3267. Count Almost Equal Pairs II
Description
Attention: In this version, the number of operations that can be performed, has been increased to twice.
You are given an array nums consisting of positive integers.
We call two integers x and y almost equal if both integers can become equal after performing the following operation at most twice:
- Choose either
xoryand swap any two digits within the chosen number.
Return the number of indices i and j in nums where i < j such that nums[i] and nums[j] are almost equal.
Note that it is allowed for an integer to have leading zeros after performing an operation.
Example 1:
Input: nums = [1023,2310,2130,213]
Output: 4
Explanation:
The almost equal pairs of elements are:
- 1023 and 2310. By swapping the digits 1 and 2, and then the digits 0 and 3 in 1023, you get 2310.
- 1023 and 213. By swapping the digits 1 and 0, and then the digits 1 and 2 in 1023, you get 0213, which is 213.
- 2310 and 213. By swapping the digits 2 and 0, and then the digits 3 and 2 in 2310, you get 0213, which is 213.
- 2310 and 2130. By swapping the digits 3 and 1 in 2310, you get 2130.
Example 2:
Input: nums = [1,10,100]
Output: 3
Explanation:
The almost equal pairs of elements are:
- 1 and 10. By swapping the digits 1 and 0 in 10, you get 01 which is 1.
- 1 and 100. By swapping the second 0 with the digit 1 in 100, you get 001, which is 1.
- 10 and 100. By swapping the first 0 with the digit 1 in 100, you get 010, which is 10.
Constraints:
2 <= nums.length <= 50001 <= nums[i] < 107
Solutions
Solution 1: Sorting + Enumeration
We can enumerate each number, and for each number, we can enumerate each pair of different digits, then swap these two digits to get a new number. Record this new number in a hash table $\textit{vis}$, representing all possible numbers after at most one swap. Then continue to enumerate each pair of different digits, swap these two digits to get a new number, and record it in the hash table $\textit{vis}$, representing all possible numbers after at most two swaps.
This enumeration may miss some pairs of numbers, such as $[100, 1]$, because the number obtained after swapping $100$ is $1$, and the previously enumerated numbers do not include $1$, so some pairs of numbers will be missed. We only need to sort the array before enumeration to solve this problem.
The time complexity is $O(n \times (\log n + \log^5 M))$, and the space complexity is $O(n + \log^4 M)$. Here, $n$ is the length of the array $\textit{nums}$, and $M$ is the maximum value in the array $\textit{nums}$.
Python3
class Solution:
def countPairs(self, nums: List[int]) -> int:
nums.sort()
ans = 0
cnt = defaultdict(int)
for x in nums:
vis = {x}
s = list(str(x))
m = len(s)
for j in range(m):
for i in range(j):
s[i], s[j] = s[j], s[i]
vis.add(int("".join(s)))
for q in range(i + 1, m):
for p in range(i + 1, q):
s[p], s[q] = s[q], s[p]
vis.add(int("".join(s)))
s[p], s[q] = s[q], s[p]
s[i], s[j] = s[j], s[i]
ans += sum(cnt[x] for x in vis)
cnt[x] += 1
return ans
Java
class Solution {
public int countPairs(int[] nums) {
Arrays.sort(nums);
int ans = 0;
Map<Integer, Integer> cnt = new HashMap<>();
for (int x : nums) {
Set<Integer> vis = new HashSet<>();
vis.add(x);
char[] s = String.valueOf(x).toCharArray();
for (int j = 0; j < s.length; ++j) {
for (int i = 0; i < j; ++i) {
swap(s, i, j);
vis.add(Integer.parseInt(String.valueOf(s)));
for (int q = i; q < s.length; ++q) {
for (int p = i; p < q; ++p) {
swap(s, p, q);
vis.add(Integer.parseInt(String.valueOf(s)));
swap(s, p, q);
}
}
swap(s, i, j);
}
}
for (int y : vis) {
ans += cnt.getOrDefault(y, 0);
}
cnt.merge(x, 1, Integer::sum);
}
return ans;
}
private void swap(char[] s, int i, int j) {
char t = s[i];
s[i] = s[j];
s[j] = t;
}
}
C++
class Solution {
public:
int countPairs(vector<int>& nums) {
sort(nums.begin(), nums.end());
int ans = 0;
unordered_map<int, int> cnt;
for (int x : nums) {
unordered_set<int> vis = {x};
string s = to_string(x);
for (int j = 0; j < s.length(); ++j) {
for (int i = 0; i < j; ++i) {
swap(s[i], s[j]);
vis.insert(stoi(s));
for (int q = i + 1; q < s.length(); ++q) {
for (int p = i + 1; p < q; ++p) {
swap(s[p], s[q]);
vis.insert(stoi(s));
swap(s[p], s[q]);
}
}
swap(s[i], s[j]);
}
}
for (int y : vis) {
ans += cnt[y];
}
cnt[x]++;
}
return ans;
}
};
Go
func countPairs(nums []int) (ans int) {
sort.Ints(nums)
cnt := make(map[int]int)
for _, x := range nums {
vis := make(map[int]struct{})
vis[x] = struct{}{}
s := []rune(strconv.Itoa(x))
for j := 0; j < len(s); j++ {
for i := 0; i < j; i++ {
s[i], s[j] = s[j], s[i]
y, _ := strconv.Atoi(string(s))
vis[y] = struct{}{}
for q := i + 1; q < len(s); q++ {
for p := i + 1; p < q; p++ {
s[p], s[q] = s[q], s[p]
z, _ := strconv.Atoi(string(s))
vis[z] = struct{}{}
s[p], s[q] = s[q], s[p]
}
}
s[i], s[j] = s[j], s[i]
}
}
for y := range vis {
ans += cnt[y]
}
cnt[x]++
}
return
}
TypeScript
function countPairs(nums: number[]): number {
nums.sort((a, b) => a - b);
let ans = 0;
const cnt = new Map<number, number>();
for (const x of nums) {
const vis = new Set<number>();
vis.add(x);
const s = x.toString().split('');
for (let j = 0; j < s.length; j++) {
for (let i = 0; i < j; i++) {
[s[i], s[j]] = [s[j], s[i]];
vis.add(+s.join(''));
for (let q = i + 1; q < s.length; ++q) {
for (let p = i + 1; p < q; ++p) {
[s[p], s[q]] = [s[q], s[p]];
vis.add(+s.join(''));
[s[p], s[q]] = [s[q], s[p]];
}
}
[s[i], s[j]] = [s[j], s[i]];
}
}
for (const y of vis) {
ans += cnt.get(y) || 0;
}
cnt.set(x, (cnt.get(x) || 0) + 1);
}
return ans;
}