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3160. Find the Number of Distinct Colors Among the Balls

Description

You are given an integer limit and a 2D array queries of size n x 2.

There are limit + 1 balls with distinct labels in the range [0, limit]. Initially, all balls are uncolored. For every query in queries that is of the form [x, y], you mark ball x with the color y. After each query, you need to find the number of distinct colors among the balls.

Return an array result of length n, where result[i] denotes the number of distinct colors after ith query.

Note that when answering a query, lack of a color will not be considered as a color.

 

Example 1:

Input: limit = 4, queries = [[1,4],[2,5],[1,3],[3,4]]

Output: [1,2,2,3]

Explanation:

  • After query 0, ball 1 has color 4.
  • After query 1, ball 1 has color 4, and ball 2 has color 5.
  • After query 2, ball 1 has color 3, and ball 2 has color 5.
  • After query 3, ball 1 has color 3, ball 2 has color 5, and ball 3 has color 4.

Example 2:

Input: limit = 4, queries = [[0,1],[1,2],[2,2],[3,4],[4,5]]

Output: [1,2,2,3,4]

Explanation:

  • After query 0, ball 0 has color 1.
  • After query 1, ball 0 has color 1, and ball 1 has color 2.
  • After query 2, ball 0 has color 1, and balls 1 and 2 have color 2.
  • After query 3, ball 0 has color 1, balls 1 and 2 have color 2, and ball 3 has color 4.
  • After query 4, ball 0 has color 1, balls 1 and 2 have color 2, ball 3 has color 4, and ball 4 has color 5.

 

Constraints:

  • 1 <= limit <= 109
  • 1 <= n == queries.length <= 105
  • queries[i].length == 2
  • 0 <= queries[i][0] <= limit
  • 1 <= queries[i][1] <= 109

Solutions

Solution 1: Double Hash Tables

We use a hash table g to record the color of each ball, and another hash table cnt to record the count of each color.

Next, we traverse the array queries. For each query $(x, y)$, we increase the count of color $y$ by $1$, then check whether ball $x$ has been colored. If it has, we decrease the count of the color of ball $x$ by $1$. If the count drops to $0$, we remove it from the hash table cnt. Then, we update the color of ball $x$ to $y$, and add the current size of the hash table cnt to the answer array.

After the traversal, we return the answer array.

The time complexity is $O(m)$, and the space complexity is $O(m)$, where $m$ is the length of the array queries.

  • class Solution {
    public int[] queryResults(int limit, int[][] queries) {
        Map<Integer, Integer> g = new HashMap<>();
        Map<Integer, Integer> cnt = new HashMap<>();
        int m = queries.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            int x = queries[i][0], y = queries[i][1];
            cnt.merge(y, 1, Integer::sum);
            if (g.containsKey(x) && cnt.merge(g.get(x), -1, Integer::sum) == 0) {
                cnt.remove(g.get(x));
            }
            g.put(x, y);
            ans[i] = cnt.size();
        }
        return ans;
    }
}

  • class Solution {
public:
    vector<int> queryResults(int limit, vector<vector<int>>& queries) {
        unordered_map<int, int> g;
        unordered_map<int, int> cnt;
        vector<int> ans;
        for (auto& q : queries) {
            int x = q[0], y = q[1];
            cnt[y]++;
            if (g.contains(x) && --cnt[g[x]] == 0) {
                cnt.erase(g[x]);
            }
            g[x] = y;
            ans.push_back(cnt.size());
        }
        return ans;
    }
};

  • class Solution:
    def queryResults(self, limit: int, queries: List[List[int]]) -> List[int]:
        g = {}
        cnt = Counter()
        ans = []
        for x, y in queries:
            cnt[y] += 1
            if x in g:
                cnt[g[x]] -= 1
                if cnt[g[x]] == 0:
                    cnt.pop(g[x])
            g[x] = y
            ans.append(len(cnt))
        return ans


  • func queryResults(limit int, queries [][]int) (ans []int) {
	g := map[int]int{}
	cnt := map[int]int{}
	for _, q := range queries {
		x, y := q[0], q[1]
		cnt[y]++
		if v, ok := g[x]; ok {
			cnt[v]--
			if cnt[v] == 0 {
				delete(cnt, v)
			}
		}
		g[x] = y
		ans = append(ans, len(cnt))
	}
	return
}

  • function queryResults(limit: number, queries: number[][]): number[] {
    const g = new Map<number, number>();
    const cnt = new Map<number, number>();
    const ans: number[] = [];
    for (const [x, y] of queries) {
        cnt.set(y, (cnt.get(y) ?? 0) + 1);
        if (g.has(x)) {
            const v = g.get(x)!;
            cnt.set(v, cnt.get(v)! - 1);
            if (cnt.get(v) === 0) {
                cnt.delete(v);
            }
        }
        g.set(x, y);
        ans.push(cnt.size);
    }
    return ans;
}

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