Range and Update Sum Queries with Factorial
Given an array arr[] of N integers and number of queries Q. The task is to answer three types of queries.
- Update [l, r] – for every i in range [l, r] increment arr[i] by 1.
- Update [l, val] – change the value of arr[l] to val.
- Query [l, r] – calculate the sum of arr[i]! % 109 for all i in range [l, r] where arr[i]! is the factorial of arr[i].
Prerequisite :Binary Indexed Trees | Segment Trees Examples:
Input: Q = 6, arr[] = { 1, 2, 1, 4, 5 } 3 1 5 1 1 3 2 2 4 3 2 4 1 2 5 3 1 5 Output: 148 50 968 1st query, the required sum is (1! + 2! + 1! + 4! + 5!) % 109 = 148 2nd query, the array becomes arr[] = { 2, 3, 2, 4, 5 } 3rd query, array becomes arr[] = { 2, 4, 2, 4, 5 } 4th query, the required sum is (4! + 2! + 4!) % 109 = 50 5th query, the array becomes arr[] = { 2, 5, 3, 5, 6 } 6th query, the required sum is (2! + 5! + 3! + 5! + 6!) % 109 = 968
Naive Approach: A simple solution is to run a loop from l to r and calculate sum of factorial of elements (pre-computed) in the given range for the 3rd query. For the 2nd query, to update a value, simply replace arr[i] with the given value i.e. arr[i] = val. For the 1st type query, increment the value of arr[i] i.e. arr[i] = arr[i] + 1. Efficient Approach: It can be observed from careful analysis that 40! is divisible by 109, that means factorial of every number greater than 40 will be divisible by 109. Hence, that adds zero to our answer for the 3rd query. 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 of the 1st type, the key observation is that the number in given range can at max be updated to 40, since after that it won’t matter as it will add zero to our final answer. We will use a set to store the index of only those numbers which are lesser than 10 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] becomes greater than or equal to 40 after incrementing by 1, remove it from the set as it will not affect our answer of sum query even after any next update query.
- For the update operation of the 2nd type, call the update function with the given value. Also, the given value is < 40, insert the index of the element to be replaced with into the set and if the given value is ? 40, remove it from the set since it will have no importance in sum query.
- For the sum query of the 3rd type, simply do query(r) – query(l – 1).
Below is the implementation of the above approach:
C++
// CPP program to calculate sum of // factorials in an interval and update // with two types of operations #include <bits/stdc++.h> using namespace std; // Modulus const int MOD = 1e9; // Maximum size of input array const int MAX = 100; // Size for factorial array const int SZ = 40; int BIT[MAX + 1], fact[SZ + 1]; // structure for queries with members type, // leftIndex, rightIndex of the query struct queries { int type, l, r; }; // function for updating the value void 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 indexes int sum( int x) { int s = 0; for (x; x > 0; x -= x & -x) s += BIT[x]; return s; } // function to return answer to queries void answerQueries( int arr[], queries que[], int n, int q) { // Precomputing factorials fact[0] = 1; for ( int i = 1; i < 41; i++) fact[i] = (fact[i - 1] * i) % MOD; // Declaring a Set set< int > s; for ( int i = 1; i < n; i++) { // inserting indexes of those // numbers which are lesser // than 40 if (arr[i] < 40) { s.insert(i); update(i, fact[arr[i]], n); } else update(i, 0, n); } for ( int i = 0; i < q; i++) { // update query of the 1st type 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; int val = arr[*it] + 1; // update the value of arr[i] to // its new value update(*it, fact[val] - fact[arr[*it]], n); arr[*it]++; // if updated value becomes greater // than or equal to 40 remove it from // the set if (arr[*it] >= 40) s.erase(*it); // increment the index que[i].l++; } } // update query of the 2nd type else if (que[i].type == 2) { int idx = que[i].l; int val = que[i].r; // update the value to its new value update(idx, fact[val] - fact[arr[idx]], n); arr[idx] = val; // If the value is less than 40, insert // it into set, otherwise remove it if (val < 40) s.insert(idx); else s.erase(idx); } // sum query of the 3rd type else cout << (sum(que[i].r) - sum(que[i].l - 1)) << endl; } } // Driver Code to test above functions int main() { int q = 6; // input array using 1-based indexing int arr[] = { 0, 1, 2, 1, 4, 5 }; int n = sizeof (arr) / sizeof (arr[0]); // declaring array of structure of type queries queries que[q + 1]; que[0].type = 3, que[0].l = 1, que[0].r = 5; que[1].type = 1, que[1].l = 1, que[1].r = 3; que[2].type = 2, que[2].l = 2, que[2].r = 4; que[3].type = 3, que[3].l = 2, que[3].r = 4; que[4].type = 1, que[4].l = 2, que[4].r = 5; que[5].type = 3, que[5].l = 1, que[5].r = 5; // answer the Queries answerQueries(arr, que, n, q); return 0; } |
Python3
# Python3 program to calculate sum of # factorials in an interval and update # with two types of operations from bisect import bisect_left as lower_bound # Modulus MOD = 1e9 # Maximum size of input array MAX = 100 # Size for factorial array SZ = 40 BIT = [ 0 ] * ( MAX + 1 ) fact = [ 0 ] * (SZ + 1 ) # structure for queries with members type, # leftIndex, rightIndex of the query class queries: def __init__( self , tpe, l, r): self . type = tpe self .l = l self .r = r # function for updating the value def update(x, val, n): global BIT while x < = n: BIT[x] + = val x + = x & - x # function for calculating the required # sum between two indexes def summ(x): global BIT s = 0 while x > 0 : s + = BIT[x] x - = x & - x return s # function to return answer to queries def answerQueries(arr: list , que: list , n: int , q: int ): global fact # Precomputing factorials fact[ 0 ] = 1 for i in range ( 1 , 41 ): fact[i] = int ((fact[i - 1 ] * i) % MOD) # Declaring a Set s = set () for i in range ( 1 , n): # inserting indexes of those # numbers which are lesser # than 40 if arr[i] < 40 : s.add(i) update(i, fact[arr[i]], n) else : update(i, 0 , n) for i in range (q): # update query of the 1st type if que[i]. type = = 1 : while True : s = list (s) s.sort() # find the left index of query in # the set using binary search it = lower_bound(s, que[i].l) # if it crosses the right index of # query or end of set, then break if it = = len (s) or s[it] > que[i].r: break que[i].l = s[it] val = arr[s[it]] + 1 # update the value of arr[i] to # its new value update(s[it], fact[val] - fact[arr[s[it]]], n) arr[s[it]] + = 1 # if updated value becomes greater # than or equal to 40 remove it from # the set if arr[s[it]] > = 40 : s.remove(it) # increment the index que[i].l + = 1 # update query of the 2nd type elif que[i]. type = = 2 : s = set (s) idx = que[i].l val = que[i].r # update the value to its new value update(idx, fact[val] - fact[arr[idx]], n) arr[idx] = val # If the value is less than 40, insert # it into set, otherwise remove it if val < 40 : s.add(idx) else : s.remove(idx) # sum query of the 3rd type else : print ((summ(que[i].r) - summ(que[i].l - 1 ))) # Driver Code if __name__ = = "__main__" : q = 6 # input array using 1-based indexing arr = [ 0 , 1 , 2 , 1 , 4 , 5 ] n = len (arr) # declaring array of structure of type queries que = [ queries( 3 , 1 , 5 ), queries( 1 , 1 , 3 ), queries( 2 , 2 , 4 ), queries( 3 , 2 , 4 ), queries( 1 , 2 , 5 ), queries( 3 , 1 , 5 ) ] # answer the Queries answerQueries(arr, que, n, q) # This code is contributed by # sanjeev2552 |
C#
// C# program to calculate sum of // factorials in an interval and update // with two types of operations using System; using System.Linq; using System.Collections.Generic; class Program { // Modulus private const int MOD = 1000000000; // Maximum size of input array private const int MAX = 100; // Size for factorial array private const int SZ = 40; private static int [] BIT = new int [MAX + 1]; private static int [] fact = new int [SZ + 1]; // structure for queries with members type, // leftIndex, rightIndex of the query private struct Queries { public int type; public int l; public int r; // Constructor public Queries( int x, int y, int z) { type = x; l = y; r = z; } }; // function for updating the value private 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 private 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 private static void AnswerQueries( int [] arr, Queries[] que, int n, int q) { // Precomputing factorials fact[0] = 1; for ( int i = 1; i < 41; i++) { fact[i] = (fact[i - 1] * i) % MOD; } // Declaring a Set SortedSet< int > s = new SortedSet< int >(); for ( int i = 1; i < n; i++) { // inserting indexes of those // numbers which are lesser // than 40 if (arr[i] < 40) { s.Add(i); Update(i, fact[arr[i]], n); } else { Update(i, 0, n); } } for ( int i = 0; i < q; i++) { // update query of the 1st type if (que[i].type == 1) { int l = que[i].l; int r = que[i].r; // find the left index of query in // the set using binary search var it = s.GetViewBetween(l, r).GetEnumerator(); // if it crosses the right index of // query or end of set, then break while (it.MoveNext()) { int idx = it.Current; int val = arr[idx] + 1; // update the value of arr[i] to // its new value Update(idx, fact[val] - fact[arr[idx]], n); arr[idx] = val; // if updated value becomes greater // than or equal to 40 remove it from // the set if (arr[idx] >= 40) { s.Remove(idx); } } } // update query of the 2nd type else if (que[i].type == 2) { int idx = que[i].l; int val = que[i].r; // update the value to its new value Update(idx, fact[val] - fact[arr[idx]], n); arr[idx] = val; // If the value is less than 40, insert // it into set, otherwise remove it if (val < 40) { s.Add(idx); } else { s.Remove(idx); } } // sum query of the 3rd type else { Console.WriteLine(Sum(que[i].r) - Sum(que[i].l - 1)); } } } // Driver Code to test above functions public static void Main( string [] args) { int q = 6; // input array using 1-based indexing int [] arr = { 0, 1, 2, 1, 4, 5 }; int n = arr.Length; // declaring array of structure of type queries Queries[] que = new Queries[q + 1]; que[0] = new Queries(3, 1, 5); que[1] = new Queries(1, 1, 3); que[2] = new Queries(2, 2, 4); que[3] = new Queries(3, 2, 4); que[4] = new Queries(1, 2, 5); que[5] = new Queries(3, 1, 5); // answer the Queries AnswerQueries(arr, que, n, q); } } // This code is contributed by phasing17. |
Javascript
// JavaScript program to calculate sum of // factorials in an interval and update // with two types of operations // Modulus let MOD = 10000000000; // Maximum size of input array let MAX = 100; // Size for factorial array let SZ = 40; let BIT = new Array(MAX + 1).fill(0); let fact = new Array(SZ + 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; } // structure for queries with members type, // leftIndex, rightIndex of the query class queries { constructor(tpe, l, r) { this .type = tpe; this .l = l; this .r = r; } } // function for updating the value function update(BIT, x, val, n) { while (x <= n) { BIT[x] += val; x += (x & -x); } return BIT } // function for calculating the required // sum between two indexes function summ(x) { var s = 0; while (x > 0) { s += BIT[x]; x -= x & -x; } return s; } // function to return answer to queries function answerQueries(arr, que, n, q) { // Precomputing factorials fact[0] = 1; for ( var i = 1; i <= 40; i++) fact[i] = Number((fact[i - 1] * i) % MOD); // Declaring a Set var s = new Set(); for ( var i = 1; i < n; i++) { // inserting indexes of those // numbers which are lesser // than 40 if (arr[i] < 40) { s.add(i); BIT = update(BIT, i, fact[arr[i]], n); } else BIT = update(BIT, i, 0, n); } for ( var i = 0; i < q; i++) { // update query of the 1st type if (que[i].type == 1) { while ( true ) { s = Array.from(s); s.sort(); // find the left index of query in // the set using binary search it = lower_bound(s, que[i].l); // if it crosses the right index of // query or end of set, then break if (it == s.length || s[it] > que[i].r) break ; que[i].l = s[it]; val = arr[s[it]] + 1; // update the value of arr[i] to // its new value BIT = update(BIT, s[it], fact[val] - fact[arr[s[it]]], n); arr[s[it]] += 1; // if updated value becomes greater // than or equal to 40 remove it from // the set if (arr[s[it]] >= 40) s.splice(it, 1); // increment the index que[i].l += 1; } } // update query of the 2nd type else if (que[i].type == 2) { s = new Set(s); var idx = que[i].l; var val = que[i].r; //update the value to its new value BIT = update(BIT, idx, fact[val] - fact[arr[idx]], n); arr[idx] = val; // If the value is less than 40, insert // it into set, otherwise remove it if (val < 40) s.add(idx); else s.remove(idx); } // sum query of the 3rd type else console.log((summ(que[i].r) - summ(que[i].l - 1))); } } // Driver Code let q = 6; // input array using 1-based indexing let arr = [0, 1, 2, 1, 4, 5]; let n = arr.length; // declaring array of structure of type queries let que = [ new queries(3, 1, 5), new queries(1, 1, 3), new queries(2, 2, 4), new queries(3, 2, 4), new queries(1, 2, 5), new queries(3, 1, 5) ]; // answer the Queries answerQueries(arr, que, n, q); // This code is contributed by phasing17 |
Java
// Java program to calculate sum of // factorials in an interval and update // with two types of operations import java.util.SortedSet; import java.util.TreeSet; class GFG { // Modulus private static final int MOD = 1000000000 ; // Maximum size of input array private static final int MAX = 100 ; // Size for factorial array private static final int SZ = 40 ; private static int [] BIT = new int [MAX + 1 ]; private static int [] fact = new int [SZ + 1 ]; // structure for queries with members type, // leftIndex, rightIndex of the query private static class Queries { public int type; public int l; public int r; // Constructor public Queries( int type, int l, int r) { this .type = type; this .l = l; this .r = r; } } // function for updating the value private 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 private 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 private static void answerQueries( int [] arr, Queries[] que, int n, int q) { // Precomputing factorials fact[ 0 ] = 1 ; for ( int i = 1 ; i < 41 ; i++) { fact[i] = (fact[i - 1 ] * i) % MOD; } // Declaring a Set SortedSet<Integer> s = new TreeSet<>(); for ( int i = 1 ; i < n; i++) { // inserting indexes of those // numbers which are lesser // than 40 if (arr[i] < 40 ) { s.add(i); update(i, fact[arr[i]], n); } else { update(i, 0 , n); } } for ( int i = 0 ; i < q; i++) { if (que[i].type == 1 ) { // find the left index of query in // the set using binary search SortedSet<Integer> view = s.subSet(que[i].l, que[i].r + 1 ); // if it crosses the right index of // query or end of set, then break for ( int idx : view) { int val = arr[idx] + 1 ; // update the value of arr[i] to // its new value update(idx, fact[val] - fact[arr[idx]], n); arr[idx] = val; // if updated value becomes greater // than or equal to 40 remove it from // the set if (arr[idx] >= 40 ) { s.remove(idx); } } } // update query of the 2nd type else if (que[i].type == 2 ) { int idx = que[i].l; int val = que[i].r; update(idx, fact[val] - fact[arr[idx]], n); arr[idx] = val; // If the value is less than 40, insert // it into set, otherwise remove it if (val < 40 ) { s.add(idx); } else { s.remove(idx); } } // sum query of the 3rd type else { System.out.println(sum(que[i].r) - sum(que[i].l - 1 )); } } } // Driver Code to test above functions public static void main(String[] args) { int q = 6 ; int [] arr = { 0 , 1 , 2 , 1 , 4 , 5 }; int n = arr.length; // declaring array of structure of type queries Queries[] que = new Queries[q + 1 ]; que[ 0 ] = new Queries( 3 , 1 , 5 ); que[ 1 ] = new Queries( 1 , 1 , 3 ); que[ 2 ] = new Queries( 2 , 2 , 4 ); que[ 3 ] = new Queries( 3 , 2 , 4 ); que[ 4 ] = new Queries( 1 , 2 , 5 ); que[ 5 ] = new Queries( 3 , 1 , 5 ); // answer the Queries answerQueries(arr, que, n, q); } } // This code is contributed by phasing17. |
148 50 968
Time Complexity: O(n+q) Since the set contains less than 40(constant) numbers, so its time complexity is treated as constant with respect to the large values of n and q.
Auxiliary Space: O(MAX+q)
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