# Secret Code CodeKaze June23 Round 2 Solution || India’s Biggest Hiring Challenge

Table of Contents

## Secret Code CodeKaze June23 Round 2 Solution || India’s Biggest Hiring Challenge

## Secret Code CodeKaze June23 Round 2 Solution || India’s Biggest Hiring Challenge

## Problem Statement

#### You are given an array ‘A’ of length ‘N’. Three consecutive elements are considered the ‘secret code’ if their sum is divisible by ’10’.

#### Return ‘1’ if the array contains the ‘secret code’ and return ‘0’ otherwise.

##### For Example :

```
Let 'N' = 4, 'A' = [ 10, 11, 21, 8 ].
There are two sets of three consecutive elements in the given array.
The first, second, and third elements make the first set, with a sum of '10 + 11 + 21 = 42', which is not divisible by '10'.
The second, third, and fourth elements make the second set, with a sum of '11 + 21 + 8 = 40', which is divisible by '10'. Therefore, the array does contain the 'secret code'.
Thus, the answer is '1'.
```

##### Hint:

```
Iterate through each array element and check if the last three elements are considered the 'secret code' by finding their sum and checking its divisibility by 10.
```

Detailed explanation ( Input/output format, Notes, Images )

##### Constraints :

```
1 <= 'T' <= 10
3 <= 'N' <= 10^5
1 <= 'A[i]' <= 10^8
Time Limit: 1 sec
```

##### Sample Input 1 :

```
2
5
1 2 3 4 1
4
30 10 20 10
```

##### Sample Output 1 :

```
0
1
```

##### Explanation Of Sample Input 1 :

```
First test case:-
There are three sets of three consecutive elements in the given array.
The sums of all three sets are less than '10'.
Thus, the answer is '0'.
Second test case:-
Every element is divisible by '10'.
Therefore, the sum of any three consecutive elements will be divisible by '10'.
Thus, the answer is '1'.
```

##### Sample Input 2 :

```
2
3
1082 1 1037
4
34 225 17 9
```

##### Sample Output 2 :

```
1
0
```

6 digit code appear after 69 seconds.

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## Secret Code CodeKaze June23 Round 2 Solution || India’s Biggest Hiring Challenge

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