Given an array arr[] of length N, the task is the find the length of the longest sub-array with the maximum possible GCD value.
Examples:
Input: arr[] = {1, 2, 2}
Output: 2
Here all possible sub-arrays and there GCD’s are:
1) {1} -> 1
2) {2} -> 2
3) {2} -> 2
4) {1, 2} -> 1
5) {2, 2} -> 2
6) {1, 2, 3} -> 1
Here, the maximum GCD value is 2 and longest sub-array having GCD = 2 is {2, 2}.
Thus, the answer is {2, 2}.Input: arr[] = {3, 3, 3, 3}
Output: 4
Naive approach: Generate all the possible sub-arrays and find the GCD of each of them individually in order to find the longest such sub-array. This approach will take O(N3) time to solve the problem.
Better approach: The maximum GCD value will always be equal to the largest number present in the array. Let’s say that the largest number present in the array is X. Now, the task is to find the largest sub-array having all X. The same can be done using the two-pointer approach. Below is the algorithm:
- Find the largest number in the array. Let us call this number X.
- Run a loop from i = 0
- If arr[i] != X then increment i and continue.
- Else initialize j = i.
- While j < n and arr[j] = X, increment j.
- Update the answer as ans = max(ans, j – i).
- Update i as i = j.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the length of // the largest subarray with // maximum possible GCD int findLength(int* arr, int n)
{ // To store the maximum number
// present in the array
int x = 0;
// Finding the maximum element
for (int i = 0; i < n; i++)
x = max(x, arr[i]);
// To store the final answer
int ans = 0;
// Two pointer
for (int i = 0; i < n; i++) {
if (arr[i] != x)
continue;
// Running a loop from j = i
int j = i;
// Condition for incrementing 'j'
while (arr[j] == x)
j++;
// Updating the answer
ans = max(ans, j - i);
}
return ans;
} // Driver code int main()
{ int arr[] = { 1, 2, 2 };
int n = sizeof(arr) / sizeof(int);
cout << findLength(arr, n);
return 0;
} |
// Java implementation of the approach class GFG
{ // Function to return the length of
// the largest subarray with
// maximum possible GCD
static int findLength(int []arr, int n)
{
// To store the maximum number
// present in the array
int x = 0;
// Finding the maximum element
for (int i = 0; i < n; i++)
x = Math.max(x, arr[i]);
// To store the final answer
int ans = 0;
// Two pointer
for (int i = 0; i < n; i++)
{
if (arr[i] != x)
continue;
// Running a loop from j = i
int j = i;
// Condition for incrementing 'j'
while (arr[j] == x)
{
j++;
if (j >= n )
break;
}
// Updating the answer
ans = Math.max(ans, j - i);
}
return ans;
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 1, 2, 2 };
int n = arr.length;
System.out.println(findLength(arr, n));
}
} // This code is contributed by AnkitRai01 |
# Python3 implementation of the approach # Function to return the length of # the largest subarray with # maximum possible GCD def findLength(arr, n) :
# To store the maximum number
# present in the array
x = 0;
# Finding the maximum element
for i in range(n) :
x = max(x, arr[i]);
# To store the final answer
ans = 0;
# Two pointer
for i in range(n) :
if (arr[i] != x) :
continue;
# Running a loop from j = i
j = i;
# Condition for incrementing 'j'
while (arr[j] == x) :
j += 1;
if j >= n :
break
# Updating the answer
ans = max(ans, j - i);
return ans;
# Driver code if __name__ == "__main__" :
arr = [ 1, 2, 2 ];
n = len(arr);
print(findLength(arr, n));
# This code is contributed by AnkitRai01 |
// C# implementation of the approach using System;
class GFG
{ // Function to return the length of
// the largest subarray with
// maximum possible GCD
static int findLength(int []arr, int n)
{
// To store the maximum number
// present in the array
int x = 0;
// Finding the maximum element
for (int i = 0; i < n; i++)
x = Math.Max(x, arr[i]);
// To store the final answer
int ans = 0;
// Two pointer
for (int i = 0; i < n; i++)
{
if (arr[i] != x)
continue;
// Running a loop from j = i
int j = i;
// Condition for incrementing 'j'
while (arr[j] == x)
{
j++;
if (j >= n )
break;
}
// Updating the answer
ans = Math.Max(ans, j - i);
}
return ans;
}
// Driver code
public static void Main ()
{
int []arr = { 1, 2, 2 };
int n = arr.Length;
Console.WriteLine(findLength(arr, n));
}
} // This code is contributed by AnkitRai01 |
<script> // Javascript implementation of the approach // Function to return the length of // the largest subarray with // maximum possible GCD function findLength(arr, n)
{ // To store the maximum number
// present in the array
var x = 0;
// Finding the maximum element
for (var i = 0; i < n; i++)
x = Math.max(x, arr[i]);
// To store the final answer
var ans = 0;
// Two pointer
for (var i = 0; i < n; i++) {
if (arr[i] != x)
continue;
// Running a loop from j = i
var j = i;
// Condition for incrementing 'j'
while (arr[j] == x)
j++;
// Updating the answer
ans = Math.max(ans, j - i);
}
return ans;
} // Driver code var arr = [1, 2, 2 ];
var n = arr.length;
document.write( findLength(arr, n)); </script> |
2
Time Complexity: O(n)
Auxiliary Space: O(1)
Related Topic: Subarrays, Subsequences, and Subsets in Array