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| 1 | +// Source : https://leetcode.com/problems/number-of-different-subsequences-gcds/ |
| 2 | +// Author : Hao Chen |
| 3 | +// Date : 2021-04-05 |
| 4 | + |
| 5 | +/***************************************************************************************************** |
| 6 | + * |
| 7 | + * You are given an array nums that consists of positive integers. |
| 8 | + * |
| 9 | + * The GCD of a sequence of numbers is defined as the greatest integer that divides all the numbers in |
| 10 | + * the sequence evenly. |
| 11 | + * |
| 12 | + * For example, the GCD of the sequence [4,6,16] is 2. |
| 13 | + * |
| 14 | + * A subsequence of an array is a sequence that can be formed by removing some elements (possibly |
| 15 | + * none) of the array. |
| 16 | + * |
| 17 | + * For example, [2,5,10] is a subsequence of [1,2,1,2,4,1,5,10]. |
| 18 | + * |
| 19 | + * Return the number of different GCDs among all non-empty subsequences of nums. |
| 20 | + * |
| 21 | + * Example 1: |
| 22 | + * |
| 23 | + * Input: nums = [6,10,3] |
| 24 | + * Output: 5 |
| 25 | + * Explanation: The figure shows all the non-empty subsequences and their GCDs. |
| 26 | + * The different GCDs are 6, 10, 3, 2, and 1. |
| 27 | + * |
| 28 | + * Example 2: |
| 29 | + * |
| 30 | + * Input: nums = [5,15,40,5,6] |
| 31 | + * Output: 7 |
| 32 | + * |
| 33 | + * Constraints: |
| 34 | + * |
| 35 | + * 1 <= nums.length <= 10^5 |
| 36 | + * 1 <= nums[i] <= 2 * 10^5 |
| 37 | + ******************************************************************************************************/ |
| 38 | + |
| 39 | +class Solution { |
| 40 | +private: |
| 41 | + // Euclidean algorithm |
| 42 | + // https://en.wikipedia.org/wiki/Euclidean_algorithm |
| 43 | + int gcd(int a, int b) { |
| 44 | + while ( b != 0 ) { |
| 45 | + int t = b; |
| 46 | + b = a % b; |
| 47 | + a = t; |
| 48 | + } |
| 49 | + return a; |
| 50 | + } |
| 51 | + |
| 52 | +public: |
| 53 | + int countDifferentSubsequenceGCDs(vector<int>& nums) { |
| 54 | + int len = nums.size(); |
| 55 | + vector<int> gcds(200001, 0); |
| 56 | + |
| 57 | + for(int i=0; i<len; i++) { |
| 58 | + int n = nums[i]; |
| 59 | + int m = sqrt(n); |
| 60 | + for(int g=1; g<=m; g++){ |
| 61 | + if (n % g != 0) continue; |
| 62 | + int x = g, y = n/g; |
| 63 | + if (x != y ){ |
| 64 | + gcds[x] = gcd(n, gcds[x]); |
| 65 | + gcds[y] = gcd(n, gcds[y]); |
| 66 | + }else { |
| 67 | + gcds[x] = gcd(n, gcds[x]); |
| 68 | + } |
| 69 | + } |
| 70 | + } |
| 71 | + |
| 72 | + int cnt = 0; |
| 73 | + for(int i=1; i<gcds.size(); i++){ |
| 74 | + if (gcds[i]==i) cnt++; |
| 75 | + } |
| 76 | + return cnt; |
| 77 | + } |
| 78 | +}; |
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