# Maximum houses that can be painted continuously when the painting duration and date limit is given

Given an array **arr[][]** consisting of **N** pairs such that each pair **{L, R}** represents that i^{th} house can be painted in **L** number of days before the **R ^{th}** day, the task is to find the maximum number of house that can be painted continuously.

**Examples:**

Input:N = 4, paint[ ][ ] = [[1, 19], [2, 2], [4, 17], [1, 1]]Output:3Explanation:Maximum of three houses can be painted and order is {4, 3, 1}

Input:N = 4, paint[ ][ ] = [[100, 210], [200, 1310], [1000, 1275], [2500, 3000]]Output:3

**Approach: **The problem can be solved using the greedy approach. The idea is to choose from the lowest DaysRequired within the current LastDay. Follow the steps below for the approach.

- Initialize a vector, say
**V,**to store all the pairs which can be taken. - Sort the vector
**V**by using a comparator**pair.second**less than**pair.first** - Initialize a priority queue
**pq**and push the first pair of the vector**V.** - Initialize a variable, say
**t**, to store the time. - Traverse the vector,
**V**and put the current pair in priority queue**pq**:- If
**t + DaysRequired**is less than or equal to**LastDay**, then**continue**. - else, pop from the priority queue, store in
**temp**variable and also update**t**equals**t – temp.first**

- If
- Finally, return the
**size**of the priority queue.

Below is the implementation of the above approach:

## C++14

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `typedef` `vector<vector<` `int` `> > Matrix;` `// Comparator for sorting` `static` `bool` `cmp(pair<` `int` `, ` `int` `> a, pair<` `int` `, ` `int` `> b)` `{` ` ` `return` `a.second < b.second;` `}` `// Function to count number of houses` `// that can be painted` `void` `countMaxPinted(` `int` `n, Matrix& paint)` `{` ` ` `// Vector to store the pairs` ` ` `vector<pair<` `int` `, ` `int` `> > V;` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `// If house can be painted` ` ` `if` `(paint[i][0] <= paint[i][1]) {` ` ` `V.push_back(make_pair(paint[i][0], paint[i][1]));` ` ` `}` ` ` `}` ` ` `// Sort the vector` ` ` `sort(V.begin(), V.end(), cmp);` ` ` `// If vector is empty` ` ` `if` `(V.size() == 0) {` ` ` `cout << 0 << endl;` ` ` `return` `;` ` ` `}` ` ` `// Initialize t` ` ` `int` `t = V[0].first;` ` ` `// Initialize priority queue` ` ` `priority_queue<pair<` `int` `, ` `int` `> > pq;` ` ` `pq.push(V[0]);` ` ` `// Traversing the vector` ` ` `for` `(` `int` `i = 1; i < V.size(); i++) {` ` ` `t += V[i].first;` ` ` `// Pushing in Vectors` ` ` `pq.push(V[i]);` ` ` `if` `(t <= V[i].second) {` ` ` `continue` `;` ` ` `}` ` ` `else` `{` ` ` `// Pop the top element` ` ` `auto` `temp = pq.top();` ` ` `pq.pop();` ` ` `t = t - temp.first;` ` ` `}` ` ` `}` ` ` `// Print the ans` ` ` `cout << pq.size() << endl;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given Input` ` ` `int` `n = 4;` ` ` `Matrix paint = { { 1, 19 }, { 2, 2 }, { 4, 17 }, { 1, 1 } };` ` ` `// Function Call` ` ` `countMaxPinted(n, paint);` `}` |

**Output**

3

**Time Complexity:** O(NlogN)**Auxiliary Space: **O(N)

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