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# 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"]
[[], , [2, 10], [1, 10], [3, 10], [5, 10], , [1, 20], ]
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<>());
}

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
heappush(self.numberIndexesHeap[number], index)
else:
self.numbersByIndex[index] = number
heappush(self.numberIndexesHeap[number], index)

def find(self, number: int) -> int:
while self.numberIndexesHeap[number] and self.numberIndexesHeap[number] 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] if self.numberIndexesHeap[number] else -1
``````
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