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3867. Sum of GCD of Formed Pairs
Description
You are given an integer array nums of length n.
Construct an array prefixGcd where for each index i:
- Let
mxi = max(nums[0], nums[1], ..., nums[i]). prefixGcd[i] = gcd(nums[i], mxi).
After constructing prefixGcd:
- Sort
prefixGcdin non-decreasing order. - Form pairs by taking the smallest unpaired element and the largest unpaired element.
- Repeat this process until no more pairs can be formed.
- For each formed pair, compute the
gcdof the two elements. - If
nis odd, the middle element in theprefixGcdarray remains unpaired and should be ignored.
Return an integer denoting the sum of the GCD values of all formed pairs.
The term gcd(a, b) denotes the greatest common divisor of a and b.
Example 1:
Input: nums = [2,6,4]
Output: 2
Explanation:
Construct prefixGcd:
i |
nums[i] |
mxi |
prefixGcd[i] |
|---|---|---|---|
| 0 | 2 | 2 | 2 |
| 1 | 6 | 6 | 6 |
| 2 | 4 | 6 | 2 |
prefixGcd = [2, 6, 2]. After sorting, it forms [2, 2, 6].
Pair the smallest and largest elements: gcd(2, 6) = 2. The remaining middle element 2 is ignored. Thus, the sum is 2.
Example 2:
Input: nums = [3,6,2,8]
Output: 5
Explanation:
Construct prefixGcd:
i |
nums[i] |
mxi |
prefixGcd[i] |
|---|---|---|---|
| 0 | 3 | 3 | 3 |
| 1 | 6 | 6 | 6 |
| 2 | 2 | 6 | 2 |
| 3 | 8 | 8 | 8 |
prefixGcd = [3, 6, 2, 8]. After sorting, it forms [2, 3, 6, 8].
Form pairs: gcd(2, 8) = 2 and gcd(3, 6) = 3. Thus, the sum is 2 + 3 = 5.
Constraints:
1 <= n == nums.length <= 1051 <= nums[i] <= 109
Solutions
Solution 1: Simulation
We simulate according to the problem description.
We create an array $\textit{prefixGcd}$ to store the value for each index $i$. We also maintain a variable $mx$ to track the current maximum value. For each element $nums[i]$, we update $mx$ and compute the value of $\textit{prefixGcd}[i]$. Then we sort $\textit{prefixGcd}$ and calculate the sum of GCDs of the formed pairs.
The time complexity is $O(n \log M + n \log n)$, and the space complexity is $O(n)$, where $n$ is the length of the array and $M$ is the maximum value in the array.
-
class Solution { public long gcdSum(int[] nums) { int n = nums.length; int[] prefixGcd = new int[n]; int mx = 0; for (int i = 0; i < n; i++) { int x = nums[i]; mx = Math.max(mx, x); prefixGcd[i] = gcd(x, mx); } Arrays.sort(prefixGcd); long ans = 0; for (int i = 0; i < n / 2; i++) { ans += gcd(prefixGcd[i], prefixGcd[n - i - 1]); } return ans; } private int gcd(int a, int b) { while (b != 0) { int t = a % b; a = b; b = t; } return a; } } -
class Solution { public: long long gcdSum(vector<int>& nums) { int n = nums.size(); vector<int> prefix_gcd(n); int mx = 0; for (int i = 0; i < n; i++) { int x = nums[i]; mx = max(mx, x); prefix_gcd[i] = gcd(x, mx); } sort(prefix_gcd.begin(), prefix_gcd.end()); long long ans = 0; for (int i = 0; i < n / 2; i++) { ans += gcd(prefix_gcd[i], prefix_gcd[n - i - 1]); } return ans; } }; -
class Solution: def gcdSum(self, nums: list[int]) -> int: n = len(nums) prefix_gcd = [0] * n mx = 0 for i, x in enumerate(nums): mx = max(mx, x) prefix_gcd[i] = gcd(x, mx) prefix_gcd.sort() return sum(gcd(prefix_gcd[i], prefix_gcd[-i - 1]) for i in range(n // 2)) -
func gcdSum(nums []int) int64 { n := len(nums) prefixGcd := make([]int, n) mx := 0 for i, x := range nums { if x > mx { mx = x } prefixGcd[i] = gcd(x, mx) } sort.Ints(prefixGcd) var ans int64 for i := 0; i < n/2; i++ { ans += int64(gcd(prefixGcd[i], prefixGcd[n-i-1])) } return ans } func gcd(a, b int) int { for b != 0 { a, b = b, a%b } return a } -
function gcdSum(nums: number[]): number { const n = nums.length; const prefixGcd: number[] = new Array(n); let mx = 0; for (let i = 0; i < n; i++) { const x = nums[i]; mx = Math.max(mx, x); prefixGcd[i] = gcd(x, mx); } prefixGcd.sort((a, b) => a - b); let ans = 0; for (let i = 0; i < n >> 1; i++) { ans += gcd(prefixGcd[i], prefixGcd[n - i - 1]); } return ans; } function gcd(a: number, b: number): number { while (b !== 0) { const t = a % b; a = b; b = t; } return a; }