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Najmanjši razpon z elementi s razvrščenih seznamov

Poskusite na praksi GFG ' title=

Glede na 2D celotno matriko arr [] [] vrstnega reda k * n kjer je vsaka vrstica razvrščeno v naraščajočem vrstnem redu. Vaša naloga je najti najmanjši razpon, ki vključuje vsaj en element iz vsakega od  K  sezname. Če najdemo več takšnih razponov, se vrne prva.

Primeri:  



Vnos: arr [] [] = [[4 7 9 12 15]
[0 8 10 14 20]
[6 12 16 30 50]]
Izhod: 6 8
Pojasnilo: Najmanjši razpon je oblikovan s številko 7 s prvega seznama 8 s drugega seznama in 6 s tretjega seznama.

Vnos: arr [] [] = [[2 4]
[1 7]
[20 40]]
Izhod: 4 20
Pojasnilo: Območje [4 20] vsebuje 4 7 20, ki vsebuje element iz vseh treh nizov.

Tabela vsebine



[Naivni pristop] - Uporaba K kazalcev - o (n k^2) čas in o (k) prostor

Ideja je, da K -kazalce obdržite za vsak seznam, ki se začne pri indeksu 0. Na vsakem koraku vzemite min in max trenutnih K elementov, ki tvorijo obseg. Do zmanjšati obseg Moramo Povečajte min vrednost Ker ne moremo zmanjšati maksimuma (vsi kazalci se začnejo pri 0). Torej premaknite kazalec seznama, ki ima trenutni minimum in posodobite obseg. Ponovite, dokler se en seznam ne izčrpa.

Korak za korakom Izvedba:

  • Ustvari seznam kazalcev Eno za vsak vhodni seznam, ki se začne pri indeksu 0.
  • Ponovite postopek dokler eden od kazalcev ne doseže konca svojega seznama.
  • Na vsakem koraku Izberite trenutne elemente poudarjajo vsi kazalci.
  • Poiščite minimalno in največjo med temi elementi.
  • Izračunajte območje z uporabo vrednosti min in max.
  • Če je ta razpon manjši kot prejšnja najboljša posodobitev odgovora.
  • Premaknite kazalec seznama, ki je imel minimalni element.
  • Ustavite se, ko je kateri koli seznam izčrpan in vrnite najboljši najdeni doseg.
C++
// 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]); 

Izhod
6 8

[Boljši pristop] Uporaba dveh kazalcev - o (n*k log (n*k)) čas in o (n*k) prostor

Ideja je najti najmanjšo težavo z dosegom tako, da jo spremenite v težavo z drsnim oknom nad združenim in razvrščenim seznamom vseh elementov s seznamov vhodov. Vsak element je shranjen skupaj z originalnim indeksom seznama, da sledi svojemu viru. Po razvrščanju kombiniranega seznama po vrednosti dva kazalca (leftinright) se uporabljajo za določitev okna, ki se premika po seznamu. Ko okno razširi frekvenčni zemljevid, sledi, koliko edinstvenih seznamov je predstavljenih. Ko okno vključuje vsaj eno številko s vsakega seznama, ga algoritem poskuša skrčiti z leve, da bi našli manjši veljaven obseg. Kot rezultat se vrne najmanjši takšen obseg, ki ga najdemo med tem postopkom.



C++
#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]); 

Izhod
6 8

[Učinkovit pristop] - Uporaba min -heap - o (n k log k) čas in o (k) prostor

Min-hep lahko uporabite za iskanje minimalne vrednosti v logaritmičnem času ali dnevniku K namesto v linearnem času. Če želite najti največjo vrednost, na začetku inicializiramo največjo vrednost vseh indeksov 0. Za preostale največje vrednosti v zanki preprosto primerjamo trenutno največjo vrednost z naslednjim elementom s seznama, s katerega se odstrani Min Element. Preostali pristop ostaja enak. 

Korak za korakom Izvedba:

  • Min-hep lahko uporabite za iskanje minimalne vrednosti v logaritmičnem času ali dnevniku K namesto v linearnem času. Če želite najti največjo vrednost, na začetku inicializiramo največjo vrednost vseh indeksov 0. Za preostale največje vrednosti v zanki preprosto primerjamo trenutno največjo vrednost z naslednjim elementom s seznama, s katerega se odstrani Min Element. Preostali pristop ostaja enak. 

    Ustvarite min-hep za shranjevanje elementov K enega iz vsakega matrika in spremenljivko Minrange inicializirano na največjo vrednost in ohranite tudi spremenljivko Max za shranjevanje največjega števila.

  • Min-hep lahko uporabite za iskanje minimalne vrednosti v logaritmičnem času ali dnevniku K namesto v linearnem času. Če želite najti največjo vrednost, na začetku inicializiramo največjo vrednost vseh indeksov 0. Za preostale največje vrednosti v zanki preprosto primerjamo trenutno največjo vrednost z naslednjim elementom s seznama, s katerega se odstrani Min Element. Preostali pristop ostaja enak. 

    Sprva prvi element postavite s vsakega seznama in shranite največjo vrednost v Max .

  • Min-hep lahko uporabite za iskanje minimalne vrednosti v logaritmičnem času ali dnevniku K namesto v linearnem času. Če želite najti največjo vrednost, na začetku inicializiramo največjo vrednost vseh indeksov 0. Za preostale največje vrednosti v zanki preprosto primerjamo trenutno največjo vrednost z naslednjim elementom s seznama, s katerega se odstrani Min Element. Preostali pristop ostaja enak. 

    Naslednje korake ponovite, dokler ne izpuste vsaj enega seznama: 

    • poiščite minimalno vrednost oz min Uporabite zgornji del ali koren min gomile, ki je najmanjši element.
    • Zdaj posodobite Minrange Če je tok (max-min) manjši od Minrange .
    • Odstranite zgornji ali korenski element iz čakalne vrste Priority vstavite naslednji element s seznama, ki vsebuje element min
    • Posodobite max z vstavljenim novim elementom, če je nov element večji od prejšnjega max -a.
Min-hep lahko uporabite za iskanje minimalne vrednosti v logaritmičnem času ali dnevniku K namesto v linearnem času. Če želite najti največjo vrednost, na začetku inicializiramo največjo vrednost vseh indeksov 0. Za preostale največje vrednosti v zanki preprosto primerjamo trenutno največjo vrednost z naslednjim elementom s seznama, s katerega se odstrani Min Element. Preostali pristop ostaja enak. 

C++

#include    using namespace std; // Struct to represent elements in the heap struct Node {  int val row col;  bool operator>(const Node& other) const {  return val > other.val;  } }; // Function to find the smallest range vector<int> findSmallestRange(vector<vector<int>>& arr) {  int N = arr.size(); // Number of rows  int K = arr[0].size(); // Number of columns (same for each row)  priority_queue<Node vector<Node> greater<Node>> pq;  int maxVal = INT_MIN;  // Push the first element of each list into the min-heap  for (int i = 0; i < N; i++) {  pq.push({arr[i][0] i 0});  maxVal = max(maxVal arr[i][0]);  }  int minRange = INT_MAX minEl maxEl;  while (true) {  Node curr = pq.top(); pq.pop();  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 == K) break;  // Push next element from the same list  int nextVal = arr[curr.row][curr.col + 1];  pq.push({nextVal curr.row curr.col + 1});  maxVal = max(maxVal nextVal);  }  return {minEl maxEl}; } // 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 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));  } } 
Python
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]) 
C#
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    }  // Helper to get the minimum element (first element in SortedSet)  static Node GetMin(SortedSet<Node> pq) {  foreach (var node in pq)  return node;  return null;  }  // Driver code  static void Main() {  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
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]); 

Izhod
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