Secret Code CodeKaze June23 Round 2 Solution || India’s Biggest Hiring Challenge
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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
06 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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