Given an array of integers, you need to determine if it can be partitioned into one or more valid subarrays. A subarray is valid if it meets one of three conditions:
Companies like Coinbase and Google use this to test your understanding of linear Dynamic Programming. It’s a "decision" problem where at each index, you have to decide which rule to apply. It evaluates your ability to break a large problem into overlapping subproblems and your skill in using memoization to avoid exponential time complexity.
The pattern is 1D Dynamic Programming. Let dp[i] be a boolean representing whether the prefix of the array ending at index i-1 can be validly partitioned. To calculate dp[i], you check:
dp[i-2] is true and the last 2 elements match Rule 1.dp[i-3] is true and the last 3 elements match Rule 2 or Rule 3.
Since you only ever look back 3 steps, you can optimize this to space.Array: [4, 4, 4, 5, 6]
dp[0] = True (empty prefix).dp[2]: Is [4, 4] valid? Yes. dp[2] = dp[0] = True.dp[3]: Is [4, 4, 4] valid? Yes. dp[3] = dp[0] = True.dp[5]: Can we reach here?
dp[2]: Is [4, 5, 6] valid? Yes. dp[5] = True.
Result: True.A common error is not checking the bounds correctly when looking back 2 or 3 elements. Another is failing to realize that a single index could be the end of multiple valid partitions (e.g., an index could be the end of a group of 2 AND a group of 3). Candidates also sometimes forget to handle the base cases for very short arrays.
Linear DP problems where you "look back" a fixed number of steps are very common. Practice recognizing when a global property ("can the whole array be partitioned?") can be derived from local properties ("can the last elements form a valid group?").
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Inverse Coin Change | Medium | Solve | |
| Maximize Total Cost of Alternating Subarrays | Medium | Solve | |
| Minimum Increment Operations to Make Array Beautiful | Medium | Solve | |
| Zero Array Transformation IV | Medium | Solve | |
| Best Time to Buy and Sell Stock V | Medium | Solve |