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# What is the logic behind this problem "Number of ways" ?

 0 How do I solve this problem ? Number of ways . I read the editorial here but didnt quite understand why cnt[i+2] is being added when ss==s . The editorialist's solution can be found here. asked 25 May '17, 21:38 2★chari407 448●2●7 accept rate: 0%

One Answer:
 2 I think it will become clear with an example. Let's assume the given array is a = [0 0 0 0 0]. "Let’s create an array cnt[], where cnt[i] equals 1, if the sum of elements from i-th to n-th equals S/3 and 0 — otherwise." -where S is the sum of all the elements of the array a. Constructing the cnts array, we get a = [0 0 0 0 0] cnts = [1 1 1 1 1] "Now, to calculate the answer we have to find the sum cnt[j] + cnt[j+1] + ... + cnt[n] faster then O(n). There are a lot of required ways to do this, but the easiest one is to create a new additional array sums[] where in j-th element will be cnt[j] + cnt[j+1] + ... + cnt[n]" a = [0 0 0 0 0] cnts = [1 1 1 1 1] sums = [5 4 3 2 1] "Thus, we receive very simple solution: for each prefix of initial array 1..i with the sum that equals we need to add to the answer sums[i+2]." We are required to break the given array into 3 pieces. If we are at the i-th element and the sum of elements from 0 to i is equal to S/3 then the segment from 0 to i is a valid first piece of the array. And for all the position j where cnts[j] is 1, the segment from j to the end is a valid third piece of the array. Since we require a second piece of the array, there must be a segment of non-zero length between these 2 segments, which is why the "i+2" comes in. It ensures that the second segment is of length ≥ 1. Hence if you choose the given prefix 0 to i with sum S/3 as the first piece, then the number of ways to choose the third piece is equal to the number of positions ≥ i+2 where the cnts[] value is 1. This is nothing but sums[i+2]. Getting back to the example, only 3 prefixes within acceptable range have sum S/3 marked with | and their +2 positions marked with ^. a = [0 0 0 0 0] cnts = [1 1 1 1 1] sums = [5 4 3 2 1] . |---^ . |---^ . |---^ The answer is hence 3+2+1=6, and the actual 6 ways are [0 | 0 | 0 0 0] [0 | 0 0 | 0 0] [0 | 0 0 0 | 0] [0 0 | 0 | 0 0] [0 0 | 0 0 | 0] [0 0 0 | 0 | 0] answered 26 May '17, 05:06 6★meooow ♦ 7.1k●7●18 accept rate: 48% Thanks a lot mate. (26 May '17, 07:11) chari4072★ No problem :) (27 May '17, 02:17) meooow ♦6★
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question asked: 25 May '17, 21:38

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last updated: 27 May '17, 02:17