Given an array
arr[]
, the task is to count the number of the unique arrays can be formed by updating the elements of the given array to any element in the range [1, arr[i]] such that the
of the updated array is equal to the maximum element.
Examples:
Input: arr[] = {6, 3} Output: 13 Explanation: Possible Arrays are – {[1, 1], [1, 2], [2, 1], [2, 2], [1, 3], [3, 1], [3, 3], [4, 1], [4, 2], [5, 1], [6, 1], [6, 2], [6, 3]} Input: arr[] = {1, 4, 3, 2} Output: 15
Approach:
- For the maximum element to be the LCM of the array, we need to fix the maximum element of the array.
- As, we have fixed some number as maximum, now for the LCM to be , we’ll need to ensure that every element in the array is some multiple of including
- We’ll find the factors for the number and find the number of ways to place them in the array.
- Let’s say that the factors of be . The count of factors is .
- Let’s assume that number of positions that be means there are number of positions that have number in the array which is greater than equal to and let have positions and have positions.
- Now, number of ways to distribute x in positions, in positions and in positions are and so on.
- Now, we’ll have to subtract those ways which have LCM but is not there.
- We’ll need to subtract from the ways.
- We’ll use BIT(Binary Indexed Tree) to find number of positions greater than some number .
Below is the implementation of the above approach:
C++
// C++ implementation to find the // Number of ways to change the array // such that maximum element of the // array is the LCM of the array #include <bits/stdc++.h> using namespace std; // Modulo const int MOD = 1e9 + 7; const int N = 1e5 + 5; // Fenwick tree to find number // of indexes greater than x vector< int > BIT(N, 0); // Function to compute // x ^ y % MOD int power( int x, int y) { if (x == 0) return 0; int ans = 1; // Loop to compute the // x^y % MOD while (y > 0) { if (y & 1) ans = (1LL * ans * x) % MOD; x = (1LL * x * x) % MOD; y >>= 1; } return ans; } // Function to update the binary // indexed tree void updateBIT( int idx, int val) { assert (idx > 0); while (idx < N) { BIT[idx] += val; idx += idx & -idx; } } // Function to find the prefix sum // upto the current index int queryBIT( int idx) { int ans = 0; while (idx > 0) { ans += BIT[idx]; idx -= idx & -idx; } return ans; } // Function to find the number of // ways to change the array such // that the LCM of array is // maximum element of the array int numWays( int arr[], int n) { int mx = 0; for ( int i = 0; i < n; i++) { // Updating BIT with the // frequency of elements updateBIT(arr[i], 1); // Maximum element in the array mx = max(mx, arr[i]); } // 1 is for every element // is 1 in the array; int ans = 1; for ( int i = 2; i <= mx; i++) { // Vector for storing the factors vector< int > factors; for ( int j = 1; j * j <= i; j++) { // finding factors of i if (i % j == 0) { factors.push_back(j); if (i / j != j) factors.push_back(i / j); } } // Sorting in descending order sort(factors.rbegin(), factors.rend()); int cnt = 1; // for storing number of indexex // greater than the i - 1 element int prev = 0; for ( int j = 0; j < factors.size(); j++) { // Number of remaining factors int remFactors = int (factors.size()) - j; // Number of indexes in the array // with element factor[j] and above int indexes = n - queryBIT(factors[j] - 1); // Multiplying count with // remFcators ^ (indexes - prev) cnt = (1LL * cnt * power(remFactors, indexes - prev)) % MOD; prev = max(prev, indexes); } // Remove those counts which have // lcm as i but i is not present factors.erase(factors.begin()); int toSubtract = 1; prev = 0; // Loop to find the count which have // lcm as i but i is not present for ( int j = 0; j < factors.size(); j++) { int remFactors = int (factors.size()) - j; int indexes = n - queryBIT(factors[j] - 1); toSubtract = (1LL * toSubtract * power(remFactors, indexes - prev)); prev = max(prev, indexes); } // Adding cnt - toSubtract to answer ans = (1LL * ans + cnt - toSubtract + MOD) % MOD; } return ans; } // Driver Code int main() { int arr[] = { 6, 3 }; int n = sizeof (arr) / sizeof (arr[0]); int ans = numWays(arr, n); cout << ans << endl; return 0; } |
Java
import java.util.*; public class Main { // Modulo static final int MOD = 1000000007 ; static final int N = 100005 ; // Fenwick tree to find number // of indexes greater than x static List<Integer> BIT = new ArrayList<>(Collections.nCopies(N, 0 )); // Function to compute // x ^ y % MOD static int power( int x, int y) { if (x == 0 ) return 0 ; int ans = 1 ; // Loop to compute the // x^y % MOD while (y > 0 ) { if ((y & 1 ) != 0 ) ans = ( int )((1L * ans * x) % MOD); x = ( int )((1L * x * x) % MOD); y >>= 1 ; } return ans; } // Function to update the binary // indexed tree static void updateBIT( int idx, int val) { assert idx > 0 ; while (idx < N) { BIT.set(idx, BIT.get(idx) + val); idx += idx & -idx; } } // Function to find the prefix sum // upto the current index static int queryBIT( int idx) { int ans = 0 ; while (idx > 0 ) { ans += BIT.get(idx); idx -= idx & -idx; } return ans; } // Function to find the number of // ways to change the array such // that the LCM of array is // maximum element of the array static int numWays( int [] arr, int n) { int mx = 0 ; for ( int i = 0 ; i < n; i++) { // Updating BIT with the // frequency of elements updateBIT(arr[i], 1 ); // Maximum element in the array mx = Math.max(mx, arr[i]); } // 1 is for every element // is 1 in the array; int ans = 1 ; for ( int i = 2 ; i <= mx; i++) { // Vector for storing the factors List<Integer> factors = new ArrayList<>(); for ( int j = 1 ; j * j <= i; j++) { // finding factors of i if (i % j == 0 ) { factors.add(j); if (i / j != j) factors.add(i / j); } } // Sorting in descending order factors.sort(Collections.reverseOrder()); int cnt = 1 ; // for storing number of indexex // greater than the i - 1 element int prev = 0 ; for ( int j = 0 ; j < factors.size(); j++) { // Number of remaining factors int remFactors = factors.size() - j; // Number of indexes in the array // with element factor[j] and above int indexes = n - queryBIT(factors.get(j) - 1 ); // Multiplying count with // remFcators ^ (indexes - prev) cnt = ( int )((1L * cnt * power(remFactors, indexes - prev)) % MOD); prev = Math.max(prev, indexes); } // Remove those counts which have // lcm as i but i is not present factors.remove( 0 ); int toSubtract = 1 ; prev = 0 ; // Loop to find the count which have // lcm as i but i is not present for ( int j = 0 ; j < factors.size(); j++) { int remFactors = factors.size() - j; int indexes = n - queryBIT(factors.get(j) - 1 ); toSubtract = ( int )((1L * toSubtract * power(remFactors, indexes - prev)) % MOD); prev = Math.max(prev, indexes); } // Adding cnt - toSubtract to answer ans = ( int )((1L * ans + cnt - toSubtract + MOD) % MOD); } return ans; } // Driver Code public static void main(String[] args) { int [] arr = { 6 , 3 }; int n = arr.length; int ans = numWays(arr, n); System.out.println(ans); } } // This code is contributed by Samim Hossain Mondal. |
Python3
# Python implementation to find the # Number of ways to change the array # such that maximum element of the # array is the LCM of the array # Modulo MOD = int ( 1e9 ) + 9 MAXN = int ( 1e5 ) + 5 # Fenwick tree to find number # of indexes greater than x BIT = [ 0 for _ in range (MAXN)] # Function to compute # x ^ y % MOD def power(x, y): if x = = 0 : return 0 ans = 1 # Loop to compute the # x ^ y % MOD while y > 0 : if y % 2 = = 1 : ans = (ans * x) % MOD x = (x * x) % MOD y = y / / 2 return ans # Function to update the # Binary Indexed Tree def updateBIT(idx, val): # Loop to update the BIT while idx < MAXN: BIT[idx] + = val idx + = idx & ( - idx) # Function to find # prefix sum upto idx def queryBIT(idx): ans = 0 while idx > 0 : ans + = BIT[idx] idx - = idx & ( - idx) return ans # Function to find number of ways # to change the array such that # MAX of array is same as LCM def numWays(arr): mx = 0 # Updating BIT with the # frequency of elements for i in arr: updateBIT(i, 1 ) # Maximum element # in the array mx = max (mx, i) ans = 1 for i in range ( 2 , mx + 1 ): # For storing factors of i factors = [] for j in range ( 1 , i + 1 ): if j * j > i: break # Finding factors of i if i % j = = 0 : factors.append(j) if i / / j ! = j: factors.append(i / / j) # Sorting in descending order factors.sort() factors.reverse() # For storing ans cnt = 1 # For storing number of indexes # greater than the i - 1 element prev = 0 for j in range ( len (factors)): # Number of remaining factors remFactors = len (factors) - j # Number of indexes in the array # with element factor[j] and above indexes = len (arr) - queryBIT(factors[j] - 1 ) # Multiplying count with # remFcators ^ (indexes - prev) cnt = (cnt * power(remFactors, \ indexes - prev)) % MOD prev = max (prev, indexes) # Remove those counts which have # lcm as i but i is not present factors.remove(factors[ 0 ]) toSubtract = 1 prev = 0 for j in range ( len (factors)): remFactors = len (factors) - j indexes = len (arr) - queryBIT(factors[j] - 1 ) toSubtract = (toSubtract * \ power(remFactors, indexes - prev)) prev = max (prev, indexes) # Adding cnt - toSubtract to ans; ans = (ans + cnt - toSubtract + MOD) % MOD; return ans # Driver Code if __name__ = = "__main__" : arr = [ 1 , 4 , 3 , 2 ] ans = numWays(arr); print (ans) |
C#
using System; using System.Collections.Generic; class MainClass { // Modulo const int MOD = 1000000007; const int N = 100005; // Fenwick tree to find the number // of indexes greater than x static List< int > BIT = new List< int >( new int [N]); // Function to compute x ^ y % MOD static int Power( int x, int y) { if (x == 0) return 0; int ans = 1; // Loop to compute x^y % MOD while (y > 0) { if ((y & 1) == 1) ans = ( int )((1L * ans * x) % MOD); x = ( int )((1L * x * x) % MOD); y >>= 1; } return ans; } // Function to update the binary // indexed tree static void UpdateBIT( int idx, int val) { if (idx <= 0) return ; while (idx < N) { BIT[idx] += val; idx += idx & -idx; } } // Function to find the prefix sum // up to the current index static int QueryBIT( int idx) { int ans = 0; while (idx > 0) { ans += BIT[idx]; idx -= idx & -idx; } return ans; } // Function to find the number of // ways to change the array such // that the LCM of the array is // the maximum element of the array static int NumWays( int [] arr, int n) { int mx = 0; for ( int i = 0; i < n; i++) { // Updating BIT with the frequency of elements UpdateBIT(arr[i], 1); // Maximum element in the array mx = Math.Max(mx, arr[i]); } // 1 is for every element is 1 in the array; int ans = 1; for ( int i = 2; i <= mx; i++) { // List for storing the factors List< int > factors = new List< int >(); for ( int j = 1; j * j <= i; j++) { // Finding factors of i if (i % j == 0) { factors.Add(j); if (i / j != j) factors.Add(i / j); } } // Sorting in descending order factors.Sort((a, b) => b.CompareTo(a)); int cnt = 1; // For storing the number of indexes greater than the i - 1 element int prev = 0; for ( int j = 0; j < factors.Count; j++) { // Number of remaining factors int remFactors = factors.Count - j; // Number of indexes in the array with element factor[j] and above int indexes = n - QueryBIT(factors[j] - 1); // Multiplying count with remFcators ^ (indexes - prev) cnt = ( int )( (1L * cnt * Power(remFactors, indexes - prev)) % MOD); prev = Math.Max(prev, indexes); } // Remove those counts which have lcm as i but i is not present factors.RemoveAt(0); int toSubtract = 1; prev = 0; // Loop to find the count which has lcm as i but i is not present for ( int j = 0; j < factors.Count; j++) { int remFactors = factors.Count - j; int indexes = n - QueryBIT(factors[j] - 1); toSubtract = ( int )( (1L * toSubtract * Power(remFactors, indexes - prev)) % MOD); prev = Math.Max(prev, indexes); } // Adding cnt - toSubtract to the answer ans = ( int )((1L * ans + cnt - toSubtract + MOD) % MOD); } return ans; } public static void Main( string [] args) { int [] arr = { 6, 3 }; int n = arr.Length; int ans = NumWays(arr, n); Console.WriteLine(ans); } } |
Javascript
// Modulo const MOD = 1e9 + 7; const N = 1e5 + 5; // Fenwick tree to find the number of indexes greater than x const BIT = new Array(N).fill(0); // Function to compute x^y % MOD function power(x, y) { if (x === 0) return 0; let ans = 1; // Loop to compute x^y % MOD while (y > 0) { if (y & 1) ans = (ans * x) % MOD; x = (x * x) % MOD; y >>= 1; } return ans; } // Function to update the binary indexed tree function updateBIT(idx, val) { while (idx > 0) { BIT[idx] += val; idx += idx & -idx; } } // Function to find the prefix sum up to // the current index function queryBIT(idx) { let ans = 0; while (idx > 0) { ans += BIT[idx]; idx -= idx & -idx; } return ans; } function GFG(arr, n) { let mx = 0; for (let i = 0; i < n; i++) { // Updating BIT with the frequency of elements updateBIT(arr[i], 1); // Maximum element in the array mx = Math.max(mx, arr[i]); } // 1 is for every element is 1 in the array let ans = 1; for (let i = 2; i <= mx; i++) { // Vector for storing the factors const factors = []; for (let j = 1; j * j <= i; j++) { // Finding factors of i if (i % j === 0) { factors.push(j); if (i / j !== j) factors.push(i / j); } } // Sorting in descending order factors.sort((a, b) => b - a); let cnt = 1; // For storing the number of indexes greater than // the (i - 1) element let prev = 0; for (let j = 0; j < factors.length; j++) { // Number of remaining factors const remFactors = factors.length - j; // Number of indexes in the array with an element >= factor[j] const indexes = n - queryBIT(factors[j] - 1); // Multiplying count with (remFactors ^ (indexes - prev)) cnt = (cnt * power(remFactors, indexes - prev)) % MOD; prev = Math.max(prev, indexes); } // Remove those counts which have LCM as i but i is not present factors.shift(); let toSubtract = 1; prev = 0; // Loop to find the count which have LCM as i but i is not present for (let j = 0; j < factors.length; j++) { const remFactors = factors.length - j; const indexes = n - queryBIT(factors[j] - 1); toSubtract = (toSubtract * power(remFactors, indexes - prev)) % MOD prev = Math.max(prev, indexes); } // Adding cnt - toSubtract to answer ans = (ans + cnt - toSubtract + MOD) % MOD; } return ans; } // Driver Code const arr = [6, 3]; const n = arr.length; const ans = GFG(arr, n); console.log(ans); |
13
Time Complexity:
, where
is the maximum element in the array.
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