Given an array A of N integers. You have to answer two types of queries : 1. Update [l, r] – for every i in range from l to r update Ai with D(Ai), where D(Ai) represents the number of divisors of Ai 2. Query [l, r] – calculate the sum of all numbers ranging between l and r in array A. Input is given as two integers N and Q, representing number of integers in array and number of queries respectively. Next line contains an array of n integers followed by Q queries where ith query is represented as typei, li, ri.
Prerequisite : Binary Indexed Trees | Segment Trees Examples :
Input : 7 4
6 4 1 10 3 2 4
2 1 7
2 4 5
1 3 5
2 4 4
Output : 30
13
4
Explanation : First query is to calculate the sum of numbers from A1 to A7 which is 6 + 4
+ 1 + 10 + 3 + 2 + 4 = 30. Similarly, second query results into 13. For third query,
which is update operation, hence A3 will remain 1, A4 will become 4 and A5 will become 2.
Fourth query will result into A4 = 4.
Naive Approach: A simple solution is to run a loop from l to r and calculate sum of elements in given range. To update a value, precompute the values of number of divisors of every number and simply do arr[i] = divisors[arr[i]].
Efficient Approach :
The idea is to reduce the time complexity for each query and update operation to O(logN). Use Binary Indexed Trees (BIT) or Segment Trees. Construct a BIT[] array and have two functions for query and update operation and precompute the number of divisors for each number.
Now, for each update operation, the key observation is that the numbers ‘1’ and ‘2’ will have ‘1’ and ‘2’ as their number of divisors respectively, so if it exists in the range of update query, they don’t need to be updated. We will use a set to store the index of only those numbers which are greater than 2 and use binary search to find the l index of the update query and increment the l index until every element is updated in range of that update query.
If the arr[i] has only 2 divisors then after updating it, remove it from the set as it will always be 2 even after any next update query. For sum query operation, simply do query(r) – query(l – 1).
Implementation:
C++
// CPP program to calculate sum// in an interval and update with// number of divisors#include <bits/stdc++.h>using namespace std;int divisors[100], BIT[100];// structure for queries with members type,// leftIndex, rightIndex of the querystruct queries{ int type, l, r;};// function to calculate the number// of divisors of each numbervoid calcDivisors(){ for (int i = 1; i < 100; i++) { for (int j = i; j < 100; j += i) { divisors[j]++; } }}// function for updating the valuevoid update(int x, int val, int n){ for (x; x <= n; x += x&-x) { BIT[x] += val; }}// function for calculating the required// sum between two indexesint sum(int x){ int s = 0; for (x; x > 0; x -= x&-x) { s += BIT[x]; } return s;}// function to return answer to queriesvoid answerQueries(int arr[], queries que[], int n, int q){ // Declaring a Set set<int> s; for (int i = 1; i < n; i++) { // inserting indexes of those numbers // which are greater than 2 if(arr[i] > 2) s.insert(i); update(i, arr[i], n); } for (int i = 0; i < q; i++) { // update query if (que[i].type == 1) { while (true) { // find the left index of query in // the set using binary search auto it = s.lower_bound(que[i].l); // if it crosses the right index of // query or end of set, then break if(it == s.end() || *it > que[i].r) break; que[i].l = *it; // update the value of arr[i] to // its number of divisors update(*it, divisors[arr[*it]] - arr[*it], n); arr[*it] = divisors[arr[*it]]; // if updated value becomes less than or // equal to 2 remove it from the set if(arr[*it] <= 2) s.erase(*it); // increment the index que[i].l++; } } // sum query else { cout << (sum(que[i].r) - sum(que[i].l - 1)) << endl; } }}// Driver Codeint main(){ // precompute the number of divisors for each number calcDivisors(); int q = 4; // input array int arr[] = {0, 6, 4, 1, 10, 3, 2, 4}; int n = sizeof(arr) / sizeof(arr[0]); // declaring array of structure of type queries queries que[q + 1]; que[0].type = 2, que[0].l = 1, que[0].r = 7; que[1].type = 2, que[1].l = 4, que[1].r = 5; que[2].type = 1, que[2].l = 3, que[2].r = 5; que[3].type = 2, que[3].l = 4, que[3].r = 4; // answer the Queries answerQueries(arr, que, n, q); return 0;} |
Java
import java.util.*;public class Main { static int[] divisors = new int[100]; static int[] BIT = new int[100]; // structure for queries with members type, // leftIndex, rightIndex of the query static class Query { int type, l, r; public Query(int type, int l, int r) { this.type = type; this.l = l; this.r = r; } } // function to calculate the number // of divisors of each number static void calcDivisors() { for (int i = 1; i < 100; i++) { for (int j = i; j < 100; j += i) { divisors[j]++; } } } // function for updating the value static void update(int x, int val, int n) { for (int i = x; i <= n; i += i & -i) { BIT[i] += val; } } // function for calculating the required // sum between two indexes static int sum(int x) { int s = 0; for (int i = x; i > 0; i -= i & -i) { s += BIT[i]; } return s; } // function to return answer to queries static void answerQueries(int[] arr, Query[] que, int n, int q) { // Declaring a Set Set<Integer> s = new TreeSet<>(); for (int i = 1; i < n; i++) { // inserting indexes of those numbers // which are greater than 2 if (arr[i] > 2) s.add(i); update(i, arr[i], n); } for (int i = 0; i < q; i++) { // update query if (que[i].type == 1) { while (true) { // find the left index of query in // the set using binary search Iterator<Integer> it = s.iterator(); int idx = 0; while (it.hasNext()) { idx = it.next(); if (idx >= que[i].l && idx <= que[i].r) break; } if (idx > que[i].r) break; que[i].l = idx; // update the value of arr[i] to // its number of divisors update(idx, divisors[arr[idx]] - arr[idx], n); arr[idx] = divisors[arr[idx]]; // if updated value becomes less than or // equal to 2 remove it from the set if (arr[idx] <= 2) s.remove(idx); que[i].l++; } } // sum query else { System.out.println(sum(que[i].r) - sum(que[i].l - 1)); } } } // Driver's code public static void main(String[] args) { calcDivisors(); int q = 4; int[] arr = { 0, 6, 4, 1, 10, 3, 2, 4 }; int n = arr.length; // declaring array of structure of type queries Query[] que = new Query[q + 1]; que[0] = new Query(2, 1, 7); que[1] = new Query(2, 4, 5); que[2] = new Query(1, 3, 5); que[3] = new Query(2, 4, 4); answerQueries(arr, que, n, q); }} |
Python3
# Python code for the above approach# function to calculate the number# of divisors of each numberdef calcDivisors(): divisors = [0]*100 for i in range(1, 100): for j in range(i, 100, i): divisors[j] += 1 return divisors# function for updating the valuedef update(x, val, n, BIT): while x <= n: BIT[x] += val x += (x & -x)# function for calculating the required# sum between two indexesdef sum(x, BIT): s = 0 while x > 0: s += BIT[x] x -= (x & -x) return s# function to return answer to queriesdef answerQueries(arr, que, n, q): # Declaring a Set s = set() BIT = [0] * (n + 1) divisors = calcDivisors() for i in range(1, n): # inserting indexes of those numbers # which are greater than 2 if arr[i] > 2: s.add(i) update(i, arr[i], n, BIT) for i in range(q): # update query if que[i][0] == 1: leftIndex = que[i][1] rightIndex = que[i][2] for it in set(s): if it >= leftIndex and it <= rightIndex: # update the value of arr[i] to update(it, divisors[arr[it]] - arr[it], n, BIT) arr[it] = divisors[arr[it]] if arr[it] <= 2: s.remove(it) # sum query else: print(sum(que[i][2], BIT) - sum(que[i][1] - 1, BIT))# Driver Codeif __name__ == '__main__': # precompute the number of divisors for each number # input array arr = [0, 6, 4, 1, 10, 3, 2, 4] n = len(arr) # declaring array of structure of type queries que = [(2, 1, 7), (2, 4, 5), (1, 3, 5), (2, 4, 4)] # answer the Queries answerQueries(arr, que, n, len(que)) # This code is contributed by lokeshpotta20 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;public class GFG{ static int[] divisors = new int[100]; static int[] BIT = new int[100]; // structure for queries with members type, // leftIndex, rightIndex of the query public class Query { public int type, l, r; public Query(int type, int l, int r) { this.type = type; this.l = l; this.r = r; } } // function to calculate the number // of divisors of each number static void CalcDivisors() { for (int i = 1; i < 100; i++) { for (int j = i; j < 100; j += i) { divisors[j]++; } } } // function for updating the value static void Update(int x, int val, int n) { for (int i = x; i <= n; i += i & -i) { BIT[i] += val; } } // function for calculating the required // sum between two indexes static int Sum(int x) { int s = 0; for (int i = x; i > 0; i -= i & -i) { s += BIT[i]; } return s; } // function to return answer to queries static void AnswerQueries(int[] arr, Query[] que, int n, int q) { // Declaring a Set SortedSet<int> s = new SortedSet<int>(); for (int i = 1; i < n; i++) { // inserting indexes of those numbers // which are greater than 2 if (arr[i] > 2) s.Add(i); Update(i, arr[i], n); } for (int i = 0; i < q; i++) { // update query if (que[i].type == 1) { while (true) { // find the left index of query in // the set using binary search var it = s.GetEnumerator(); int idx = 0; while (it.MoveNext()) { idx = it.Current; if (idx >= que[i].l && idx <= que[i].r) break; } if (idx > que[i].r) break; que[i].l = idx; // update the value of arr[i] to // its number of divisors Update(idx, divisors[arr[idx]] - arr[idx], n); arr[idx] = divisors[arr[idx]]; // if updated value becomes less than or // equal to 2 remove it from the set if (arr[idx] <= 2) s.Remove(idx); que[i].l++; } } // sum query else { Console.WriteLine(Sum(que[i].r) - Sum(que[i].l - 1)); } } } // Driver's code public static void Main(string[] args) { CalcDivisors(); int q = 4; int[] arr = { 0, 6, 4, 1, 10, 3, 2, 4 }; int n = arr.Length; // declaring array of structure of type queries Query[] que = new Query[q + 1]; que[0] = new Query(2, 1, 7); que[1] = new Query(2, 4, 5); que[2] = new Query(1, 3, 5); que[3] = new Query(2, 4, 4); AnswerQueries(arr, que, n, q); }}// This code is contribute by rishab |
Javascript
// function to calculate the number of divisors of each numberfunction calcDivisors() { const divisors = new Array(100).fill(0); for (let i = 1; i < 100; i++) { for (let j = i; j < 100; j += i) { divisors[j] += 1; } } return divisors;}// function for updating the valuefunction update(x, val, n, BIT) { while (x <= n) { BIT[x] += val; x += (x & -x); }}// function for calculating the required sum between two indexesfunction sum(x, BIT) { let s = 0; while (x > 0) { s += BIT[x]; x -= (x & -x); } return s;}// function to return answer to queriesfunction answerQueries(arr, que, n, q) { // Declaring a Set const s = new Set(); const BIT = new Array(n + 1).fill(0); const divisors = calcDivisors(); for (let i = 1; i < n; i++) { // inserting indexes of those numbers which are greater than 2 if (arr[i] > 2) { s.add(i); } update(i, arr[i], n, BIT); } for (let i = 0; i < q; i++) { // update query if (que[i][0] === 1) { const leftIndex = que[i][1]; const rightIndex = que[i][2]; for (const it of s.values()) { if (it >= leftIndex && it <= rightIndex) { // update the value of arr[i] update(it, divisors[arr[it]] - arr[it], n, BIT); arr[it] = divisors[arr[it]]; if (arr[it] <= 2) { s.delete(it); } } } } // sum query else { console.log(sum(que[i][2], BIT) - sum(que[i][1] - 1, BIT)); } }}// Driver Codeconst arr = [0, 6, 4, 1, 10, 3, 2, 4];const n = arr.length;const que = [ [2, 1, 7], [2, 4, 5], [1, 3, 5], [2, 4, 4]];// answer the QueriesanswerQueries(arr, que, n, que.length); |
Output
30 13 4
Time Complexity for answering Q queries will be O(Q * log(N)).
Auxiliary Space: O(N)
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