wish_me's blog

Count of subarrays whose products don’t have any repeating prime factor
By wish_me, history, 7 years ago, In English

Can any one tell me the solution of the problem.

Eg. Input: 2 3 9 Output: 3 Explanation: Total sub-array are:- {2}, {3}, {9}, {2, 3}, {3, 9}, {2, 3, 9}

Subarray which violets the property are:- {9} -> {3 * 3}, Since 3 is a repeating prime factor in prime decomposition of 9 {3, 9} -> {3 * 3 * 3}, 3 is a repeating prime factor in prime decomposition of 27 {2, 3, 9} -> {2 * 3 * 3 * 3}, 3 is repeating prime factor in prime decomposition of 54 Hence total subarray remains which satisfies our condition are 3.

Input: 2, 3, 5, 15, 7, 2 Output: 12

  • Vote: I like it
  • -9
  • Vote: I do not like it

Comments Comments (2)
| Write comment?
»
7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

You can use two pointers approach. We'll store array cnt, cnt[x] means count of prime factor x in our current subarray. So solution is O(Nsqrt(X)) or O(NlogX + X) depending on the way you'll factorize numbers.

→ Reply
»
7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Why do you do that?

→ Reply