int arr[] = { 4, 8, 15, 16, 23, 42 };
  
    int N = arr.length;
      
    // Given K
    int K = 3;
      
    // Function Call
    System.out.print(find(arr, N, K) + "\n");
}
}
  
// This code is contributed by Amit Katiyar

Python3




# Python3 program for the above approach
  
# Function to find the subarray
def find(a, n, k):
      
    v = []
      
    # Add the difference to vectors
    for i in range(1, n):
        v.append(a[i - 1] - a[i])
          
    # Sort vector to find minimum k
    v.sort()
      
    # Initialize result
    res = a[n - 1] - a[0]
      
    # Adding first k-1 values
    for i in range(k - 1):
        res += v[i]
      
    # Return the minimized sum
    return res
      
# Driver code
arr = [ 4, 8, 15, 16, 23, 42 ]
  
# Length of array
N = len(arr)
  
K = 3
  
# Function Call
print(find(arr, N, K))
  
# This code is contributed by sanjoy_62

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
  
class GFG{
  
// Function to find the subarray
static int find(int []a, int n, int k)
{
    List<int> v = new List<int>();
  
    // Add the difference to vectors
    for(int i = 1; i < n; ++i)
    {
       v.Add(a[i - 1] - a[i]);
    }
  
    // Sort vector to find minimum k
    v.Sort();
  
    // Initialize result
    int res = a[n - 1] - a[0];
  
    // Adding first k-1 values
    for(int i = 0; i < k - 1; ++i)
    {
       res += v[i];
    }
      
    // Return the minimized sum
    return res;
}
  
// Driver Code
public static void Main(String[] args)
{
      
    // Given array []arr
    int []arr = { 4, 8, 15, 16, 23, 42 };
    int N = arr.Length;
      
    // Given K
    int K = 3;
      
    // Function Call
    Console.Write(find(arr, N, K) + "\n");
}
}
  
// This code is contributed by Amit Katiyar
Output:
12

Time Complexity: O(N), where N is the number of elements in the array.
Auxiliary Space: O(N), where N is the number of elements in the array.

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