The painter’s partition problem

We have to paint n boards of length {A1, A2…An}. There are k painters available and each takes 1 unit time to paint 1 unit of board. The problem is to find the minimum time to get
this job done under the constraints that any painter will only paint continuous sections of boards, say board {2, 3, 4} or only board {1} or nothing but not board {2, 4, 5}.

Examples:

Input : k = 2, A = {10, 10, 10, 10} 
Output : 20.
Here we can divide the boards into 2
equal sized partitions, so each painter 
gets 20 units of board and the total
time taken is 20. 

Input : k = 2, A = {10, 20, 30, 40} 
Output : 60.
Here we can divide first 3 boards for
one painter and the last board for 
second painter.



From the above examples, it is obvious that the strategy of dividing the boards into k equal partitions won’t work for all the cases. We can observe that the problem can be broken down into: Given an array A of non-negative integers and a positive integer k, we have to divide A into k of fewer partitions such that the maximum sum of the elements in a partition, overall partitions is minimized. So for the second example above, possible divisions are:
* One partition: so time is 100.
* Two partitions: (10) & (20, 30, 40), so time is 90. Similarly we can put the first divider
after 20 (=> time 70) or 30 (=> time 60); so this means the minimum time: (100, 90, 70, 60) is 60.

A brute force solution is to consider all possible set of contiguous partitions and calculate the maximum sum partition in each case and return the minimum of all these cases.

1) Optimal Substructure:
We can implement the naive solution using recursion with the following optimal substructure property:
Assuming that we already have k-1 partitions in place (using k-2 dividers), we now have to put the k-1 th divider to get k partitions.
How can we do this? We can put the k-1 th divider between the i th and i+1 th element where i = 1 to n. Please note that putting it before the first element is the same as putting it after the last element.

The total cost of this arrangement can be calculated as the maximum of the following:
a) The cost of the last partition: sum(Ai..An), where the k-1 th divider is
before element i.
b) The maximum cost of any partition already formed to the left of the k-1 th divider.



Here a) can be found out using a simple helper function to calculate sum
of elements between two indices in the array. How to find out b) ?
We can observe that b) actually is to place the k-2 separators as fairly as
possible, so it is a subproblem of the given problem. Thus we can write the optimal
substructure property as the following recurrence relation:

painter-partition

Following is the implementation of the above recursive equation:

C++

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// CPP program for The painter's partition problem
#include <climits>
#include <iostream>
using namespace std;
  
// function to calculate sum between two indices 
// in array
int sum(int arr[], int from, int to)
{
    int total = 0;
    for (int i = from; i <= to; i++)
        total += arr[i];
    return total;
}
  
// for n boards and k partitions
int partition(int arr[], int n, int k)
{
    // base cases    
    if (k == 1) // one partition
        return sum(arr, 0, n - 1);    
    if (n == 1)  // one board
        return arr[0];
  
    int best = INT_MAX;
  
    // find minimum of all possible maximum
    // k-1 partitions to the left of arr[i],
    // with i elements, put k-1 th divider 
    // between arr[i-1] & arr[i] to get k-th 
    // partition
    for (int i = 1; i <= n; i++)
        best = min(best, max(partition(arr, i, k - 1), 
                                sum(arr, i, n - 1)));
  
    return best;
}
  
int main()
{
    int arr[] = { 10, 20, 60, 50, 30, 40 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    cout << partition(arr, n, k) << endl;
  
    return 0;
}

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Java

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// Java Program for The painter's partition problem
import java.util.*;
import java.io.*;
  
class GFG
{
// function to calculate sum between two indices 
// in array
static int sum(int arr[], int from, int to)
{
    int total = 0;
    for (int i = from; i <= to; i++)
        total += arr[i];
    return total;
}
   
// for n boards and k partitions
static int partition(int arr[], int n, int k)
{
    // base cases    
    if (k == 1) // one partition
        return sum(arr, 0, n - 1);    
    if (n == 1// one board
        return arr[0];
   
    int best = Integer.MAX_VALUE;
   
    // find minimum of all possible maximum
    // k-1 partitions to the left of arr[i],
    // with i elements, put k-1 th divider 
    // between arr[i-1] & arr[i] to get k-th 
    // partition
    for (int i = 1; i <= n; i++)
        best = Math.min(best, Math.max(partition(arr, i, k - 1), 
                                sum(arr, i, n - 1)));
   
    return best;
}
  
// Driver code
public static void main(String args[])
{
 int arr[] = { 10, 20, 60, 50, 30, 40 };
   
    // Calculate size of array.
    int n = arr.length;
        int k = 3;
 System.out.println(partition(arr, n, k));
}
}
  
// This code is contributed by Sahil_Bansall

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Python3

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# Python program for The painter's 
# partition problem function to 
# calculate sum between two indices 
# in array
def sum(arr, frm, to): 
    total = 0;
    for i in range(frm, to + 1):
        total += arr[i]
    return total
      
# for n boards and k partitions 
def partition(arr, n, k):
      
    # base cases
    if k == 1: # one partition
        return sum(arr, 0, n - 1)
    if n == 1: # one board
        return arr[0]
    best = 100000000
      
    # find minimum of all possible  
    # maximum k-1 partitions to  
    # the left of arr[i], with i 
    # elements, put k-1 th divider 
    # between arr[i-1] & arr[i] to 
    # get k-th partition 
    for i in range(1, n + 1):
        best = min(best, 
               max(partition(arr, i, k - 1), 
                         sum(arr, i, n - 1)))
    return best
      
# Driver Code
arr = [10, 20, 60, 50, 30, 40 ]
n = len(arr)
k = 3
print(partition(arr, n, k))
  
# This code is contributed
# by sahilshelangia

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C#

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// C# Program for The painter's partition problem
using System;
  
class GFG {
      
// function to calculate sum 
// between two indices in array
static int sum(int []arr, int from, int to)
{
    int total = 0;
    for (int i = from; i <= to; i++)
        total += arr[i];
    return total;
}
  
// for n boards and k partitions
static int partition(int []arr, int n, int k)
{
    // base cases 
    if (k == 1) // one partition
        return sum(arr, 0, n - 1); 
          
    if (n == 1) // one board
        return arr[0];
  
    int best = int.MaxValue;
  
    // find minimum of all possible maximum
    // k-1 partitions to the left of arr[i],
    // with i elements, put k-1 th divider 
    // between arr[i-1] & arr[i] to get k-th 
    // partition
    for (int i = 1; i <= n; i++)
        best = Math.Min(best, Math.Max(partition(arr, i, k - 1), 
                                           sum(arr, i, n - 1)));
  
    return best;
}
  
// Driver code
public static void Main()
{
    int []arr = {10, 20, 60, 50, 30, 40};
  
    // Calculate size of array.
    int n = arr.Length;
    int k = 3;
      
    // Function calling
    Console.WriteLine(partition(arr, n, k));
}
}
  
// This code is contributed by vt_m

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PHP

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<?php
// PHP program for The 
// painter's partition problem
  
// function to calculate sum 
// between two indices in array
function sum($arr, $from, $to)
{
    $total = 0;
    for ($i = $from; $i <= $to; $i++)
        $total += $arr[$i];
    return $total;
}
  
// for n boards 
// and k partitions
function partition($arr, $n, $k)
{
    // base cases 
    if ($k == 1) // one partition
        return sum($arr, 0, $n - 1); 
    if ($n == 1) // one board
        return $arr[0];
  
    $best = PHP_INT_MAX;
  
    // find minimum of all possible 
    // maximum k-1 partitions to the 
    // left of arr[i], with i elements,
    // put k-1 th divider between 
    // arr[i-1] & arr[i] to get k-th 
    // partition
    for ($i = 1; $i <= $n; $i++)
        $best = min($best
                max(partition($arr, $i, $k - 1), 
                          sum($arr, $i, $n - 1)));
  
    return $best;
}
// Driver Code
$arr = array(10, 20, 60, 
             50, 30, 40);
$n = sizeof($arr);
$k = 3;
echo partition($arr, $n, $k), "\n";
  
// This code is contributed by ajit
?>

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Output :

90

The time complexity of the above solution is exponential.

2) Overlapping subproblems:
Following is the partial recursion tree for T(4, 3) in above equation.

      T(4, 3)
     /    /    \ ..         
T(1, 2)  T(2, 2) T(3, 2) 
          /..      /..     
      T(1, 1)    T(1, 1) 

We can observe that many subproblems like T(1, 1) in the above problem are being solved again and again. Because of these two properties of this problem, we can solve it using dynamic programming, either by top down memoized method or bottom up
tabular method. Following is the bottom up tabular implementation:

C++

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// A DP based CPP program for painter's partition problem
#include <climits>
#include <iostream>
using namespace std;
  
// function to calculate sum between two indices
// in array
int sum(int arr[], int from, int to)
{
    int total = 0;
    for (int i = from; i <= to; i++)
        total += arr[i];
    return total;
}
  
// bottom up tabular dp
int findMax(int arr[], int n, int k)
{
    // initialize table
    int dp[k + 1][n + 1] = { 0 };
  
    // base cases
    // k=1
    for (int i = 1; i <= n; i++)
        dp[1][i] = sum(arr, 0, i - 1);
  
    // n=1
    for (int i = 1; i <= k; i++)
        dp[i][1] = arr[0];
  
    // 2 to k partitions
    for (int i = 2; i <= k; i++) { // 2 to n boards
        for (int j = 2; j <= n; j++) {
  
            // track minimum
            int best = INT_MAX;
  
            // i-1 th separator before position arr[p=1..j]
            for (int p = 1; p <= j; p++) 
                best = min(best, max(dp[i - 1][p],
                              sum(arr, p, j - 1)));       
  
            dp[i][j] = best;
        }
    }
  
    // required
    return dp[k][n];
}
  
// driver function
int main()
{
    int arr[] = { 10, 20, 60, 50, 30, 40 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    cout << findMax(arr, n, k) << endl;
    return 0;
}

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Java

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// A DP based Java program for
// painter's partition problem
import java.util.*;
import java.io.*;
  
class GFG
{
// function to calculate sum between two indices
// in array
static int sum(int arr[], int from, int to)
{
    int total = 0;
    for (int i = from; i <= to; i++)
        total += arr[i];
    return total;
}
   
// bottom up tabular dp
static int findMax(int arr[], int n, int k)
{
    // initialize table
    int dp[][] = new int[k+1][n+1];
   
    // base cases
    // k=1
    for (int i = 1; i <= n; i++)
        dp[1][i] = sum(arr, 0, i - 1);
   
    // n=1
    for (int i = 1; i <= k; i++)
        dp[i][1] = arr[0];
   
    // 2 to k partitions
    for (int i = 2; i <= k; i++) { // 2 to n boards
        for (int j = 2; j <= n; j++) {
   
            // track minimum
            int best = Integer.MAX_VALUE;
   
            // i-1 th separator before position arr[p=1..j]
            for (int p = 1; p <= j; p++) 
                best = Math.min(best, Math.max(dp[i - 1][p],
                              sum(arr, p, j - 1)));       
   
            dp[i][j] = best;
        }
    }
   
    // required
    return dp[k][n];
}
  
// Driver code
public static void main(String args[])
{
 int arr[] = { 10, 20, 60, 50, 30, 40 };
   
    // Calculate size of array.
    int n = arr.length;
        int k = 3;
 System.out.println(findMax(arr, n, k));
}
}
  
// This code is contributed by Sahil_Bansall

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C#

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// A DP based C# program for
// painter's partition problem
using System;
  
class GFG {
      
// function to calculate sum between 
// two indices in array
static int sum(int []arr, int from, int to)
{
    int total = 0;
    for (int i = from; i <= to; i++)
        total += arr[i];
    return total;
}
  
// bottom up tabular dp
static int findMax(int []arr, int n, int k)
{
    // initialize table
    int [,]dp = new int[k+1,n+1];
  
    // base cases
    // k=1
    for (int i = 1; i <= n; i++)
        dp[1,i] = sum(arr, 0, i - 1);
  
    // n=1
    for (int i = 1; i <= k; i++)
        dp[i,1] = arr[0];
  
    // 2 to k partitions
    for (int i = 2; i <= k; i++) { // 2 to n boards
        for (int j = 2; j <= n; j++) {
  
            // track minimum
            int best = int.MaxValue;
  
            // i-1 th separator before position arr[p=1..j]
            for (int p = 1; p <= j; p++) 
                best = Math.Min(best, Math.Max(dp[i - 1,p],
                                      sum(arr, p, j - 1))); 
  
            dp[i,j] = best;
        }
    }
  
    // required
    return dp[k,n];
}
  
// Driver code
public static void Main()
{
    int []arr = {10, 20, 60, 50, 30, 40};
  
    // Calculate size of array.
    int n = arr.Length;
    int k = 3;
    Console.WriteLine(findMax(arr, n, k));
}
}
  
// This code is contributed by vt_m

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PHP

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<?php
// A DP based PHP program for
// painter's partition problem
  
// function to calculate sum 
// between two indices in array
function sum($arr, $from, $to)
{
    $total = 0;
    for ($i = $from; $i <= $to; $i++)
        $total += $arr[$i];
    return $total;
}
  
// bottom up tabular dp
function findMax($arr, $n, $k)
{
    // initialize table
    $dp[$k + 1][$n + 1] = array( 0 );
  
    // base cases
    // k=1
    for ($i = 1; $i <= $n; $i++)
        $dp[1][$i] = sum($arr, 0, 
                         $i - 1);
  
    // n=1
    for ($i = 1; $i <= $k; $i++)
        $dp[$i][1] = $arr[0];
  
    // 2 to k partitions
    for ($i = 2; $i <= $k; $i++) 
    
        // 2 to n boards
        for ($j = 2; $j <= $n; $j++) 
        {
  
            // track minimum
            $best = PHP_INT_MAX;
  
            // i-1 th separator before
            // position arr[p=1..j]
            for ($p = 1; $p <= $j; $p++) 
                $best = min($best, max($dp[$i - 1][$p],
                               sum($arr, $p, $j - 1)));     
  
            $dp[$i][$j] = $best;
        }
    }
  
    // required
    return $dp[$k][$n];
}
  
// Driver Code
$arr = array (10, 20, 60, 
              50, 30, 40 );
$n = sizeof($arr);
$k = 3;
echo findMax($arr, $n, $k) ,"\n";
  
// This code is contribted by m_kit
?>

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Output:



90

Optimizations:
1) The time complexity of the above program is O(k*N^3). It can be easily brought down to O(k*N^2) by precomputing the cumulative sums in an array thus avoiding repeated calls to the sum function:

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 int sum[n+1] = {0};
  
 // sum from 1 to i elements of arr
 for (int i = 1; i <= n; i++)
   sum[i] = sum[i-1] + arr[i-1];
  
 for (int i = 1; i <= n; i++)
   dp[1][i] = sum[i];
 
and using it to calculate the result as:
best = min(best, max(dp[i-1][p], sum[j] - sum[p]));

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2) Though here we consider to divide A into k or fewer partitions, we can observe that
the optimal case always occurs when we divide A into exactly k partitions. So we can use:

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for (int i = k-1; i <= n; i++)
    best = min(best, max( partition(arr, i, k-1),
                            sum(arr, i, n-1)));

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and modify the other implementations accordingly.

Exercise:
1) Can you come up with a solution using binary search which runs in O(N lg N) time? Prerequisite for this: https://www.topcoder.com/community/data-science/data-science-tutorials/binary-search/

References:
https://articles.leetcode.com/the-painters-partition-problem/

Asked in: Google, CodeNation



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Improved By : vt_m, jit_t, sahilshelangia