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This code is actually quite efficient. So for the following values, n=5 (number of elements in the array), d=2 (number of rotations)and assuming i goes from 0-4.
(i + n - d)%n with the iteration of i from 0-4:
(0+5-2)%5=3
(1+5-2)%5=4
.
.
.
(4+5-2)%5=2
So essentially, you are changing the locations of the elements in the array. How the "%" works is that it takes the remainder. So for example, 1%7=1 since 1 cannot be divided by 7 which means 1 is the remainder.

## Left Rotation

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This code is actually quite efficient. So for the following values, n=5 (number of elements in the array), d=2 (number of rotations)and assuming i goes from 0-4. (i + n - d)%n with the iteration of i from 0-4: (0+5-2)%5=3 (1+5-2)%5=4 . . . (4+5-2)%5=2 So essentially, you are changing the locations of the elements in the array. How the "%" works is that it takes the remainder. So for example, 1%7=1 since 1 cannot be divided by 7 which means 1 is the remainder.

that's great, can you provide any mathematical illustration to arrive (i+n-d)%n ?