Clustering/Partitioning an array such that sum of square differences is minimum

Given an array of n numbers and a number k. We need to divide the array into k partitions (clusters) of same or different length. For a given k, there can be one or more ways to make clusters (partitions). We define a function Cost(i) for the $i_{th}$ cluster, as the square of the difference between its first and last element. If the current cluster is $S_x,S_{x+1},S_{x+2}.........,S_{x+len_i-1}$ , where $len_{i}$ is the length of current cluster, then $Cost_{i}={(S_{x+len_i-1}-S_x)}^{2}$ .
Amongst all the possible kinds of partitions, we have to find the partition that will minimize the function,

$f(x)=\sum_{i=1}^{k} Cost_{i}$

Examples :

Input : arr[] = {1, 5, 8, 10}
            k = 2
Output : 20  
Explanation : 
Consider clustering 4 elements 1, 5, 8, 10
into 2 clusters. There are three options:
1. S1 = 1, S2 = 5, 8, 10, with total cost
    +  = 25.
2. S1 = 1, 5, S2 = 8, 10, with total cost 
    +  = 20.
3. S1 = 1, 5, 8, S2 = 10, with total cost 
    +  = 49.
So, the optimal clustering is the second one, 
so the output of the above problem is 20.

Input : arr[] = {5, 8, 1, 10}
            k = 3
Output : 20 
Explanation :
The three partitions are {5, 8}, {1} and {10}

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

To solve the problem, we assume that we have k slabs. We have to insert them in some k different positions in the array, which will give us the required partition scheme, and the one having minimum value for f(x) will be the answer.

Naive solution:

If we solve the above problem by the naive method, we would simply take all the possibilities and compute the minimum.

C++

// C++ program to find minimum cost k partitions 
// of array. 
#include<iostream> 
using namespace std; 
  
// Initialize answer as infinite. 
const int inf = 1000000000; 
int ans = inf; 
  
// function to generate all possible answers. 
// and comute minimum of all costs. 
// i   --> is index of previous partition 
// par --> is current number of partitions 
// a[] and n --> Input array and its size 
// current_ans --> Cost of partitions made so far. 
void solve(int i, int par, int a[], int n, 
                  int k, int current_ans) 
{ 
    // If number of partitions is more than k 
    if (par > k) 
        return; 
  
    // If we have mad k partitions and have 
    // reached last element 
    if (par==k && i==n-1) 
    { 
        ans = min(ans, current_ans); 
        return; 
    } 
  
    // 1) Partition array at different points 
    // 2) For every point, increase count of  
    //    partitions, "par" by 1. 
    // 3) Before recursive call, add cost of  
    //    the partition to current_ans 
    for (int j=i+1; j<n; j++) 
        solve(j, par+1, a, n, k, current_ans + 
                  (a[j]-a[i+1])*(a[j]-a[i+1])); 
} 
  
// Driver code 
int main() 
{ 
    int k = 2; 
    int a[] = {1, 5, 8, 10}; 
    int n = sizeof(a)/sizeof(a[0]); 
    solve(-1, 0, a, n, k, 0); 
    cout << ans << endl; 
    return 0; 
} 

Java

// Java program to find minimum cost k partitions 
// of array. 
import java.io.*; 
  
class GFG 
{ 
    // Initialize answer as infinite. 
    static int inf = 1000000000; 
    static int ans = inf; 
      
    // function to generate all possible answers. 
    // and comute minimum of all costs. 
    // i --> is index of previous partition 
    // par --> is current number of partitions 
    // a[] and n --> Input array and its size 
    // current_ans --> Cost of partitions made so far. 
    static void solve(int i, int par, int a[], int n, 
                               int k, int current_ans) 
    { 
        // If number of partitions is more than k 
        if (par > k) 
            return; 
      
        // If we have mad k partitions and have 
        // reached last element 
        if (par == k && i == n - 1) 
        { 
            ans = Math.min(ans, current_ans); 
            return; 
        } 
      
        // 1) Partition array at different points 
        // 2) For every point, increase count of  
        // partitions, "par" by 1. 
        // 3) Before recursive call, add cost of  
        // the partition to current_ans 
        for (int j = i + 1; j < n; j++) 
            solve(j, par + 1, a, n, k, current_ans + 
                 (a[j] - a[i + 1]) * (a[j] - a[i + 1])); 
    } 
      
    // Driver code 
    public static void main (String[] args)  
    { 
        int k = 2; 
        int a[] = {1, 5, 8, 10}; 
        int n = a.length; 
        solve(-1, 0, a, n, k, 0); 
        System.out.println(ans); 
          
    } 
}  
  
// This code is contributed by vt_m. 

Python3

# Python3 program to find minimum  
# cost k partitions of array. 
  
# Initialize answer as infinite.  
inf = 1000000000
ans = inf 
  
# function to generate all possible answers.  
# and comute minimum of all costs.  
# i --> is index of previous partition  
# par --> is current number of partitions  
# a[] and n --> Input array and its size  
# current_ans --> Cost of partitions made so far. 
def solve(i, par, a, n, k, current_ans): 
      
    # If number of partitions is more than k 
    if (par > k): 
        return 0
          
    # If we have mad k partitions and  
    # have reached last element 
    global ans 
    if (par == k and i == n - 1): 
        ans = min(ans, current_ans) 
        return 0
  
    # 1) Partition array at different points  
    # 2) For every point, increase count of  
    # partitions, "par" by 1.  
    # 3) Before recursive call, add cost of  
    # the partition to current_ans  
    for j in range(i + 1, n): 
        solve(j, par + 1, a, n, k, current_ans + 
             (a[j] - a[i + 1]) * (a[j] - a[i + 1])) 
  
# Driver code  
k = 2
a = [1, 5, 8, 10] 
n = len(a) 
solve(-1, 0, a, n, k, 0) 
print(ans) 
  
# This code is contributed by sahilshelangia 

C#

// C# program to find minimum 
// cost k partitions of array. 
using System; 
  
class GFG 
{ 
    // Initialize answer as infinite. 
    static int inf = 1000000000; 
    static int ans = inf; 
      
    // function to generate all possible answers. 
    // and comute minimum of all costs. 
    // i --> is index of previous partition 
    // par --> is current number of partitions 
    // a[] and n --> Input array and its size 
    // current_ans --> Cost of partitions made so far. 
    static void solve(int i, int par, int []a, 
                      int n, int k, int current_ans) 
    { 
        // If number of partitions is more than k 
        if (par > k) 
            return; 
      
        // If we have mad k partitions and 
        // have reached last element 
        if (par == k && i == n - 1) 
        { 
            ans = Math.Min(ans, current_ans); 
            return; 
        } 
      
        // 1) Partition array at different points 
        // 2) For every point, increase count of  
        // partitions, "par" by 1. 
        // 3) Before recursive call, add cost of  
        // the partition to current_ans 
        for (int j = i + 1; j < n; j++) 
            solve(j, par + 1, a, n, k, current_ans + 
                 (a[j] - a[i + 1]) * (a[j] - a[i + 1])); 
    } 
      
    // Driver code 
    public static void Main ()  
    { 
        int k = 2; 
        int []a = {1, 5, 8, 10}; 
        int n = a.Length; 
        solve(-1, 0, a, n, k, 0); 
        Console.Write(ans); 
    } 
}  
  
// This code is contributed by nitin mittal. 

PHP

<?php  
// PHP program to find minimum cost k  
// partitions of array. 
  
// Initialize answer as infinite. 
$inf = 1000000000; 
$ans = $inf; 
  
// function to generate all possible answers. 
// and compute minimum of all costs. 
// i --> is index of previous partition 
// par --> is current number of partitions 
// a[] and n --> Input array and its size 
// current_ans --> Cost of partitions made so far. 
function solve($i, $par, &$a, $n, $k, $current_ans) 
{ 
    global $inf, $ans; 
      
    // If number of partitions is  
    // more than k 
    if ($par > $k) 
        return; 
  
    // If we have mad k partitions and  
    // have reached last element 
    if ($par == $k && $i == $n - 1) 
    { 
        $ans = min($ans, $current_ans); 
        return; 
    } 
  
    // 1) Partition array at different points 
    // 2) For every point, increase count of  
    //    partitions, "par" by 1. 
    // 3) Before recursive call, add cost of  
    //    the partition to current_ans 
    for ($j = $i + 1; $j < $n; $j++) 
        solve($j, $par + 1, $a, $n, $k, $current_ans + 
                           ($a[$j] - $a[$i + 1]) *  
                           ($a[$j] - $a[$i + 1])); 
} 
  
// Driver code 
$k = 2; 
$a = array(1, 5, 8, 10); 
$n = sizeof($a); 
solve(-1, 0, $a, $n, $k, 0); 
echo $ans . "\n"; 
  
// This code is contributed by ita_c 
?> 

Output:

Time Complexity: Its clear that the above algorithm has Time Complexity of $O(2^n)$ .

Dynamic Programming:

We create a table dp[n+1][k+1] table and initialize all values as infinite.

dp[i][j] stores optimal partition cost 
         for arr[0..i-1] and j partitions.

Let us compute the value of dp[i][j]. we take an index m, such that m < i, and put a partition next to that position such that there is no slab in between the indices i and m. It can be seen simply that answer to the current scenario is dp[m][j-1] + (a[i-1]-a[m])*(a[i-1]-a[m]), where the first term signifies the minimum f(x) till the $m_{th)$ element with j-1 partitions and the second one signifies the cost of current cluster. So we will take the minimum of all the possible indices m and dp[i][j] will be assigned the minimum amongst them.

C++

// C++ program to find minimum cost k partitions 
// of array. 
#include<iostream> 
using namespace std; 
const int inf = 1000000000; 
  
// Returns minimum cost of partitioning a[] in 
// k clusters. 
int minCost(int a[], int n, int k) 
{ 
    // Create a dp[][] table and initialize 
    // all values as infinite. dp[i][j] is 
    // going to store optimal partition cost 
    // for arr[0..i-1] and j partitions 
    int dp[n+1][k+1]; 
    for (int i=0; i<=n; i++) 
        for (int j=0;j<=k;j++) 
            dp[i][j] = inf; 
  
    // Fill dp[][] in bottom up manner 
    dp[0][0] = 0; 
  
    // Current ending position (After i-th 
    // iteration result for a[0..i-1] is computed. 
    for (int i=1;i<=n;i++) 
  
        // j is number of partitions 
        for (int j=1;j<=k;j++) 
  
            // Picking previous partition for 
            // current i. 
            for (int m=i-1;m>=0;m--) 
                dp[i][j] = min(dp[i][j], dp[m][j-1] + 
                          (a[i-1]-a[m])*(a[i-1]-a[m])); 
  
  
    return dp[n][k]; 
} 
  
// Driver code 
int main() 
{ 
    int k = 2; 
    int a[] = {1, 5, 8, 10}; 
    int n = sizeof(a)/sizeof(a[0]); 
    cout << minCost(a, n, k) << endl; 
    return 0; 
} 

Java

// Java program to find minimum cost  
// k partitions of array. 
import java.io.*; 
  
class GFG  
{ 
    static int inf = 1000000000; 
      
    // Returns minimum cost of partitioning  
    // a[] in k clusters. 
    static int minCost(int a[], int n, int k) 
    { 
        // Create a dp[][] table and initialize 
        // all values as infinite. dp[i][j] is 
        // going to store optimal partition cost 
        // for arr[0..i-1] and j partitions 
        int dp[][] = new int[n + 1][k + 1]; 
        for (int i = 0; i <= n; i++) 
            for (int j = 0; j <= k; j++) 
                dp[i][j] = inf; 
      
        // Fill dp[][] in bottom up manner 
        dp[0][0] = 0; 
      
        // Current ending position (After i-th 
        // iteration result for a[0..i-1] is computed. 
        for (int i = 1; i <= n; i++) 
      
            // j is number of partitions 
            for (int j = 1; j <= k; j++) 
      
                // Picking previous partition for 
                // current i. 
                for (int m = i - 1; m >= 0; m--) 
                    dp[i][j] = Math.min(dp[i][j], dp[m][j - 1] + 
                              (a[i - 1] - a[m]) * (a[i - 1] - a[m])); 
      
      
        return dp[n][k]; 
    } 
      
    // Driver code 
    public static void main (String[] args)  
    { 
        int k = 2; 
        int a[] = {1, 5, 8, 10}; 
        int n = a.length; 
        System.out.println(minCost(a, n, k)); 
              
    } 
} 
  
// This code is contributed by vt_m. 

Python3

# Python3 program to find minimum cost k partitions 
# of array. 
inf = 1000000000; 
  
# Returns minimum cost of partitioning a[] in 
# k clusters. 
def minCost(a, n, k): 
  
    # Create a dp[][] table and initialize 
    # all values as infinite. dp[i][j] is 
    # going to store optimal partition cost 
    # for arr[0..i-1] and j partitions 
    dp = [[inf for i in range(k + 1)]  
               for j in range(n + 1)]; 
  
    # Fill dp[][] in bottom up manner 
    dp[0][0] = 0; 
  
    # Current ending position (After i-th 
    # iteration result for a[0..i-1] is computed. 
    for i in range(1, n + 1): 
  
        # j is number of partitions 
        for j in range(1, k + 1): 
  
            # Picking previous partition for 
            # current i. 
            for m in range(i - 1, -1, -1): 
                dp[i][j] = min(dp[i][j], dp[m][j - 1] +
                                    (a[i - 1] - a[m]) * 
                                    (a[i - 1] - a[m])); 
  
    return dp[n][k]; 
  
# Driver code 
if __name__ == '__main__': 
    k = 2; 
    a = [1, 5, 8, 10]; 
    n = len(a); 
    print(minCost(a, n, k)); 
  
# This code is contributed by 29AjayKumar 

C#

// C# program to find minimum cost  
// k partitions of array. 
using System; 
  
class GFG { 
      
    static int inf = 1000000000; 
      
    // Returns minimum cost of partitioning  
    // a[] in k clusters. 
    static int minCost(int []a, int n, int k) 
    { 
          
        // Create a dp[][] table and initialize 
        // all values as infinite. dp[i][j] is 
        // going to store optimal partition cost 
        // for arr[0..i-1] and j partitions 
        int [,]dp = new int[n + 1,k + 1]; 
        for (int i = 0; i <= n; i++) 
            for (int j = 0; j <= k; j++) 
                dp[i,j] = inf; 
      
        // Fill dp[][] in bottom 
        // up manner 
        dp[0,0] = 0; 
      
        // Current ending position  
        // (After i-th iteration  
        // result for a[0..i-1] 
        // is computed. 
        for (int i = 1; i <= n; i++) 
      
            // j is number of partitions 
            for (int j = 1; j <= k; j++) 
      
                // Picking previous 
                // partition for 
                // current i. 
                for (int m = i - 1; m >= 0; m--) 
                    dp[i,j] = Math.Min(dp[i,j], 
                                 dp[m,j - 1] + 
                               (a[i - 1] - a[m]) *  
                               (a[i - 1] - a[m])); 
      
      
        return dp[n,k]; 
    } 
      
    // Driver code 
    public static void Main ()  
    { 
        int k = 2; 
        int []a = {1, 5, 8, 10}; 
        int n = a.Length; 
        Console.Write(minCost(a, n, k)); 
              
    } 
} 
  
// This code is contributed by nitin mittal 

Output:

Time Complexity: Having the three simple loops, the complexity of the above algorithm is $O(n^2k)$ .

This article is contributed by Amritya Vagmi and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

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