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Medium
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Weekly Contest 412 Q2
Array
Hash Table
Counting
Enumeration
Sorting

3265. Count Almost Equal Pairs I

中文文档

Description

You are given an array nums consisting of positive integers.

We call two integers x and y in this problem almost equal if both integers can become equal after performing the following operation at most once:

  • Choose either x or y and 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 = [3,12,30,17,21]

Output: 2

Explanation:

The almost equal pairs of elements are:

  • 3 and 30. By swapping 3 and 0 in 30, you get 3.
  • 12 and 21. By swapping 1 and 2 in 12, you get 21.

Example 2:

Input: nums = [1,1,1,1,1]

Output: 10

Explanation:

Every two elements in the array are almost equal.

Example 3:

Input: nums = [123,231]

Output: 0

Explanation:

We cannot swap any two digits of 123 or 231 to reach the other.

 

Constraints:

  • 2 <= nums.length <= 100
  • 1 <= nums[i] <= 106

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. We record this new number in a hash table $s$, representing all possible numbers after at most one swap. Then, we count how many numbers previously enumerated are in the hash table $s$ and add this count to the answer. Next, we add the currently enumerated number to the hash table $\textit{cnt}$, representing the count of the current number.

This enumeration method may miss some pairs, such as $[100, 1]$, because the number obtained by swapping digits in $100$ is $1$, and previously enumerated numbers do not include $1$, thus missing some pairs. We can solve this problem by sorting the array before enumeration.

The time complexity is $O(n \times (\log n + \log^3 M))$, and the space complexity is $O(n + \log^2 M)$. 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))
            for j in range(len(s)):
                for i in range(j):
                    s[i], s[j] = s[j], s[i]
                    vis.add(int("".join(s)))
                    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)));
                    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));
                    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{}{}
				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(''));
                [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;
}