Maximum length subarray with LCM equal to product
Last Updated :
11 Jul, 2025
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 condition
int 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 code
int main()
{
int arr[] = { 6, 10, 21 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxLengthSubArray(arr, n);
return 0;
}
// Java implementation of the above approach
import 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 condition
static 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 code
public 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 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 condition
def 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 code
arr = [6, 10, 21 ]
n = len(arr)
print(maxLengthSubArray(arr, n))
# This code is contributed by
# mohit kumar 29
// C# implementation of the above approach
using 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 condition
static 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 code
public 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 implementation of the above approach
// Function to calculate gcd(a, b)
function gcd($a, $b)
{
if ($b == 0)
return $a;
return gcd($b, $a % $b);
}
// Function to return max length of subarray
// that satisfies the condition
function maxLengthSubArray(&$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
?>
<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 condition
function 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 code
let arr=[6, 10, 21 ];
let n = arr.length;
document.write(maxLengthSubArray(arr, n));
// This code is contributed by unknown2108
</script>
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 element
vector<int> v[N];
// Stores the position of a prime in the subarray
// in two pointer technique
int f[MAX];
// Function to store smallest prime factor of numbers
void 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 = product
int 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 code
int main()
{
sieve();
int arr[] = { 6, 10, 21 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxLengthSubArray(arr, n);
return 0;
}
// Java implementation of the above approach
import 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 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 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 implementation of the above approach
N = 100005
MAX = 1000002
prime = [0 for i in range(MAX + 1)]
# Stores array of primes for every element
v = [[] for i in range(N)]
# Stores the position of a prime in the subarray
# in two pointer technique
f = [-1 for i in range(MAX)]
# Function to store smallest prime factor of numbers
def 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 = product
def 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 code
sieve()
arr = [6, 10, 21]
n = len(arr)
print(maxLengthSubArray(arr, n))
# This code is contributed by mohit kumar
// C# implementation of the above approach
using 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
<script>
// Javascript implementation of the above approach
let 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 element
let v = [];
// Stores the position of a prime in the
// subarray in two pointer technique
let f = new Array(MAX);
// Function to store smallest prime
// factor of numbers
function 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 = product
function 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 code
sieve();
let arr=[ 6, 10, 21];
let n = arr.length;
document.write(maxLengthSubArray(arr, n));
// This code is contributed by patel2127
</script>
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Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
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