The Painter's Partition Problem using Binary Search
Last Updated :
29 Apr, 2025
Given an array arr[] and k, where the array represents the boards and each element of the given array represents the length of each board. k numbers of painters are available to paint these boards. Consider that each unit of a board takes 1 unit of time to paint.
The task is to find the minimum time to get this job done by painting all the boards under the constraint that any painter will only paint the continuous sections of boards. say board [2, 3, 4] or only board [1] or nothing but not board [2, 4, 5].
Examples:
Input: arr[] = [5, 10, 30, 20, 15], k = 3
Output: 35
Explanation: The most optimal way will be:
Painter 1 allocation : [5,10]
Painter 2 allocation : [30]
Painter 3 allocation : [20, 15]
Job will be done when all painters finish i.e. at time = max(5 + 10, 30, 20 + 15) = 35
Input: arr[] = [10, 20, 30, 40], k = 2
Output: 60
Explanation: The most optimal way to paint:
Painter 1 allocation : [10, 20, 30]
Painter 2 allocation : [40]
Job will be complete at time = 60
Input: arr[] = [100, 200, 300, 400], k = 1
Output: 1000
In the Painter's Partition Problem, we've previously explored a dynamic programming based approach with a time complexity of O(k*(n^2)) and an extra space of O(k*n). In this article, we present a more optimized solution using Binary Search, significantly improving the overall efficiency.
Approach:
The idea is based on the observation that if we fix a time limit (say mid), we can check whether it's possible to divide the boards among k painters such that no one paints for more than this time. We apply Binary Search on the answer space to minimize the maximum time taken by any painter.
- The highest possible value in this range is the sum of all elements in the array - this represents the case when one painter paints all the boards.
- The lowest possible value is the maximum element of the array - this ensures at least one painter paints the largest board, and others are assigned workloads not exceeding this value.
- For each mid value, we simulate the partitioning and check if it's feasible using the given number of painters.
- As the time limit increases, the number of painters required decreases, and vice versa. This monotonic property ensures the search space is ordered, making it ideal for binary search.
C++
// C++ program to find the minimum time for
// painter's partition problem using Binary Search.
#include <bits/stdc++.h>
using namespace std;
// Function to check if it is possible to
// paint all boards by k painters such that no
// painter paints more than maxTime
bool isPossible(int maxTime, vector<int> &arr, int k) {
int painters = 1;
int currSum = 0;
for (int i = 0; i < arr.size(); i++) {
// If a single board exceeds maxTime,
// return false
if (arr[i] > maxTime) {
return false;
}
// If adding current board exceeds
// limit, assign to next painter
if (currSum + arr[i] > maxTime) {
painters++;
currSum = arr[i];
} else {
currSum += arr[i];
}
}
// Return true if total painters used is <= k
return painters <= k;
}
int minTime(vector<int> &arr, int k) {
int n = arr.size();
// Find the maximum board length manually
int maxBoard = arr[0];
for (int i = 1; i < n; i++) {
if (arr[i] > maxBoard) {
maxBoard = arr[i];
}
}
// Find the sum of all board lengths manually
int totalSum = 0;
for (int i = 0; i < n; i++) {
totalSum += arr[i];
}
// Apply Binary Search between
// maxBoard and totalSum
int low = maxBoard;
int high = totalSum;
int res = totalSum;
while (low <= high) {
int mid = (low + high) / 2;
// If it is possible with max workload as mid
if (isPossible(mid, arr, k)) {
res = mid;
// try to minimize further
high = mid - 1;
} else {
// need more time
low = mid + 1;
}
}
return res;
}
int main() {
vector<int> arr = {5, 10, 30, 20, 15};
int k = 3;
int res = minTime(arr, k);
cout << res << endl;
return 0;
}
// Java program to find the minimum time for
// painter's partition problem using Binary Search.
class GfG {
// Function to check if it is possible to
// paint all boards by k painters such that no
// painter paints more than maxTime
static boolean isPossible(int maxTime, int[] arr, int k) {
int painters = 1;
int currSum = 0;
for (int i = 0; i < arr.length; i++) {
// If a single board exceeds maxTime,
// return false
if (arr[i] > maxTime) {
return false;
}
// If adding current board exceeds
// limit, assign to next painter
if (currSum + arr[i] > maxTime) {
painters++;
currSum = arr[i];
} else {
currSum += arr[i];
}
}
// Return true if total painters used is <= k
return painters <= k;
}
static int minTime(int[] arr, int k) {
int n = arr.length;
// Find the maximum board length manually
int maxBoard = arr[0];
for (int i = 1; i < n; i++) {
if (arr[i] > maxBoard) {
maxBoard = arr[i];
}
}
// Find the sum of all board lengths manually
int totalSum = 0;
for (int i = 0; i < n; i++) {
totalSum += arr[i];
}
// Apply Binary Search between
// maxBoard and totalSum
int low = maxBoard;
int high = totalSum;
int res = totalSum;
while (low <= high) {
int mid = (low + high) / 2;
// If it is possible with max workload as mid
if (isPossible(mid, arr, k)) {
res = mid;
// try to minimize further
high = mid - 1;
} else {
// need more time
low = mid + 1;
}
}
return res;
}
public static void main(String[] args) {
int[] arr = {5, 10, 30, 20, 15};
int k = 3;
int res = minTime(arr, k);
System.out.println(res);
}
}
# Python program to find the minimum time for
# painter's partition problem using Binary Search.
def isPossible(maxTime, arr, k):
painters = 1
currSum = 0
for i in range(len(arr)):
# If a single board exceeds maxTime,
# return false
if arr[i] > maxTime:
return False
# If adding current board exceeds
# limit, assign to next painter
if currSum + arr[i] > maxTime:
painters += 1
currSum = arr[i]
else:
currSum += arr[i]
# Return true if total painters used is <= k
return painters <= k
def minTime(arr, k):
n = len(arr)
# Find the maximum board length manually
maxBoard = arr[0]
for i in range(1, n):
if arr[i] > maxBoard:
maxBoard = arr[i]
# Find the sum of all board lengths manually
totalSum = 0
for i in range(n):
totalSum += arr[i]
# Apply Binary Search between
# maxBoard and totalSum
low = maxBoard
high = totalSum
res = totalSum
while low <= high:
mid = (low + high) // 2
# If it is possible with max workload as mid
if isPossible(mid, arr, k):
res = mid
# try to minimize further
high = mid - 1
else:
# need more time
low = mid + 1
return res
if __name__ == "__main__":
arr = [5, 10, 30, 20, 15]
k = 3
res = minTime(arr, k)
print(res)
// C# program to find the minimum time for
// painter's partition problem using Binary Search.
using System;
class GfG {
// Function to check if it is possible to
// paint all boards by k painters such that no
// painter paints more than maxTime
static bool isPossible(int maxTime, int[] arr, int k) {
int painters = 1;
int currSum = 0;
for (int i = 0; i < arr.Length; i++) {
// If a single board exceeds maxTime,
// return false
if (arr[i] > maxTime) {
return false;
}
// If adding current board exceeds
// limit, assign to next painter
if (currSum + arr[i] > maxTime) {
painters++;
currSum = arr[i];
} else {
currSum += arr[i];
}
}
// Return true if total painters used is <= k
return painters <= k;
}
static int minTime(int[] arr, int k) {
int n = arr.Length;
// Find the maximum board length manually
int maxBoard = arr[0];
for (int i = 1; i < n; i++) {
if (arr[i] > maxBoard) {
maxBoard = arr[i];
}
}
// Find the sum of all board lengths manually
int totalSum = 0;
for (int i = 0; i < n; i++) {
totalSum += arr[i];
}
// Apply Binary Search between
// maxBoard and totalSum
int low = maxBoard;
int high = totalSum;
int res = totalSum;
while (low <= high) {
int mid = (low + high) / 2;
// If it is possible with max workload as mid
if (isPossible(mid, arr, k)) {
res = mid;
// try to minimize further
high = mid - 1;
} else {
// need more time
low = mid + 1;
}
}
return res;
}
static void Main() {
int[] arr = {5, 10, 30, 20, 15};
int k = 3;
int res = minTime(arr, k);
Console.WriteLine(res);
}
}
// JavaScript program to find the minimum time for
// painter's partition problem using Binary Search.
function isPossible(maxTime, arr, k) {
let painters = 1;
let currSum = 0;
for (let i = 0; i < arr.length; i++) {
// If a single board exceeds maxTime,
// return false
if (arr[i] > maxTime) {
return false;
}
// If adding current board exceeds
// limit, assign to next painter
if (currSum + arr[i] > maxTime) {
painters++;
currSum = arr[i];
} else {
currSum += arr[i];
}
}
// Return true if total painters used is <= k
return painters <= k;
}
function minTime(arr, k) {
let n = arr.length;
// Find the maximum board length manually
let maxBoard = arr[0];
for (let i = 1; i < n; i++) {
if (arr[i] > maxBoard) {
maxBoard = arr[i];
}
}
// Find the sum of all board lengths manually
let totalSum = 0;
for (let i = 0; i < n; i++) {
totalSum += arr[i];
}
// Apply Binary Search between
// maxBoard and totalSum
let low = maxBoard;
let high = totalSum;
let res = totalSum;
while (low <= high) {
let mid = Math.floor((low + high) / 2);
// If it is possible with max workload as mid
if (isPossible(mid, arr, k)) {
res = mid;
// try to minimize further
high = mid - 1;
} else {
// need more time
low = mid + 1;
}
}
return res;
}
// Driver Code
let arr = [5, 10, 30, 20, 15];
let k = 3;
let res = minTime(arr, k);
console.log(res);
Time Complexity: O(n*log(sum(arr) - MAX)), as we perform Binary Search on the answer space from maximum board length to total sum, and for each mid value, we iterate over all n boards to check feasibility.
Space Complexity: O(1), as the approach uses constant space.
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