Rectangle Area II in C++

Suppose we have a list of (axis-aligned) rectangles. Here each rectangle[i] = {x1, y1, x2, y2}, where (x1, y1) is the point of the bottom-left corner, and (x2, y2) are the point of the top-right corner of the ith rectangle.

We have to find the total area covered by all rectangles in the plane. The answer may be very , so we can use modulo 10^9 + 7.

So, if the input is like

then the output will be 6.

To solve this, we will follow these steps −

• m = 10^9 + 7

• Define a function add(), this will take a, b,

• return ((a mod m) + (b mod m) mod m)

• Define one function compress this will take 2d matrix v

• Define an array temp

• for initialize i := 0, when i < size of v, update (increase i by 1), do −

  • insert v[i, 0] at the end of temp

  • insert v[i, 2] at the end of temp

• sort the array temp

• Define one map, ret

• idx := 0

• for initialize i := 0, when i < size of temp, update (increase i by 1), do

  • if temp[i] is not member of ret, then −

    • ret[temp[i]] := idx

    • (increase idx by 1)

• return ret

• From the main method do the following −

• Define an array xv

• insert { 0 } at the end of xv

• for initialize i := 0, when i < size of v, update (increase i by 1), do −

  • insert v[i, 0] at the end of xv

  • insert v[i, 2] at the end of xv

• sort the array xv

• uniItr = first element of a list with unique elements of xv

• delete uniItr, from xv

• Define one map index

• idx := 0

• for initialize i := 0, when i < size of xv, update (increase i by 1), do −

  • index[xv[i]] := i

• Define an array count of size same as index size

• Define one 2D array x

• for initialize i := 0, when i < size of v, update (increase i by 1), do −

  • x1 := v[i, 0], y1 := v[i, 1]

  • x2 := v[i, 2], y2 := v[i, 3]

  • insert { y1, x1, x2, 1 } at the end of x

  • insert { y2, x1, x2, -1 } at the end of x

• sort the array x

• ret := 0

• sum := 0, currentY := 0

• for initialize i := 0, when i < size of x, update (increase i by 1), do −

  • y := x[i, 0]

  • x1 := x[i, 1], x2 := x[i, 2]

  • sig := x[i, 3]

  • ret := add(ret, (y - currentY) * sum)

  • currentY := y

  • for initialize i := index[x1], when i < index[x2], update (increase i by 1), do −

    • count[i] := count[i] + sig

  • sum := 0

  • for initialize i := 0, when i < size of count, update (increase i by 1), do −

    • if count[i] > 0, then

      • sum := sum + (xv[i + 1] - xv[i])

• return ret mod m

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h> using namespace std; typedef long long int lli; const int m = 1e9 + 7; class Solution {    public:    lli add(lli a, lli b){       return ((a % m) + (b % m) % m);    }    map<int, int> compress(vector<vector<int> >& v){       vector<int> temp;       for (int i = 0; i < v.size(); i++) {          temp.push_back(v[i][0]);          temp.push_back(v[i][2]);       }       sort(temp.begin(), temp.end());       map<int, int> ret;       int idx = 0;       for (int i = 0; i < temp.size(); i++) {          if (!ret.count(temp[i])) {             ret[temp[i]] = idx;             idx++;          }       }       return ret;    }    int rectangleArea(vector<vector<int> >& v){       vector<int> xv;       xv.push_back({ 0 });       for (int i = 0; i < v.size(); i++) {          xv.push_back(v[i][0]);          xv.push_back(v[i][2]);       }       sort(xv.begin(), xv.end());       vector<int>::iterator uniItr = unique(xv.begin(), xv.end());       xv.erase(uniItr, xv.end());       map<int, int> index;       int idx = 0;       for (int i = 0; i < xv.size(); i++) {          index[xv[i]] = i;       }       vector<int> count(index.size());       vector<vector<int> > x;       int x1, x2, y1, y2;       for (int i = 0; i < v.size(); i++) {          x1 = v[i][0];          y1 = v[i][1];          x2 = v[i][2];          y2 = v[i][3];          x.push_back({ y1, x1, x2, 1 });          x.push_back({ y2, x1, x2, -1 });       }       sort(x.begin(), x.end());       lli ret = 0;       lli sum = 0, currentY = 0;       for (int i = 0; i < x.size(); i++) {          lli y = x[i][0];          x1 = x[i][1];          x2 = x[i][2];          int sig = x[i][3];          ret = add(ret, (y - currentY) * sum);          currentY = y;          for (int i = index[x1]; i < index[x2]; i++) {             count[i] += sig;          }          sum = 0;          for (int i = 0; i < count.size(); i++) {             if (count[i] > 0) {                sum += (xv[i + 1] - xv[i]);             }          }       }       return ret % m;    } }; main(){    Solution ob;    vector<vector<int>> v = {{0,0,2,2},{1,0,2,3},{1,0,3,1}};    cout << (ob.rectangleArea(v)); }

Input

{{0,0,2,2},{1,0,2,3},{1,0,3,1}}

Output

6