Given an array arr consisting of N elements and Q queries of the following two types:
- 1 K: For this type of query, the array needs to be rotated by K indices anticlockwise from its current state.
- 2 L R: For this query, the sum of the array elements present in the indices [L, R] needs to be calculated.
Example:
Input: arr = { 1, 2, 3, 4, 5, 6 }, query = { {2, 1, 3}, {1, 3}, {2, 0, 3}, {1, 4}, {2, 3, 5} }
Output:
9
16
12
Explanation:
For the 1st query {2, 1, 3} -> Sum of the elements in the indices [1, 3] = 2 + 3 + 4 = 9.
For the 2nd query {1, 3} -> Modified array after anti-clockwise rotation by 3 places is { 4, 5, 6, 1, 2, 3 }
For the 3rd query {2, 0, 3} -> Sum of the elements in the indices [0, 3] = 4 + 5 + 6 + 1 = 16.
For the 4th query {1, 4} -> Modified array after anti-clockwise rotation by 4 places is { 2, 3, 4, 5, 6, 1 }
For the 5th query {2, 3, 5} -> Sum of the elements in the indices [3, 5] = 5 + 6 + 1 = 12.
Approach:
- Create a prefix array which is double the size of the arr and copy the element at the ith index of arr to ith and N + ith index of prefix for all i in [0, N).
- Precompute the prefix sum for every index of that array and store in prefix.
- Set the pointer start at 0 to denote the starting index of the initial array.
- For query of type 1, shift start to
((start + K) % N)th position
- For query of type 2, calculate
prefix[start + R] - prefix[start + L- 1 ]
- if start + L >= 1 or print the value of
prefix[start + R]
- otherwise.
Below code is the implementation of the above approach:
C++
// C++ Program to calculate range sum// queries for anticlockwise// rotations of array by K#include <bits/stdc++.h>using namespace std;// Function to execute the queriesvoid rotatedSumQuery( int arr[], int n, vector<vector<int> >& query, int Q){ // Construct a new array // of size 2*N to store // prefix sum of every index int prefix[2 * n]; // Copy elements to the new array for (int i = 0; i < n; i++) { prefix[i] = arr[i]; prefix[i + n] = arr[i]; } // Calculate the prefix sum // for every index for (int i = 1; i < 2 * n; i++) prefix[i] += prefix[i - 1]; // Set start pointer as 0 int start = 0; for (int q = 0; q < Q; q++) { // Query to perform // anticlockwise rotation if (query[q][0] == 1) { int k = query[q][1]; start = (start + k) % n; } // Query to answer range sum else if (query[q][0] == 2) { int L, R; L = query[q][1]; R = query[q][2]; // If pointing to 1st index if (start + L == 0) // Display the sum upto start + R cout << prefix[start + R] << endl; else // Subtract sum upto start + L - 1 // from sum upto start + R cout << prefix[start + R] - prefix[start + L - 1] << endl; } }}// Driver codeint main(){ int arr[] = { 1, 2, 3, 4, 5, 6 }; // Number of query int Q = 5; // Store all the queries vector<vector<int> > query = { { 2, 1, 3 }, { 1, 3 }, { 2, 0, 3 }, { 1, 4 }, { 2, 3, 5 } }; int n = sizeof(arr) / sizeof(arr[0]); rotatedSumQuery(arr, n, query, Q); return 0;} |
Java
// Java program to calculate range sum// queries for anticlockwise// rotations of array by Kclass GFG{// Function to execute the queriesstatic void rotatedSumQuery(int arr[], int n, int [][]query, int Q){ // Construct a new array // of size 2*N to store // prefix sum of every index int []prefix = new int[2 * n]; // Copy elements to the new array for(int i = 0; i < n; i++) { prefix[i] = arr[i]; prefix[i + n] = arr[i]; } // Calculate the prefix sum // for every index for(int i = 1; i < 2 * n; i++) prefix[i] += prefix[i - 1]; // Set start pointer as 0 int start = 0; for(int q = 0; q < Q; q++) { // Query to perform // anticlockwise rotation if (query[q][0] == 1) { int k = query[q][1]; start = (start + k) % n; } // Query to answer range sum else if (query[q][0] == 2) { int L, R; L = query[q][1]; R = query[q][2]; // If pointing to 1st index if (start + L == 0) // Display the sum upto start + R System.out.print(prefix[start + R] + "\n"); else // Subtract sum upto start + L - 1 // from sum upto start + R System.out.print(prefix[start + R] - prefix[start + L - 1] + "\n"); } }}// Driver codepublic static void main(String[] args){ int arr[] = { 1, 2, 3, 4, 5, 6 }; // Number of query int Q = 5; // Store all the queries int [][]query = { { 2, 1, 3 }, { 1, 3 }, { 2, 0, 3 }, { 1, 4 }, { 2, 3, 5 } }; int n = arr.length; rotatedSumQuery(arr, n, query, Q);}}// This code is contributed by Rohit_ranjan |
Python3
# Python3 program to calculate range sum# queries for anticlockwise# rotations of the array by K# Function to execute the queriesdef rotatedSumQuery(arr, n, query, Q): # Construct a new array # of size 2*N to store # prefix sum of every index prefix = [0] * (2 * n) # Copy elements to the new array for i in range(n): prefix[i] = arr[i] prefix[i + n] = arr[i] # Calculate the prefix sum # for every index for i in range(1, 2 * n): prefix[i] += prefix[i - 1]; # Set start pointer as 0 start = 0; for q in range(Q): # Query to perform # anticlockwise rotation if (query[q][0] == 1): k = query[q][1] start = (start + k) % n; # Query to answer range sum elif (query[q][0] == 2): L = query[q][1] R = query[q][2] # If pointing to 1st index if (start + L == 0): # Display the sum upto start + R print(prefix[start + R]) else: # Subtract sum upto start + L - 1 # from sum upto start + R print(prefix[start + R]- prefix[start + L - 1]) # Driver codearr = [ 1, 2, 3, 4, 5, 6 ];# Number of queryQ = 5# Store all the queriesquery= [ [ 2, 1, 3 ], [ 1, 3 ], [ 2, 0, 3 ], [ 1, 4 ], [ 2, 3, 5 ] ]n = len(arr);rotatedSumQuery(arr, n, query, Q);# This code is contributed by ankitkumar34 |
C#
// C# program to calculate range sum// queries for anticlockwise// rotations of array by Kusing System;class GFG{// Function to execute the queriesstatic void rotatedSumQuery(int[] arr, int n, int[,] query, int Q){ // Construct a new array // of size 2*N to store // prefix sum of every index int[] prefix = new int[2 * n]; // Copy elements to the new array for(int i = 0; i < n; i++) { prefix[i] = arr[i]; prefix[i + n] = arr[i]; } // Calculate the prefix sum // for every index for(int i = 1; i < 2 * n; i++) prefix[i] += prefix[i - 1]; // Set start pointer as 0 int start = 0; for(int q = 0; q < Q; q++) { // Query to perform // anticlockwise rotation if (query[q, 0] == 1) { int k = query[q, 1]; start = (start + k) % n; } // Query to answer range sum else if (query[q, 0] == 2) { int L, R; L = query[q, 1]; R = query[q, 2]; // If pointing to 1st index if (start + L == 0) // Display the sum upto start + R Console.Write(prefix[start + R] + "\n"); else // Subtract sum upto start + L - 1 // from sum upto start + R Console.Write(prefix[start + R] - prefix[start + L - 1] + "\n"); } }}// Driver codepublic static void Main(){ int[] arr = new int[] { 1, 2, 3, 4, 5, 6 }; // Number of query int Q = 5; // Store all the queries int[,] query = new int[,] { { 2, 1, 3 }, { 1, 3, 0 }, { 2, 0, 3 }, { 1, 4, 0 }, { 2, 3, 5 } }; int n = arr.Length; rotatedSumQuery(arr, n, query, Q);}}// This code is contributed by sanjoy_62 |
Javascript
<script> // Javascript program to calculate range sum// queries for anticlockwise// rotations of array by K// Function to execute the queriesfunction rotatedSumQuery(arr, n, query, Q){ // Construct a new array // of size 2*N to store // prefix sum of every index let prefix = []; // Copy elements to the new array for(let i = 0; i < n; i++) { prefix[i] = arr[i]; prefix[i + n] = arr[i]; } // Calculate the prefix sum // for every index for(let i = 1; i < 2 * n; i++) prefix[i] += prefix[i - 1]; // Set start pointer as 0 let start = 0; for(let q = 0; q < Q; q++) { // Query to perform // anticlockwise rotation if (query[q][0] == 1) { let k = query[q][1]; start = (start + k) % n; } // Query to answer range sum else if (query[q][0] == 2) { let L, R; L = query[q][1]; R = query[q][2]; // If pointing to 1st index if (start + L == 0) // Display the sum upto start + R document.write(prefix[start + R] + "<br/>"); else // Subtract sum upto start + L - 1 // from sum upto start + R document.write(prefix[start + R] - prefix[start + L - 1] + "<br/>"); } }} // Driver code let arr = [ 1, 2, 3, 4, 5, 6 ]; // Number of query let Q = 5; // Store all the queries let query = [[ 2, 1, 3 ], [ 1, 3 ], [ 2, 0, 3 ], [ 1, 4 ], [ 2, 3, 5 ]]; let n = arr.length; rotatedSumQuery(arr, n, query, Q); // This code is contributed by susmitakundugoaldanga.</script> |
Output:
9 16 12
Time Complexity: O(N+Q), where Q is the number of queries, and as each query will cost O (1) time for Q queries time complexity would be O(N+Q).
Auxiliary Space: O(N), as we are using extra space for prefix.
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