Minimum Processing Time for OS Tasks - Expedia Online Assessment IIT Roorkee

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"Pattern : 2 DSA + 1 SQL An operating system has `n` tasks numbered from `1` to `n`. You are given: * `high_priority`: 1-based indices of high-priority tasks. * `t_normal`: processing time for a normal task. * `t_high`: processing time for a high-priority task. The tasks must be divided into **exactly two contiguous parts**: ```text Prefix: tasks 1 ... k Suffix: tasks k+1 ... n ``` where `0 <= k <= n`. * Core 1 processes the prefix. * Core 2 processes the suffix. * Both cores work in parallel. Therefore, the total processing time for a split is: ```text max(prefix_time, suffix_time) ``` Find the minimum possible total processing time. ### Constraints ```text 1 <= n <= 10^5 0 <= high_priority.length <= n 1 <= high_priority[i] <= n 1 <= t_normal < t_high <= 1000 high_priority contains unique values ``` ### Example **Input:** ```text n = 5 high_priority = [2, 4] t_normal = 2 t_high = 5 ``` The task processing times are: ```text [2, 5, 2, 5, 2] ``` ### Split Evaluation ```text k = 0 → max(0, 16) = 16 k = 1 → max(2, 14) = 14 k = 2 → max(7, 9) = 9 k = 3 → max(9, 7) = 9 k = 4 → max(14, 2) = 14 k = 5 → max(16, 0) = 16 ``` The minimum processing time is: ```text 9 ``` It is achieved by splitting at: ```text k = 2 k = 3 ``` **Expected result:** ```text 9 ```"