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Capital one|OA| Research Scientist – DS OA – OA代写 – 面试代面 – 面试辅助 – Capital One OA Experience and Solutions for Memory Allocation and Query Operations

Recently, We took an Online Assessment (OA) from Capital One, consisting of four questions. The first two questions were straightforward and completed within minutes. The challenging parts were the last two questions, which I'll discuss in detail, including my approach and thoughts.

First Part

You are given numbers, an array of non-negative integers. Your task is to perform the following algorithm on this array:

Step 1:

Find the index i of the leftmost non-zero element numbers[i] = x ≠ 0. If there is no such element, finish the algorithm.

Step 2:

Starting at index i and going to the right, attempt to subtract x from each element. If the element is strictly less than x, move on to step 3; otherwise, subtract x from the element and move on to the next element. If you reach the end of the array, move on to step 3.

Step 3:

Add x to the final result.

Step 4:

Go back to step 1.

Return the resulting sum obtained from step 3. It is guaranteed that the algorithm is finite and will finish at some point.

Note: You are not expected to provide the most optimal solution, but a solution with time complexity not worse than O(numbers.length × NUMBER_OF_STEPS) will fit within the execution time limit.

Second Part

You are given an array of integers memory consisting of 0s and 1s - whether the corresponding memory unit is free or not. memory[i] = 0 means that the ith memory unit is free, and memory[i] = 1 means it's occupied.

Your task is to perform two types of queries:

alloc X:

Find the left-most memory block of X consecutive free memory units and mark these units as occupied (i.e.: find the left-most contiguous subarray of 0s, and replace them all with 1s). If there are no blocks with X consecutive free units, return -1; otherwise return the index of the first position of the allocated block segment.

erase index:

If there exists an allocated memory block starting at position index, free all its memory units. Otherwise, if the memory cell at position index was occupied before the very first operation ever applied to the memory, free this cell only. Return the length of the deleted memory block. If there is no such allocated block starting at the position index, return -1.

The queries are given in the following format:

queries is an array of 2-elements arrays; if queries[i][0] = 0 then this is an alloc type query, where X = queries[i][1]; if queries[i][0] = 1 then this is an erase type query, where index = queries[i][1].

Return an array containing the results of all the queries.

Note: You are not expected to provide the most optimal solution, but a solution with time complexity not worse than O(memory.length^2 * queries.length) will fit within the execution time limit.




找到左起第一个非零元素的索引 i,即 numbers[i] = x ≠ 0。如果没有这样的元素,算法结束。


从索引 i 开始向右尝试从每个元素中减去 x。如果元素严格小于 x,则转到步骤3;否则,从该元素中减去 x 并继续下一个元素。如果到达数组的末尾,则转到步骤3。


x 加到最终结果中。




注意:你不需要提供最优解,但时间复杂度不应超过 O(numbers.length × NUMBER_OF_STEPS),以便在执行时间限制内完成。


  1. 对于分配操作 (queries[i][0] = 0):
    • 使用滑动窗口找到第一个长度为 X 的连续 0 子数组。
    • 将这些单元标记为 1 并返回起始位置。
    • 维护一个映射记录已分配块的起始位置和长度。
  2. 对于擦除操作 (queries[i][0] = 1):
    • 检查索引是否有记录分配。
    • 如果匹配记录分配,则释放内存;否则处理初始占用状态。
    • 相应更新映射。


  • 理解擦除操作的确切行为,尤其是初始占用状态的条件,比较复杂。
  • 在给定的时间约束内实现高效解决方案,处理所有边界情况。




你被给定了一个由0和1组成的整数数组 memory,表示相应的内存单元是否空闲。memory[i] = 0 表示第 i 个内存单元空闲,memory[i] = 1 表示被占用。


分配X (alloc X):

找到最左边的 X 个连续空闲内存单元并将这些单元标记为已占用(即:找到最左边的连续的0子数组,并将它们全部替换为1)。如果没有 X 个连续的空闲单元,则返回 -1;否则返回第一个内存块段的起始位置。

删除索引 (erase index):

如果在位置 index 处存在分配的内存块,则释放其所有内存单元。否则,如果位置 index 处的内存单元在最初从未应用过任何操作,则仅释放此单元。返回删除的内存块长度。如果在该位置没有分配的内存块,则返回 -1


queries 是一个由2元素数组组成的数组; 如果 queries[i][0] = 0,则这是一个分配类型查询,其中 X = queries[i][1]; 如果 queries[i][0] = 1,则这是一个删除类型查询,其中 index = queries[i][1]


注意:你不需要提供最优解,但时间复杂度不应超过 O(memory.length^2 * queries.length),以便在执行时间限制内完成。


  1. 使用映射存储 ab 中元素的计数。
  2. 对于更新查询 (queries[i][0] = 0),调整 b 的映射以反映新计数。
  3. 对于和查询 (queries[i][0] = 1),遍历一个映射的键并检查另一个映射中是否存在相应值,乘以它们的计数得到总对数。


  • 初始实现可能因为更新或计数逻辑错误而失败。
  • 确保解决方案处理所有边界情况并优化性能。




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