LeetCode #2945 — HARD

Find Maximum Non-decreasing Array Length

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given a 0-indexed integer array nums.

You can perform any number of operations, where each operation involves selecting a subarray of the array and replacing it with the sum of its elements. For example, if the given array is [1,3,5,6] and you select subarray [3,5] the array will convert to [1,8,6].

Return the maximum length of a non-decreasing array that can be made after applying operations.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [5,2,2]
Output: 1
Explanation: This array with length 3 is not non-decreasing.
We have two ways to make the array length two.
First, choosing subarray [2,2] converts the array to [5,4].
Second, choosing subarray [5,2] converts the array to [7,2].
In these two ways the array is not non-decreasing.
And if we choose subarray [5,2,2] and replace it with [9] it becomes non-decreasing. 
So the answer is 1.

Example 2:

Input: nums = [1,2,3,4]
Output: 4
Explanation: The array is non-decreasing. So the answer is 4.

Example 3:

Input: nums = [4,3,2,6]
Output: 3
Explanation: Replacing [3,2] with [5] converts the given array to [4,5,6] that is non-decreasing.
Because the given array is not non-decreasing, the maximum possible answer is 3.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums. You can perform any number of operations, where each operation involves selecting a subarray of the array and replacing it with the sum of its elements. For example, if the given array is [1,3,5,6] and you select subarray [3,5] the array will convert to [1,8,6]. Return the maximum length of a non-decreasing array that can be made after applying operations. A subarray is a contiguous non-empty sequence of elements within an array.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Dynamic Programming · Stack · Monotonic Queue

Example 1

[5,2,2]

Example 2

[1,2,3,4]

Example 3

[4,3,2,6]

Step 02

Core Insight

What unlocks the optimal approach

Let <code>dp[i]</code> be the maximum number of elements in the increasing sequence after processing the first <code>i</code> elements of the original array.
We have <code>dp[0] = 0</code>. <code>dp[i + 1] >= dp[i]</code> (since if we have the solution for the first <code>i</code> elements, we can always merge the last one of the first <code>i + 1</code> elements which is <code>nums[i]</code> into the solution of the first <code>i</code> elements.
For <code>i > 0</code>, we want to <code>dp[i] = max(dp[j] + 1)</code> where <code>sum(nums[i - 1] + nums[i - 2] +… + nums[j]) >= v[j]</code> and <code>v[j]</code> is the last element of the solution ending with <code>nums[j - 1]</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2945: Find Maximum Non-decreasing Array Length
class Solution {
    public int findMaximumLength(int[] nums) {
        int n = nums.length;
        long[] s = new long[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        int[] f = new int[n + 1];
        int[] pre = new int[n + 2];
        for (int i = 1; i <= n; ++i) {
            pre[i] = Math.max(pre[i], pre[i - 1]);
            f[i] = f[pre[i]] + 1;
            int j = Arrays.binarySearch(s, s[i] * 2 - s[pre[i]]);
            pre[j < 0 ? -j - 1 : j] = i;
        }
        return f[n];
    }
}

// Accepted solution for LeetCode #2945: Find Maximum Non-decreasing Array Length
func findMaximumLength(nums []int) int {
	n := len(nums)
	f := make([]int, n+1)
	pre := make([]int, n+2)
	s := make([]int, n+1)
	for i, x := range nums {
		s[i+1] = s[i] + x
	}
	for i := 1; i <= n; i++ {
		pre[i] = max(pre[i], pre[i-1])
		f[i] = f[pre[i]] + 1
		j := sort.SearchInts(s, s[i]*2-s[pre[i]])
		pre[j] = max(pre[j], i)
	}
	return f[n]
}

# Accepted solution for LeetCode #2945: Find Maximum Non-decreasing Array Length
class Solution:
    def findMaximumLength(self, nums: List[int]) -> int:
        n = len(nums)
        s = list(accumulate(nums, initial=0))
        f = [0] * (n + 1)
        pre = [0] * (n + 2)
        for i in range(1, n + 1):
            pre[i] = max(pre[i], pre[i - 1])
            f[i] = f[pre[i]] + 1
            j = bisect_left(s, s[i] * 2 - s[pre[i]])
            pre[j] = i
        return f[n]

// Accepted solution for LeetCode #2945: Find Maximum Non-decreasing Array Length
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2945: Find Maximum Non-decreasing Array Length
// class Solution {
//     public int findMaximumLength(int[] nums) {
//         int n = nums.length;
//         long[] s = new long[n + 1];
//         for (int i = 0; i < n; ++i) {
//             s[i + 1] = s[i] + nums[i];
//         }
//         int[] f = new int[n + 1];
//         int[] pre = new int[n + 2];
//         for (int i = 1; i <= n; ++i) {
//             pre[i] = Math.max(pre[i], pre[i - 1]);
//             f[i] = f[pre[i]] + 1;
//             int j = Arrays.binarySearch(s, s[i] * 2 - s[pre[i]]);
//             pre[j < 0 ? -j - 1 : j] = i;
//         }
//         return f[n];
//     }
// }

// Accepted solution for LeetCode #2945: Find Maximum Non-decreasing Array Length
function findMaximumLength(nums: number[]): number {
    const n = nums.length;
    const f: number[] = Array(n + 1).fill(0);
    const pre: number[] = Array(n + 2).fill(0);
    const s: number[] = Array(n + 1).fill(0);
    for (let i = 1; i <= n; ++i) {
        s[i] = s[i - 1] + nums[i - 1];
    }
    const search = (nums: number[], x: number): number => {
        let [l, r] = [0, nums.length];
        while (l < r) {
            const mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    for (let i = 1; i <= n; ++i) {
        pre[i] = Math.max(pre[i], pre[i - 1]);
        f[i] = f[pre[i]] + 1;
        const j = search(s, s[i] * 2 - s[pre[i]]);
        pre[j] = i;
    }
    return f[n];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.