Array range queries over range queries
Given an array of size n and a give set of commands of size m. The commands are enumerated from 1 to m. These commands can be of the following two types of commands:
- Type 1 [l r (1 <= l <= r <= n)] : Increase all elements of the array by one, whose indices belongs to the range [l, r]. In these queries of the index is inclusive in the range.
- Type 2 [l r (1 <= l <= r <= m)] : Execute all the commands whose indices are in the range [l, r]. In these queries of the index is inclusive in the range. It's guaranteed that r is strictly less than the enumeration/number of the current command.
Note : The array indexing is from 1 as per the problem statement.
Example 1
Input : 5 5 1 1 2 1 4 5 2 1 2 2 1 3 2 3 4 Output : 7 7 0 7 7
Explanation of Example 1 :
Our array initially is of size 5 whose each element has been initialized to 0.
So now the question states that we have 5 queries for the above example.
- Query 1 is of type 1 : As stated above we will simply increment the array indices by 1 the given indices are 1 and 2 so after the execution of the first our array turns down to be 1 1 0 0 0 .
- Query 2 is of type 1 : As stated above we will simply increment the array indices by 1
the given indices are 4 and 5 so after the execution of the first our array turns down to be 1 1 0 1 1 . - Query 3 is of type 2 : As stated in the definition of this type of query we will execute the queries stated in the range i.e. we will operate the queries instead of the array. The range given is 1 and 2 so we will execute queries 1 and 2 again i.e. we will use repetitive approach for the type 2 queries so we will execute query 1 again and our array will be 2 2 0 1 1. Now when we execute the query we will execute query 2 and our resultant array will be 2 2 0 2 2 .
- Query 4 is of type 2 : As stated in the definition of this type of query we will execute the queries stated in the range i.e. we will operate the queries instead of the array. The range given is 1 and 3 so we will execute queries 1, 2 and 3 again i.e. using repetitive approach queries 1, 2 and 3 will be executed. After the execution of the query 1 again the array will be 3 3 0 2 2 . After the execution of the query 2 again the array will be 3 3 0 3 3 . Now due to query 3 inclusive in the range we will execute query 3 the resultant array will be 4 4 0 4 4 . As explained above.
- Query 5 is of type 2 : The last query will execute the 3rd and 4th query which has been explained above. After the execution of the 3rd query our array will be 5 5 0 5 5 . And after the execution of the 4th query i.e. execution of query 1, 2 and 3 our array will be 7 7 0 7 7 The above is the desired result
Example 2
Input : 1 2 1 1 1 1 1 1 Output : 2
Explanation of the example 2:
Our array initially is of size 1 whose each element has been initialized to 0.
So now the question states that we have 2 queries for the above example.
- Query 1 is of type 1 : As stated above we will simply increment the array indices by 1 the given indices are 1 and 1 so after the execution of the first our array turns down to be 1 .
- Query 2 is of type 1 : As stated above we will simply increment the array indices by 1 the given indices are 1 and 1 so after the execution of the first our array turns down to be 2 . This gives us the desired result
Method 1 :
This method is the brute force method where by simple recursion is applied on the type 2 queries and for type 1 queries simple increment in the array index is performed.
C++
// CPP program to perform range queries over range // queries. #include <bits/stdc++.h> using namespace std; // Function to execute type 1 query void type1( int arr[], int start, int limit) { // incrementing the array by 1 for type // 1 queries for ( int i = start; i <= limit; i++) arr[i]++; } // Function to execute type 2 query void type2( int arr[], int query[][3], int start, int limit) { for ( int i = start; i <= limit; i++) { // If the query is of type 1 function // call to type 1 query if (query[i][0] == 1) type1(arr, query[i][1], query[i][2]); // If the query is of type 2 recursive call // to type 2 query else if (query[i][0] == 2) type2(arr, query, query[i][1], query[i][2]); } } // Driver code int main() { // Input size of array amd number of queries int n = 5, m = 5; int arr[n + 1]; for ( int i = 1; i <= n; i++) arr[i] = 0; // Build query matrix int temp[15] = { 1, 1, 2, 1, 4, 5, 2, 1, 2, 2, 1, 3, 2, 3, 4 }; int query[5][3]; int j = 0; for ( int i = 1; i <= m; i++) { query[i][0] = temp[j++]; query[i][1] = temp[j++]; query[i][2] = temp[j++]; } // Perform queries for ( int i = 1; i <= m; i++) if (query[i][0] == 1) type1(arr, query[i][1], query[i][2]); else if (query[i][0] == 2) type2(arr, query, query[i][1], query[i][2]); // printing the result for ( int i = 1; i <= n; i++) cout << arr[i] << " " ; return 0; } |
Java
// Java program to perform range queries // over range queries. import java.util.*; class GFG { // Function to execute type 1 query static void type1( int [] arr, int start, int limit) { // incrementing the array by 1 for type // 1 queries for ( int i = start; i <= limit; i++) arr[i]++; } // Function to execute type 2 query static void type2( int [] arr, int [][] query, int start, int limit) { for ( int i = start; i <= limit; i++) { // If the query is of type 1 function // call to type 1 query if (query[i][ 0 ] == 1 ) type1(arr, query[i][ 1 ], query[i][ 2 ]); // If the query is of type 2 recursive call // to type 2 query else if (query[i][ 0 ] == 2 ) type2(arr, query, query[i][ 1 ], query[i][ 2 ]); } } // Driver Code public static void main(String[] args) { // Input size of array amd number of queries int n = 5 , m = 5 ; int [] arr = new int [n + 1 ]; // Build query matrix int [] temp = { 1 , 1 , 2 , 1 , 4 , 5 , 2 , 1 , 2 , 2 , 1 , 3 , 2 , 3 , 4 }; int [][] query = new int [ 6 ][ 4 ]; int j = 0 ; for ( int i = 1 ; i <= m; i++) { query[i][ 0 ] = temp[j++]; query[i][ 1 ] = temp[j++]; query[i][ 2 ] = temp[j++]; } // Perform queries for ( int i = 1 ; i <= m; i++) if (query[i][ 0 ] == 1 ) type1(arr, query[i][ 1 ], query[i][ 2 ]); else if (query[i][ 0 ] == 2 ) type2(arr, query, query[i][ 1 ], query[i][ 2 ]); // printing the result for ( int i = 1 ; i <= n; i++) System.out.print(arr[i] + " " ); System.out.println(); } } // This code is contributed by // sanjeev2552 |
Python3
# Python3 program to perform range # queries over range queries. # Function to execute type 1 query def type1(arr, start, limit): # incrementing the array by 1 # for type 1 queries for i in range (start, limit + 1 ): arr[i] + = 1 # Function to execute type 2 query def type2(arr, query, start, limit): for i in range (start, limit + 1 ): # If the query is of type 1 # function call to type 1 query if (query[i][ 0 ] = = 1 ): type1(arr, query[i][ 1 ], query[i][ 2 ]) # If the query is of type 2 # recursive call to type 2 query elif (query[i][ 0 ] = = 2 ): type2(arr, query, query[i][ 1 ], query[i][ 2 ]) # Driver code # Input size of array amd # number of queries n = 5 m = 5 arr = [ 0 for i in range (n + 1 )] # Build query matrix temp = [ 1 , 1 , 2 , 1 , 4 , 5 , 2 , 1 , 2 , 2 , 1 , 3 , 2 , 3 , 4 ] query = [[ 0 for i in range ( 3 )] for j in range ( 6 )] j = 0 for i in range ( 1 , m + 1 ): query[i][ 0 ] = temp[j] j + = 1 query[i][ 1 ] = temp[j] j + = 1 query[i][ 2 ] = temp[j] j + = 1 # Perform queries for i in range ( 1 , m + 1 ): if (query[i][ 0 ] = = 1 ): type1(arr, query[i][ 1 ], query[i][ 2 ]) elif (query[i][ 0 ] = = 2 ): type2(arr, query, query[i][ 1 ], query[i][ 2 ]) # printing the result for i in range ( 1 , n + 1 ): print (arr[i], end = " " ) # This code is contributed # by mohit kumar |
Output:
7 7 0 7 7
The Time complexity of the above code is O(2 ^ m)
Method 2 :
In this method we use an extra array for creating the record array to find the number of time a particular query is being executed and after creating the record array we simply execute the queries of type 1 and the contains of the record array is simply added to the main array the and this would give us the resultant array.
C++
// CPP program to perform range queries over range // queries. #include <bits/stdc++.h> using namespace std; // Function to create the record array void record_sum( int record[], int l, int r, int n, int adder) { for ( int i = l; i <= r; i++) record[i] += adder; } // Driver Code int main() { int n = 5, m = 5; int arr[n]; // Build query matrix memset (arr, 0, sizeof arr); int query[5][3] = { { 1, 1, 2 }, { 1, 4, 5 }, { 2, 1, 2 }, { 2, 1, 3 }, { 2, 3, 4 } }; int record[m]; memset (record, 0, sizeof record); for ( int i = m - 1; i >= 0; i--) { // If query is of type 2 then function // call to record_sum if (query[i][0] == 2) record_sum(record, query[i][1] - 1, query[i][2] - 1, m, record[i] + 1); // If query is of type 1 then simply add // 1 to the record array else record_sum(record, i, i, m, 1); } // for type 1 queries adding the contains of // record array to the main array record array for ( int i = 0; i < m; i++) { if (query[i][0] == 1) record_sum(arr, query[i][1] - 1, query[i][2] - 1, n, record[i]); } // printing the array for ( int i = 0; i < n; i++) cout << arr[i] << ' ' ; return 0; } |
Java
// Java program to perform range queries // over range queries. import java.util.Arrays; class GFG { // Function to create the record array static void record_sum( int record[], int l, int r, int n, int adder) { for ( int i = l; i <= r; i++) { record[i] += adder; } } // Driver Code public static void main(String[] args) { int n = 5 , m = 5 ; int arr[] = new int [n]; // Build query matrix Arrays.fill(arr, 0 ); int query[][] = {{ 1 , 1 , 2 }, { 1 , 4 , 5 }, { 2 , 1 , 2 }, { 2 , 1 , 3 }, { 2 , 3 , 4 }}; int record[] = new int [m]; Arrays.fill(record, 0 ); for ( int i = m - 1 ; i >= 0 ; i--) { // If query is of type 2 then function // call to record_sum if (query[i][ 0 ] == 2 ) { record_sum(record, query[i][ 1 ] - 1 , query[i][ 2 ] - 1 , m, record[i] + 1 ); } // If query is of type 1 then // simply add 1 to the record array else { record_sum(record, i, i, m, 1 ); } } // for type 1 queries adding the contains of // record array to the main array record array for ( int i = 0 ; i < m; i++) { if (query[i][ 0 ] == 1 ) { record_sum(arr, query[i][ 1 ] - 1 , query[i][ 2 ] - 1 , n, record[i]); } } // printing the array for ( int i = 0 ; i < n; i++) { System.out.print(arr[i] + " " ); } } } // This code is contributed // by Princi Singh |
C#
// C# program to perform range queries // over range queries. using System; class GFG { // Function to create the record array static void record_sum( int []record, int l, int r, int n, int adder) { for ( int i = l; i <= r; i++) { record[i] += adder; } } // Driver Code public static void Main(String[] args) { int n = 5, m = 5; int []arr = new int [n]; // Build query matrix int [,]query = {{1, 1, 2}, {1, 4, 5}, {2, 1, 2}, {2, 1, 3}, {2, 3, 4}}; int []record = new int [m]; for ( int i = m - 1; i >= 0; i--) { // If query is of type 2 then function // call to record_sum if (query[i,0] == 2) { record_sum(record, query[i,1] - 1, query[i,2] - 1, m, record[i] + 1); } // If query is of type 1 then // simply add 1 to the record array else { record_sum(record, i, i, m, 1); } } // for type 1 queries adding the contains of // record array to the main array record array for ( int i = 0; i < m; i++) { if (query[i, 0] == 1) { record_sum(arr, query[i, 1] - 1, query[i, 2] - 1, n, record[i]); } } // printing the array for ( int i = 0; i < n; i++) { Console.Write(arr[i] + " " ); } } } // This code is contributed by Rajput-Ji |
Output :
7 7 0 7 7
The Time complexity of the above code is O(n^2)
Method 3 :
This method has been made more efficient by applying square root decomposition to the record array.
C++
// CPP program to perform range queries over range // queries. #include <bits/stdc++.h> #define max 10000 using namespace std; // For prefix sum array void update( int arr[], int l) { arr[l] += arr[l - 1]; } // This function is used to apply square root // decomposition in the record array void record_func( int block_size, int block[], int record[], int l, int r, int value) { // traversing first block in range while (l < r && l % block_size != 0 && l != 0) { record[l] += value; l++; } // traversing completely overlapped blocks in range while (l + block_size <= r + 1) { block[l / block_size] += value; l += block_size; } // traversing last block in range while (l <= r) { record[l] += value; l++; } } // Function to print the resultant array void print( int arr[], int n) { for ( int i = 0; i < n; i++) cout << arr[i] << " " ; } // Driver code int main() { int n = 5, m = 5; int arr[n], record[m]; int block_size = sqrt (m); int block[max]; int command[5][3] = { { 1, 1, 2 }, { 1, 4, 5 }, { 2, 1, 2 }, { 2, 1, 3 }, { 2, 3, 4 } }; memset (arr, 0, sizeof arr); memset (record, 0, sizeof record); memset (block, 0, sizeof block); for ( int i = m - 1; i >= 0; i--) { // If query is of type 2 then function // call to record_func if (command[i][0] == 2) { int x = i / (block_size); record_func(block_size, block, record, command[i][1] - 1, command[i][2] - 1, (block[x] + record[i] + 1)); } // If query is of type 1 then simply add // 1 to the record array else record[i]++; } // Merging the value of the block in the record array for ( int i = 0; i < m; i++) { int check = (i / block_size); record[i] += block[check]; } for ( int i = 0; i < m; i++) { // If query is of type 1 then the array // elements are over-written by the record // array if (command[i][0] == 1) { arr[command[i][1] - 1] += record[i]; if ((command[i][2] - 1) < n - 1) arr[(command[i][2])] -= record[i]; } } // The prefix sum of the array for ( int i = 1; i < n; i++) update(arr, i); // Printing the resultant array print(arr, n); return 0; } |
Java
// Java program to perform range queries over range // queries. public class GFG { static final int max = 10000 ; // For prefix sum array static void update( int arr[], int l) { arr[l] += arr[l - 1 ]; } // This function is used to apply square root // decomposition in the record array static void record_func( int block_size, int block[], int record[], int l, int r, int value) { // traversing first block in range while (l < r && l % block_size != 0 && l != 0 ) { record[l] += value; l++; } // traversing completely overlapped blocks in range while (l + block_size <= r + 1 ) { block[l / block_size] += value; l += block_size; } // traversing last block in range while (l <= r) { record[l] += value; l++; } } // Function to print the resultant array static void print( int arr[], int n) { for ( int i = 0 ; i < n; i++) { System.out.print(arr[i] + " " ); } } // Driver code public static void main(String[] args) { int n = 5 , m = 5 ; int arr[] = new int [n], record[] = new int [m]; int block_size = ( int ) Math.sqrt(m); int block[] = new int [max]; int command[][] = {{ 1 , 1 , 2 }, { 1 , 4 , 5 }, { 2 , 1 , 2 }, { 2 , 1 , 3 }, { 2 , 3 , 4 }}; for ( int i = m - 1 ; i >= 0 ; i--) { // If query is of type 2 then function // call to record_func if (command[i][ 0 ] == 2 ) { int x = i / (block_size); record_func(block_size, block, record, command[i][ 1 ] - 1 , command[i][ 2 ] - 1 , (block[x] + record[i] + 1 )); } // If query is of type 1 then simply add // 1 to the record array else { record[i]++; } } // Merging the value of the block in the record array for ( int i = 0 ; i < m; i++) { int check = (i / block_size); record[i] += block[check]; } for ( int i = 0 ; i < m; i++) { // If query is of type 1 then the array // elements are over-written by the record // array if (command[i][ 0 ] == 1 ) { arr[command[i][ 1 ] - 1 ] += record[i]; if ((command[i][ 2 ] - 1 ) < n - 1 ) { arr[(command[i][ 2 ])] -= record[i]; } } } // The prefix sum of the array for ( int i = 1 ; i < n; i++) { update(arr, i); } // Printing the resultant array print(arr, n); } } // This code is contributed by 29AjayKumar |
C#
// C# program to perform range queries over range // queries. using System; public class GFG { static readonly int max = 10000; // For prefix sum array static void update( int []arr, int l) { arr[l] += arr[l - 1]; } // This function is used to apply square root // decomposition in the record array static void record_func( int block_size, int []block, int []record, int l, int r, int value) { // traversing first block in range while (l < r && l % block_size != 0 && l != 0) { record[l] += value; l++; } // traversing completely overlapped blocks in range while (l + block_size <= r + 1) { block[l / block_size] += value; l += block_size; } // traversing last block in range while (l <= r) { record[l] += value; l++; } } // Function to print the resultant array static void print( int []arr, int n) { for ( int i = 0; i < n; i++) { Console.Write(arr[i] + " " ); } } // Driver code public static void Main() { int n = 5, m = 5; int []arr = new int [n]; int []record = new int [m]; int block_size = ( int ) Math.Sqrt(m); int []block = new int [max]; int [,]command= {{1, 1, 2}, {1, 4, 5}, {2, 1, 2}, {2, 1, 3}, {2, 3, 4}}; for ( int i = m - 1; i >= 0; i--) { // If query is of type 2 then function // call to record_func if (command[i,0] == 2) { int x = i / (block_size); record_func(block_size, block, record, command[i,1] - 1, command[i,2] - 1, (block[x] + record[i] + 1)); } // If query is of type 1 then simply add // 1 to the record array else { record[i]++; } } // Merging the value of the block in the record array for ( int i = 0; i < m; i++) { int check = (i / block_size); record[i] += block[check]; } for ( int i = 0; i < m; i++) { // If query is of type 1 then the array // elements are over-written by the record // array if (command[i,0] == 1) { arr[command[i,1] - 1] += record[i]; if ((command[i,2] - 1) < n - 1) { arr[(command[i,2])] -= record[i]; } } } // The prefix sum of the array for ( int i = 1; i < n; i++) { update(arr, i); } // Printing the resultant array print(arr, n); } } // This code is contributed by 29AjayKumar |
Output :
7 7 0 7 7
Method 4 :
This method has been made more efficient by applying Binary Indexed Tree or Fenwick Tree by creating two binary indexed tree for query 1 and query 2 respectively.
// CPP program to perform range queries over range // queries. #include <bits/stdc++.h> 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 + BITree[index]); // Update index to that of parent in update View index = (index + (index & (-index))); } return ; } // Constructs and returns a Binary Indexed Tree for given // array of size n. int * constructBITree( int n) { // Create and initialize BITree[] as 0 int * BITree = new int [n + 1]; for ( int i = 1; i <= n; i++) BITree[i] = 0; return BITree; } // 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; // index in BITree[] is 1 more than the index in arr[] index = index + 1; // Traverse ancestors of BITree[index] while (index > 0) { // Add element of BITree to sum sum = (sum + BITree[index]); // Move index to parent node in getSum View index -= index & (-index); } return sum; } // Function to update the BITree void update( int BITree[], int l, int r, int n, int val) { updateBIT(BITree, n, l, val); updateBIT(BITree, n, r + 1, -val); return ; } // Driver code int main() { int n = 5, m = 5; int temp[15] = { 1, 1, 2, 1, 4, 5, 2, 1, 2, 2, 1, 3, 2, 3, 4 }; int q[5][3]; int j = 0; for ( int i = 1; i <= m; i++) { q[i][0] = temp[j++]; q[i][1] = temp[j++]; q[i][2] = temp[j++]; } // BITree for query of type 2 int * BITree = constructBITree(m); // BITree for query of type 1 int * BITree2 = constructBITree(n); // Input the queries in a 2D matrix for ( int i = 1; i <= m; i++) cin >> q[i][0] >> q[i][1] >> q[i][2]; // If query is of type 2 then function call // to update with BITree for ( int i = m; i >= 1; i--) if (q[i][0] == 2) update(BITree, q[i][1] - 1, q[i][2] - 1, m, 1); for ( int i = m; i >= 1; i--) { if (q[i][0] == 2) { long int val = getSum(BITree, i - 1); update(BITree, q[i][1] - 1, q[i][2] - 1, m, val); } } // If query is of type 1 then function call // to update with BITree2 for ( int i = m; i >= 1; i--) { if (q[i][0] == 1) { long int val = getSum(BITree, i - 1); update(BITree2, q[i][1] - 1, q[i][2] - 1, n, (val + 1)); } } for ( int i = 1; i <= n; i++) cout << (getSum(BITree2, i - 1)) << " " ; return 0; } |
Output :
7 7 0 7 7
The Time complexity of Method 3 and Method 4 is O(log n) .
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