I came across this question on a programming contest although I was not able to solve it. The editorial was released later but it was super unhelpful, so I am asking it here.

You are given an array, in which you have to calculate sum of all possible subarrays' OR.

For example: $a = [1, 2, 3, 4, 5]$, then the answer would be $71$.

Click here to see how

How would I go about solving such a problem in less than quadratic time?


closed as off-topic by Namaste, Adrian Keister, Theoretical Economist, José Carlos Santos, user91500 Sep 16 '18 at 9:19

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  • $\begingroup$ You should define that a subarray is a contiguous set of elements of the array. Are you guaranteed that the elements are a run of numbers, or could the array be $[1,4,9,14,26]?$ Are you guaranteed that the elements are sorted? Please state the whole problem. $\endgroup$ – Ross Millikan Sep 15 '18 at 20:52
  • $\begingroup$ They are not a run of consecutive numbers and they are not sorted either. $\endgroup$ – Andrew Scott Sep 16 '18 at 7:23

One idea is to compute how much is contributed by each bit in the sum. If the array is length $n$ there are $\frac 12n(n+1)$ subarrays because you choose two positions with replacement, with the earlier being the start and the later being the end. For each bit position the OR of a subarray will be $1$ unless all the numbers in it have a $0$ in that location, so we want to count the runs of $0$s. A run of $k\ 0$'s will generate $\frac 12k(k+1)$ subarrays that have a zero in that bit. Add up the number of subarrays that have a $0$ in the position, subtract from the total number of subarrays, and you have the number of subarrays that have a $1$ in that position.

As an example I will use the array $[1,2,3,4,5,6,7,8,9,10]$. There are $\frac 12\cdot 10 \cdot 11=55$ subarrays. In the ones bit there are five runs of $0$'s, each of length $1$. Each of those contributes one subarray with a $0$ in the ones bit, so there are $50$ subarrays with a $1$ in the ones place, starting our sum with $50$. In the twos bit there is a run of $1$ from $1$, a run of $2$ from $4,5$, and a run of $2$ from $9,10$. These give $1+3+3=7$ subarrays that have a $0$ in the twos bit, so there are $48$ that have a $1$, contributing $96$ to the sum. In the fours bit there is a run of three $0$s to start off and a run of three at the end, so there are $12$ subarrays with a $0$ in the fours place, giving $43$ with a $1$ and contributing $172$ to the sum. Finally for the eights there is a run of $7$ numbers at the start with a $0$ which gives $28$ subarrays with a $0$ in the eights place, so $27$ with a $1$, contributing $216$ to the sum. The sum is then $50+96+172+216=534$

This approach is linear in the length of the array. The operations count also grows with the number of bits in the largest number. If you had a short array with large numbers it would be faster to compute each subarray, which is actually an $n^3$ process-there are about $\frac 12n^2$ subarrays of average length $\frac n2$. If you had a thousand numbers of billions of bits and you can do a bitwise OR as a single operation it would be more efficient to just compute each subarray.

  • $\begingroup$ I am not sure I understand. I asked for the bitwise OR of all subarrays but the answer discusses about XOR? $\endgroup$ – Andrew Scott Sep 16 '18 at 7:22
  • $\begingroup$ @AndrewScott: I meant OR everywhere. I don't know why I typed XOR. The argument is correct. I'll fix it. Thanks. It does not depend on the numbers being sorted or contiguous. $\endgroup$ – Ross Millikan Sep 16 '18 at 13:40

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