Given an array A of N integers and number of queries Q. You have to answer two types of queries.
- Update [l, r] – for every i in range from l to r update Ai with sqrt(Ai), where sqrt(Ai) represents the square root of Ai in integral form.
- Query [l, r] – calculate the sum of all numbers ranging between l and r in array A.
Prerequisite: Binary Indexed Trees | Segment Trees
Examples:
Input: A[] = { 4, 5, 1, 2, 4 }, Q = {{2, 1, 5}, {1, 1, 2}, {1, 2, 4}, {2, 1, 5}}
Output: 16
9
Considering 1-based indexing, first query is to calculate sum of numbers from A1 to A5
which is 4 + 5 + 1 + 2 + 4 = 16.
Second query is to update A1 to A2 with its square root. Now, array becomes A[] = { 2, 2, 1, 2, 4 }.
Similarly, third query is to update A2 to A4 with its square root. Now, array becomes A[] = { 2, 1, 1, 1, 4 }.
Fourth query is to calculate sum of numbers from A1 to A5 which is 2 + 1 + 1 + 1 + 4 = 9.Input: A[] = { 4, 9, 25, 36 }, Q = {{1, 2, 4}, {2, 1, 4}}
Output: 18
Naive Approach: A simple solution is to run a loop from l to r and calculate sum of elements in the given range. To update a value, simple replace arr[i] with its square root, i.e., arr[i] = sqrt[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. Now, for each update operation the key observation is that the number 1 will have 1 as its square root, so if it exists in the range of update query, it doesn’t need to be updated. We will use a set to store the index of only those numbers which are greater than 1 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 1 as its square root then after updating it, remove it from the set as it will always be 1 even after any next update query. For sum query operation, simply do query(r) – query(l – 1).
Below is the implementation of the above approach:
CPP
// CPP program to calculate sum// in an interval and update with// square root#include <bits/stdc++.h>using namespace std;// Maximum size of input arrayconst int MAX = 100;int BIT[MAX + 1];// structure for queries with members type,// leftIndex, rightIndex of the querystruct queries { int type, l, r;};// 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 1 if (arr[i] > 1) 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 square root update(*it, (int)sqrt(arr[*it]) - arr[*it], n); arr[*it] = (int)sqrt(arr[*it]); // if updated value becomes equal to 1 // remove it from the set if (arr[*it] == 1) 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(){ int q = 4; // input array using 1-based indexing int arr[] = { 0, 4, 5, 1, 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 = 5; que[1].type = 1, que[1].l = 1, que[1].r = 2; que[2].type = 1, que[2].l = 2, que[2].r = 4; que[3].type = 2, que[3].l = 1, que[3].r = 5; // answer the Queries answerQueries(arr, que, n, q); return 0;} |
Java
import java.util.*;import java.util.stream.*;class Program {// Maximum size of input arraystatic int MAX = 100;static int[] BIT = new int[MAX + 1];// structure for queries with members type,// leftIndex, rightIndex of the querystatic class queries { public int type, l, r; public queries(int t_val, int l_val, int r_val) { type = t_val; l = l_val; r = r_val; } }// function for updating the valuestatic void update(int x, int val, int n) { for (; x <= n; x += x & -x) { BIT[x] += val; }}// function for calculating the required// sum between two indexesstatic int sum(int x) { int s = 0; for (; x > 0; x -= x & -x) { s += BIT[x]; } return s;}// function to return answer to queriesstatic void answerQueries(int[] arr, queries[] que, int n, int q) { // Declaring a Set HashSet<Integer> s = new HashSet<Integer>(); for (int i = 1; i < n; i++) { // inserting indexes of those numbers // which are greater than 1 if (arr[i] > 1) 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 final var val = que[i].l; Stream<Integer> st = s.stream().filter(x -> x >= val); int it = st.findFirst().orElse(0); // if it crosses the right index of // query or end of set, then break if (it == 0 || it > que[i].r) break; que[i].l = it; // update the value of arr[i] to // its square root update(it, (int)Math.sqrt(arr[it]) - arr[it], n); arr[it] = (int)Math.sqrt(arr[it]); // if updated value becomes equal to 1 // remove it from the set if (arr[it] == 1) s.remove(it); // increment the index que[i].l++; } } // sum query else { System.out.println(sum(que[i].r) - sum(que[i].l - 1)); } }}// Driver Codepublic static void main(String[] args) { int q = 4; // input array using 1-based indexing int[] arr = { 0, 4, 5, 1, 2, 4 }; int n = arr.length; // declaring array of structure of type queries queries[] que = new queries[q + 1]; que[0] = new queries(2, 1, 5); que[1] = new queries(1, 1, 2); que[2] = new queries(1, 2, 4); que[3] = new queries(2, 1, 5); // answer the Queries answerQueries(arr, que, n, q); }} |
Python3
# Python program to calculate sum# in an interval and update with# square rootfrom typing import Listimport bisectfrom math import sqrt, floor# Maximum size of input arrayMAX = 100BIT = [0 for _ in range(MAX + 1)]# structure for queries with members type,# leftIndex, rightIndex of the queryclass queries: def __init__(self, type: int = 0, l: int = 0, r: int = 0) -> None: self.type = type self.l = l self.r = r# function for updating the valuedef update(x: int, val: int, n: int) -> None: a = x while a <= n: BIT[a] += val a += a & -a# function for calculating the required# sum between two indexesdef sum(x: int) -> int: s = 0 a = x while a: s += BIT[a] a -= a & -a return s# function to return answer to queriesdef answerQueries(arr: List[int], que: List[queries], n: int, q: int) -> None: # Declaring a Set s = set() for i in range(1, n): # inserting indexes of those numbers # which are greater than 1 if (arr[i] > 1): s.add(i) update(i, arr[i], n) for i in range(q): # update query if (que[i].type == 1): while True: ss = list(sorted(s)) # find the left index of query in # the set using binary search # auto it = s.lower_bound(que[i].l); it = bisect.bisect_left(ss, que[i].l) # if it crosses the right index of # query or end of set, then break if it == len(s) or ss[it] > que[i].r: break que[i].l = ss[it] # update the value of arr[i] to # its square root update(ss[it], floor(sqrt(arr[ss[it]]) - arr[ss[it]]), n) arr[ss[it]] = floor(sqrt(arr[ss[it]])) # if updated value becomes equal to 1 # remove it from the set if (arr[ss[it]] == 1): s.remove(ss[it]) # increment the index que[i].l += 1 # sum query else: print(sum(que[i].r) - sum(que[i].l - 1))# Driver Codeif __name__ == "__main__": q = 4 # input array using 1-based indexing arr = [0, 4, 5, 1, 2, 4] n = len(arr) # declaring array of structure of type queries que = [queries() for _ in range(q + 1)] que[0].type, que[0].l, que[0].r = 2, 1, 5 que[1].type, que[1].l, que[1].r = 1, 1, 2 que[2].type, que[2].l, que[2].r = 1, 2, 4 que[3].type, que[3].l, que[3].r = 2, 1, 5 # answer the Queries answerQueries(arr, que, n, q)# This code is contributed by sanjeev2552 |
C#
using System;using System.Linq;using System.Collections.Generic;class Program { // Maximum size of input array const int MAX = 100; static int[] BIT = new int[MAX + 1]; // structure for queries with members type, // leftIndex, rightIndex of the query struct queries { public int type, l, r; } // function for updating the value static void update(int x, int val, int n) { for (; x <= n; x += x & -x) { BIT[x] += val; } } // function for calculating the required // sum between two indexes static int sum(int x) { int s = 0; for (; x > 0; x -= x & -x) { s += BIT[x]; } return s; } // function to return answer to queries static void answerQueries(int[] arr, queries[] que, int n, int q) { // Declaring a Set var s = new HashSet<int>(); for (int i = 1; i < n; i++) { // inserting indexes of those numbers // which are greater than 1 if (arr[i] > 1) 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.Where(x => x >= que[i].l) .FirstOrDefault(); // if it crosses the right index of // query or end of set, then break if (it == 0 || it > que[i].r) break; que[i].l = it; // update the value of arr[i] to // its square root update(it, (int)Math.Sqrt(arr[it]) - arr[it], n); arr[it] = (int)Math.Sqrt(arr[it]); // if updated value becomes equal to 1 // remove it from the set if (arr[it] == 1) s.Remove(it); // increment the index que[i].l++; } } // sum query else { Console.WriteLine(sum(que[i].r) - sum(que[i].l - 1)); } } } // Driver Code static void Main(string[] args) { int q = 4; // input array using 1-based indexing int[] arr = { 0, 4, 5, 1, 2, 4 }; int n = arr.Length; // declaring array of structure of type queries var que = new queries[q + 1]; que[0].type = 2; que[0].l = 1; que[0].r = 5; que[1].type = 1; que[1].l = 1; que[1].r = 2; que[2].type = 1; que[2].l = 2; que[2].r = 4; que[3].type = 2; que[3].l = 1; que[3].r = 5; // answer the Queries answerQueries(arr, que, n, q); }}// This code is contributed by phasing17 |
Javascript
// JavaScript program to calculate sum// in an interval and update with// square root// Maximum size of input arraylet MAX = 100;let BIT = new Array(MAX + 1).fill(0);function lower_bound(arr, ele){ for (var i = 0; i < arr.length; i++) { if (arr[i] >= ele) return i; } return arr.length - 1;}// structure for queries with members type,// leftIndex, rightIndex of the queryclass queries{ constructor(type, l, r) { this.type = type; this.l = l; this.r = r; }}// function for updating the valuefunction update(BIT, x, val, n){ var a = x; while (a <= n) { BIT[a] += val; a += (a & -a); } return BIT;}// function for calculating the required// sum between two indexesfunction sum(x){ var s = 0; var a = x; while (a > 0) { s += BIT[a]; a -= (a & -a); } return s;}// function to return answer to queriesfunction answerQueries(arr, que, n, q){ // Declaring a Set let s = new Set(); for (var i = 1; i < n; i++) { // inserting indexes of those numbers // which are greater than 1 if (arr[i] > 1) s.add(i); BIT = update(BIT, i, arr[i], n); } for (var i = 0; i < q; i++) { // update query if (que[i].type == 1) { while (true) { var ss = Array.from(s); ss.sort(); // find the left index of query in // the set using binary search // auto it = s.lower_bound(que[i].l); let it = lower_bound(ss, que[i].l); // if it crosses the right index of // query or end of set, then break if (it == s.length || ss[it] > que[i].r) break; que[i].l = ss[it]; // update the value of arr[i] to // its square root BIT = update(BIT, ss[it], (Math.pow(arr[ss[it]], 0.5)) - arr[ss[it]], n); arr[ss[it]] = (Math.pow(arr[ss[it]], 0.5)); // if updated value becomes equal to 1 // remove it from the set if (arr[ss[it]] == 1) arr.splice(ss[it], 1); // increment the index que[i].l += 1; } } // sum query else console.log(Math.floor(sum(que[i].r) - sum(que[i].l - 1))); }}// Driver Codelet q = 4;// input array using 1-based indexinglet arr = [0, 4, 5, 1, 2, 4];let n = arr.length;// declaring array of structure of type querieslet que = [new queries(2, 1, 5), new queries(1, 1, 2), new queries(1, 2, 4),new queries(2, 1, 5)];// answer the QueriesanswerQueries(arr, que, n, q);// This code is contributed by phasing17 |
Output
16 9
Complexity Analysis:
- Time Complexity: O(logN) per query
- Auxiliary Space: O(N)
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