## Similar Problems

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# Plates Between Candles - Leetcode Solution

LeetCode: Plates Between Candles Leetcode Solution

Difficulty: Medium

Topics: string prefix-sum binary-search array

Problem Statement:

Given a number of candles and a number of plates, divide the candles between the plates as evenly as possible. Each plate must have at least one candle and the difference between the number of candles on any two plates must not exceed one.

Solution Approach:

The problem requires dividing a given number of candles between a given number of plates. It is also mentioned that each plate must have at least one candle. A brute-force approach can be to generate all possible combinations of candles on plates and find the one that meets the criteria. However, this approach will be inefficient for large inputs.

To make the solution more efficient, we can make some observations. If we have only one plate, we can place all candles on that plate. If we have more plates than candles, we can still place only one candle on each plate, and the remaining plates will have to be left empty.

The most interesting case is where we have more candles than plates. In this case, we can start by placing one candle on each plate. Then, we can distribute the remaining candles among the plates such that the difference between the number of candles on any two plates does not exceed one. For example, if we have 7 candles and 3 plates, we can distribute the candles as follows:

Plate 1: 2 candles Plate 2: 2 candles Plate 3: 3 candles

We can see that the difference between the number of candles on any two plates is at most one. If we have more candles than plates, we can continue this process until all candles are distributed.

Code Implementation:

We can implement the above approach as follows:

```
def distribute_candles(candles, plates):
# each plate has at least one candle
base_candles = (candles // plates)
extra_candles = candles % plates
result = [base_candles] * plates
for i in range(extra_candles):
result[i] += 1
return result
```

The above function takes two arguments, candles, and plates, and returns a list of the number of candles on each plate. It first calculates the base number of candles that will be distributed on each plate. Then, it calculates the number of extra candles that need to be distributed. Finally, it distributes the extra candles evenly among the plates.

Test Cases:

We can test the function with the following test cases:

```
assert distribute_candles(5, 2) == [3, 2]
assert distribute_candles(6, 2) == [3, 3]
assert distribute_candles(7, 3) == [2, 2, 3]
assert distribute_candles(8, 3) == [3, 3, 2]
assert distribute_candles(9, 3) == [3, 3, 3]
```

The first test case checks the case when there are more candles than plates and both plates have a different number of candles. The second test case checks the same scenario, but the plates have the same number of candles. The third, fourth, and fifth test cases check different scenarios when the number of plates is greater than or equal to the number of candles.

Conclusion:

We have seen how to solve the Plates Between Candles problem efficiently using some observations. The time complexity of the proposed solution is O(N), where N is the number of plates. The space complexity is also O(N) as we need to store the number of candles on each plate.

## Plates Between Candles Solution Code

`1`