Find Maximum Non-decreasing Array Length

Find Maximum Non-decreasing Array Length

What is this problem about?

The Find Maximum Non-decreasing Array Length coding problem asks you to take an array of integers and transform it into a non-decreasing array by merging adjacent elements. When you merge two adjacent elements, they are replaced by their sum. Your goal is to maximize the length of the resulting non-decreasing array. For example, in [4, 3, 2, 6], you could merge 4 and 3 to get [7, 2, 6], but that's not non-decreasing. If you merge 3 and 2, you get [4, 5, 6], which has length 3.

Why is this asked in interviews?

Companies like Amazon and Google use this "Hard" problem to test a candidate's mastery of Dynamic Programming (DP) and optimization. It evaluation your ability to manage two competing goals: keeping the last element of the sequence as small as possible (to allow for more subsequent elements) while maximizing the total count. It’s a sophisticated test of Monotonic Queue or Monotonic Stack optimizations applied to DP state transitions.

Algorithmic pattern used

This problem is solved using Dynamic Programming with Monotonic Queue Optimization.

1. Define dp[i] as the maximum length of a non-decreasing array formed from the prefix nums[0...i-1].
2. Define last[i] as the value of the last element in that optimal non-decreasing array.
3. To maximize the length, you want to find the largest $j < i$ such that the sum of elements from $j$ to $i-1$ is greater than or equal to last[j].
4. Among all such $j$, you pick the one that maximizes dp[j] + 1. If there are ties, pick the one that minimizes the sum of the new last element.
5. Using a monotonic queue and prefix sums, you can find the optimal $j$ in $O(1)$ amortized time.

Example explanation

nums = [1, 2, 3, 4]

1. dp[0]=0, last[0]=0.
2. Index 1 (val 1): Sum(0,1)=1 $\ge$ last[0]. dp[1]=1, last[1]=1.
3. Index 2 (val 2): Sum(1,2)=2 $\ge$ last[1]. dp[2]=2, last[2]=2. ... Result: 4. (No merges needed).

nums = [4, 3, 2]

1. Index 1: dp[1]=1, last[1]=4.
2. Index 2 (val 3): 3 < 4. Must merge. Sum(0,2)=7. dp[2]=1, last[2]=7.
3. Index 3 (val 2): 2 < 7. Must merge. Sum(0,3)=9. dp[3]=1, last[3]=9. Wait, if we merged differently: [4, 5] has length 2. The DP finds this.

Common mistakes candidates make

• Greedy without DP: Trying to merge only when necessary, which might lead to a larger "last" element than necessary, blocking future additions.
• $O(n^2)$ DP: Failing to use a monotonic queue to find the optimal previous state, which results in Time Limit Exceeded.
• Sum calculation: Re-calculating range sums instead of using a Prefix Sum array.

Interview preparation tip

When a DP state depends on finding the "best" previous index based on a sum condition, think about Monotonic Queues. It's the standard way to optimize $O(n^2)$ DP to $O(n)$.