The answer is the longest subarray that contains digit $$$7$$$ only

Submission

Do some casework, and found out some of the followings:

• If $$$n$$$ is less than $$$3$$$ then the answer is $$$0$$$
• If $$$n$$$ is divisible by $$$3$$$ then the answer is $$$\displaystyle \frac{n}{3} \times k$$$
• If $$$n = 4$$$, you can notice the following pattern $$$[1, 2, 4, 5, 6, 8, 9, 10, 12, \dots]$$$
• If $$$n = 5$$$, you can notice the following pattern $$$[1, 3, 5, 6, 8, 10, 11, 13, 15, \dots]$$$
• Then for every other $$$n$$$ we can separate the points into $$$X$$$ and $$$Y$$$ $$$(X + Y = n)$$$ where $$$X$$$ is divisible by $$$3$$$ and $$$Y$$$ is either $$$4$$$ or $$$5$$$ and use the above patterns

Submission

Use Sweep Line, basically store events in the form of (time, type, index), for inserting we use type = 0, for the query use type = 1, and for deletion use type = 2.

See my submission

Use DP from back to beginning, where $$$dp[i] = (x, y)$$$ such that $$$x$$$ is the first player's point and $$$y$$$ is the second player's point, if we chose to take the $$$i$$$-th character.

To get the recursion, $$$dp[i] = (cnt[i] + dp[j][1], dp[j][0])$$$ we want the difference between the second player and the first player to be as large as possible (Since the second player is the one controlling the $$$j$$$ index).

Submission

The number of distinct possible value of $$$\displaystyle\left\lfloor \frac{n}{i} \right\rfloor$$$ is around $$$O(\sqrt{N})$$$.

To get all the values, we can use the following code:

for (int x = 1; x <= n; ) {
   int val = n / x;
   x = n / val + 1;
}

Submission

Let's store the following value for each zombie:

• $$$p_i = (dist_i, t_i)$$$
• Where $$$dist_i$$$ is the distance of the zombie to our location
• And $$$t_i$$$ is the required time to kill the zombie

Now we need to kill the nearest zombie first, so we can sort $$$p$$$.

Iterate over all the zombies, and let $$$sum$$$ be the summation of $$$t_i$$$ that we take:

• If $$$sum \leq dist_i$$$ then we continue
• If not we need to take the largest $$$t$$$ that we have take (use priority queue) and replace it with $$$1$$$ (since it needs exactly $$$1$$$ second to use the high damage bullet)
• If it turns out that $$$sum$$$ is always $$$> dist_i$$$ then return $$$-1$$$
• Otherwise return the number of time we replace values of $$$t$$$

Submission