Split a given array into K subarrays minimizing the difference between their maximum and minimum

Last Updated : 21 Aug, 2021

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 – 4 = 12
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:

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

Javascript

   <script>
// javascript program for the above approach
 
    // Function to find the subarray
    function find(a , n , k) {
        var v = [];
 
        // Add the difference to vectors
        for (i = 1; i < n; ++i) {
            v.push(a[i - 1] - a[i]);
        }
 
        // Sort vector to find minimum k
        v.sort((a,b)=>a-b);
 
        // Initialize result
        var res = a[n - 1] - a[0];
 
        // Adding first k-1 values
        for (i = 0; i < k - 1; ++i) {
            res += v[i];
        }
 
        // Return the minimized sum
        return res;
    }
 
    // Driver Code
     
 
        // Given array arr
        var arr = [ 4, 8, 15, 16, 23, 42 ];
 
        var N = arr.length;
 
        // Given K
        var K = 3;
 
        // Function Call
        document.write(find(arr, N, K) + "\n");
 
// This code contributed by aashish1995
</script>
 
  

Output:

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.

Min difference between maximum and minimum element in all Y size subarrays

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Split a given array into K subarrays minimizing the difference between their maximum and minimum

C++

Java

Python3

C#

Javascript

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