Design a Number Container System LeetCode Solution

Problem – Design a Number Container System LeetCode Solution

Design a number container system that can do the following:

  • Insert or Replace a number at the given index in the system.
  • Return the smallest index for the given number in the system.

Implement the NumberContainers class:

  • NumberContainers() Initializes the number container system.
  • void change(int index, int number) Fills the container at index with the number. If there is already a number at that index, replace it.
  • int find(int number) Returns the smallest index for the given number, or -1 if there is no index that is filled by number in the system.

Example 1:

Input
["NumberContainers", "find", "change", "change", "change", "change", "find", "change", "find"]
[[], [10], [2, 10], [1, 10], [3, 10], [5, 10], [10], [1, 20], [10]]
Output
[null, -1, null, null, null, null, 1, null, 2]

Explanation
NumberContainers nc = new NumberContainers();
nc.find(10); // There is no index that is filled with number 10. Therefore, we return -1.
nc.change(2, 10); // Your container at index 2 will be filled with number 10.
nc.change(1, 10); // Your container at index 1 will be filled with number 10.
nc.change(3, 10); // Your container at index 3 will be filled with number 10.
nc.change(5, 10); // Your container at index 5 will be filled with number 10.
nc.find(10); // Number 10 is at the indices 1, 2, 3, and 5. Since the smallest index that is filled with 10 is 1, we return 1.
nc.change(1, 20); // Your container at index 1 will be filled with number 20. Note that index 1 was filled with 10 and then replaced with 20. 
nc.find(10); // Number 10 is at the indices 2, 3, and 5. The smallest index that is filled with 10 is 2. Therefore, we return 2.

Constraints:

  • 1 <= index, number <= 109
  • At most 105 calls will be made in total to change and find.

Design a Number Container System LeetCode Solution in Java

class NumberContainers {
    
    Map<Integer,TreeSet<Integer>> map;
    Map<Integer,Integer> m;
    public NumberContainers() {
        map=new HashMap<>();
        m=new HashMap<>();
        
    }
    public void change(int index, int number) {
        m.put(index,number);
        if(!map.containsKey(number)) map.put(number,new TreeSet<>());
        map.get(number).add(index);
    }
    
    public int find(int number) {
        if(!map.containsKey(number)) return -1;
        for(Integer a:map.get(number)){
            if(m.get(a)==number) return a;
        }
        return -1;
    }
}

Design a Number Container System LeetCode Solution in C++

class NumberContainers {
public:
    unordered_map<int, int> ind_num;
    unordered_map<int, set<int>> num_inds;
    void change(int index, int number) {
        auto it = ind_num.find(index);
        if (it != end(ind_num))
            num_inds[it->second].erase(index);
        ind_num[index] = number;
        num_inds[number].insert(index);
    }
    int find(int number) {
        auto it = num_inds.find(number);
        return it == end(num_inds) || it->second.empty() ? -1 : *begin(it->second);
    }
};

Design a Number Container System LeetCode Solution in Python

class NumberContainers:
    def __init__(self):
        self.numbersByIndex = {}
        self.numberIndexes = defaultdict(set)
        self.numberIndexesHeap = defaultdict(list)

    def change(self, index: int, number: int) -> None:
        if index in self.numbersByIndex:
            if number != self.numbersByIndex[index]:
                self.numberIndexes[self.numbersByIndex[index]].remove(index)
                self.numbersByIndex[index] = number
                self.numberIndexes[number].add(index)
                heappush(self.numberIndexesHeap[number], index)
        else:
            self.numbersByIndex[index] = number
            self.numberIndexes[number].add(index)
            heappush(self.numberIndexesHeap[number], index)

    def find(self, number: int) -> int:
        while self.numberIndexesHeap[number] and self.numberIndexesHeap[number][0] not in self.numberIndexes[number]:
                heappop(self.numberIndexesHeap[number])  # make sure the smallest index in heap is still an index for number
        return self.numberIndexesHeap[number][0] if self.numberIndexesHeap[number] else -1
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