In divisible subarray problem why in case of negative sum
we are taking modulus two times? For ex: if sum is -13 and N=5 so if we simply |sum%N| ++ then counter at position 3 will be increased but on taking mod two times we’re increasing counter of (sum%N + N)%N =2
Can someone provide the reason behind this please!
It’s calculated exactly like the mod of a positive numbr. In arithmetic modulo c, we seek to express any x as qc+r, where r must be a non-negative integer.
now try to print -22%5 in your compiler it will give -2, but this is not correct, so the correct aproach is (-2+5)%5 = 3. why , because 5*-5+3=-22;
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