Solution

Approach #1: Brute Force [Memory Limit Exceeded]

Intuition and Algorithm

Create the multiplication table and sort it, then take the $$k^{th}$$ element.

Complexity Analysis

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  Time Complexity: $$O(m*n)$$ to create the table, and $$O(m*n\\log(m*n))$$ to sort it.

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  Space Complexity: $$O(m*n)$$ to store the table.

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Approach #2: Next Heap [Time Limit Exceeded]

Intuition

Maintain a heap of the smallest unused element of each row. Then, finding the next element is a pop operation on the heap.

Algorithm

Our heap is going to consist of elements $$\\text{(val, root)}$$ , where $$\\text{val}$$ is the next unused value of that row, and $$\\text{root}$$ was the starting value of that row.

We will repeatedly find the next lowest element in the table. To do this, we pop from the heap. Then, if there\'s a next lowest element in that row, we\'ll put that element back on the heap.

Complexity Analysis

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  Time Complexity: $$O(k * m \\log m) = O(m^2 n \\log m)$$ . Our initial heapify operation is $$O(m)$$ . Afterwards, each pop and push is $$O(m \\log m)$$ , and our outer loop is $$O(k) = O(m*n)$$ \n

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  Space Complexity: $$O(m)$$ . Our heap is implemented as an array with $$m$$ elements.

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Approach #3: Binary Search [Accepted]

Intuition

As $$\\text{k}$$ and $$\\text{m*n}$$ are up to $$9 * 10^8$$ , linear solutions will not work. This motivates solutions with $$\\log$$ complexity, such as binary search.

Algorithm

Let\'s do the binary search for the answer $$\\text{A}$$ .

Say enough(x) is true if and only if there are $$\\text{k}$$ or more values in the multiplication table that are less than or equal to $$\\text{x}$$ . Colloquially, enough describes whether $$\\text{x}$$ is large enough to be the $$k^{th}$$ value in the multiplication table.

Then (for our answer $$\\text{A}$$ ), whenever $$\\text{x &geq; A}$$ , enough(x) is True; and whenever $$\\text{x < A}$$ , enough(x) is False.

In our binary search, our loop invariant is enough(hi) = True. At the beginning, enough(m*n) = True, and whenever hi is set, it is set to a value that is "enough" (enough(mi) = True). That means hi will be the lowest such value at the end of our binary search.

This leaves us with the task of counting how many values are less than or equal to $$\\text{x}$$ . For each of $$\\text{m}$$ rows, the $$i^{th}$$ row looks like $$\\text{[i, 2*i, 3*i, ..., n*i]}$$ . The largest possible $$\\text{k*i &leq; x}$$ that could appear is $$\\text{k = x // i}$$ . However, if $$\\text{x}$$ is really big, then perhaps $$\\text{k > n}$$ , so in total there are $$\\text{min(k, n) = min(x // i, n)}$$ values in that row that are less than or equal to $$\\text{x}$$ .

After we have the count of how many values in the table are less than or equal to $$\\text{x}$$ , by the definition of enough(x), we want to know if that count is greater than or equal to $$\\text{k}$$ .

Complexity Analysis

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  Time Complexity: $$O(m * \\log (m*n))$$ . Our binary search divides the interval $$\\text{[lo, hi]}$$ into half at each step. At each step, we call enough which requires $$O(m)$$ time.

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  Space Complexity: $$O(1)$$ . We only keep integers in memory during our intermediate calculations.

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Analysis written by: @awice