LeetCode has started publishing problems numbered 4000 and above. At the time of writing this post, there are already 4055 problems.
In this post, I’ll take a look at the first of them, from 4000 to 4010 (except 4004, 4005), including the difficulty level of each problem and the patterns that can be used to solve it.
There are no solutions here, so you can try to solve these problems on your own using the suggested patterns as hints before looking at the actual solutions.
Here:
Easy level: 4000, 4006, 4010
Medium level: 4001, 4002, 4008
Hard level: 4003, 4007, 4009
4000. Largest Integer With Given Digit Sum
Easy level
This is a numbers (digit manipulation) problem. We need to construct the largest number with at most n digits whose digit sum is equal to s.
To make the number as large as possible, we need to place the largest possible digits on the left - first 9, then the remainder, and then 0s.
If s > 9 * n, it is impossible to construct such a number.
Complexity should be:
Time complexity: O(n)
Space complexity: O(n)
Patterns: Greedy, Digit Construction (Math)
4001. Aggregate Two Time Series
Medium level
This is an array (merge two sorted arrays) problem. We need to iterate through two time series and, for each timestamp, calculate the sum of the values from both series. Since both arrays are already sorted by timestamp, the best approach is to move two pointers through series1 and series2 simultaneously.
Complexity should be:
Time complexity: O(n+m)
Space complexity: O(n+m)
Patterns: Two Pointers, Merge Sorted Arrays, Forward Fill or Next Available Value
Medium level
This is a combinatorics (integer compositions problem). We need to count the number of sequences of k positive integers with sum n whose product is even.
The key idea is to count all sequences with sum n and then subtract those where all numbers are odd, because this is the only case where the product is odd.
Complexity for the standard solution using factorials:
Time complexity: O(n)
Space complexity: O(n)
Patterns: Combinatorics, Stars and Bars, Parity Counting, Modular Arithmetic
4003. Minimum Cost Path with Alternating Directions III
Hard level
This is a graph (shortest path in a grid) problem. The state depends not only on the cell (i, j), but also on the parity of the next action.
So, each cell is effectively represented by two states: (i, j, odd) and (i, j, even).
Complexity should be:
Time complexity: O(m*n * log(m*n))
Space complexity: O(m*n)
Patterns: Dijkstra, State Graph, Grid Shortest Path, Parity State
Easy level
This is a strings (prefix counting) problem. For each prefix, we need to check whether its characters can be rearranged to form an alternating string.
The key condition is that, for an alternating binary string, the number of 0s and 1s can differ by at most 1.
Complexity should be:
Time complexity: O(n)
Space complexity: O(1)
Patterns: Prefix Counting, Frequency Counting, Parity / Balance Check
Hard level
This is an arrays (counting) problem. We need to find a height h that allows us to obtain the maximum number of boards, either by using existing boards of height h or by combining pairs of boards whose heights sum to h.
For each pair of different heights x and y, the contribution to height x + y is min(count[x], count[y]). For equal heights x + x, we can create count[x] / 2 boards. Existing boards of height x are also added directly to the result for that height.
Complexity in worst case:
Time complexity: O(n2)
Space complexity: O(n2)
Patterns: Frequency Counting, Hash Map, Pair Sum / Two Sum, Enumeration
4008. Minimum Initial Strength to Defeat All Monsters
Medium level
This is an arrays (range updates, prefix sums) problem. First, we need to calculate the total bonus for each monster, and then determine the minimum initial strength required to defeat all monsters from left to right.
Using a Difference Array pattern lets us apply all boosts [l, r, v] in O(n + boosts.length) time without processing each range separately.
For each i, we effectively determine the minimum initial strength that would have been required, taking into account the strength already spent and the temporary bonus.
Complexity should be:
Time complexity: O(n+b)
Space complexity: O(n)
Patterns: Difference Array, Prefix Sum, Prefix Maximum / Greedy
4009. Minimum Possible Maximum Waiting Time
Hard level
This is a dynamic programming (scheduling with two resources) problem. We need to assign cars to one of two dispensers so that we first maximize the number of cars served and, among those solutions, minimize the maximum waiting time.
The DP state needs to track how much fuel remains in both dispensers, when each dispenser becomes available, and the waiting time accumulated so far.
Because of the small constraints fuel[j] <= 50 and demand[i] <= 20, we can build a DP over the possible states of the two dispensers.
Complexity should be:
Time complexity: O(n*F2)
Space complexity: O(F2)
Here, F <= 50 is the maximum fuel capacity of a single dispenser.
Patterns: Dynamic Programming, State Compression, Scheduling, Minimax Optimization
4010. Maximize Pair Strength Using GCD
Easy level
This is an arrays (number theory) problem. We need to iterate over pairs of numbers and maximize a value that depends on their product and gcd.
For each pair (i, j), we calculate gcd(nums[i], nums[j]) and the corresponding strength.
Complexity should be:
Time complexity: O(n2 log M)
Space complexity: O(1)
Here, M = max(nums[i]).
Patterns: Greatest Common Divisor (GCD), Number Theory, Pair Enumeration
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