LeetCode #2771 — MEDIUM

Longest Non-decreasing Subarray From Two Arrays

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given two 0-indexed integer arrays nums1 and nums2 of length n.

Let's define another 0-indexed integer array, nums3, of length n. For each index i in the range [0, n - 1], you can assign either nums1[i] or nums2[i] to nums3[i].

Your task is to maximize the length of the longest non-decreasing subarray in nums3 by choosing its values optimally.

Return an integer representing the length of the longest non-decreasing subarray in nums3.

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

Example 1:

Input: nums1 = [2,3,1], nums2 = [1,2,1]
Output: 2
Explanation: One way to construct nums3 is: 
nums3 = [nums1[0], nums2[1], nums2[2]] => [2,2,1]. 
The subarray starting from index 0 and ending at index 1, [2,2], forms a non-decreasing subarray of length 2. 
We can show that 2 is the maximum achievable length.

Example 2:

Input: nums1 = [1,3,2,1], nums2 = [2,2,3,4]
Output: 4
Explanation: One way to construct nums3 is: 
nums3 = [nums1[0], nums2[1], nums2[2], nums2[3]] => [1,2,3,4]. 
The entire array forms a non-decreasing subarray of length 4, making it the maximum achievable length.

Example 3:

Input: nums1 = [1,1], nums2 = [2,2]
Output: 2
Explanation: One way to construct nums3 is: 
nums3 = [nums1[0], nums1[1]] => [1,1]. 
The entire array forms a non-decreasing subarray of length 2, making it the maximum achievable length.

Constraints:

1 <= nums1.length == nums2.length == n <= 10⁵
1 <= nums1[i], nums2[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given two 0-indexed integer arrays nums1 and nums2 of length n. Let's define another 0-indexed integer array, nums3, of length n. For each index i in the range [0, n - 1], you can assign either nums1[i] or nums2[i] to nums3[i]. Your task is to maximize the length of the longest non-decreasing subarray in nums3 by choosing its values optimally. Return an integer representing the length of the longest non-decreasing subarray in nums3. Note: 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 · Dynamic Programming

Example 1

[2,3,1]
[1,2,1]

Example 2

[1,3,2,1]
[2,2,3,4]

Example 3

[1,1]
[2,2]

Core Insight

What unlocks the optimal approach

Consider using dynamic programming.
Let dp[i][0] (dp[i][1]) be the length of the longest non-decreasing ending with nums1[i] (nums2[i]).
Initialize dp[i][0] to 1. If nums1[i] >= nums1[i - 1] then dp[i][0] may be dp[i - 1][0] + 1. If nums1[i] >= nums2[i - 1] then dp[i][0] may be dp[i - 1][1] + 1. Perform a similar calculation for nums2[i] and dp[i][1].

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

Upper-end input sizes

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 #2771: Longest Non-decreasing Subarray From Two Arrays
class Solution {
    public int maxNonDecreasingLength(int[] nums1, int[] nums2) {
        int n = nums1.length;
        int f = 1, g = 1;
        int ans = 1;
        for (int i = 1; i < n; ++i) {
            int ff = 1, gg = 1;
            if (nums1[i] >= nums1[i - 1]) {
                ff = Math.max(ff, f + 1);
            }
            if (nums1[i] >= nums2[i - 1]) {
                ff = Math.max(ff, g + 1);
            }
            if (nums2[i] >= nums1[i - 1]) {
                gg = Math.max(gg, f + 1);
            }
            if (nums2[i] >= nums2[i - 1]) {
                gg = Math.max(gg, g + 1);
            }
            f = ff;
            g = gg;
            ans = Math.max(ans, Math.max(f, g));
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2771: Longest Non-decreasing Subarray From Two Arrays
func maxNonDecreasingLength(nums1 []int, nums2 []int) int {
	n := len(nums1)
	f, g, ans := 1, 1, 1
	for i := 1; i < n; i++ {
		ff, gg := 1, 1
		if nums1[i] >= nums1[i-1] {
			ff = max(ff, f+1)
		}
		if nums1[i] >= nums2[i-1] {
			ff = max(ff, g+1)
		}
		if nums2[i] >= nums1[i-1] {
			gg = max(gg, f+1)
		}
		if nums2[i] >= nums2[i-1] {
			gg = max(gg, g+1)
		}
		f, g = ff, gg
		ans = max(ans, max(f, g))
	}
	return ans
}

# Accepted solution for LeetCode #2771: Longest Non-decreasing Subarray From Two Arrays
class Solution:
    def maxNonDecreasingLength(self, nums1: List[int], nums2: List[int]) -> int:
        n = len(nums1)
        f = g = 1
        ans = 1
        for i in range(1, n):
            ff = gg = 1
            if nums1[i] >= nums1[i - 1]:
                ff = max(ff, f + 1)
            if nums1[i] >= nums2[i - 1]:
                ff = max(ff, g + 1)
            if nums2[i] >= nums1[i - 1]:
                gg = max(gg, f + 1)
            if nums2[i] >= nums2[i - 1]:
                gg = max(gg, g + 1)
            f, g = ff, gg
            ans = max(ans, f, g)
        return ans

// Accepted solution for LeetCode #2771: Longest Non-decreasing Subarray From Two Arrays
// 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 #2771: Longest Non-decreasing Subarray From Two Arrays
// class Solution {
//     public int maxNonDecreasingLength(int[] nums1, int[] nums2) {
//         int n = nums1.length;
//         int f = 1, g = 1;
//         int ans = 1;
//         for (int i = 1; i < n; ++i) {
//             int ff = 1, gg = 1;
//             if (nums1[i] >= nums1[i - 1]) {
//                 ff = Math.max(ff, f + 1);
//             }
//             if (nums1[i] >= nums2[i - 1]) {
//                 ff = Math.max(ff, g + 1);
//             }
//             if (nums2[i] >= nums1[i - 1]) {
//                 gg = Math.max(gg, f + 1);
//             }
//             if (nums2[i] >= nums2[i - 1]) {
//                 gg = Math.max(gg, g + 1);
//             }
//             f = ff;
//             g = gg;
//             ans = Math.max(ans, Math.max(f, g));
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2771: Longest Non-decreasing Subarray From Two Arrays
function maxNonDecreasingLength(nums1: number[], nums2: number[]): number {
    const n = nums1.length;
    let [f, g, ans] = [1, 1, 1];
    for (let i = 1; i < n; ++i) {
        let [ff, gg] = [1, 1];
        if (nums1[i] >= nums1[i - 1]) {
            ff = Math.max(ff, f + 1);
        }
        if (nums1[i] >= nums2[i - 1]) {
            ff = Math.max(ff, g + 1);
        }
        if (nums2[i] >= nums1[i - 1]) {
            gg = Math.max(gg, f + 1);
        }
        if (nums2[i] >= nums2[i - 1]) {
            gg = Math.max(gg, g + 1);
        }
        f = ff;
        g = gg;
        ans = Math.max(ans, f, g);
    }
    return ans;
}

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(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

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.