The Skyline Problem using Divide and Conquer algorithm
Given n rectangular buildings in a 2-dimensional city, computes the skyline of these buildings, eliminating hidden lines. The main task is to view buildings from a side and remove all sections that are not visible. All buildings share common bottom and every building is represented by triplet (left, ht, right)
- ‘left’: is x coordinated of left side (or wall).
- ‘right’: is x coordinate of right side
- ‘ht’: is height of building.
A skyline is a collection of rectangular strips. A rectangular strip is represented as a pair (left, ht) where left is x coordinate of left side of strip and ht is height of strip. Examples:
Input: Array of buildings { (1, 11, 5), (2, 6, 7), (3, 13, 9), (12, 7, 16), (14, 3, 25), (19, 18, 22), (23, 13, 29), (24, 4, 28) } Output: Skyline (an array of rectangular strips) A strip has x coordinate of left side and height (1, 11), (3, 13), (9, 0), (12, 7), (16, 3), (19, 18), (22, 3), (25, 0) Below image is for input 1 :
Consider following as another example when there is only one building Input: {(1, 11, 5)} Output: (1, 11), (5, 0)
A Simple Solution is to initialize skyline or result as empty, then one by one add buildings to skyline. A building is added by first finding the overlapping strip(s). If there are no overlapping strips, the new building adds new strip(s). If overlapping strip is found, then height of the existing strip may increase. Time complexity of this solution is O(n2) We can find Skyline in Θ(nLogn) time using Divide and Conquer. The idea is similar to Merge Sort, divide the given set of buildings in two subsets. Recursively construct skyline for two halves and finally merge the two skylines. How to Merge two Skylines? The idea is similar to merge of merge sort, start from first strips of two skylines, compare x coordinates. Pick the strip with smaller x coordinate and add it to result. The height of added strip is considered as maximum of current heights from skyline1 and skyline2. Example to show working of merge:
Height of new Strip is always obtained by takin maximum of following (a) Current height from skyline1, say 'h1'. (b) Current height from skyline2, say 'h2' h1 and h2 are initialized as 0. h1 is updated when a strip from SkyLine1 is added to result and h2 is updated when a strip from SkyLine2 is added. Skyline1 = {(1, 11), (3, 13), (9, 0), (12, 7), (16, 0)} Skyline2 = {(14, 3), (19, 18), (22, 3), (23, 13), (29, 0)} Result = {} h1 = 0, h2 = 0 Compare (1, 11) and (14, 3). Since first strip has smaller left x, add it to result and increment index for Skyline1. h1 = 11, New Height = max(11, 0) Result = {(1, 11)} Compare (3, 13) and (14, 3). Since first strip has smaller left x, add it to result and increment index for Skyline1 h1 = 13, New Height = max(13, 0) Result = {(1, 11), (3, 13)} Similarly (9, 0) and (12, 7) are added. h1 = 7, New Height = max(7, 0) = 7 Result = {(1, 11), (3, 13), (9, 0), (12, 7)} Compare (16, 0) and (14, 3). Since second strip has smaller left x, it is added to result. h2 = 3, New Height = max(7, 3) = 7 Result = {(1, 11), (3, 13), (9, 0), (12, 7), (14, 7)} Compare (16, 0) and (19, 18). Since first strip has smaller left x, it is added to result. h1 = 0, New Height = max(0, 3) = 3 Result = {(1, 11), (3, 13), (9, 0), (12, 7), (14, 7), (16, 3)} Since Skyline1 has no more items, all remaining items of Skyline2 are added Result = {(1, 11), (3, 13), (9, 0), (12, 7), (14, 7), (16, 3), (19, 18), (22, 3), (23, 13), (29, 0)} One observation about above output is, the strip (14, 7) is redundant (There is already an strip of same height). We remove all redundant strips. Result = {(1, 11), (3, 13), (9, 0), (12, 7), (16, 3), (19, 18), (22, 3), (23, 13), (29, 0)} In below code, redundancy is handled by not appending a strip if the previous strip in result has same height.
Below is C++ implementation of above idea.
C++
// A divide and conquer based C++ // program to find skyline of given buildings #include <iostream> using namespace std; // A structure for building struct Building { // x coordinate of left side int left; // height int ht; // x coordinate of right side int right; }; // A strip in skyline class Strip { // x coordinate of left side int left; // height int ht; public : Strip( int l = 0, int h = 0) { left = l; ht = h; } friend class SkyLine; }; // Skyline: To represent Output(An array of strips) class SkyLine { // Array of strips Strip* arr; // Capacity of strip array int capacity; // Actual number of strips in array int n; public : ~SkyLine() { delete [] arr; } int count() { return n; } // A function to merge another skyline // to this skyline SkyLine* Merge(SkyLine* other); // Constructor SkyLine( int cap) { capacity = cap; arr = new Strip[cap]; n = 0; } // Function to add a strip 'st' to array void append(Strip* st) { // Check for redundant strip, a strip is // redundant if it has same height or left as previous if (n > 0 && arr[n - 1].ht == st->ht) return ; if (n > 0 && arr[n - 1].left == st->left) { arr[n - 1].ht = max(arr[n - 1].ht, st->ht); return ; } arr[n] = *st; n++; } // A utility function to print all strips of // skyline void print() { for ( int i = 0; i < n; i++) { cout << " (" << arr[i].left << ", " << arr[i].ht << "), " ; } } }; // This function returns skyline for a // given array of buildings arr[l..h]. // This function is similar to mergeSort(). SkyLine* findSkyline(Building arr[], int l, int h) { if (l == h) { SkyLine* res = new SkyLine(2); res->append( new Strip( arr[l].left, arr[l].ht)); res->append( new Strip( arr[l].right, 0)); return res; } int mid = (l + h) / 2; // Recur for left and right halves // and merge the two results SkyLine* sl = findSkyline( arr, l, mid); SkyLine* sr = findSkyline( arr, mid + 1, h); SkyLine* res = sl->Merge(sr); // To avoid memory leak delete sl; delete sr; // Return merged skyline return res; } // Similar to merge() in MergeSort // This function merges another skyline // 'other' to the skyline for which it is called. // The function returns pointer to the // resultant skyline SkyLine* SkyLine::Merge(SkyLine* other) { // Create a resultant skyline with // capacity as sum of two skylines SkyLine* res = new SkyLine( this ->n + other->n); // To store current heights of two skylines int h1 = 0, h2 = 0; // Indexes of strips in two skylines int i = 0, j = 0; while (i < this ->n && j < other->n) { // Compare x coordinates of left sides of two // skylines and put the smaller one in result if ( this ->arr[i].left < other->arr[j].left) { int x1 = this ->arr[i].left; h1 = this ->arr[i].ht; // Choose height as max of two heights int maxh = max(h1, h2); res->append( new Strip(x1, maxh)); i++; } else { int x2 = other->arr[j].left; h2 = other->arr[j].ht; int maxh = max(h1, h2); res->append( new Strip(x2, maxh)); j++; } } // If there are strips left in this // skyline or other skyline while (i < this ->n) { res->append(&arr[i]); i++; } while (j < other->n) { res->append(&other->arr[j]); j++; } return res; } // Driver Function int main() { Building arr[] = { { 1, 11, 5 }, { 2, 6, 7 }, { 3, 13, 9 }, { 12, 7, 16 }, { 14, 3, 25 }, { 19, 18, 22 }, { 23, 13, 29 }, { 24, 4, 28 } }; int n = sizeof (arr) / sizeof (arr[0]); // Find skyline for given buildings // and print the skyline SkyLine* ptr = findSkyline(arr, 0, n - 1); cout << " Skyline for given buildings is \n" ; ptr->print(); return 0; } |
Java
// A divide and conquer based Java // program to find skyline of given buildings import java.util.*; // A class for building class Building { int left, ht, right; public Building( int left, int ht, int right) { this .left = left; this .ht = ht; this .right = right; } } // A strip in skyline class Strip { int left, ht; public Strip( int left, int ht) { this .left = left; this .ht = ht; } } // Skyline: To represent Output(An array of strips) class SkyLine { List<Strip> arr; int capacity, n; public SkyLine( int cap) { this .arr = new ArrayList<>(); this .capacity = cap; this .n = 0 ; } public int count() { return this .n; } // A function to merge another skyline // to this skyline public SkyLine merge(SkyLine other) { SkyLine res = new SkyLine( this .n + other.n); int h1 = 0 , h2 = 0 , i = 0 , j = 0 ; while (i < this .n && j < other.n) { if ( this .arr.get(i).left < other.arr.get(j).left) { int x1 = this .arr.get(i).left; h1 = this .arr.get(i).ht; int maxh = Math.max(h1, h2); res.append( new Strip(x1, maxh)); i++; } else { int x2 = other.arr.get(j).left; h2 = other.arr.get(j).ht; int maxh = Math.max(h1, h2); res.append( new Strip(x2, maxh)); j++; } } while (i < this .n) { res.append( this .arr.get(i)); i++; } while (j < other.n) { res.append(other.arr.get(j)); j++; } return res; } // Function to add a strip 'st' to array public void append(Strip st) { if ( this .n > 0 && this .arr.get( this .n- 1 ).ht == st.ht) { return ; } if ( this .n > 0 && this .arr.get( this .n- 1 ).left == st.left) { this .arr.get( this .n- 1 ).ht = Math.max( this .arr.get( this .n- 1 ).ht, st.ht); return ; } this .arr.add(st); this .n++; } // A utility function to print all strips of // skyline public void printSkyline() { System.out.println( "Skyline for given buildings is:" ); for ( int i = 0 ; i < this .n; i++) { System.out.print( "(" + this .arr.get(i).left + ", " + this .arr.get(i).ht + "), " ); } System.out.println(); } } // This function returns skyline for a // given array of buildings arr[l..h]. // This function is similar to mergeSort(). class SkylineProblem { public static SkyLine findSkyline(Building[] arr, int l, int h) { if (l == h) { SkyLine res = new SkyLine( 2 ); res.append( new Strip(arr[l].left, arr[l].ht)); res.append( new Strip(arr[l].right, 0 )); return res; } int mid = (l + h) / 2 ; // Recur for left and right halves // and merge the two results SkyLine sl = findSkyline(arr, l, mid); SkyLine sr = findSkyline(arr, mid+ 1 , h); SkyLine res = sl.merge(sr); return res; } // Driver Code public static void main(String[] args) { Building[] arr = { new Building( 1 , 11 , 5 ), new Building( 2 , 6 , 7 ), new Building( 3 , 13 , 9 ), new Building( 12 , 7 , 16 ), new Building( 14 , 3 , 25 ), new Building( 19 , 18 , 22 ), new Building( 23 , 13 , 29 ), new Building( 24 , 4 , 28 )}; // Find skyline for given buildings // and print the skyline SkyLine res = findSkyline(arr, 0 , arr.length- 1 ); res.printSkyline(); } } /* Output: Skyline for given buildings is: (1, 11), (3, 13), (9, 0), (12, 18), (22, 3), (25, 0), (28, 4), (29, 0), */ |
Python3
class Building: def __init__( self , left, ht, right): self .left = left self .ht = ht self .right = right class Strip: def __init__( self , left = 0 , ht = 0 ): self .left = left self .ht = ht class SkyLine: def __init__( self , cap): self .arr = [] self .capacity = cap self .n = 0 def count( self ): return self .n def merge( self , other): res = SkyLine( self .n + other.n) h1, h2, i, j = 0 , 0 , 0 , 0 while i < self .n and j < other.n: if self .arr[i].left < other.arr[j].left: x1, h1 = self .arr[i].left, self .arr[i].ht maxh = max (h1, h2) res.append(Strip(x1, maxh)) i + = 1 else : x2, h2 = other.arr[j].left, other.arr[j].ht maxh = max (h1, h2) res.append(Strip(x2, maxh)) j + = 1 while i < self .n: res.append( self .arr[i]) i + = 1 while j < other.n: res.append(other.arr[j]) j + = 1 return res def append( self , st): if self .n > 0 and self .arr[ self .n - 1 ].ht = = st.ht: return if self .n > 0 and self .arr[ self .n - 1 ].left = = st.left: self .arr[ self .n - 1 ].ht = max ( self .arr[ self .n - 1 ].ht, st.ht) return self .arr.append(st) self .n + = 1 def print_skyline( self ): print ( "Skyline for given buildings is" ) for i in range ( self .n): print ( " ({}, {})," . format ( self .arr[i].left, self .arr[i].ht), end = "") print () def find_skyline(arr, l, h): if l = = h: res = SkyLine( 2 ) res.append(Strip(arr[l].left, arr[l].ht)) res.append(Strip(arr[l].right, 0 )) return res mid = (l + h) / / 2 sl = find_skyline(arr, l, mid) sr = find_skyline(arr, mid + 1 , h) res = sl.merge(sr) return res arr = [Building( 1 , 11 , 5 ), Building( 2 , 6 , 7 ), Building( 3 , 13 , 9 ), Building( 12 , 7 , 16 ), Building( 14 , 3 , 25 ), Building( 19 , 18 , 22 ), Building( 23 , 13 , 29 ), Building( 24 , 4 , 28 )] n = len (arr) skyline = find_skyline(arr, 0 , n - 1 ) skyline.print_skyline() |
C#
// A divide and conquer based Java // program to find skyline of given buildings using System; using System.Collections.Generic; // A class for building public class Building { public int left, ht, right; public Building( int left, int ht, int right) { this .left = left; this .ht = ht; this .right = right; } } // A strip in skyline public class Strip { public int left, ht; public Strip( int left, int ht) { this .left = left; this .ht = ht; } } // Skyline: To represent Output(An array of strips) public class SkyLine { public List<Strip> arr; public int capacity, n; public SkyLine( int cap) { this .arr = new List<Strip>(); this .capacity = cap; this .n = 0; } public int count() { return this .n; } // A function to merge another skyline // to this skyline public SkyLine merge(SkyLine other) { SkyLine res = new SkyLine( this .n + other.n); int h1 = 0, h2 = 0, i = 0, j = 0; while (i < this .n && j < other.n) { if ( this .arr[i].left < other.arr[j].left) { int x1 = this .arr[i].left; h1 = this .arr[i].ht; int maxh = Math.Max(h1, h2); res.append( new Strip(x1, maxh)); i++; } else { int x2 = other.arr[j].left; h2 = other.arr[j].ht; int maxh = Math.Max(h1, h2); res.append( new Strip(x2, maxh)); j++; } } while (i < this .n) { res.append( this .arr[i]); i++; } while (j < other.n) { res.append(other.arr[j]); j++; } return res; } // Function to add a strip 'st' to array public void append(Strip st) { if ( this .n > 0 && this .arr[ this .n-1].ht == st.ht) { return ; } if ( this .n > 0 && this .arr[ this .n-1].left == st.left) { this .arr[ this .n-1].ht = Math.Max( this .arr[ this .n-1].ht, st.ht); return ; } this .arr.Add(st); this .n++; } // A utility function to print all strips of // skyline public void printSkyline() { Console.WriteLine( "Skyline for given buildings is:" ); for ( int i = 0; i < this .n; i++) { Console.Write( "(" + this .arr[i].left + ", " + this .arr[i].ht + "), " ); } Console.WriteLine(); } } // This function returns skyline for a // given array of buildings arr[l..h]. // This function is similar to mergeSort(). public class SkylineProblem { public static SkyLine findSkyline(Building[] arr, int l, int h) { if (l == h) { SkyLine res2 = new SkyLine(2); res2.append( new Strip(arr[l].left, arr[l].ht)); res2.append( new Strip(arr[l].right, 0)); return res2; } int mid = (l + h) / 2; // Recur for left and right halves // and merge the two results SkyLine sl = findSkyline(arr, l, mid); SkyLine sr = findSkyline(arr, mid+1, h); SkyLine res3 = sl.merge(sr); return res3; } // Driver Code public static void Main( string [] args) { Building[] arr = { new Building(1, 11, 5), new Building(2, 6, 7), new Building(3, 13, 9), new Building(12, 7, 16), new Building(14, 3, 25), new Building(19, 18, 22), new Building(23, 13, 29), new Building(24, 4, 28)}; // Find skyline for given buildings and print the skyline SkyLine res1 = findSkyline(arr, 0, arr.Length-1); res1.printSkyline(); } } |
Javascript
// Js program to find skyline of given buildings // A structure for building const Building = { left: Number, ht: Number, right: Number, }; class Strip { constructor(l = 0, h = 0) { this .left = l; this .ht = h; } } class SkyLine { constructor(cap) { this .capacity = cap; this .arr = new Array(cap); this .n = 0; } count() { return this .n; } Merge(other) { let res = new SkyLine( this .n + other.n); let h1 = 0, h2 = 0; let i = 0, j = 0; // Merge two skylines by comparing the left coordinates of the strips while (i < this .n && j < other.n) { if ( this .arr[i].left < other.arr[j].left) { let x1 = this .arr[i].left; h1 = this .arr[i].ht; let maxh = Math.max(h1, h2); res.append( new Strip(x1, maxh)); i++; } else { let x2 = other.arr[j].left; h2 = other.arr[j].ht; let maxh = Math.max(h1, h2); res.append( new Strip(x2, maxh)); j++; } } // Append remaining strips from skyline 1 and 2 while (i < this .n) { res.append( this .arr[i]); i++; } while (j < other.n) { res.append(other.arr[j]); j++; } return res; } // Append a strip to skyline if it is not redundant append(st) { if ( this .n > 0 && this .arr[ this .n - 1].ht == st.ht) { return ; } if ( this .n > 0 && this .arr[ this .n - 1].left == st.left) { this .arr[ this .n - 1].ht = Math.max( this .arr[ this .n - 1].ht, st.ht); return ; } this .arr[ this .n] = st; this .n++; } // Print the skyline strips print() { let str = '' ; for (let i = 0; i < this .n; i++) { str += `(${ this .arr[i].left}, ${ this .arr[i].ht}), `; } console.log(`Skyline for given buildings is \n${str}`); } } function findSkyline(arr, l, h) { if (l == h) { // Base case: when a single building is left let res = new SkyLine(2); res.append( new Strip(arr[l].left, arr[l].ht)); res.append( new Strip(arr[l].right, 0)); return res; } // Divide and Conquer approach let mid = Math.floor((l + h) / 2); let sl = findSkyline(arr, l, mid); let sr = findSkyline(arr, mid + 1, h); let res = sl.Merge(sr); delete sl; delete sr; return res; } const main = () => { const arr = [ { left: 1, ht: 11, right: 5 }, { left: 2, ht: 6, right: 7 }, { left: 3, ht: 13, right: 9 }, { left: 12, ht: 7, right: 16 }, { left: 14, ht: 3, right: 25 }, { left: 19, ht: 18, right: 22 }, { left: 23, ht: 13, right: 29 }, { left: 24, ht: 4, right: 28 }, ]; const n = arr.length; const ptr = findSkyline(arr, 0, n - 1); console.log( "Skyline for given buildings is" ); ptr.print(); }; main(); |
Skyline for given buildings is (1, 11), (3, 13), (9, 0), (12, 7), (16, 3), (19, 18), (22, 3), (23, 13), (29, 0),
Time complexity of above recursive implementation is same as Merge Sort. T(n) = T(n/2) + Θ(n) Solution of above recurrence is Θ(nLogn) References:
- http://faculty.kfupm.edu.sa/ics/darwish/stuff/ics353handouts/Ch4Ch5.pdf
- www.cs.ucf.edu/~sarahb/COP3503/Lectures/DivideAndConquer.ppt
This article is contributed Abhay Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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