LeetCode #2871 — MEDIUM

Split Array Into Maximum Number of Subarrays

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

The Problem

Problem Statement

You are given an array nums consisting of non-negative integers.

We define the score of subarray nums[l..r] such that l <= r as nums[l] AND nums[l + 1] AND ... AND nums[r] where AND is the bitwise AND operation.

Consider splitting the array into one or more subarrays such that the following conditions are satisfied:

Each element of the array belongs to exactly one subarray.
The sum of scores of the subarrays is the minimum possible.

Return the maximum number of subarrays in a split that satisfies the conditions above.

A subarray is a contiguous part of an array.

Example 1:

Input: nums = [1,0,2,0,1,2]
Output: 3
Explanation: We can split the array into the following subarrays:
- [1,0]. The score of this subarray is 1 AND 0 = 0.
- [2,0]. The score of this subarray is 2 AND 0 = 0.
- [1,2]. The score of this subarray is 1 AND 2 = 0.
The sum of scores is 0 + 0 + 0 = 0, which is the minimum possible score that we can obtain.
It can be shown that we cannot split the array into more than 3 subarrays with a total score of 0. So we return 3.

Example 2:

Input: nums = [5,7,1,3]
Output: 1
Explanation: We can split the array into one subarray: [5,7,1,3] with a score of 1, which is the minimum possible score that we can obtain.
It can be shown that we cannot split the array into more than 1 subarray with a total score of 1. So we return 1.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given an array nums consisting of non-negative integers. We define the score of subarray nums[l..r] such that l <= r as nums[l] AND nums[l + 1] AND ... AND nums[r] where AND is the bitwise AND operation. Consider splitting the array into one or more subarrays such that the following conditions are satisfied: Each element of the array belongs to exactly one subarray. The sum of scores of the subarrays is the minimum possible. Return the maximum number of subarrays in a split that satisfies the conditions above. A subarray is a contiguous part of an array.

Baseline thinking

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

Pattern signal: Array · Greedy · Bit Manipulation

Example 1

[1,0,2,0,1,2]

Example 2

[5,7,1,3]

Step 02

Core Insight

What unlocks the optimal approach

The minimum score will always be the bitwise <code>AND</code> of all elements of the array.
If the minimum score is not equal to <code>0</code>, the only possible split will be to keep all elements in one subarray.
Otherwise, all of the subarrays should have a score of <code>0</code>, we can greedily split the array while trying to make each subarray as small as possible.

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 #2871: Split Array Into Maximum Number of Subarrays
class Solution {
    public int maxSubarrays(int[] nums) {
        int score = -1;
        int ans = 1;
        for (int num : nums) {
            score &= num;
            if (score == 0) {
                ans++;
                score = -1;
            }
        }
        return ans == 1 ? 1 : ans - 1;
    }
}

// Accepted solution for LeetCode #2871: Split Array Into Maximum Number of Subarrays
func maxSubarrays(nums []int) int {
	ans, score := 1, -1
	for _, num := range nums {
		score &= num
		if score == 0 {
			score--
			ans++
		}
	}
	if ans == 1 {
		return 1
	}
	return ans - 1
}

# Accepted solution for LeetCode #2871: Split Array Into Maximum Number of Subarrays
class Solution:
    def maxSubarrays(self, nums: List[int]) -> int:
        score, ans = -1, 1
        for num in nums:
            score &= num
            if score == 0:
                score = -1
                ans += 1
        return 1 if ans == 1 else ans - 1

// Accepted solution for LeetCode #2871: Split Array Into Maximum Number of Subarrays
// 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 #2871: Split Array Into Maximum Number of Subarrays
// class Solution {
//     public int maxSubarrays(int[] nums) {
//         int score = -1;
//         int ans = 1;
//         for (int num : nums) {
//             score &= num;
//             if (score == 0) {
//                 ans++;
//                 score = -1;
//             }
//         }
//         return ans == 1 ? 1 : ans - 1;
//     }
// }

// Accepted solution for LeetCode #2871: Split Array Into Maximum Number of Subarrays
function maxSubarrays(nums: number[]): number {
    let [ans, score] = [1, -1];
    for (const num of nums) {
        score &= num;
        if (score === 0) {
            --score;
            ++ans;
        }
    }
    return ans == 1 ? 1 : ans - 1;
}

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 log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.