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Binaire geïndexeerde boom: bereikupdates en puntquery's

Gegeven een array arr[0..n-1]. De volgende handelingen moeten worden uitgevoerd.

    update(l r val): Voeg ‘val’ toe aan alle elementen in de array van [l r].getElement(i): Zoek element in de array geïndexeerd op 'i'.

Aanvankelijk zijn alle elementen in de array 0. Zoekopdrachten kunnen in elke volgorde staan, dat wil zeggen dat er veel updates kunnen plaatsvinden vóór de puntzoekopdracht.



Voorbeeld:

Linux hernoemt map
Input: arr = {0 0 0 0 0} Queries: update : l = 0 r = 4 val = 2 getElement : i = 3 update : l = 3 r = 4 val = 3 getElement : i = 3 Output: Element at 3 is 2 Element at 3 is 5 Explanation: Array after first update becomes {2 2 2 2 2} Array after second update becomes {2 2 2 5 5}

Methode 1 [update: O(n) getElement(): O(1)]

    update(l r val):Herhaal de subarray van l tot r en verhoog alle elementen met val.getElement(i):Om het element op de i-de index te krijgen, retourneert u eenvoudigweg arr[i].

De tijdscomplexiteit in het ergste geval is O(q*n), waarbij q het aantal queries is en n het aantal elementen.  



Methode 2 [update: O(1) getElement(): O(n)]

We kunnen voorkomen dat alle elementen worden bijgewerkt en kunnen slechts 2 indexen van de array bijwerken!

    update(l r val) :Voeg ‘val’ toe aan de leelement en trek ‘val’ af van de (r+1)eelement doe dit voor alle updatequery's.
 arr[l] = arr[l] + val arr[r+1] = arr[r+1] - val
    getElement(i): Om ik te krijgeneelement in de array vindt de som van alle gehele getallen in de array van 0 tot i. (Prefix Sum).

Laten we de updatequery analyseren. Waarom val toevoegen aan leindex? Val toevoegen aan leindex betekent dat alle elementen na l worden verhoogd met val, omdat we voor elk element de prefixsom zullen berekenen. Waarom val aftrekken van (r+1)eindex? Er was een bereikupdate nodig van [lr], maar wat we hebben bijgewerkt is [l n-1], dus we moeten val verwijderen van alle elementen na r, d.w.z. val aftrekken van (r+1)eindex. De val wordt dus toegevoegd aan bereik [lr]. Hieronder vindt u de implementatie van bovenstaande aanpak. 



lexicografische volgorde
C++
// C++ program to demonstrate Range Update // and Point Queries Without using BIT #include    using namespace std; // Updates such that getElement() gets an increased // value when queried from l to r. void update(int arr[] int l int r int val) {  arr[l] += val;  arr[r+1] -= val; } // Get the element indexed at i int getElement(int arr[] int i) {  // To get ith element sum of all the elements  // from 0 to i need to be computed  int res = 0;  for (int j = 0 ; j <= i; j++)  res += arr[j];  return res; } // Driver program to test above function int main() {  int arr[] = {0 0 0 0 0};  int n = sizeof(arr) / sizeof(arr[0]);  int l = 2 r = 4 val = 2;  update(arr l r val);  //Find the element at Index 4  int index = 4;  cout << 'Element at index ' << index << ' is ' <<  getElement(arr index) << endl;  l = 0 r = 3 val = 4;  update(arrlrval);  //Find the element at Index 3  index = 3;  cout << 'Element at index ' << index << ' is ' <<  getElement(arr index) << endl;  return 0; } 
Java
// Java program to demonstrate Range Update  // and Point Queries Without using BIT  class GfG {  // Updates such that getElement() gets an increased  // value when queried from l to r.  static void update(int arr[] int l int r int val)  {   arr[l] += val;  if(r + 1 < arr.length)  arr[r+1] -= val;  }  // Get the element indexed at i  static int getElement(int arr[] int i)  {   // To get ith element sum of all the elements   // from 0 to i need to be computed   int res = 0;   for (int j = 0 ; j <= i; j++)   res += arr[j];   return res;  }  // Driver program to test above function  public static void main(String[] args)  {   int arr[] = {0 0 0 0 0};   int n = arr.length;   int l = 2 r = 4 val = 2;   update(arr l r val);   //Find the element at Index 4   int index = 4;   System.out.println('Element at index ' + index + ' is ' +getElement(arr index));   l = 0;  r = 3;  val = 4;   update(arrlrval);   //Find the element at Index 3   index = 3;   System.out.println('Element at index ' + index + ' is ' +getElement(arr index));  } }  
Python3
# Python3 program to demonstrate Range  # Update and PoQueries Without using BIT  # Updates such that getElement() gets an  # increased value when queried from l to r.  def update(arr l r val): arr[l] += val if r + 1 < len(arr): arr[r + 1] -= val # Get the element indexed at i  def getElement(arr i): # To get ith element sum of all the elements  # from 0 to i need to be computed  res = 0 for j in range(i + 1): res += arr[j] return res # Driver Code if __name__ == '__main__': arr = [0 0 0 0 0] n = len(arr) l = 2 r = 4 val = 2 update(arr l r val) # Find the element at Index 4  index = 4 print('Element at index' index 'is' getElement(arr index)) l = 0 r = 3 val = 4 update(arr l r val) # Find the element at Index 3  index = 3 print('Element at index' index 'is' getElement(arr index)) # This code is contributed by PranchalK 
C#
// C# program to demonstrate Range Update  // and Point Queries Without using BIT  using System; class GfG  {  // Updates such that getElement()  // gets an increased value when // queried from l to r.  static void update(int []arr int l   int r int val)  {   arr[l] += val;   if(r + 1 < arr.Length)   arr[r + 1] -= val;  }  // Get the element indexed at i  static int getElement(int []arr int i)  {   // To get ith element sum of all the elements   // from 0 to i need to be computed   int res = 0;   for (int j = 0 ; j <= i; j++)   res += arr[j];   return res;  }  // Driver code  public static void Main(String[] args)  {   int []arr = {0 0 0 0 0};   int n = arr.Length;   int l = 2 r = 4 val = 2;   update(arr l r val);   //Find the element at Index 4   int index = 4;   Console.WriteLine('Element at index ' +   index + ' is ' +  getElement(arr index));   l = 0;   r = 3;   val = 4;   update(arrlrval);   //Find the element at Index 3   index = 3;   Console.WriteLine('Element at index ' +   index + ' is ' +  getElement(arr index));  }  }  // This code is contributed by PrinciRaj1992 
PHP
 // PHP program to demonstrate Range Update  // and Point Queries Without using BIT  // Updates such that getElement() gets an  // increased value when queried from l to r.  function update(&$arr $l $r $val) { $arr[$l] += $val; if($r + 1 < sizeof($arr)) $arr[$r + 1] -= $val; } // Get the element indexed at i  function getElement(&$arr $i) { // To get ith element sum of all the elements  // from 0 to i need to be computed  $res = 0; for ($j = 0 ; $j <= $i; $j++) $res += $arr[$j]; return $res; } // Driver Code $arr = array(0 0 0 0 0); $n = sizeof($arr); $l = 2; $r = 4; $val = 2; update($arr $l $r $val); // Find the element at Index 4  $index = 4; echo('Element at index ' . $index . ' is ' . getElement($arr $index) . 'n'); $l = 0; $r = 3; $val = 4; update($arr $l $r $val); // Find the element at Index 3  $index = 3; echo('Element at index ' . $index . ' is ' . getElement($arr $index)); // This code is contributed by Code_Mech ?> 
JavaScript
//JavaScript program to demonstrate Range Update // and Point Queries Without using BIT // Updates such that getElement() gets an increased // value when queried from l to r. function update(arr l r val) {  arr[l] += val;  arr[r+1] -= val; } // Get the element indexed at i function getElement(rr i) {  // To get ith element sum of all the elements  // from 0 to i need to be computed  let res = 0;  for (let j = 0 ; j <= i; j++)  res += arr[j];  return res; } // Driver program to test above function  let arr = [0 0 0 0 0];  let n = arr.length;  let l = 2 r = 4 val = 2;  update(arr l r val);  // Find the element at Index 4  let index = 4;  console.log('Element at index 'index' is 'getElement(arr index));  l = 0 r = 3 val = 4;  update(arrlrval);  // Find the element at Index 3  index = 3;  console.log('Element at index 'index' is 'getElement(arr index)); // This code is contributed by vikkycirus 

Uitgang:

Element at index 4 is 2 Element at index 3 is 6

Tijdcomplexiteit : O(q*n) waarbij q het aantal zoekopdrachten is.  

Hulpruimte: Op)

Methode 3 (binair geïndexeerde boom gebruiken)

Bij methode 2 hebben we gezien dat het probleem kan worden beperkt tot het bijwerken en prefixen van somquery's. Dat hebben wij gezien BIT kan worden gebruikt voor het uitvoeren van update- en prefix-query's in O(Logn)-tijd. Hieronder vindt u de implementatie. 

C++
// C++ code to demonstrate Range Update and // Point Queries on a Binary Index Tree #include    using namespace std; // Updates a node in Binary Index Tree (BITree) at given index // in BITree. The given value 'val' is added to BITree[i] and // all of its ancestors in tree. void updateBIT(int BITree[] int n int index int val) {  // index in BITree[] is 1 more than the index in arr[]  index = index + 1;  // Traverse all ancestors and add 'val'  while (index <= n)  {  // Add 'val' to current node of BI Tree  BITree[index] += val;  // Update index to that of parent in update View  index += index & (-index);  } } // Constructs and returns a Binary Indexed Tree for given // array of size n. int *constructBITree(int arr[] int n) {  // Create and initialize BITree[] as 0  int *BITree = new int[n+1];  for (int i=1; i<=n; i++)  BITree[i] = 0;  // Store the actual values in BITree[] using update()  for (int i=0; i<n; i++)  updateBIT(BITree n i arr[i]);  // Uncomment below lines to see contents of BITree[]  //for (int i=1; i<=n; i++)  // cout << BITree[i] << ' ';  return BITree; } // SERVES THE PURPOSE OF getElement() // Returns sum of arr[0..index]. This function assumes // that the array is preprocessed and partial sums of // array elements are stored in BITree[] int getSum(int BITree[] int index) {  int sum = 0; // Initialize result  // index in BITree[] is 1 more than the index in arr[]  index = index + 1;  // Traverse ancestors of BITree[index]  while (index>0)  {  // Add current element of BITree to sum  sum += BITree[index];  // Move index to parent node in getSum View  index -= index & (-index);  }  return sum; } // Updates such that getElement() gets an increased // value when queried from l to r. void update(int BITree[] int l int r int n int val) {  // Increase value at 'l' by 'val'  updateBIT(BITree n l val);  // Decrease value at 'r+1' by 'val'  updateBIT(BITree n r+1 -val); } // Driver program to test above function int main() {  int arr[] = {0 0 0 0 0};  int n = sizeof(arr)/sizeof(arr[0]);  int *BITree = constructBITree(arr n);  // Add 2 to all the element from [24]  int l = 2 r = 4 val = 2;  update(BITree l r n val);  // Find the element at Index 4  int index = 4;  cout << 'Element at index ' << index << ' is ' <<  getSum(BITreeindex) << 'n';  // Add 2 to all the element from [03]  l = 0 r = 3 val = 4;  update(BITree l r n val);  // Find the element at Index 3  index = 3;  cout << 'Element at index ' << index << ' is ' <<  getSum(BITreeindex) << 'n' ;  return 0; } 
Java
/* Java code to demonstrate Range Update and * Point Queries on a Binary Index Tree. * This method only works when all array * values are initially 0.*/ class GFG {  // Max tree size  final static int MAX = 1000;  static int BITree[] = new int[MAX];  // Updates a node in Binary Index  // Tree (BITree) at given index  // in BITree. The given value 'val'  // is added to BITree[i] and  // all of its ancestors in tree.  public static void updateBIT(int n   int index   int val)  {  // index in BITree[] is 1   // more than the index in arr[]  index = index + 1;  // Traverse all ancestors   // and add 'val'  while (index <= n)  {  // Add 'val' to current   // node of BITree  BITree[index] += val;  // Update index to that   // of parent in update View  index += index & (-index);  }  }  // Constructs Binary Indexed Tree   // for given array of size n.  public static void constructBITree(int arr[]  int n)  {  // Initialize BITree[] as 0  for(int i = 1; i <= n; i++)  BITree[i] = 0;  // Store the actual values   // in BITree[] using update()  for(int i = 0; i < n; i++)  updateBIT(n i arr[i]);  // Uncomment below lines to   // see contents of BITree[]  // for (int i=1; i<=n; i++)  // cout << BITree[i] << ' ';  }  // SERVES THE PURPOSE OF getElement()  // Returns sum of arr[0..index]. This   // function assumes that the array is  // preprocessed and partial sums of  // array elements are stored in BITree[]  public static int getSum(int index)  {  int sum = 0; //Initialize result  // index in BITree[] is 1 more   // than the index in arr[]  index = index + 1;  // Traverse ancestors  // of BITree[index]  while (index > 0)  {  // Add current element   // of BITree to sum  sum += BITree[index];  // Move index to parent   // node in getSum View  index -= index & (-index);  }  // Return the sum  return sum;  }  // Updates such that getElement()   // gets an increased value when   // queried from l to r.  public static void update(int l int r   int n int val)  {  // Increase value at   // 'l' by 'val'  updateBIT(n l val);  // Decrease value at  // 'r+1' by 'val'  updateBIT(n r + 1 -val);  }  // Driver Code  public static void main(String args[])  {  int arr[] = {0 0 0 0 0};  int n = arr.length;  constructBITree(arrn);  // Add 2 to all the  // element from [24]  int l = 2 r = 4 val = 2;  update(l r n val);  int index = 4;  System.out.println('Element at index '+   index + ' is '+   getSum(index));  // Add 2 to all the   // element from [03]  l = 0; r = 3; val = 4;  update(l r n val);  // Find the element  // at Index 3  index = 3;  System.out.println('Element at index '+   index + ' is '+   getSum(index));  } } // This code is contributed // by Puneet Kumar. 
Python3
# Python3 code to demonstrate Range Update and # PoQueries on a Binary Index Tree # Updates a node in Binary Index Tree (BITree) at given index # in BITree. The given value 'val' is added to BITree[i] and # all of its ancestors in tree. def updateBIT(BITree n index val): # index in BITree[] is 1 more than the index in arr[] index = index + 1 # Traverse all ancestors and add 'val' while (index <= n): # Add 'val' to current node of BI Tree BITree[index] += val # Update index to that of parent in update View index += index & (-index) # Constructs and returns a Binary Indexed Tree for given # array of size n. def constructBITree(arr n): # Create and initialize BITree[] as 0 BITree = [0]*(n+1) # Store the actual values in BITree[] using update() for i in range(n): updateBIT(BITree n i arr[i]) return BITree # SERVES THE PURPOSE OF getElement() # Returns sum of arr[0..index]. This function assumes # that the array is preprocessed and partial sums of # array elements are stored in BITree[] def getSum(BITree index): sum = 0 # Initialize result # index in BITree[] is 1 more than the index in arr[] index = index + 1 # Traverse ancestors of BITree[index] while (index > 0): # Add current element of BITree to sum sum += BITree[index] # Move index to parent node in getSum View index -= index & (-index) return sum # Updates such that getElement() gets an increased # value when queried from l to r. def update(BITree l r n val): # Increase value at 'l' by 'val' updateBIT(BITree n l val) # Decrease value at 'r+1' by 'val' updateBIT(BITree n r+1 -val) # Driver code arr = [0 0 0 0 0] n = len(arr) BITree = constructBITree(arr n) # Add 2 to all the element from [24] l = 2 r = 4 val = 2 update(BITree l r n val) # Find the element at Index 4 index = 4 print('Element at index' index 'is' getSum(BITree index)) # Add 2 to all the element from [03] l = 0 r = 3 val = 4 update(BITree l r n val) # Find the element at Index 3 index = 3 print('Element at index' index 'is' getSum(BITreeindex)) # This code is contributed by mohit kumar 29 
C#
using System; /* C# code to demonstrate Range Update and  * Point Queries on a Binary Index Tree.  * This method only works when all array  * values are initially 0.*/ public class GFG {  // Max tree size   public const int MAX = 1000;  public static int[] BITree = new int[MAX];  // Updates a node in Binary Index   // Tree (BITree) at given index   // in BITree. The given value 'val'   // is added to BITree[i] and   // all of its ancestors in tree.   public static void updateBIT(int n int index int val)  {  // index in BITree[] is 1   // more than the index in arr[]   index = index + 1;  // Traverse all ancestors   // and add 'val'   while (index <= n)  {  // Add 'val' to current   // node of BITree   BITree[index] += val;  // Update index to that   // of parent in update View   index += index & (-index);  }  }  // Constructs Binary Indexed Tree   // for given array of size n.   public static void constructBITree(int[] arr int n)  {  // Initialize BITree[] as 0   for (int i = 1; i <= n; i++)  {  BITree[i] = 0;  }  // Store the actual values   // in BITree[] using update()   for (int i = 0; i < n; i++)  {  updateBIT(n i arr[i]);  }  // Uncomment below lines to   // see contents of BITree[]   // for (int i=1; i<=n; i++)   // cout << BITree[i] << ' ';   }  // SERVES THE PURPOSE OF getElement()   // Returns sum of arr[0..index]. This   // function assumes that the array is   // preprocessed and partial sums of   // array elements are stored in BITree[]   public static int getSum(int index)  {  int sum = 0; //Initialize result  // index in BITree[] is 1 more   // than the index in arr[]   index = index + 1;  // Traverse ancestors   // of BITree[index]   while (index > 0)  {  // Add current element   // of BITree to sum   sum += BITree[index];  // Move index to parent   // node in getSum View   index -= index & (-index);  }  // Return the sum   return sum;  }  // Updates such that getElement()   // gets an increased value when   // queried from l to r.   public static void update(int l int r int n int val)  {  // Increase value at   // 'l' by 'val'   updateBIT(n l val);  // Decrease value at   // 'r+1' by 'val'   updateBIT(n r + 1 -val);  }  // Driver Code   public static void Main(string[] args)  {  int[] arr = new int[] {0 0 0 0 0};  int n = arr.Length;  constructBITree(arrn);  // Add 2 to all the   // element from [24]   int l = 2 r = 4 val = 2;  update(l r n val);  int index = 4;  Console.WriteLine('Element at index ' + index + ' is ' + getSum(index));  // Add 2 to all the   // element from [03]   l = 0;  r = 3;  val = 4;  update(l r n val);  // Find the element   // at Index 3   index = 3;  Console.WriteLine('Element at index ' + index + ' is ' + getSum(index));  } }  // This code is contributed by Shrikant13 
JavaScript
// Updates a node in Binary Index Tree (BITree) at given index // in BITree. The given value 'val' is added to BITree[i] and // all of its ancestors in tree. function updateBIT(BITree n index val) {  // index in BITree[] is 1 more than the index in arr[]  index = index + 1;  // Traverse all ancestors and add 'val'  while (index <= n) {  // Add 'val' to current node of BI Tree  BITree[index] += val;  // Update index to that of parent in update View  index += index & (-index);  } } // Constructs and returns a Binary Indexed Tree for given // array of size n. function constructBITree(arr n) {  // Create and initialize BITree[] as 0  let BITree = new Array(n+1).fill(0);  // Store the actual values in BITree[] using update()  for (let i = 0; i < n; i++) {  updateBIT(BITree n i arr[i]);  }  return BITree; } // SERVES THE PURPOSE OF getElement() // Returns sum of arr[0..index]. This function assumes // that the array is preprocessed and partial sums of // array elements are stored in BITree[] function getSum(BITree index) {  let sum = 0; // Initialize result  // index in BITree[] is 1 more than the index in arr[]  index = index + 1;  // Traverse ancestors of BITree[index]  while (index > 0) {  // Add current element of BITree to sum  sum += BITree[index];  // Move index to parent node in getSum View  index -= index & (-index);  }  return sum; } // Updates such that getElement() gets an increased // value when queried from l to r. function update(BITree l r n val) {  // Increase value at 'l' by 'val'  updateBIT(BITree n l val);  // Decrease value at 'r+1' by 'val'  updateBIT(BITree n r+1 -val); } // Test the functions let arr = [0 0 0 0 0]; let n = arr.length; let BITree = constructBITree(arr n); // Add 2 to all the element from [24] let l = 2 r = 4 val = 2; update(BITree l r n val); // Find the element at Index 4 let index = 4; console.log(`Element at index ${index} is ${getSum(BITreeindex)}`); // Add 2 to all the element from [03] l = 0 r = 3 val = 4; update(BITree l r n val); // Find the element at Index 3 index = 3; console.log(`Element at index ${index} is ${getSum(BITreeindex)}`); 

Uitgang:

tostring in Java
Element at index 4 is 2 Element at index 3 is 6

Tijdcomplexiteit: O(q * log n) + O(n * log n) waarbij q het aantal zoekopdrachten is. 

Hulpruimte: Op)

Methode 1 is efficiënt als de meeste query's getElement() zijn. Methode 2 is efficiënt als de meeste query's updates() zijn en methode 3 heeft de voorkeur als er een combinatie van beide query's is.