# Find number of Polygons lying inside each given Polygons on Cartesian Plane

Given N non-intersecting nested polygons, and a 2D array arr[][] of pairs, where each cell of the array represents the coordinates of the vertices of a polygon. The task is to find the count of polygons lying inside each polygon.

Examples:

Input: N = 3, arr[][][] = {{{-2, 2}, {-1, 1}, {2, 2}, {2, -1}, {1, -2}, {-2, -2}}, {{-1, -1}, {1, -1}, {1, 1}}, {{3, 3}, {-3, 3}, {-3, -3}, {3, -3}}}
Output: 1 0 2
Explanation:

1. The polygon represented at index 0, is the green-colored polygon shown in the picture above.
2. The polygon represented at index 1, is the blue-colored polygon shown in the picture above.
3. The polygon represented at index 2, is the yellow-colored polygon shown in the picture above.

Therefore, there is 1 polygon inside the polygon at index 0, and 0 polygons inside the polygon at index 1, and 2 polygons inside the polygon at index 2.

Approach: The above problem can be solved based on the following observations:

1. It can be observed that the polygon lying outside will have the largest X- coordinates. The maximum X-coordinates of the polygon will decrease, as one will go towards the inside, as it is given that polygons are nested and non-intersecting.
2. Therefore, sorting the polygons by the maximum X-coordinate will give the correct order of the polygons lying.

Follow the steps below to solve the problem:

• Initialize a 2D array say maximumX[][] of size N*2 to store the pair of maximum X-coordinates of a polygon and the index value of the polygon.
• Iterate over the array arr[][] using the variable, i perform the following steps:
• Initialize a variable, say maxX, to store the maximum X-coordinate of the current polygon.
• Iterate over the array arr[i] using the variable j and in each iteration update the maxX as maxX = max(maxX, arr[i][j][0]).
• Assign the pair {maxX, i} to maximumX[i].
• Sort the array of pairs in descending order by the first element using a custom comparator.
• Initialize an array, say res[], to store the count of polygons lying inside the current polygon.
• Iterate over the range [0, N-1] using the variable i and in each iteration assign N-i-1 to res[maximumX[i][1]], as there are N-i-1 polygons lying inside the maximumX[i][1]th polygon.
• Finally, after completing the above steps, print the array res[] as the answer.

Below is the implementation of the above approach:

## C++

 `// C++ program for above approach` `#include` `using` `namespace` `std;`   `// Function to calculate maximum sum` `void` `findAllPolygons(``int` `N,` `                     ``vector > > arr)` `{` `  `  `    ``// Stores the maximum X co-` `    ``// ordinate of every polygon` `    ``vector > maximumX(N);`   `    ``// Traverse the array arr[][]` `    ``for` `(``int` `i = 0; i < N; i++) {`   `        ``// Stores the max X co-` `        ``// ordinate of current` `        ``// polygon` `        ``int` `maxX = INT_MIN;`   `        ``// Traverse over the vertices` `        ``// of the current polygon`   `        ``for` `(``int` `j = 0; j < arr[i].size(); j++) {` `            ``// Update maxX` `            ``maxX = max(maxX, arr[i][j][0]);` `        ``}`   `        ``// Assign maxX to maximumX[i][0]` `        ``maximumX[i].first = maxX;`   `        ``// Assign i to maximumX[i][1]` `        ``maximumX[i].second = i;` `    ``}`   `    ``// Sort the array of pairs` `    ``// maximumX in descending` `    ``// order by first element` `    ``sort(maximumX.rbegin(), maximumX.rend());`   `    ``// Stores the count of` `    ``// polygons lying inside` `    ``// a polygon` `    ``vector<``int``> res(N, 0);`   `    ``// Traverse the maximumX array` `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// Update value at` `        ``// maximumX[i][1] in` `        ``// res` `        ``res[maximumX[i].second] = N - 1 - i;` `    ``}`   `    ``// Print the array res` `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``cout << res[i] << ``" "``;` `        ``// System.out.print(res[i] + " ");` `    ``}` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 3;` `    ``vector > > arr = {` `        ``{ { -2, 2 },` `          ``{ -1, 1 },` `          ``{ 2, 2 },` `          ``{ 2, -1 },` `          ``{ 1, -2 },` `          ``{ -2, -2 } },` `        ``{ { -1, -1 }, { 1, -1 }, { 1, 1 } },` `        ``{ { 3, 3 }, { -3, 3 }, { -3, -3 }, { 3, -3 } }` `    ``};`   `    ``findAllPolygons(N, arr);` `    ``return` `0;` `}`   `// The code is contributed by Gautam goel (gautamgoel962)`

## Java

 `// Java program for above approach` `import` `java.util.*;`   `class` `GFG {`   `    ``// Function to sort a 2D` `    ``// array using custom a comparator` `    ``public` `static` `void` `sort2d(``int``[][] arr)` `    ``{` `        ``Arrays.sort(arr, ``new` `Comparator<``int``[]>() {` `            ``public` `int` `compare(``final` `int``[] a,` `                               ``final` `int``[] b)` `            ``{` `                ``if` `(a[``0``] < b[``0``])` `                    ``return` `1``;` `                ``else` `                    ``return` `-``1``;` `            ``}` `        ``});` `    ``}`   `    ``// Function to calculate maximum sum` `    ``static` `void` `findAllPolygons(``int` `N,` `                                ``int``[][][] arr)` `    ``{` `        ``// Stores the maximum X co-` `        ``// ordinate of every polygon` `        ``int``[][] maximumX = ``new` `int``[N][``2``];`   `        ``// Traverse the array arr[][]` `        ``for` `(``int` `i = ``0``; i < N; i++) {`   `            ``// Stores the max X co-` `            ``// ordinate of current` `            ``// polygon` `            ``int` `maxX = Integer.MIN_VALUE;`   `            ``// Traverse over the vertices` `            ``// of the current polygon`   `            ``for` `(``int` `j = ``0``; j < arr[i].length; j++) {` `                ``// Update maxX` `                ``maxX = Math.max(maxX, arr[i][j][``0``]);` `            ``}`   `            ``// Assign maxX to maximumX[i][0]` `            ``maximumX[i][``0``] = maxX;`   `            ``// Assign i to maximumX[i][1]` `            ``maximumX[i][``1``] = i;` `        ``}`   `        ``// Sort the array of pairs` `        ``// maximumX in descending` `        ``// order by first element` `        ``sort2d(maximumX);`   `        ``// Stores the count of` `        ``// polygons lying inside` `        ``// a polygon` `        ``int``[] res = ``new` `int``[N];`   `        ``// Traverse the maximumX array` `        ``for` `(``int` `i = ``0``; i < N; i++) {` `            ``// Update value at` `            ``// maximumX[i][1] in` `            ``// res` `            ``res[maximumX[i][``1``]] = N - ``1` `- i;` `        ``}`   `        ``// Print the array res` `        ``for` `(``int` `i = ``0``; i < N; i++) {` `            ``System.out.print(res[i] + ``" "``);` `        ``}` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``int` `N = ``3``;` `        ``int``[][][] arr = {` `            ``{ { -``2``, ``2` `}, { -``1``, ``1` `}, { ``2``, ``2` `}, { ``2``, -``1` `}, { ``1``, -``2` `}, { -``2``, -``2` `} },` `            ``{ { -``1``, -``1` `}, { ``1``, -``1` `}, { ``1``, ``1` `} },` `            ``{ { ``3``, ``3` `}, { -``3``, ``3` `}, { -``3``, -``3` `}, { ``3``, -``3` `} }` `        ``};`   `        ``findAllPolygons(N, arr);` `    ``}` `}`

## Python3

 `# Python3 program for the above approach`   `# Function to sort a 2D array using custom a comparator` `def` `sort2d(arr):` `    ``arr.sort(key``=``lambda` `x: x[``0``], reverse``=``True``)`   `# Function to calculate maximum sum` `def` `findAllPolygons(N, arr):` `    ``# Stores the maximum X co-ordinate of every polygon` `    ``maximumX ``=` `[]`   `    ``# Traverse the array arr[][]` `    ``for` `i ``in` `range``(N):` `        ``# Stores the max X co-ordinate of current polygon` `        ``maxX ``=` `float``(``'-inf'``)`   `        ``# Traverse over the vertices of the current polygon` `        ``for` `j ``in` `range``(``len``(arr[i])):` `            ``# Update maxX` `            ``maxX ``=` `max``(maxX, arr[i][j][``0``])`   `        ``# Assign maxX to maximumX[i][0]` `        ``maximumX.append([maxX, i])`   `    ``# Sort the array of pairs maximumX in descending order by first element` `    ``sort2d(maximumX)`   `    ``# Stores the count of polygons lying inside a polygon` `    ``res ``=` `[``0``] ``*` `N`   `    ``# Traverse the maximumX array` `    ``for` `i ``in` `range``(N):` `        ``# Update value at maximumX[i][1] in res` `        ``res[maximumX[i][``1``]] ``=` `N ``-` `1` `-` `i`   `    ``# Print the array res` `    ``print``(``" "``.join(``map``(``str``, res)))`   `# Driver Code` `N ``=` `3` `arr ``=` `[` `    ``[` `        ``[``-``2``, ``2``],` `        ``[``-``1``, ``1``],` `        ``[``2``, ``2``],` `        ``[``2``, ``-``1``],` `        ``[``1``, ``-``2``],` `        ``[``-``2``, ``-``2``]` `    ``],` `    ``[` `        ``[``-``1``, ``-``1``],` `        ``[``1``, ``-``1``],` `        ``[``1``, ``1``]` `    ``],` `    ``[` `        ``[``3``, ``3``],` `        ``[``-``3``, ``3``],` `        ``[``-``3``, ``-``3``],` `        ``[``3``, ``-``3``]` `    ``]` `]`   `findAllPolygons(N, arr)`

## C#

 `using` `System;` `using` `System.Linq;`   `class` `GFG` `{` `  ``// Function to sort a 2D array using a custom comparator` `  ``public` `static` `void` `Sort2d(``int``[][] arr)` `  ``{` `    ``Array.Sort(arr, ``new` `Comparison<``int``[]>(` `      ``(a, b) => a[0] < b[0] ? 1 : -1` `    ``));` `  ``}`   `  ``// Function to calculate the maximum sum` `  ``static` `void` `FindAllPolygons(``int` `N, ``int``[][][] arr)` `  ``{` `    ``// Stores the maximum X coordinate of every polygon` `    ``int``[][] maximumX = ``new` `int``[N][];`   `    ``// Traverse the array arr[][]` `    ``for` `(``int` `i = 0; i < N; i++)` `    ``{` `      ``// Stores the max X coordinate of the current polygon` `      ``int` `maxX = ``int``.MinValue;`   `      ``// Traverse over the vertices of the current polygon` `      ``for` `(``int` `j = 0; j < arr[i].Length; j++)` `      ``{` `        ``// Update maxX` `        ``maxX = Math.Max(maxX, arr[i][j][0]);` `      ``}`   `      ``// Assign maxX to maximumX[i][0] and i to maximumX[i][1]` `      ``maximumX[i] = ``new` `int``[] { maxX, i };` `    ``}`   `    ``// Sort the array of pairs maximumX in descending order by the first element` `    ``Sort2d(maximumX);`   `    ``// Stores the count of polygons lying inside a polygon` `    ``int``[] res = ``new` `int``[N];`   `    ``// Traverse the maximumX array` `    ``for` `(``int` `i = 0; i < N; i++)` `    ``{` `      ``// Update value at maximumX[i][1] in res` `      ``res[maximumX[i][1]] = N - 1 - i;` `    ``}`   `    ``// Print the array res` `    ``for` `(``int` `i = 0; i < N; i++)` `    ``{` `      ``Console.Write(res[i] + ``" "``);` `    ``}` `  ``}`   `  ``// Driver Code` `  ``static` `void` `Main(``string``[] args)` `  ``{` `    ``int` `N = 3;` `    ``int``[][][] arr = {` `      ``new` `int``[][] { ``new` `int``[] { -2, 2 }, ``new` `int``[] { -1, 1 }, ``new` `int``[] { 2, 2 }, ``new` `int``[] { 2, -1 }, ``new` `int``[] { 1, -2 }, ``new` `int``[] { -2, -2 } },` `      ``new` `int``[][] { ``new` `int``[] { -1, -1 }, ``new` `int``[] { 1, -1 }, ``new` `int``[] { 1, 1 } },` `      ``new` `int``[][] { ``new` `int``[] { 3, 3 }, ``new` `int``[] { -3, 3 }, ``new` `int``[] { -3, -3 }, ``new` `int``[] { 3, -3 } }` `    ``};`   `    ``FindAllPolygons(N, arr);` `  ``}` `}`   `// This code is contributed by phasing17.`

## Javascript

 `// JavaScript program for the above approach`   `// Function to sort a 2D array using custom a comparator` `function` `sort2d(arr) {` `    ``arr.sort(``function``(a, b) {` `        ``if` `(a[0] < b[0]) {` `            ``return` `1;` `        ``} ``else` `{` `            ``return` `-1;` `        ``}` `    ``});` `}`   `// Function to calculate maximum sum` `function` `findAllPolygons(N, arr) {` `    ``// Stores the maximum X co-ordinate of every polygon` `    ``let maximumX = [];`   `    ``// Traverse the array arr[][]` `    ``for` `(let i = 0; i < N; i++) {` `        ``// Stores the max X co-ordinate of current polygon` `        ``let maxX = Number.MIN_SAFE_INTEGER;`   `        ``// Traverse over the vertices of the current polygon` `        ``for` `(let j = 0; j < arr[i].length; j++) {` `            ``// Update maxX` `            ``maxX = Math.max(maxX, arr[i][j][0]);` `        ``}`   `        ``// Assign maxX to maximumX[i][0]` `        ``maximumX[i] = [maxX, i];` `    ``}`   `    ``// Sort the array of pairs maximumX in descending order by first element` `    ``sort2d(maximumX);`   `    ``// Stores the count of polygons lying inside a polygon` `    ``let res = [];`   `    ``// Traverse the maximumX array` `    ``for` `(let i = 0; i < N; i++) {` `        ``// Update value at maximumX[i][1] in res` `        ``res[maximumX[i][1]] = N - 1 - i;` `    ``}`   `    ``// Print the array res` `    ``console.log(res.join(``" "``));` `}`   `// Driver Code` `let N = 3;` `let arr = [` `    ``[` `        ``[-2, 2],` `        ``[-1, 1],` `        ``[2, 2],` `        ``[2, -1],` `        ``[1, -2],` `        ``[-2, -2]` `    ``],` `    ``[` `        ``[-1, -1],` `        ``[1, -1],` `        ``[1, 1]` `    ``],` `    ``[` `        ``[3, 3],` `        ``[-3, 3],` `        ``[-3, -3],` `        ``[3, -3]` `    ``]` `];`   `findAllPolygons(N, arr);`   `// This code is contributed by phasing17.`

Output

`1 0 2 `

Time Complexity: O(M + N log N), Where M maximum size of a polygon
Auxiliary Space: O(N)

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