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 cluster, as the square of the difference between its first and last element. If the current cluster is , where is the length of current cluster, then .

Amongst all the possible kinds of partitions, we have to find the partition that will minimize the function,

Example:

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 =3Output : 20 Explanation : The three partitions are {5, 8}, {1} and {10}

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++ 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; }

Output:

20

**Time Complexity:** Its clear that the above algorithm has Time Complexity of .

**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 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++ 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; }

Output:

20

**Time Complexity:** Having the three simple loops, the complexity of the above algorithm is .

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