The longest sub-array with switching elements - arrays

Question

An array is called "switching" if the odd and even elements are equal.
Example:
[2,4,2,4] is a switching array because the members in even positions (indexes 0 and 2) and odd positions (indexes 1 and 3) are equal.
If A = [3,7,3,7, 2, 1, 2], the switching sub-arrays are:
== > [3,7,3,7] and [2,1,2]
Therefore, the longest switching sub-array is [3,7,3,7] with length = 4.
As another example if A = [1,5,6,0,1,0], the the only switching sub-array is [0,1,0].
Another example: A= [7,-5,-5,-5,7,-1,7], the switching sub-arrays are [7,-1,7] and [-5,-5,-5].
Question:
Write a function that receives an array and find its longest switching sub-array.
I would like to know how you solve this problem and which strategies you use to solve this with a good time complexity?

Answer 1

I am assuming that the array is zero indexed .
if arr.size <= 2
return arr.size
else
ans = 2
temp_ans = 2 // We will update ans when temp_ans > ans;
for i = 2; i < arr.size ; ++i
if arr[i] = arr[i-2]
temp_ans = temp_ans + 1;
else
temp_ans = 2;
ans = max(temp_ans , ans);
return ans;
I think this should work and I don't think it needs any kind of explanation .
Example Code

Answer 2

private static int solve(int[] arr){
if(arr.length == 1) return 1;
int even = arr[0],odd = arr[1];
int start = 0,max_len = 0;
for(int i=2;i<arr.length;++i){
if(i%2 == 0 && arr[i] != even || i%2 == 1 && arr[i] != odd){
max_len = Math.max(max_len,i - start);
start = i-1;
if(i%2 == 0){
even = arr[i];
odd = arr[i-1];
}else{
even = arr[i-1];
odd = arr[i];
}
}
}
return Math.max(max_len,arr.length - start);
}
It's like a sliding window problem.
We keep track of even and odd equality with 2 variables, even and odd.
Whenever we come across a unmet condition, like index even but not equal with even variable and same goes for odd, we first
Record the length till now in max_len.
Reset start to i-1 as this is need incase of all elements equal.
Reset even and odd according to current index i to arr[i] and arr[i-1] respectively.
Demo: https://ideone.com/iUQti7

Answer 3

I didn't analyse the time complexity, just wrote a solution that uses recursion and it works (I think):
public class Main
{
public static int switching(int[] arr, int index, int end)
{
try
{
if (arr[index] == arr[index+2])
{
end = index+2;
return switching(arr, index+1, end);
}
} catch (Exception e) {}
return end;
}
public static void main(String[] args)
{
//int[] arr = {3,2,3,2,3};
//int[] arr = {3,2,3};
//int[] arr = {4,4,4};
int[] arr = {1,2,3,4,5,4,4,7,9,8,10};
int best = -1;
for (int i = 0; i < arr.length; i++)
best = Math.max(best, (switching(arr, i, 0) - i));
System.out.println(best+1); // It returns, in this example, 3
}
}

Answer 4

int switchingSubarray(vector<int> &arr, int n) {
if(n==1||n==2) return n;
int i=0;
int ans=2;
int j=2;
while(j<n)
{
if(arr[j]==arr[j-2]) j++;
else
{
ans=max(ans,j-i);
i=j-1;
j++;
}
}
ans=max(ans,j-i);
return ans;
}
Just using sliding window technique to solve this problems as element at j and j-2 need to be same.
Try to dry run on paper u will surely get it .

Answer 5

# Switching if numbers in even positions equal to odd positions find length of longest switch in continuos sub array
def check(t):
even = []
odd = []
i = 0
while i < len(t):
if i % 2 == 0:
even.append(t[i])
else:
odd.append(t[i])
i += 1
if len(set(even)) == 1 and len(set(odd)) == 1:
return True
else:
return False
def solution(A):
maxval = 0
if len(A) == 1:
return 1
for i in range(0, len(A)):
for j in range(0, len(A)):
if check(A[i:j+1]) == True:
val = len(A[i:j+1])
print(A[i:j+1])
if val > maxval:
maxval = val
return maxval
A = [3,2,3,2,3]
A = [7,4,-2,4,-2,-9]
A=[4]
A = [7,-5,-5,-5,7,-1,7]
print(solution(A))