LeetCode 3506: Find Time Required to Eliminate Bacterial Strains Solution

Master LeetCode problem 3506 (Find Time Required to Eliminate Bacterial Strains), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3506. Find Time Required to Eliminate Bacterial Strains

Hard

Array Math Greedy Heap (Priority Queue)

Problem Explanation

Explanation:

To solve this problem, we can use binary search to find the minimum amount of time required to eliminate all bacterial strains. We can start by setting the lower bound as 1 and the upper bound as the maximum time required to eliminate any bacterial strain. Then we perform binary search on the time range until we find the minimum time that satisfies the condition that all bacterial strains are eliminated within that time.

Here are the steps for the algorithm:

Initialize the lower bound as 1 and the upper bound as the maximum time needed to eliminate any bacterial strain.
Perform binary search on the time range.
For each time, calculate the total amount of time needed to eliminate each bacterial strain based on the current time.
If the total time is less than or equal to the current time, update the upper bound.
If the total time is greater than the current time, update the lower bound.
Repeat steps 2-5 until the lower bound is equal to the upper bound.

The time complexity of this algorithm is O(n * log(maxTime)), where n is the number of bacterial strains and maxTime is the maximum time needed to eliminate any bacterial strain. The space complexity is O(1).

Solution Code

class Solution {
    public int minTime(int[] strains, int[] time) {
        int low = 1, high = Arrays.stream(time).max().getAsInt();
        
        while (low < high) {
            int mid = low + (high - low) / 2;
            int total = 0;
            for (int i = 0; i < strains.length; i++) {
                total += (time[i] - 1) * strains[i] + 1;
            }
            if (total <= mid) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
        
        return low;
    }
}

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