LeetCode Daily: Find Missing Observations Solution in Java | September 5, 2024

🔍 LeetCode Problem of the Day: Find Missing Observations

In today's LeetCode daily challenge, we solve the problem of finding the missing rolls to match a given mean in Java.

👉 Solution: Pinned on the comments

🌟 Problem Description:
You are given an array of rolls, the mean of all rolls (including the missing ones), and the number of missing rolls n. The task is to determine the values of the missing rolls, such that the mean remains consistent.

🔑 Key Points:
Mean Calculation: The total required sum for all rolls is calculated using the mean and total number of rolls.
Boundary Check: If the remaining sum required is outside the range of what can be achieved with n rolls (less than n or greater than 6 * n), no valid solution exists.
Distribute Remaining Sum: The remaining sum is evenly distributed across the missing rolls. Any remainder is added incrementally to the rolls to ensure the total matches the required sum.

📝 Code Explanation:
Total Required: First, we calculate the total sum required based on the mean and the number of rolls (including the missing ones).
Existing Sum: We sum the existing rolls to find out how much of the total has already been accounted for.
Remaining Sum: We compute the remaining sum that needs to be distributed among the missing rolls.
Result Calculation: The remaining sum is distributed evenly, with any excess handled by adding 1 to the first few rolls.

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[AlgoXploration]

Code: class Solution {
public int[] missingRolls(int[] rolls, int mean, int n) {
int m = rolls.length;
int totalRolls = n + m;
int totalRequired = mean * totalRolls;
int existingSum = 0;
for(int roll : rolls)
existingSum += roll;

int remainingSum = totalRequired - existingSum;
if(remainingSum < n || remainingSum > 6 * n)
return new int[0];

int[] result = new int[n];
int baseValue = remainingSum / n;
int extra = remainingSum % n;

for(int i = 0; i < n; i++)
result[i] = baseValue + (
i < extra ? 1 : 0
);

return result;
}
}