Given an arr[], the task is to find the maximum length of the sub-array such that the LCM of the sub-array is equal to the product of numbers in the sub-array. If no valid sub-array exists, then print -1.
Note: The length of the sub-array must be ? 2.
Examples:
Input: arr[] = { 6, 10, 21}
Output: 2
The sub-array { 10, 21 } satisfies the condition.Input: arr[] = { 2, 2, 4}
Output: -1
No sub-array satisfies the condition. Hence, the output is -1.
Naive Approach: Run nested loops to check the condition for every possible sub-array of length ? 2. If the sub-array satisfies the condition, then update ans = max(ans, length(sub-array)). Print the ans in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;#define ll long long// Function to calculate gcd(a, b)int gcd(int a, int b){ if (b == 0) return a; return gcd(b, a % b);}// Function to return max length of subarray// that satisfies the conditionint maxLengthSubArray(const int* arr, int n){ int maxLen = -1; for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { ll lcm = 1LL * arr[i]; ll product = 1LL * arr[i]; // Update LCM and product of the // sub-array for (int k = i + 1; k <= j; k++) { lcm = (((arr[k] * lcm)) / (gcd(arr[k], lcm))); product = product * arr[k]; } // If the current sub-array satisfies // the condition if (lcm == product) { // Choose the maximum maxLen = max(maxLen, j - i + 1); } } } return maxLen;}// Driver codeint main(){ int arr[] = { 6, 10, 21 }; int n = sizeof(arr) / sizeof(arr[0]); cout << maxLengthSubArray(arr, n); return 0;} |
Java
// Java implementation of the above approachimport java.util.*;class GFG{// Function to calculate gcd(a, b)static int gcd(int a, int b){ if (b == 0) return a; return gcd(b, a % b);}// Function to return max length of subarray// that satisfies the conditionstatic int maxLengthSubArray(int arr[], int n){ int maxLen = -1; for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { int lcm = 1 * arr[i]; int product = 1 * arr[i]; // Update LCM and product of the // sub-array for (int k = i + 1; k <= j; k++) { lcm = (((arr[k] * lcm)) / (gcd(arr[k], lcm))); product = product * arr[k]; } // If the current sub-array satisfies // the condition if (lcm == product) { // Choose the maximum maxLen = Math.max(maxLen, j - i + 1); } } } return maxLen;}// Driver codepublic static void main(String args[]){ int arr[] = { 6, 10, 21 }; int n = arr.length; System.out.println(maxLengthSubArray(arr, n));}}// This code is contributed by// Shashank_Sharma |
Python3
# Python3 implementation of the # above approach# Function to calculate gcd(a, b)def gcd(a, b): if (b == 0): return a return gcd(b, a % b)# Function to return max length of # subarray that satisfies the conditiondef maxLengthSubArray(arr, n): maxLen = -1 for i in range(n - 1): for j in range(n): lcm = arr[i] product = arr[i] # Update LCM and product of the # sub-array for k in range(i + 1, j + 1): lcm = (((arr[k] * lcm)) // (gcd(arr[k], lcm))) product = product * arr[k] # If the current sub-array satisfies # the condition if (lcm == product): # Choose the maximum maxLen = max(maxLen, j - i + 1) return maxLen# Driver codearr = [6, 10, 21 ]n = len(arr)print(maxLengthSubArray(arr, n))# This code is contributed by # mohit kumar 29 |
C#
// C# implementation of the above approachusing System;class GFG{// Function to calculate gcd(a, b)static int gcd(int a, int b){ if (b == 0) return a; return gcd(b, a % b);}// Function to return max length of subarray// that satisfies the conditionstatic int maxLengthSubArray(int[] arr, int n){ int maxLen = -1; for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { int lcm = 1 * arr[i]; int product = 1 * arr[i]; // Update LCM and product of the // sub-array for (int k = i + 1; k <= j; k++) { lcm = (((arr[k] * lcm)) / (gcd(arr[k], lcm))); product = product * arr[k]; } // If the current sub-array satisfies // the condition if (lcm == product) { // Choose the maximum maxLen = Math.Max(maxLen, j - i + 1); } } } return maxLen;}// Driver codepublic static void Main(){ int[] arr = { 6, 10, 21 }; int n = arr.Length; Console.Write(maxLengthSubArray(arr, n));}}// This code is contributed by ita_c |
PHP
<?php// PHP implementation of the above approach// Function to calculate gcd(a, b)function gcd($a, $b){ if ($b == 0) return $a; return gcd($b, $a % $b);}// Function to return max length of subarray// that satisfies the conditionfunction maxLengthSubArray(&$arr, $n){ $maxLen = -1; for ($i = 0; $i < $n - 1; $i++) { for ($j = $i + 1; $j < $n; $j++) { $lcm = $arr[$i]; $product = $arr[$i]; // Update LCM and product of the // sub-array for ($k = $i + 1; $k <= $j; $k++) { $lcm = ((($arr[$k] * $lcm)) / (gcd($arr[$k], $lcm))); $product = $product * $arr[$k]; } // If the current sub-array satisfies // the condition if ($lcm == $product) { // Choose the maximum $maxLen = max($maxLen, $j - $i + 1); } } } return $maxLen;}// Driver code$arr = array(6, 10, 21 );$n = sizeof($arr);echo(maxLengthSubArray($arr, $n));// This code is contributed by Shivi_Aggarwal?> |
Javascript
<script>// Javascript implementation of the above approach// Function to calculate gcd(a, b)function gcd(a,b){ if (b == 0) return a; return gcd(b, a % b);}// Function to return max length of subarray// that satisfies the conditionfunction maxLengthSubArray(arr,n){ let maxLen = -1; for (let i = 0; i < n - 1; i++) { for (let j = i + 1; j < n; j++) { let lcm = 1 * arr[i]; let product = 1 * arr[i]; // Update LCM and product of the // sub-array for (let k = i + 1; k <= j; k++) { lcm = (((arr[k] * lcm)) / (gcd(arr[k], lcm))); product = product * arr[k]; } // If the current sub-array satisfies // the condition if (lcm == product) { // Choose the maximum maxLen = Math.max(maxLen, j - i + 1); } } } return maxLen;}// Driver codelet arr=[6, 10, 21 ];let n = arr.length;document.write(maxLengthSubArray(arr, n)); // This code is contributed by unknown2108</script> |
Output:
2
Efficient Approach: A sub-array will have its LCM equal to its product when no two elements in the sub-array have any common factor.
For example:
arr[] = { 6, 10, 21 }
Prime factorization yields:
arr[] = { 2 * 3, 2 * 5, 3 * 7 }
[6, 10] has 2 as a common factor.
[6, 10, 21] has 2 as a common factor between 6 and 10.
Sub-array [10, 21] has no common factor between any 2 elements. Therefore, answer = 2.
Firstly, prime factorization of numbers is done to deal with factors. To calculate the sub-array in which no 2 elements have a common factor, we use the two-pointer technique.
Two pointers run, both from the right and they represent the current sub-array. We add elements in the sub-array from the right. Now there are two scenarios:
- An element is added in the current sub-array if it has no factor in common with the current elements in the sub-array. If a common factor is found, then starting from the left, elements are subsequently eliminated until no common factor is found with the newly added element.
- If there are no common factors between the newly added element and existing elements, then update ans = max(ans, length of sub-array)
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>#define pb push_back#define N 100005#define MAX 1000002#define mem(a, b) memset(a, b, sizeof(a))using namespace std;int prime[MAX];// Stores array of primes for every elementvector<int> v[N];// Stores the position of a prime in the subarray// in two pointer techniqueint f[MAX];// Function to store smallest prime factor of numbersvoid sieve(){ prime[0] = prime[1] = 1; for (int i = 2; i < MAX; i++) { if (!prime[i]) { for (int j = i * 2; j < MAX; j += i) { if (!prime[j]) prime[j] = i; } } } for (int i = 2; i < MAX; i++) { // If number is prime, // then smallest prime factor is the // number itself if (!prime[i]) prime[i] = i; }}// Function to return maximum length of subarray// with LCM = productint maxLengthSubArray(int* arr, int n){ // Initialize f with -1 mem(f, -1); for (int i = 0; i < n; ++i) { // Prime factorization of numbers // Store primes in a vector for every element while (arr[i] > 1) { int p = prime[arr[i]]; arr[i] /= p; v[i].pb(p); } } // Two pointers l and r // denoting left and right of subarray int l = 0, r = 1, ans = -1; // f is a mapping. // prime -> index in the current subarray // With the help of f, // we can detect whether a prime has // already occurred in the subarray for (int i : v[0]) { f[i] = 0; } while (l <= r && r < n) { int flag = 0; for (int i = 0; i < v[r].size(); i++) { // Map the prime to the index if (f[v[r][i]] == -1 || f[v[r][i]] == r) { f[v[r][i]] = r; } // If already occurred then, // start removing elements from the left else { flag = 1; break; } } // Remove elements if flag = 1 if (flag) { // Nullify entries of element at index 'l' for (int i : v[l]) { f[i] = -1; } // Increment 'l' l++; } else { // Maximize the answer when // no common factor is found ans = max(ans, r - l + 1); r++; } } // One length subarray is discarded if (ans == 1) ans = -1; return ans;}// Driver codeint main(){ sieve(); int arr[] = { 6, 10, 21 }; int n = sizeof(arr) / sizeof(arr[0]); cout << maxLengthSubArray(arr, n); return 0;} |
Java
// Java implementation of the above approachimport java.io.*;import java.util.*;class GFG{ static int N = 100005;static int MAX = 1000002;static int[] prime = new int[MAX];// Stores array of primes for every element static ArrayList< ArrayList<Integer>> v = new ArrayList< ArrayList<Integer>>();// Stores the position of a prime in the // subarray in two pointer techniquestatic int[] f = new int[MAX];// Function to store smallest prime // factor of numbers static void sieve(){ for(int i = 0; i < N; i++) { v.add(new ArrayList<Integer>()); } prime[0] = prime[1] = 1; for(int i = 2; i < MAX; i++) { if (prime[i] == 0) { for(int j = i * 2; j < MAX; j += i) { if (prime[j] == 0) { prime[j] = i; } } } } for(int i = 2; i < MAX; i++) { // If number is prime, then // smallest prime factor is // the number itself if (prime[i] == 0) { prime[i] = i; } }}// Function to return maximum length of// subarray with LCM = product static int maxLengthSubArray(int[] arr, int n) { // Initialize f with -1 Arrays.fill(f, -1); for(int i = 0; i < n; ++i) { // Prime factorization of numbers // Store primes in a vector for // every element while (arr[i] > 1) { int p = prime[arr[i]]; arr[i] /= p; v.get(i).add(p); } } // Two pointers l and r denoting // left and right of subarray int l = 0, r = 1, ans = -1; // f is a mapping. // prime -> index in the current subarray // With the help of f, // we can detect whether a prime has // already occurred in the subarray for(int i : v.get(0)) { f[i] = 0; } while (l <= r && r < n) { int flag = 0; for(int i = 0; i < v.get(r).size(); i++) { // Map the prime to the index if (f[v.get(r).get(i)] == -1 || f[v.get(r).get(i)] == r) { f[v.get(r).get(i)] = r; } // If already occurred then, // start removing elements // from the left else { flag = 1; break; } } // Remove elements if flag = 1 if (flag != 0) { // Nullify entries of element // at index 'l' for(int i : v.get(l)) { f[i] = -1; } // Increment 'l' l++; } else { // Maximize the answer when // no common factor is found ans = Math.max(ans, r - l + 1); r++; } } // One length subarray is discarded if (ans == 1) { ans = -1; } return ans; }// Driver code public static void main(String[] args) { sieve(); int arr[] = { 6, 10, 21 }; int n = arr.length; System.out.println(maxLengthSubArray(arr, n));}}// This code is contributed by avanitrachhadiya2155 |
Python3
# Python3 implementation of the above approachN = 100005MAX = 1000002prime = [0 for i in range(MAX + 1)]# Stores array of primes for every elementv = [[] for i in range(N)]# Stores the position of a prime in the subarray# in two pointer techniquef = [-1 for i in range(MAX)]# Function to store smallest prime factor of numbersdef sieve(): prime[0], prime[1] = 1, 1 for i in range(2, MAX + 1): if (prime[i] == 0): for j in range(i * 2, MAX, i): if (prime[j] == 0): prime[j] = i for i in range(2, MAX): # If number is prime, # then smallest prime factor is the # number itself if (prime[i] == 0): prime[i] = i# Function to return maximum length of subarray# with LCM = productdef maxLengthSubArray(arr, n): # Initialize f with -1 for i in range(n): f[i] = -1 for i in range(n): # Prime factorization of numbers # Store primes in a vector for every element while (arr[i] > 1): p = prime[arr[i]] arr[i] //= p v[i].append(p) # Two pointers l and r # denoting left and right of subarray l, r, ans = 0, 1, -1 # f is a mapping. # prime -> index in the current subarray # With the help of f, # we can detect whether a prime has # already occurred in the subarray for i in v[0]: f[i] = 0 while (l <= r and r < n): flag = 0 for i in range(len(v[r])): # Map the prime to the index if (f[v[r][i]] == -1 or f[v[r][i]] == r): f[v[r][i]] = r # If already occurred then, # start removing elements from the left else: flag = 1 break # Remove elements if flag = 1 if (flag): # Nullify entries of element at index 'l' for i in v[l]: f[i] = -1 # Increment 'l' l += 1 else : # Maximize the answer when # no common factor is found ans = max(ans, r - l + 1) r += 1 # One length subarray is discarded if (ans == 1): ans = -1 return ans# Driver codesieve()arr = [6, 10, 21]n = len(arr)print(maxLengthSubArray(arr, n))# This code is contributed by mohit kumar |
C#
// C# implementation of the above approachusing System;using System.Collections.Generic;class GFG{ static int N = 100005; static int MAX = 1000002; static int[] prime = new int[MAX]; // Stores array of primes for every element static List<List<int>> v = new List<List<int>>(); // Stores the position of a prime in the // subarray in two pointer technique static int[] f = new int[MAX]; // Function to store smallest prime // factor of numbers static void sieve() { for(int i = 0; i < N; i++) { v.Add(new List<int>()); } prime[0] = prime[1] = 1; for(int i = 2; i < MAX; i++) { if (prime[i] == 0) { for(int j = i * 2; j < MAX; j += i) { if (prime[j] == 0) { prime[j] = i; } } } } for(int i = 2; i < MAX; i++) { // If number is prime, then // smallest prime factor is // the number itself if (prime[i] == 0) { prime[i] = i; } } } // Function to return maximum length of // subarray with LCM = product static int maxLengthSubArray(int[] arr, int n) { // Initialize f with -1 Array.Fill(f, -1); for(int i = 0; i < n; ++i) { // Prime factorization of numbers // Store primes in a vector for // every element while (arr[i] > 1) { int p = prime[arr[i]]; arr[i] /= p; v[i].Add(p); } } // Two pointers l and r denoting // left and right of subarray int l = 0, r = 1, ans = -1; // f is a mapping. // prime -> index in the current subarray // With the help of f, // we can detect whether a prime has // already occurred in the subarray foreach(int i in v[0]) { f[i] = 0; } while (l <= r && r < n) { int flag = 0; for(int i = 0; i < v[r].Count; i++) { // Map the prime to the index if (f[v[r][i]] == -1 || f[v[r][i]] == r) { f[v[r][i]] = r; } // If already occurred then, // start removing elements // from the left else { flag = 1; break; } } // Remove elements if flag = 1 if (flag != 0) { // Nullify entries of element // at index 'l' foreach(int i in v[l]) { f[i] = -1; } // Increment 'l' l++; } else { // Maximize the answer when // no common factor is found ans = Math.Max(ans, r - l + 1); r++; } } // One length subarray is discarded if (ans == 1) { ans = -1; } return ans; } // Driver code static public void Main () { sieve(); int[] arr = { 6, 10, 21 }; int n = arr.Length; Console.WriteLine(maxLengthSubArray(arr, n)); }}// This code is contributed by rag2127 |
Javascript
<script>// Javascript implementation of the above approachlet N = 100005;let MAX = 1000002;let prime = new Array(MAX);for(let i=0;i<prime.length;i++){ prime[i]=0;}// Stores array of primes for every elementlet v = [];// Stores the position of a prime in the// subarray in two pointer techniquelet f = new Array(MAX);// Function to store smallest prime// factor of numbersfunction sieve(){ for(let i = 0; i < N; i++) { v.push([]); } prime[0] = prime[1] = 1; for(let i = 2; i < MAX; i++) { if (prime[i] == 0) { for(let j = i * 2; j < MAX; j += i) { if (prime[j] == 0) { prime[j] = i; } } } } for(let i = 2; i < MAX; i++) { // If number is prime, then // smallest prime factor is // the number itself if (prime[i] == 0) { prime[i] = i; } }}// Function to return maximum length of// subarray with LCM = productfunction maxLengthSubArray(arr,n){ // Initialize f with -1 for(let i=0;i<f.length;i++) { f[i]=-1; } for(let i = 0; i < n; ++i) { // Prime factorization of numbers // Store primes in a vector for // every element while (arr[i] > 1) { let p = prime[arr[i]]; arr[i] /= p; v[i].push(p); } } // Two pointers l and r denoting // left and right of subarray let l = 0, r = 1, ans = -1; // f is a mapping. // prime -> index in the current subarray // With the help of f, // we can detect whether a prime has // already occurred in the subarray for(let i=0;i< v[0].length;i++) { f[v[0][i]] = 0; } while (l <= r && r < n) { let flag = 0; for(let i = 0; i < v[r].length; i++) { // Map the prime to the index if (f[v[r][i]] == -1 || f[v[r][i]] == r) { f[v[r][i]] = r; } // If already occurred then, // start removing elements // from the left else { flag = 1; break; } } // Remove elements if flag = 1 if (flag != 0) { // Nullify entries of element // at index 'l' for(let i=0;i<v[l].length;i++) { f[v[l][i]] = -1; } // Increment 'l' l++; } else { // Maximize the answer when // no common factor is found ans = Math.max(ans, r - l + 1); r++; } } // One length subarray is discarded if (ans == 1) { ans = -1; } return ans;}// Driver codesieve();let arr=[ 6, 10, 21];let n = arr.length;document.write(maxLengthSubArray(arr, n));// This code is contributed by patel2127</script> |
Output:
2
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