LeetCode #659 — MEDIUM

Split Array into Consecutive Subsequences

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

The Problem

Problem Statement

You are given an integer array nums that is sorted in non-decreasing order.

Determine if it is possible to split nums into one or more subsequences such that both of the following conditions are true:

Each subsequence is a consecutive increasing sequence (i.e. each integer is exactly one more than the previous integer).
All subsequences have a length of 3 or more.

Return true if you can split nums according to the above conditions, or false otherwise.

A subsequence of an array is a new array that is formed from the original array by deleting some (can be none) of the elements without disturbing the relative positions of the remaining elements. (i.e., [1,3,5] is a subsequence of [1,2,3,4,5] while [1,3,2] is not).

Example 1:

Input: nums = [1,2,3,3,4,5]
Output: true
Explanation: nums can be split into the following subsequences:
[1,2,3,3,4,5] --> 1, 2, 3
[1,2,3,3,4,5] --> 3, 4, 5

Example 2:

Input: nums = [1,2,3,3,4,4,5,5]
Output: true
Explanation: nums can be split into the following subsequences:
[1,2,3,3,4,4,5,5] --> 1, 2, 3, 4, 5
[1,2,3,3,4,4,5,5] --> 3, 4, 5

Example 3:

Input: nums = [1,2,3,4,4,5]
Output: false
Explanation: It is impossible to split nums into consecutive increasing subsequences of length 3 or more.

Constraints:

1 <= nums.length <= 10⁴
-1000 <= nums[i] <= 1000
nums is sorted in non-decreasing order.

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums that is sorted in non-decreasing order. Determine if it is possible to split nums into one or more subsequences such that both of the following conditions are true: Each subsequence is a consecutive increasing sequence (i.e. each integer is exactly one more than the previous integer). All subsequences have a length of 3 or more. Return true if you can split nums according to the above conditions, or false otherwise. A subsequence of an array is a new array that is formed from the original array by deleting some (can be none) of the elements without disturbing the relative positions of the remaining elements. (i.e., [1,3,5] is a subsequence of [1,2,3,4,5] while [1,3,2] is not).

Baseline thinking

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

Pattern signal: Array · Hash Map · Greedy

Example 1

[1,2,3,3,4,5]

Example 2

[1,2,3,3,4,4,5,5]

Example 3

[1,2,3,4,4,5]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #659: Split Array into Consecutive Subsequences
class Solution {
    public boolean isPossible(int[] nums) {
        Map<Integer, PriorityQueue<Integer>> d = new HashMap<>();
        for (int v : nums) {
            if (d.containsKey(v - 1)) {
                var q = d.get(v - 1);
                d.computeIfAbsent(v, k -> new PriorityQueue<>()).offer(q.poll() + 1);
                if (q.isEmpty()) {
                    d.remove(v - 1);
                }
            } else {
                d.computeIfAbsent(v, k -> new PriorityQueue<>()).offer(1);
            }
        }
        for (var v : d.values()) {
            if (v.peek() < 3) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #659: Split Array into Consecutive Subsequences
func isPossible(nums []int) bool {
	d := map[int]*hp{}
	for _, v := range nums {
		if d[v] == nil {
			d[v] = new(hp)
		}
		if h := d[v-1]; h != nil {
			heap.Push(d[v], heap.Pop(h).(int)+1)
			if h.Len() == 0 {
				delete(d, v-1)
			}
		} else {
			heap.Push(d[v], 1)
		}
	}
	for _, q := range d {
		if q.IntSlice[0] < 3 {
			return false
		}
	}
	return true
}

type hp struct{ sort.IntSlice }

func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}

# Accepted solution for LeetCode #659: Split Array into Consecutive Subsequences
class Solution:
    def isPossible(self, nums: List[int]) -> bool:
        d = defaultdict(list)
        for v in nums:
            if h := d[v - 1]:
                heappush(d[v], heappop(h) + 1)
            else:
                heappush(d[v], 1)
        return all(not v or v and v[0] > 2 for v in d.values())

// Accepted solution for LeetCode #659: Split Array into Consecutive Subsequences
struct Solution;
use std::collections::HashMap;

impl Solution {
    fn is_possible(nums: Vec<i32>) -> bool {
        let mut left: HashMap<i32, usize> = HashMap::new();
        let mut end: HashMap<i32, usize> = HashMap::new();
        for &x in &nums {
            *left.entry(x).or_default() += 1;
        }

        for &x in &nums {
            if *left.entry(x).or_default() == 0 {
                continue;
            }
            if *end.entry(x - 1).or_default() > 0 {
                *left.entry(x).or_default() -= 1;
                *end.entry(x - 1).or_default() -= 1;
                *end.entry(x).or_default() += 1;
                continue;
            }
            if *left.entry(x + 1).or_default() > 0 && *left.entry(x + 2).or_default() > 0 {
                *left.entry(x).or_default() -= 1;
                *left.entry(x + 1).or_default() -= 1;
                *left.entry(x + 2).or_default() -= 1;
                *end.entry(x + 2).or_default() += 1;
                continue;
            }
            return false;
        }
        true
    }
}

#[test]
fn test() {
    let nums = vec![1, 2, 3, 3, 4, 5];
    let res = true;
    assert_eq!(Solution::is_possible(nums), res);
    let nums = vec![1, 2, 3, 3, 4, 4, 5, 5];
    let res = true;
    assert_eq!(Solution::is_possible(nums), res);
    let nums = vec![1, 2, 3, 4, 4, 5];
    let res = false;
    assert_eq!(Solution::is_possible(nums), res);
}

// Accepted solution for LeetCode #659: Split Array into Consecutive Subsequences
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #659: Split Array into Consecutive Subsequences
// class Solution {
//     public boolean isPossible(int[] nums) {
//         Map<Integer, PriorityQueue<Integer>> d = new HashMap<>();
//         for (int v : nums) {
//             if (d.containsKey(v - 1)) {
//                 var q = d.get(v - 1);
//                 d.computeIfAbsent(v, k -> new PriorityQueue<>()).offer(q.poll() + 1);
//                 if (q.isEmpty()) {
//                     d.remove(v - 1);
//                 }
//             } else {
//                 d.computeIfAbsent(v, k -> new PriorityQueue<>()).offer(1);
//             }
//         }
//         for (var v : d.values()) {
//             if (v.peek() < 3) {
//                 return false;
//             }
//         }
//         return true;
//     }
// }

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.