Remainder When Divided By 11
You are given a number N. You need to find the remainder when N is divided by 11.

Note :

Number N may be very large.

Input format :

The first line contains an integer 'T' denoting the number of test cases to be run. The 'T' test cases follow. The first line of each test case contains a single number 'N' in the form of a string.

Output Format :

For each test case, print an integer denoting the remainder when 'N' is divided by 11.

Constraints :

1

Question

Remainder When Divided By 11

You are given a number N. You need to find the remainder when N is divided by 11.

Note :

Number N may be very large.

Input format :

The first line contains an integer 'T' denoting the number of test cases to be run. The 'T' test cases follow. 

The first line of each test case contains a single number 'N' in the form of a string.

Output Format :

For each test case, print an integer denoting the remainder when 'N' is divided by 11.

Constraints :

1 <= T <= 5
1 <= length of N <= 10 ^ 6

Time Limit: 1 sec

Accepted Answer

Using Modulo Property

Since the number is very large we can not simply calculate N % 11.
Now, any number say 256 % 11 can be calculated as (25*10 + 6)%11. Which in turn can be calculated as (( 25*10)%11+ 6)%11 or ( (25%11)*10 + 6)%11. Similarly 25%11 can be calculated as (2*10+5)%11. Therefore , 256 %11 can be finally calculated as ( ( (2*10 + 5)%11 )*10 + 6)%11.
So, the idea is to store the number N in string Str.
Take a variable Num, initialize it with 0.
Now iterate character by character over the string and convert it into a digit, starting from 0th index.
Multiply your current Num by 10 and add the current digit, say b, in it and then take it modulus with 11. So Num becomes Num = (Num*10 + b)%11.
The final value of Num will be the required remainder.

Space Complexity: O(1)Explanation:

O(1) per test case.

In the worst case, we are taking constant extra space.

Time Complexity: O(n)Explanation:

O(L) per test case, where ‘L’ is the number of digits in ‘N’.

In the worst case, we are traversing ‘N’ once, digit by digit.

You are given a number N. You need to find the remainder when N is divided by 11.

Note :

Input format :

Output Format :

Constraints :

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