Problem 1: Profitable Project Pairs

Description

Given two integer lists Profit and Cost, count the number of index pairs (i, j) such that i < j and (Profit[i] - Cost[i]) + (Profit[j] - Cost[j]) > 0.

A brute-force O(n²) approach would time out, so we use the following optimized approach:

1. Compute net_profit[i] = Profit[i] - Cost[i]
2. Sort the net_profit list
3. For each index i, use binary search to find how many j > i satisfy net_profit[i] + net_profit[j] > 0

Function Signature

def count_profitable_pairs(Profit: List[int], Cost: List[int]) -> int:

Example

Profit = [5, 3, 2]
Cost = [2, 1, 4]
# net_profit = [3, 2, -2]
# Sorted = [-2, 2, 3]
# Valid pairs: (1, 2), (0, 1), (0, 2)
# Output: 3

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Problem 2: Maximum Swaps

Description

Given two lists a and b, and an integer k (maximum number of swaps allowed), return the maximum number of unique elements that can be in list a after at most k swaps from list b.

Steps:

• Compute elements in b not in a: set(b) - set(a)
• Compute the number of duplicate elements in a: len(a) - len(set(a))
• The maximum possible unique elements we can add is min(k, len(b_not_in_a), duplicates_in_a)
• Result is len(set(a)) + swaps_used

Function Signature

def max_unique_after_swaps(a: List[int], b: List[int], k: int) -> int:

Example

a = [1, 2, 2, 3]
b = [3, 4, 5]
k = 2
# set(a) = {1, 2, 3}, set(b) - set(a) = {4, 5}
# duplicates = 1, max usable swaps = min(2, 2, 1) = 1
# result = 3 + 1 = 4

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Problem 3: Binary Circuit

Description

Given a binary string (composed of '0' and '1'), calculate the total cost to move all 1s to the rightmost part of the string.

Rules:

• Only the leftmost block of 1s can be moved
• Move the rightmost 1 from the leftmost group to immediately before the next 1
• Cost = 1 + distance_moved

Use dynamic programming or greedy counting:

• Traverse left to right, for each 0, add current number of 1s before it to cost
• If a 0 is the first one after a 1, add an extra 1 to cost

Function Signature

def binary_circuit_cost(s: str) -> int:

Example

s = "1100101"
# Steps:
# 1 -> count_ones = 2
# cost += count_ones at each '0'
# total cost = 2 (first 0) + 2 (second 0) + 3 (third 0 after more 1s) = 7
# Output: 7

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