Given are N boards of with length of each given in the form of array, and K painters, such that each painter takes 1 unit of time to paint 1 unit of the board. The task is to find the minimum time to paint all boards 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: N = 4, A = {10, 10, 10, 10}, K = 2
Output : 20
Explanation: Here we can divide the boards into 2 equal sized partitions (Painter 1 will paint boards A1 and A2, and Painter 2 will paint boards A3 and A4). So each painter gets 20 units of board and the total time taken is 20.
Input: N = 4, A = {10, 20, 30, 40}, K = 2
Output : 60
Explanation: Since there are only 2 painters, therefore divide first 3 boards to painter 1 (A1, A2 and A3) with time = 60, and last board to painter 2 (A4) with time = 40. Therefore total time taken = 60, which is the minimum possible.
Please note the combination A1 and A4 to Painter 1 with time 50, and A2 and A3 to Painter 2 with time 50, will yield a smaller time (50 units). But this cant be considered due to the constraint that a painter cannot paint non-continuos series of boards.
In the Painter’s Programming, we have discussed a dynamic programming based approach having time complexity of O(k∗N2) and O(k∗N) extra space. In this post we will look into a more efficient approach using binary search.
Efficient Approach for Painter’s Problem using Binary Search
The idea to apply Binary Search Approach to Painter’s Problem is based on following observations:
We know that the invariant of binary search has three main parts:
- the target value would always be in the searching range.
- the searching range will decrease in each loop so that the termination can be reached.
- We also know that the values in this range must be in sorted order.
Here our target value is the maximum sum of a contiguous section in the optimal allocation of boards.
Now how can we apply binary search for this?
We can fix the possible low to high range for the target value and narrow down our search to get the optimal allocation.
- We can see that the highest possible value in this range is the sum of all the elements in the array and this happens when we allot 1 painter all the sections of the board.
- The lowest possible value of this range is the maximum value of the array max, as in this allocation we can allot max to one painter and divide the other sections such that the cost of them is less than or equal to max and as close as possible to max.
- Now if we use x painters in the above scenarios, it is obvious that as the value in the range increases, the value of x decreases and vice-versa.
From this we can find the target value when x=k and use a helper function to find x, the minimum number of painters required when the maximum length of section a painter can paint is given.
C++
// CPP program for painter's partition problem#include <climits>#include <iostream>using namespace std;// return the maximum element from the arrayint getMax(int arr[], int n){ int max = INT_MIN; for (int i = 0; i < n; i++) if (arr[i] > max) max = arr[i]; return max;}// return the sum of the elements in the arrayint getSum(int arr[], int n){ int total = 0; for (int i = 0; i < n; i++) total += arr[i]; return total;}// find minimum required painters for given maxlen// which is the maximum length a painter can paintint numberOfPainters(int arr[], int n, int maxLen){ int total = 0, numPainters = 1; for (int i = 0; i < n; i++) { total += arr[i]; if (total > maxLen) { // for next count total = arr[i]; numPainters++; } } return numPainters;}int partition(int arr[], int n, int k){ int lo = getMax(arr, n); int hi = getSum(arr, n); while (lo < hi) { int mid = lo + (hi - lo) / 2; int requiredPainters = numberOfPainters(arr, n, mid); // find better optimum in lower half // here mid is included because we // may not get anything better if (requiredPainters <= k) hi = mid; // find better optimum in upper half // here mid is excluded because it gives // required Painters > k, which is invalid else lo = mid + 1; } // required return lo;}// driver functionint main(){ int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 3; cout << partition(arr, n, k) << endl; return 0;} |
Java
// Java Program for painter's partition problemimport java.io.*;import java.util.*;class GFG { // return the maximum element from the array static int getMax(int arr[], int n) { int max = Integer.MIN_VALUE; for (int i = 0; i < n; i++) if (arr[i] > max) max = arr[i]; return max; } // return the sum of the elements in the array static int getSum(int arr[], int n) { int total = 0; for (int i = 0; i < n; i++) total += arr[i]; return total; } // find minimum required painters for given maxlen // which is the maximum length a painter can paint static int numberOfPainters(int arr[], int n, int maxLen) { int total = 0, numPainters = 1; for (int i = 0; i < n; i++) { total += arr[i]; if (total > maxLen) { // for next count total = arr[i]; numPainters++; } } return numPainters; } static int partition(int arr[], int n, int k) { int lo = getMax(arr, n); int hi = getSum(arr, n); while (lo < hi) { int mid = lo + (hi - lo) / 2; int requiredPainters = numberOfPainters(arr, n, mid); // find better optimum in lower half // here mid is included because we // may not get anything better if (requiredPainters <= k) hi = mid; // find better optimum in upper half // here mid is excluded because it gives // required Painters > k, which is invalid else lo = mid + 1; } // required return lo; } // Driver code public static void main(String args[]) { int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; // 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 |
Python3
# Python program for painter's partition problem# Find minimum required painters for given maxlen# which is the maximum length a painter can paintdef numberOfPainters(arr, n, maxLen): total = 0 numPainters = 1 for i in arr: total += i if (total > maxLen): # for next count total = i numPainters += 1 return numPaintersdef partition(arr, n, k): lo = max(arr) hi = sum(arr) while (lo < hi): mid = lo + (hi - lo) // 2 requiredPainters = numberOfPainters(arr, n, mid) # find better optimum in lower half # here mid is included because we # may not get anything better if (requiredPainters <= k): hi = mid # find better optimum in upper half # here mid is excluded because it gives # required Painters > k, which is invalid else: lo = mid + 1 # required return lo# Driver codearr = [1, 2, 3, 4, 5, 6, 7, 8, 9]n = len(arr)k = 3print(int(partition(arr, n, k))) |
C#
// C# Program for painter's// partition problemusing System;class GFG { // return the maximum // element from the array static int getMax(int[] arr, int n) { int max = int.MinValue; for (int i = 0; i < n; i++) if (arr[i] > max) max = arr[i]; return max; } // return the sum of the // elements in the array static int getSum(int[] arr, int n) { int total = 0; for (int i = 0; i < n; i++) total += arr[i]; return total; } // find minimum required // painters for given // maxlen which is the // maximum length a painter // can paint static int numberOfPainters(int[] arr, int n, int maxLen) { int total = 0, numPainters = 1; for (int i = 0; i < n; i++) { total += arr[i]; if (total > maxLen) { // for next count total = arr[i]; numPainters++; } } return numPainters; } static int partition(int[] arr, int n, int k) { int lo = getMax(arr, n); int hi = getSum(arr, n); while (lo < hi) { int mid = lo + (hi - lo) / 2; int requiredPainters = numberOfPainters(arr, n, mid); // find better optimum in lower // half here mid is included // because we may not get // anything better if (requiredPainters <= k) hi = mid; // find better optimum in upper // half here mid is excluded // because it gives required // Painters > k, which is invalid else lo = mid + 1; } // required return lo; } // Driver code static public void Main() { int[] arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; // Calculate size of array. int n = arr.Length; int k = 3; Console.WriteLine(partition(arr, n, k)); }}// This code is contributed by ajit |
PHP
<?php// PHP program for painter's // partition problem// return the maximum // element from the arrayfunction getMax($arr, $n){ $max = PHP_INT_MIN; for ($i = 0; $i < $n; $i++) if ($arr[$i] > $max) $max = $arr[$i]; return $max;}// return the sum of the// elements in the arrayfunction getSum($arr, $n){ $total = 0; for ($i = 0; $i < $n; $i++) $total += $arr[$i]; return $total;}// find minimum required painters // for given maxlen which is the// maximum length a painter can paintfunction numberOfPainters($arr, $n, $maxLen){ $total = 0; $numPainters = 1; for ($i = 0; $i < $n; $i++) { $total += $arr[$i]; if ($total > $maxLen) { // for next count $total = $arr[$i]; $numPainters++; } } return $numPainters;}function partition($arr, $n, $k){ $lo = getMax($arr, $n); $hi = getSum($arr, $n); while ($lo < $hi) { $mid = $lo + ($hi - $lo) / 2; $requiredPainters = numberOfPainters($arr, $n, $mid); // find better optimum in // lower half here mid is // included because we may // not get anything better if ($requiredPainters <= $k) $hi = $mid; // find better optimum in // upper half here mid is // excluded because it // gives required Painters > k, // which is invalid else $lo = $mid + 1; } // required return floor($lo);}// Driver Code$arr = array(1, 2, 3, 4, 5, 6, 7, 8, 9);$n = sizeof($arr);$k = 3;echo partition($arr, $n, $k), "\n";// This code is contributed by ajit?> |
Javascript
<script>// Javascript Program for painter's partition problem // Return the maximum element from the arrayfunction getMax(arr, n){ let max = Number.MIN_VALUE; for(let i = 0; i < n; i++) if (arr[i] > max) max = arr[i]; return max;}// Return the sum of the elements in the arrayfunction getSum(arr, n){ let total = 0; for(let i = 0; i < n; i++) total += arr[i]; return total;}// Find minimum required paleters for given maxlen// which is the maximum length a paleter can paletfunction numberOfPaleters(arr, n, maxLen){ let total = 0, numPaleters = 1; for(let i = 0; i < n; i++) { total += arr[i]; if (total > maxLen) { // For next count total = arr[i]; numPaleters++; } } return numPaleters;}function partition(arr, n, k){ let lo = getMax(arr, n); let hi = getSum(arr, n); while (lo < hi) { let mid = lo + (hi - lo) / 2; let requiredPaleters = numberOfPaleters( arr, n, mid); // Find better optimum in lower half // here mid is included because we // may not get anything better if (requiredPaleters <= k) hi = mid; // find better optimum in upper half // here mid is excluded because it gives // required Paleters > k, which is invalid else lo = mid + 1; } // Required return lo;}// Driver codelet arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ];// Calculate size of array.let n = arr.length;let k = 3;document.write(Math.round(partition(arr, n, k)));// This code is contributed by sanjoy_62</script> |
Output
17
Time Complexity: O(N*log(sum(arr[])))
Auxiliary Space: O(1)
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