LeetCode #2811 — MEDIUM

Check if it is Possible to Split Array

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

The Problem

Problem Statement

You are given an array nums of length n and an integer m. You need to determine if it is possible to split the array into n arrays of size 1 by performing a series of steps.

An array is called good if:

The length of the array is one, or
The sum of the elements of the array is greater than or equal to m.

In each step, you can select an existing array (which may be the result of previous steps) with a length of at least two and split it into two arrays, if both resulting arrays are good.

Return true if you can split the given array into n arrays, otherwise return false.

Example 1:

Input: nums = [2, 2, 1], m = 4

Output: true

Explanation:

Split [2, 2, 1] to [2, 2] and [1]. The array [1] has a length of one, and the array [2, 2] has the sum of its elements equal to 4 >= m, so both are good arrays.
Split [2, 2] to [2] and [2]. both arrays have the length of one, so both are good arrays.

Example 2:

Input: nums = [2, 1, 3], m = 5

Output: false

Explanation:

The first move has to be either of the following:

Split [2, 1, 3] to [2, 1] and [3]. The array [2, 1] has neither length of one nor sum of elements greater than or equal to m.
Split [2, 1, 3] to [2] and [1, 3]. The array [1, 3] has neither length of one nor sum of elements greater than or equal to m.

So as both moves are invalid (they do not divide the array into two good arrays), we are unable to split nums into n arrays of size 1.

Example 3:

Input: nums = [2, 3, 3, 2, 3], m = 6

Output: true

Explanation:

Split [2, 3, 3, 2, 3] to [2] and [3, 3, 2, 3].
Split [3, 3, 2, 3] to [3, 3, 2] and [3].
Split [3, 3, 2] to [3, 3] and [2].
Split [3, 3] to [3] and [3].

Constraints:

1 <= n == nums.length <= 100
1 <= nums[i] <= 100
1 <= m <= 200

Step 01

Brute Force Baseline

Problem summary: You are given an array nums of length n and an integer m. You need to determine if it is possible to split the array into n arrays of size 1 by performing a series of steps. An array is called good if: The length of the array is one, or The sum of the elements of the array is greater than or equal to m. In each step, you can select an existing array (which may be the result of previous steps) with a length of at least two and split it into two arrays, if both resulting arrays are good. Return true if you can split the given array into n arrays, otherwise return false.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming · Greedy

Example 1

[2, 2, 1]
4

Example 2

[2, 1, 3]
5

Example 3

[2, 3, 3, 2, 3]
6

Step 02

Core Insight

What unlocks the optimal approach

It can be proven that if you can split more than one element as a subarray, then you can split exactly one element.

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 #2811: Check if it is Possible to Split Array
class Solution {
    private Boolean[][] f;
    private int[] s;
    private int m;

    public boolean canSplitArray(List<Integer> nums, int m) {
        int n = nums.size();
        f = new Boolean[n][n];
        s = new int[n + 1];
        for (int i = 1; i <= n; ++i) {
            s[i] = s[i - 1] + nums.get(i - 1);
        }
        this.m = m;
        return dfs(0, n - 1);
    }

    private boolean dfs(int i, int j) {
        if (i == j) {
            return true;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        for (int k = i; k < j; ++k) {
            boolean a = k == i || s[k + 1] - s[i] >= m;
            boolean b = k == j - 1 || s[j + 1] - s[k + 1] >= m;
            if (a && b && dfs(i, k) && dfs(k + 1, j)) {
                return f[i][j] = true;
            }
        }
        return f[i][j] = false;
    }
}

// Accepted solution for LeetCode #2811: Check if it is Possible to Split Array
func canSplitArray(nums []int, m int) bool {
	n := len(nums)
	f := make([][]int, n)
	s := make([]int, n+1)
	for i, x := range nums {
		s[i+1] = s[i] + x
	}
	for i := range f {
		f[i] = make([]int, n)
	}
	var dfs func(i, j int) bool
	dfs = func(i, j int) bool {
		if i == j {
			return true
		}
		if f[i][j] != 0 {
			return f[i][j] == 1
		}
		for k := i; k < j; k++ {
			a := k == i || s[k+1]-s[i] >= m
			b := k == j-1 || s[j+1]-s[k+1] >= m
			if a && b && dfs(i, k) && dfs(k+1, j) {
				f[i][j] = 1
				return true
			}
		}
		f[i][j] = -1
		return false
	}
	return dfs(0, n-1)
}

# Accepted solution for LeetCode #2811: Check if it is Possible to Split Array
class Solution:
    def canSplitArray(self, nums: List[int], m: int) -> bool:
        @cache
        def dfs(i: int, j: int) -> bool:
            if i == j:
                return True
            for k in range(i, j):
                a = k == i or s[k + 1] - s[i] >= m
                b = k == j - 1 or s[j + 1] - s[k + 1] >= m
                if a and b and dfs(i, k) and dfs(k + 1, j):
                    return True
            return False

        s = list(accumulate(nums, initial=0))
        return dfs(0, len(nums) - 1)

// Accepted solution for LeetCode #2811: Check if it is Possible to Split Array
impl Solution {
    pub fn can_split_array(nums: Vec<i32>, m: i32) -> bool {
        let n = nums.len();
        if n <= 2 {
            return true;
        }
        for i in 1..n {
            if nums[i - 1] + nums[i] >= m {
                return true;
            }
        }
        false
    }
}

// Accepted solution for LeetCode #2811: Check if it is Possible to Split Array
function canSplitArray(nums: number[], m: number): boolean {
    const n = nums.length;
    const s: number[] = new Array(n + 1).fill(0);
    for (let i = 1; i <= n; ++i) {
        s[i] = s[i - 1] + nums[i - 1];
    }
    const f: number[][] = Array(n)
        .fill(0)
        .map(() => Array(n).fill(-1));
    const dfs = (i: number, j: number): boolean => {
        if (i === j) {
            return true;
        }
        if (f[i][j] !== -1) {
            return f[i][j] === 1;
        }
        for (let k = i; k < j; ++k) {
            const a = k === i || s[k + 1] - s[i] >= m;
            const b = k === j - 1 || s[j + 1] - s[k + 1] >= m;
            if (a && b && dfs(i, k) && dfs(k + 1, j)) {
                f[i][j] = 1;
                return true;
            }
        }
        f[i][j] = 0;
        return false;
    };
    return dfs(0, n - 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^3)

Space

O(n^2)

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.

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.