Given an array arr[] and an integer K. The task is to divide the array into K parts ( subarray ) such that the sum of the values of all subarray is minimum.
The value of every subarray is defined as:
- Take the maximum from that subarray.
- Subtract each element of the subarray with the maximum.
- Take the sum of all the values after subtraction.
The task is to minimize the sum of the values after dividing the array into K parts.
Examples:
Input: arr[] = { 2, 9, 5, 4, 8, 3, 6 }, K = 2
Output: 19
Explanation:
The two groups are : {2} with max = 2 and {9, 5, 4, 8, 3, 6} with max=9,
sum of difference of first group = 2 – 2 = 0,
sum of difference of second group = (9-9) + (9-5) + (9-4) + (9-8) + (9-3) + (9-6) = 19Input: arr[] = { 12, 20, 30, 14, 25}, K = 3
Output: 19
Approach:
The brute-force solution will be to try all the possible partitions and take the minimum overall. Although this solution is exponential in time. In the recursive solution, there are many overlapping sub-problems that can be optimised using dynamic programming.
So, We can form a basic recursive formula and that computes every possible solution and finds the best possible solution. We can see that the recursive solution has many overlapping sub-problems we can reduce the complexity using Dynamic programming.
Recursive formula:
F(i, K) = { min of all values such that j < i [ max(Arr[i..j]) * (i – j + 1) – Sum(A[i…j] ] } + F(j, K-1)
The bottom-up approach can be used to compute the values of sub-problems first and store them.
Here dp[i][j] defines the minimum value that can be obtained if the array is starting from index i and have j partition.
So, the answer to the problems will be dp[0][K], array starting at 0 and having K partitions.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to divide an array into k// parts such that the sum of difference// of every element with the maximum element// of that part is minimumint divideArray(int arr[], int n, int k){ // Dp to store the values int dp[500][500] = { 0 }; k -= 1; // Fill up the dp table for (int i = n - 1; i >= 0; i--) { for (int j = 0; j <= k; j++) { // Intitilize maximum value dp[i][j] = INT_MAX; // Max element and the sum int max_ = -1, sum = 0; // Run a loop from i to n for (int l = i; l < n; l++) { // Find the maximum number // from i to l and the sum // from i to l max_ = max(max_, arr[l]); sum += arr[l]; // Find the sum of difference // of every element with the // maximum element int diff = (l - i + 1) * max_ - sum; // If the array can be divided if (j > 0) dp[i][j] = min(dp[i][j], diff + dp[l + 1][j - 1]); else dp[i][j] = diff; } } } // Returns the minimum sum // in K parts return dp[0][k];}// Driver codeint main(){ int arr[] = { 2, 9, 5, 4, 8, 3, 6 }; int n = sizeof(arr) / sizeof(int); int k = 2; cout << divideArray(arr, n, k) << "\n"; return 0;} |
Java
// Java implementation of the above approach class GFG{ // Function to divide an array into k // parts such that the sum of difference // of every element with the maximum element // of that part is minimum static int divideArray(int arr[], int n, int k) { // Dp to store the values int dp[][] = new int[500][500]; int i, j; for(i = 0; i < 500; i++) for(j = 0; j < 500; j++) dp[i][j] = 0; k -= 1; // Fill up the dp table for (i = n - 1; i >= 0; i--) { for (j = 0; j <= k; j++) { // Intitilize maximum value dp[i][j] = Integer.MAX_VALUE; // Max element and the sum int max_ = -1, sum = 0; // Run a loop from i to n for (int l = i; l < n; l++) { // Find the maximum number // from i to l and the sum // from i to l max_ = Math.max(max_, arr[l]); sum += arr[l]; // Find the sum of difference // of every element with the // maximum element int diff = (l - i + 1) * max_ - sum; // If the array can be divided if (j > 0) dp[i][j] = Math.min(dp[i][j], diff + dp[l + 1][j - 1]); else dp[i][j] = diff; } } } // Returns the minimum sum // in K parts return dp[0][k]; } // Driver code public static void main (String[] args) { int arr[] = { 2, 9, 5, 4, 8, 3, 6 }; int n = arr.length; int k = 2; System.out.println(divideArray(arr, n, k)); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the above approach# Function to divide an array into k# parts such that the summ of difference# of every element with the maximum element# of that part is minimumdef divideArray(arr, n, k): # Dp to store the values dp = [[0 for i in range(500)] for i in range(500)] k -= 1 # Fill up the dp table for i in range(n - 1, -1, -1): for j in range(0, k + 1): # Intitilize maximum value dp[i][j] = 10**9 # Max element and the summ max_ = -1 summ = 0 # Run a loop from i to n for l in range(i, n): # Find the maximum number # from i to l and the summ # from i to l max_ = max(max_, arr[l]) summ += arr[l] # Find the summ of difference # of every element with the # maximum element diff = (l - i + 1) * max_ - summ # If the array can be divided if (j > 0): dp[i][j]= min(dp[i][j], diff + dp[l + 1][j - 1]) else: dp[i][j] = diff # Returns the minimum summ # in K parts return dp[0][k]# Driver codearr = [2, 9, 5, 4, 8, 3, 6]n = len(arr)k = 2print(divideArray(arr, n, k))# This code is contributed by Mohit Kumar |
C#
// C# implementation of above approachusing System;class GFG{ // Function to divide an array into k // parts such that the sum of difference // of every element with the maximum element // of that part is minimum static int divideArray(int []arr, int n, int k) { // Dp to store the values int [,]dp = new int[500, 500]; int i, j; for(i = 0; i < 500; i++) for(j = 0; j < 500; j++) dp[i, j] = 0; k -= 1; // Fill up the dp table for (i = n - 1; i >= 0; i--) { for (j = 0; j <= k; j++) { // Intitilize maximum value dp[i, j] = int.MaxValue; // Max element and the sum int max_ = -1, sum = 0; // Run a loop from i to n for (int l = i; l < n; l++) { // Find the maximum number // from i to l and the sum // from i to l max_ = Math.Max(max_, arr[l]); sum += arr[l]; // Find the sum of difference // of every element with the // maximum element int diff = (l - i + 1) * max_ - sum; // If the array can be divided if (j > 0) dp[i, j] = Math.Min(dp[i, j], diff + dp[l + 1, j - 1]); else dp[i, j] = diff; } } } // Returns the minimum sum // in K parts return dp[0, k]; } // Driver code public static void Main (String[] args) { int []arr = { 2, 9, 5, 4, 8, 3, 6 }; int n = arr.Length; int k = 2; Console.WriteLine(divideArray(arr, n, k)); } }// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the above approach// Function to divide an array into k// parts such that the sum of difference// of every element with the maximum element// of that part is minimumfunction divideArray(arr, n, k){ // Dp to store the values var dp = Array.from(Array(500), ()=> Array(500).fill(0)); k -= 1; // Fill up the dp table for (var i = n - 1; i >= 0; i--) { for (var j = 0; j <= k; j++) { // Intitilize maximum value dp[i][j] = 1000000000; // Max element and the sum var max_ = -1, sum = 0; // Run a loop from i to n for (var l = i; l < n; l++) { // Find the maximum number // from i to l and the sum // from i to l max_ = Math.max(max_, arr[l]); sum += arr[l]; // Find the sum of difference // of every element with the // maximum element var diff = (l - i + 1) * max_ - sum; // If the array can be divided if (j > 0) dp[i][j] = Math.min(dp[i][j], diff + dp[l + 1][j - 1]); else dp[i][j] = diff; } } } // Returns the minimum sum // in K parts return dp[0][k];}// Driver codevar arr = [2, 9, 5, 4, 8, 3, 6 ];var n = arr.length;var k = 2;document.write( divideArray(arr, n, k) + "<br>");</script> |
19
Time Complexity: O(n*n*k) , n is the size of array
Auxiliary Space: O(n*k)
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
