Painters Partition Problem

Lecture 63 :- Painters Partition Problem

The Painter's Partition Problem is another classic optimization problem related to book allocation. In this problem, there are N boards arranged in a row, each having some length. You have to paint these boards using M painters. Each painter takes 1 unit of time to paint 1 unit of board length. The goal is to minimize the time it takes to paint all the boards.

To solve this problem, we can use binary search to find the minimum time required to paint all the boards. We will set the search space to be the range from the maximum length of the boards to the sum of all board lengths. Then, we will check if it is possible to complete the painting using M painters in the mid time value (the value we are currently searching for). If it is possible, we try to minimize this time by moving to the left half of the search space. Otherwise, we move to the right half.

Here's the C++ implementation of the Painter's Partition Problem:

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#include <iostream> #include <vector> #include <algorithm> bool isPossible(std::vector<int>& boards, int numPainters, int maxTime) {    int painters = 1;    int timeTaken = 0;    for (int i = 0; i < boards.size(); i++) {        if (boards[i] > maxTime) {            return false;        }        if (timeTaken + boards[i] > maxTime) {            painters++;            timeTaken = boards[i];        } else {            timeTaken += boards[i];        }    }    return painters <= numPainters; } int partitionBoards(std::vector<int>& boards, int numPainters) {    int n = boards.size();    int totalLength = 0;    int maxLength = 0;    for (int i = 0; i < n; i++) {        totalLength += boards[i];        maxLength = std::max(maxLength, boards[i]);    }    int left = maxLength;    int right = totalLength;    int result = -1;    while (left <= right) {        int mid = left + (right - left) / 2;        if (isPossible(boards, numPainters, mid)) {            result = mid;            right = mid - 1;        } else {            left = mid + 1;        }    }    return result; } int main() {    std::vector<int> boards = {10, 20, 30, 40};    int numPainters = 2;    int minTime = partitionBoards(boards, numPainters);    std::cout << "Minimum time required to paint all boards: " << minTime << std::endl;    return 0; }

Example Output:

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Minimum time required to paint all boards: 60

In this code, the isPossible function checks if it is possible to complete the painting using M painters in the maxTime value. The partitionBoards function uses binary search to find the minimum time required to paint all the boards. The time complexity of this algorithm is O(N * log(sum)), where N is the number of boards and sum is the sum of all board lengths.