Mazākais diapazons ar elementiem no K sakārtotiem sarakstiem
Dots 2D vesels skaitļa masīvs arr [] [] kārtība k * n kur katra rinda ir sakārtots Augošā secībā. Jūsu uzdevums ir atrast mazāko diapazonu, kas ietver vismaz vienu elementu no katra no Kandidāts saraksti. Ja tiek atrasti vairāki šādi diapazoni, atgriežas pirmais.
Piemēri:
Ievade: arr [] [] = [[4 7 9 12 15]
[0 8 10 14 20]
[6 12 16 30 50]]
Izlaide: 6 8
Paskaidrojums: Mazākais diapazons tiek veidots pēc 7. numura no pirmā 8. saraksta no otrā saraksta un 6 no trešā saraksta.Ievade: arr [] [] = [[2 4]
[1 7]
[20 40]]
Izlaide: 4 20
Paskaidrojums: Diapazons [4 20] satur 4 7 20, kas satur elementu no visiem trim blokiem.
Satura rādītājs
- [Naivā pieeja] - izmantojot k pointers - o (n k^2) Laiks un O (k) telpa
- [Labāka pieeja] Izmantojot divus rādītāju - o (n*k log (n*k)) laiku un o (n*k) telpa
- [Efektīva pieeja] - Min Heap - O (n K log K) laika un O (k) telpas izmantošana
[Naivā pieeja] - izmantojot k pointers - o (n k^2) Laiks un O (k) telpa
Ideja ir saglabāt k norādes vienu katram sarakstam, sākot ar indeksu 0. Katrā solī veiciet min un max no pašreizējiem K elementiem, lai veidotu diapazonu. Līdz samazināt diapazonu Mums ir jābūt Palieliniet min vērtību Tā kā mēs nevaram samazināt maksimumu (visi rādītāji sākas ar 0). Tāpēc pārvietojiet saraksta rādītāju, kurā ir pašreizējais minimums un atjauniniet diapazonu. Atkārtojiet, līdz viens saraksts ir izsmelts.
Soli pa solim ieviešana:
- Izveidojiet norādījumu sarakstu pa vienam katram ievades sarakstam, sākot ar indeksu 0.
- Atkārtojiet procesu līdz viena no norādēm sasniedz sava saraksta beigas.
- Katrā solī Izvēlieties pašreizējos elementus norāda uz visiem norādēm.
- Atrast Minimālais un maksimālais Starp šiem elementiem.
- Aprēķiniet diapazonu izmantojot min un max vērtības.
- Ja šis diapazons ir mazāks nekā iepriekšējais labākais atbildes atjauninājums.
- Virzieties uz priekšu rādītāju no saraksta, kurā bija minimālais elements.
- Apstājieties, kad jebkurš saraksts ir izsmelts un atgriezt labāko atrasto diapazonu.
// C++ program to find the smallest range // that includes at least one element from // each of the k sorted lists using k pointers #include #include #include using namespace std ; vector < int > findSmallestRange ( vector < vector < int >>& arr ) { int k = arr . size (); int n = arr [ 0 ]. size (); // Pointers for each of the k rows vector < int > ptr ( k 0 ); int minRange = INT_MAX ; int start = -1 end = -1 ; while ( true ) { int minVal = INT_MAX ; int maxVal = INT_MIN ; int minRow = -1 ; // Traverse all k rows to get current min and max for ( int i = 0 ; i < k ; i ++ ) { // If any list is exhausted stop the loop if ( ptr [ i ] == n ) { return { start end }; } // Track min value and its row index if ( arr [ i ][ ptr [ i ]] < minVal ) { minVal = arr [ i ][ ptr [ i ]]; minRow = i ; } // Track current max value if ( arr [ i ][ ptr [ i ]] > maxVal ) { maxVal = arr [ i ][ ptr [ i ]]; } } // Update the result range if a // smaller range is found if ( maxVal - minVal < minRange ) { minRange = maxVal - minVal ; start = minVal ; end = maxVal ; } // Move the pointer of the // row with minimum value ptr [ minRow ] ++ ; } return { start end }; } int main () { vector < vector < int >> arr = { { 4 7 9 12 15 } { 0 8 10 14 20 } { 6 12 16 30 50 } }; vector < int > res = findSmallestRange ( arr ); cout < < res [ 0 ] < < ' ' < < res [ 1 ]; return 0 ; }
Java // Java program to find the smallest range import java.util.* ; class GfG { static ArrayList < Integer > findSmallestRange ( int [][] arr ) { int k = arr . length ; int n = arr [ 0 ] . length ; // Pointers for each of the k rows int [] ptr = new int [ k ] ; int minRange = Integer . MAX_VALUE ; int start = - 1 end = - 1 ; while ( true ) { int minVal = Integer . MAX_VALUE ; int maxVal = Integer . MIN_VALUE ; int minRow = - 1 ; // Traverse all k rows to get current min and max for ( int i = 0 ; i < k ; i ++ ) { // If any list is exhausted stop the loop if ( ptr [ i ] == n ) { ArrayList < Integer > result = new ArrayList <> (); result . add ( start ); result . add ( end ); return result ; } // Track min value and its row index if ( arr [ i ][ ptr [ i ]] < minVal ) { minVal = arr [ i ][ ptr [ i ]] ; minRow = i ; } // Track current max value if ( arr [ i ][ ptr [ i ]] > maxVal ) { maxVal = arr [ i ][ ptr [ i ]] ; } } // Update the result range if a smaller range is found if ( maxVal - minVal < minRange ) { minRange = maxVal - minVal ; start = minVal ; end = maxVal ; } // Move the pointer of the row with minimum value ptr [ minRow ]++ ; } } public static void main ( String [] args ) { int [][] arr = { { 4 7 9 12 15 } { 0 8 10 14 20 } { 6 12 16 30 50 } }; ArrayList < Integer > res = findSmallestRange ( arr ); System . out . println ( res . get ( 0 ) + ' ' + res . get ( 1 )); } }
Python # Python program to find the smallest range def findSmallestRange ( arr ): k = len ( arr ) n = len ( arr [ 0 ]) # Pointers for each of the k rows ptr = [ 0 ] * k min_range = float ( 'inf' ) start = - 1 end = - 1 while True : min_val = float ( 'inf' ) max_val = float ( '-inf' ) min_row = - 1 # Traverse all k rows to get current min and max for i in range ( k ): # If any list is exhausted stop the loop if ptr [ i ] == n : return [ start end ] # Track min value and its row index if arr [ i ][ ptr [ i ]] < min_val : min_val = arr [ i ][ ptr [ i ]] min_row = i # Track current max value if arr [ i ][ ptr [ i ]] > max_val : max_val = arr [ i ][ ptr [ i ]] # Update the result range if a smaller range is found if max_val - min_val < min_range : min_range = max_val - min_val start = min_val end = max_val # Move the pointer of the row with minimum value ptr [ min_row ] += 1 if __name__ == '__main__' : arr = [ [ 4 7 9 12 15 ] [ 0 8 10 14 20 ] [ 6 12 16 30 50 ] ] res = findSmallestRange ( arr ) print ( res [ 0 ] res [ 1 ])
C# using System ; using System.Collections.Generic ; class GfG { static List < int > findSmallestRange ( int [] arr ) { int k = arr . GetLength ( 0 ); int n = arr . GetLength ( 1 ); // Pointers for each of the k rows int [] ptr = new int [ k ]; int minRange = int . MaxValue ; int start = - 1 end = - 1 ; while ( true ) { int minVal = int . MaxValue ; int maxVal = int . MinValue ; int minRow = - 1 ; // Traverse all k rows to get current min and max for ( int i = 0 ; i < k ; i ++ ) { // If any list is exhausted stop the loop if ( ptr [ i ] == n ) { return new List < int > { start end }; } int current = arr [ i ptr [ i ]]; if ( current < minVal ) { minVal = current ; minRow = i ; } if ( current > maxVal ) { maxVal = current ; } } // Update the result range if a smaller range is found if ( maxVal - minVal < minRange ) { minRange = maxVal - minVal ; start = minVal ; end = maxVal ; } // Move the pointer of the row with minimum value ptr [ minRow ] ++ ; } } public static void Main ( string [] args ) { int [] arr = { { 4 7 9 12 15 } { 0 8 10 14 20 } { 6 12 16 30 50 } }; List < int > res = findSmallestRange ( arr ); Console . WriteLine ( res [ 0 ] + ' ' + res [ 1 ]); } }
JavaScript // JavaScript program to find the smallest range function findSmallestRange ( arr ) { let k = arr . length ; let n = arr [ 0 ]. length ; // Pointers for each of the k rows let ptr = new Array ( k ). fill ( 0 ); let minRange = Infinity ; let start = - 1 end = - 1 ; while ( true ) { let minVal = Infinity ; let maxVal = - Infinity ; let minRow = - 1 ; // Traverse all k rows to get current min and max for ( let i = 0 ; i < k ; i ++ ) { // If any list is exhausted stop the loop if ( ptr [ i ] === n ) { return [ start end ]; } // Track min value and its row index if ( arr [ i ][ ptr [ i ]] < minVal ) { minVal = arr [ i ][ ptr [ i ]]; minRow = i ; } // Track current max value if ( arr [ i ][ ptr [ i ]] > maxVal ) { maxVal = arr [ i ][ ptr [ i ]]; } } // Update the result range if a smaller range is found if ( maxVal - minVal < minRange ) { minRange = maxVal - minVal ; start = minVal ; end = maxVal ; } // Move the pointer of the row with minimum value ptr [ minRow ] ++ ; } } const arr = [ [ 4 7 9 12 15 ] [ 0 8 10 14 20 ] [ 6 12 16 30 50 ] ]; const res = findSmallestRange ( arr ); console . log ( res [ 0 ] + ' ' + res [ 1 ]);
Izvade
6 8
[Labāka pieeja] Izmantojot divus rādītāju - o (n*k log (n*k)) laiku un o (n*k) telpa
C++Ideja ir atrast mazāko diapazona problēmu, pārveidojot to par bīdāmā loga problēmu salīdzinājumā ar apvienotu un sakārtotu visu elementu sarakstu no ievades sarakstiem. Katrs elements tiek saglabāts kopā ar sākotnējo saraksta indeksu, lai izsekotu tā avotu. Pēc kombinētā saraksta šķirošanas pēc vērtības diviem norādēm (
leftunright) tiek izmantoti, lai definētu logu, kas pārvietojas pa sarakstu. Tā kā logs paplašina frekvences kartes celiņus, cik unikālo sarakstu tiek pārstāvēti. Kad logā ir vismaz viens skaitlis no katra saraksta, algoritms mēģina to samazināt no kreisās puses, lai atrastu mazāku derīgu diapazonu. Rezultātā tiek atgriezts mazākais šāds diapazons, kas atrodams šajā procesā.
#include using namespace std ; vector < int > findSmallestRange ( vector < vector < int >>& arr ) { int k = arr . size (); // Stores the current index for each list vector < int > pointers ( k 0 ); // Stores the current smallest range vector < int > smallestRange = { 0 INT_MAX }; while ( true ) { int currentMin = INT_MAX currentMax = INT_MIN ; int minListIndex = -1 ; // Find the minimum and maximum among current elements of all lists for ( int i = 0 ; i < k ; i ++ ) { int value = arr [ i ][ pointers [ i ]]; if ( value < currentMin ) { currentMin = value ; minListIndex = i ; } if ( value > currentMax ) { currentMax = value ; } } // Update the smallest range if this one is smaller if ( currentMax - currentMin < smallestRange [ 1 ] - smallestRange [ 0 ]) { smallestRange [ 0 ] = currentMin ; smallestRange [ 1 ] = currentMax ; } // Move the pointer in the list that had the minimum value pointers [ minListIndex ] ++ ; // If that list is exhausted break the loop if ( pointers [ minListIndex ] == arr [ minListIndex ]. size ()) break ; } return smallestRange ; } // Driver code int main () { vector < vector < int >> arr = { { 4 7 9 12 15 } { 0 8 10 14 20 } { 6 12 16 30 50 } }; vector < int > result = findSmallestRange ( arr ); cout < < result [ 0 ] < < ' ' < < result [ 1 ]; return 0 ; }
Java import java.util.* ; class GfG { // Function to find the smallest range public static ArrayList < Integer > findSmallestRange ( int [][] arr ) { int k = arr . length ; // Number of lists // Stores the current index for each list int [] pointers = new int [ k ] ; // Stores the current smallest range ArrayList < Integer > smallestRange = new ArrayList <> ( Arrays . asList ( 0 Integer . MAX_VALUE )); // Continue the loop until one list is exhausted while ( true ) { int currentMin = Integer . MAX_VALUE currentMax = Integer . MIN_VALUE ; int minListIndex = - 1 ; // Find the minimum and maximum among current elements of all lists for ( int i = 0 ; i < k ; i ++ ) { int value = arr [ i ][ pointers [ i ]] ; // Update the current minimum if ( value < currentMin ) { currentMin = value ; minListIndex = i ; } // Update the current maximum if ( value > currentMax ) { currentMax = value ; } } // Update the smallest range if this one is smaller if ( currentMax - currentMin < smallestRange . get ( 1 ) - smallestRange . get ( 0 )) { smallestRange . set ( 0 currentMin ); smallestRange . set ( 1 currentMax ); } // Move the pointer in the list that had the minimum value pointers [ minListIndex ]++ ; // If that list is exhausted break the loop if ( pointers [ minListIndex ] == arr [ minListIndex ] . length ) break ; } return smallestRange ; // Return the result as ArrayList } // Driver code public static void main ( String [] args ) { int [][] arr = { { 4 7 9 12 15 } { 0 8 10 14 20 } { 6 12 16 30 50 } }; ArrayList < Integer > result = findSmallestRange ( arr ); System . out . println ( result . get ( 0 ) + ' ' + result . get ( 1 )); } }
Python def findSmallestRange ( arr ): k = len ( arr ) # Number of lists # Stores the current index for each list pointers = [ 0 ] * k # Stores the current smallest range smallestRange = [ 0 float ( 'inf' )] # Continue the loop until one list is exhausted while True : currentMin = float ( 'inf' ) currentMax = - float ( 'inf' ) minListIndex = - 1 # Find the minimum and maximum among current elements of all lists for i in range ( k ): value = arr [ i ][ pointers [ i ]] # Update the current minimum if value < currentMin : currentMin = value minListIndex = i # Update the current maximum if value > currentMax : currentMax = value # Update the smallest range if this one is smaller if currentMax - currentMin < smallestRange [ 1 ] - smallestRange [ 0 ]: smallestRange [ 0 ] = currentMin smallestRange [ 1 ] = currentMax # Move the pointer in the list that had the minimum value pointers [ minListIndex ] += 1 # If that list is exhausted break the loop if pointers [ minListIndex ] == len ( arr [ minListIndex ]): break return smallestRange # Return the result as a list # Driver code if __name__ == '__main__' : arr = [ [ 4 7 9 12 15 ] [ 0 8 10 14 20 ] [ 6 12 16 30 50 ] ] result = findSmallestRange ( arr ) print ( result [ 0 ] result [ 1 ])
C# using System ; using System.Collections.Generic ; class GfG { // Function to find the smallest range public static List < int > findSmallestRange ( int [] arr ) { int k = arr . GetLength ( 0 ); // Number of lists (rows) // Stores the current index for each list (row) int [] pointers = new int [ k ]; // Stores the current smallest range List < int > smallestRange = new List < int > { 0 int . MaxValue }; // Continue the loop until one list is exhausted while ( true ) { int currentMin = int . MaxValue currentMax = int . MinValue ; int minListIndex = - 1 ; // Find the minimum and maximum among current elements // of all lists for ( int i = 0 ; i < k ; i ++ ) { int value = arr [ i pointers [ i ]]; // Update the current minimum if ( value < currentMin ) { currentMin = value ; minListIndex = i ; } // Update the current maximum if ( value > currentMax ) { currentMax = value ; } } // Update the smallest range if this one is smaller if ( currentMax - currentMin < smallestRange [ 1 ] - smallestRange [ 0 ]) { smallestRange [ 0 ] = currentMin ; smallestRange [ 1 ] = currentMax ; } // Move the pointer in the list that had the minimum value pointers [ minListIndex ] ++ ; // If that list is exhausted break the loop if ( pointers [ minListIndex ] == arr . GetLength ( 1 )) break ; } return smallestRange ; // Return the result as List } // Driver code public static void Main ( string [] args ) { int [] arr = { { 4 7 9 12 15 } { 0 8 10 14 20 } { 6 12 16 30 50 } }; List < int > result = findSmallestRange ( arr ); Console . WriteLine ( result [ 0 ] + ' ' + result [ 1 ]); } }
JavaScript function findSmallestRange ( arr ) { const k = arr . length ; // Number of lists // Stores the current index for each list let pointers = new Array ( k ). fill ( 0 ); // Stores the current smallest range let smallestRange = [ 0 Number . MAX_VALUE ]; // Continue the loop until one list is exhausted while ( true ) { let currentMin = Number . MAX_VALUE currentMax = - Number . MAX_VALUE ; let minListIndex = - 1 ; // Find the minimum and maximum among current elements of all lists for ( let i = 0 ; i < k ; i ++ ) { const value = arr [ i ][ pointers [ i ]]; // Update the current minimum if ( value < currentMin ) { currentMin = value ; minListIndex = i ; } // Update the current maximum if ( value > currentMax ) { currentMax = value ; } } // Update the smallest range if this one is smaller if ( currentMax - currentMin < smallestRange [ 1 ] - smallestRange [ 0 ]) { smallestRange [ 0 ] = currentMin ; smallestRange [ 1 ] = currentMax ; } // Move the pointer in the list that had the minimum value pointers [ minListIndex ] ++ ; // If that list is exhausted break the loop if ( pointers [ minListIndex ] === arr [ minListIndex ]. length ) break ; } return smallestRange ; // Return the result as an array } // Driver code const arr = [ [ 4 7 9 12 15 ] [ 0 8 10 14 20 ] [ 6 12 16 30 50 ] ]; const result = findSmallestRange ( arr ); console . log ( result [ 0 ] result [ 1 ]);
Izvade
6 8
[Efektīva pieeja] - Min Heap - O (n K log K) laika un O (k) telpas izmantošana
Mīnētāja Var izmantot, lai lineārā laika vietā atrastu minimālo vērtību logaritmiskajā laikā vai log k laiku. Lai atrastu maksimālo vērtību, mēs sākotnēji inicializējam visu 0 indeksu maksimālo vērtību. Pārējām maksimālajām vērtībām cilpā mēs vienkārši salīdzinām pašreizējo maksimālo vērtību ar nākamo vienumu no saraksta, no kura tiek noņemts min. Pārējā pieeja paliek tāda pati.
Soli pa solim ieviešana:
- Mīnētāja Var izmantot, lai lineārā laika vietā atrastu minimālo vērtību logaritmiskajā laikā vai log k laiku. Lai atrastu maksimālo vērtību, mēs sākotnēji inicializējam visu 0 indeksu maksimālo vērtību. Pārējām maksimālajām vērtībām cilpā mēs vienkārši salīdzinām pašreizējo maksimālo vērtību ar nākamo vienumu no saraksta, no kura tiek noņemts min. Pārējā pieeja paliek tāda pati.
Izveidojiet mīnieti, lai saglabātu K elementus no katra masīva un mainīgā lieluma montēt inicializēts līdz maksimālai vērtībai un arī saglabājiet mainīgo maksimums Lai saglabātu maksimālo veselu skaitli.
- Mīnētāja Var izmantot, lai lineārā laika vietā atrastu minimālo vērtību logaritmiskajā laikā vai log k laiku. Lai atrastu maksimālo vērtību, mēs sākotnēji inicializējam visu 0 indeksu maksimālo vērtību. Pārējām maksimālajām vērtībām cilpā mēs vienkārši salīdzinām pašreizējo maksimālo vērtību ar nākamo vienumu no saraksta, no kura tiek noņemts min. Pārējā pieeja paliek tāda pati.
Sākotnēji ievietojiet pirmo elementu no katra saraksta un saglabājiet maksimālo vērtību maksimums Apvidū
- Mīnētāja Var izmantot, lai lineārā laika vietā atrastu minimālo vērtību logaritmiskajā laikā vai log k laiku. Lai atrastu maksimālo vērtību, mēs sākotnēji inicializējam visu 0 indeksu maksimālo vērtību. Pārējām maksimālajām vērtībām cilpā mēs vienkārši salīdzinām pašreizējo maksimālo vērtību ar nākamo vienumu no saraksta, no kura tiek noņemts min. Pārējā pieeja paliek tāda pati.
Atkārtojiet šīs darbības, līdz vismaz viens saraksts ir izsmelts:
- Atrodiet minimālo vērtību vai minimāls Izmantojiet minimālo kaudzes augšdaļu vai sakni, kas ir minimālais elements.
- Tagad atjauniniet montēt Ja strāva (max-min) ir mazāka par montēt Apvidū
- Noņemiet augšējo vai saknes elementu no prioritārās rindas. Ievietojiet nākamo elementu no saraksta, kurā ir MIN elements
- Atjauniniet maksimumu ar jauno elementu, kas ievietots, ja jaunais elements ir lielāks nekā iepriekšējais MAX.
C++ Java #include
Python import java.util.* ; // Class to represent elements in the heap class Node implements Comparable < Node > { int val row col ; Node ( int val int row int col ) { this . val = val ; this . row = row ; this . col = col ; } // For min-heap based on value public int compareTo ( Node other ) { return this . val - other . val ; } } class GfG { // Function to find the smallest range static ArrayList < Integer > findSmallestRange ( int [][] arr ) { int k = arr . length ; int n = arr [ 0 ] . length ; PriorityQueue < Node > pq = new PriorityQueue <> (); int maxVal = Integer . MIN_VALUE ; // Push the first element of each list into the min-heap for ( int i = 0 ; i < k ; i ++ ) { pq . add ( new Node ( arr [ i ][ 0 ] i 0 )); maxVal = Math . max ( maxVal arr [ i ][ 0 ] ); } int minRange = Integer . MAX_VALUE minEl = - 1 maxEl = - 1 ; while ( true ) { Node curr = pq . poll (); int minVal = curr . val ; // Update range if better if ( maxVal - minVal < minRange ) { minRange = maxVal - minVal ; minEl = minVal ; maxEl = maxVal ; } // If we've reached the end of a list break if ( curr . col + 1 == n ) break ; // Push next element from the same list int nextVal = arr [ curr . row ][ curr . col + 1 ] ; pq . add ( new Node ( nextVal curr . row curr . col + 1 )); maxVal = Math . max ( maxVal nextVal ); } // Return result as ArrayList ArrayList < Integer > result = new ArrayList <> (); result . add ( minEl ); result . add ( maxEl ); return result ; } // Driver code public static void main ( String [] args ) { int [][] arr = { { 4 7 9 12 15 } { 0 8 10 14 20 } { 6 12 16 30 50 } }; ArrayList < Integer > res = findSmallestRange ( arr ); System . out . println ( res . get ( 0 ) + ' ' + res . get ( 1 )); } }
C# import heapq # Function to find the smallest range def findSmallestRange ( arr ): k = len ( arr ) n = len ( arr [ 0 ]) heap = [] maxVal = float ( '-inf' ) # Push the first element of each # list into the min-heap for i in range ( k ): heapq . heappush ( heap ( arr [ i ][ 0 ] i 0 )) maxVal = max ( maxVal arr [ i ][ 0 ]) minRange = float ( 'inf' ) minEl = maxEl = - 1 while True : minVal row col = heapq . heappop ( heap ) # Update range if better if maxVal - minVal < minRange : minRange = maxVal - minVal minEl = minVal maxEl = maxVal # If we've reached the end of a list break if col + 1 == n : break # Push next element from the same list nextVal = arr [ row ][ col + 1 ] heapq . heappush ( heap ( nextVal row col + 1 )) maxVal = max ( maxVal nextVal ) return [ minEl maxEl ] # Driver code if __name__ == '__main__' : arr = [ [ 4 7 9 12 15 ] [ 0 8 10 14 20 ] [ 6 12 16 30 50 ] ] res = findSmallestRange ( arr ) print ( res [ 0 ] res [ 1 ])
JavaScript using System ; using System.Collections.Generic ; // Class to represent elements in the heap class Node : IComparable < Node > { public int val row col ; public Node ( int val int row int col ) { this . val = val ; this . row = row ; this . col = col ; } // For min-heap based on value public int CompareTo ( Node other ) { if ( this . val != other . val ) return this . val . CompareTo ( other . val ); // To avoid duplicate keys in SortedSet if ( this . row != other . row ) return this . row . CompareTo ( other . row ); return this . col . CompareTo ( other . col ); } } class GfG { // Function to find the smallest range static List < int > findSmallestRange ( int [] arr ) { int k = arr . GetLength ( 0 ); int n = arr . GetLength ( 1 ); var pq = new SortedSet < Node > (); int maxVal = int . MinValue ; // Push the first element of each list into the min-heap for ( int i = 0 ; i < k ; i ++ ) { var node = new Node ( arr [ i 0 ] i 0 ); pq . Add ( node ); maxVal = Math . Max ( maxVal arr [ i 0 ]); } int minRange = int . MaxValue minEl = - 1 maxEl = - 1 ; while ( true ) { var curr = GetMin ( pq ); pq . Remove ( curr ); int minVal = curr . val ; // Update range if better if ( maxVal - minVal < minRange ) { minRange = maxVal - minVal ; minEl = minVal ; maxEl = maxVal ; } // If we've reached the end of a list break if ( curr . col + 1 == n ) break ; // Push next element from the same list int nextVal = arr [ curr . row curr . col + 1 ]; var nextNode = new Node ( nextVal curr . row curr . col + 1 ); pq . Add ( nextNode ); maxVal = Math . Max ( maxVal nextVal ); } return new List < int > { minEl maxEl }; // Return result as List
class Node { constructor ( val row col ) { this . val = val ; this . row = row ; this . col = col ; } } // Function to find the smallest range function findSmallestRange ( arr ) { const k = arr . length ; const n = arr [ 0 ]. length ; const heap = new MinHeap (); let maxVal = - Infinity ; // Push the first element of each list into the min-heap for ( let i = 0 ; i < k ; i ++ ) { heap . push ( new Node ( arr [ i ][ 0 ] i 0 )); maxVal = Math . max ( maxVal arr [ i ][ 0 ]); } let minRange = Infinity ; let minEl = - 1 maxEl = - 1 ; while ( true ) { const curr = heap . pop (); const minVal = curr . val ; // Update range if better if ( maxVal - minVal < minRange ) { minRange = maxVal - minVal ; minEl = minVal ; maxEl = maxVal ; } // If we've reached the end of a list break if ( curr . col + 1 === n ) break ; // Push next element from the same list const nextVal = arr [ curr . row ][ curr . col + 1 ]; heap . push ( new Node ( nextVal curr . row curr . col + 1 )); maxVal = Math . max ( maxVal nextVal ); } return [ minEl maxEl ]; } // Min-heap comparator class MinHeap { constructor () { this . heap = []; } push ( node ) { this . heap . push ( node ); this . _heapifyUp (); } pop () { if ( this . size () === 1 ) return this . heap . pop (); const top = this . heap [ 0 ]; this . heap [ 0 ] = this . heap . pop (); this . _heapifyDown (); return top ; } top () { return this . heap [ 0 ]; } size () { return this . heap . length ; } _heapifyUp () { let idx = this . size () - 1 ; while ( idx > 0 ) { let parent = Math . floor (( idx - 1 ) / 2 ); if ( this . heap [ parent ]. val <= this . heap [ idx ]. val ) break ; [ this . heap [ parent ] this . heap [ idx ]] = [ this . heap [ idx ] this . heap [ parent ]]; idx = parent ; } } _heapifyDown () { let idx = 0 ; const n = this . size (); while ( true ) { let left = 2 * idx + 1 ; let right = 2 * idx + 2 ; let smallest = idx ; if ( left < n && this . heap [ left ]. val < this . heap [ smallest ]. val ) { smallest = left ; } if ( right < n && this . heap [ right ]. val < this . heap [ smallest ]. val ) { smallest = right ; } if ( smallest === idx ) break ; [ this . heap [ smallest ] this . heap [ idx ]] = [ this . heap [ idx ] this . heap [ smallest ]]; idx = smallest ; } } } // Driver code const arr = [ [ 4 7 9 12 15 ] [ 0 8 10 14 20 ] [ 6 12 16 30 50 ] ]; const res = findSmallestRange ( arr ); console . log ( res [ 0 ] + ' ' + res [ 1 ]);
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