Problem
You are given a 0-indexed integer array nums, and you are allowed to traverse between its indices. You can traverse between index i and index j, i != j, if and only if gcd(nums[i], nums[j]) > 1, where gcd is the greatest common divisor.
Your task is to determine if for every pair of indices i and j in nums, where i < j, there exists a sequence of traversals that can take us from i to j.
Return true** if it is possible to traverse between all such pairs of indices,**** or false otherwise.**
Example 1:
Input: nums = [2,3,6]
Output: true
Explanation: In this example, there are 3 possible pairs of indices: (0, 1), (0, 2), and (1, 2).
To go from index 0 to index 1, we can use the sequence of traversals 0 -> 2 -> 1, where we move from index 0 to index 2 because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 > 1, and then move from index 2 to index 1 because gcd(nums[2], nums[1]) = gcd(6, 3) = 3 > 1.
To go from index 0 to index 2, we can just go directly because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 > 1. Likewise, to go from index 1 to index 2, we can just go directly because gcd(nums[1], nums[2]) = gcd(3, 6) = 3 > 1.
Example 2:
Input: nums = [3,9,5]
Output: false
Explanation: No sequence of traversals can take us from index 0 to index 2 in this example. So, we return false.
Example 3:
Input: nums = [4,3,12,8]
Output: true
Explanation: There are 6 possible pairs of indices to traverse between: (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), and (2, 3). A valid sequence of traversals exists for each pair, so we return true.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 105
Solution (Java)
class Solution {
private int getf(int[] f, int x) {
return f[x] == x ? x : (f[x] = getf(f, f[x]));
}
private void merge(int[] f, int[] num, int x, int y) {
x = getf(f, x);
y = getf(f, y);
if (x == y) {
return;
}
if (num[x] < num[y]) {
int t = x;
x = y;
y = t;
}
f[y] = x;
num[x] += num[y];
}
public boolean canTraverseAllPairs(int[] nums) {
final int n = nums.length;
if (n == 1) {
return true;
}
int[] f = new int[n], num = new int[n];
for (int i = 0; i < n; ++i) {
f[i] = i;
num[i] = 1;
}
Map<Integer, Integer> have = new HashMap<>();
for (int i = 0; i < n; ++i) {
int x = nums[i];
if (x == 1) {
return false;
}
for (int d = 2; d * d <= x; ++d) {
if (x % d == 0) {
if (have.containsKey(d)) {
merge(f, num, i, have.get(d));
} else {
have.put(d, i);
}
while (x % d == 0) {
x /= d;
}
}
}
if (x > 1) {
if (have.containsKey(x)) {
merge(f, num, i, have.get(x));
} else {
have.put(x, i);
}
}
}
return num[getf(f, 0)] == n;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).