What is the time complexity of Split Array into Consecutive Subsequences?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Split Array into Consecutive Subsequences?

This problem uses Array, Hash Table, Greedy, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Split Array into Consecutive Subsequences?

Insight 1: The core idea is to greedily extend existing subsequences whenever possible to minimize the creation of new ones, as subsequences must have a length of at least 3. Insight 2: Using two hash maps (or frequency counters) is crucial. One (`freq`) tracks the remaining frequency of each number in the input array. The other (`tails`) tracks the number of subsequences ending with a particular number.

Which companies ask LeetCode 659. Split Array into Consecutive Subsequences in interviews?

This problem is frequently asked at PhonePe.

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Split Array into Consecutive Subsequences - LeetCode 659 Solution

Split Array into Consecutive Subsequences - Complete Solution Guide

Split Array into Consecutive Subsequences is LeetCode problem 659, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem asks us to determine if a given sorted (non-decreasing order) integer array `nums` can be split into one or more subsequences. Each subsequence must be a consecutive increasing sequence (each element is exactly one greater than the previous) and must have a length of at least 3. The function should return `true` if such a split is possible, and `false` otherwise. A subsequence is formed by deleting some (or none) elements from the original array without changing the order of the remaining elements.

Solution Approach

The solution uses a greedy approach. It iterates through the sorted array `nums`, processing each number `x`. For each `x`, it checks if it can extend an existing subsequence (ending in `x-1`). If so, it extends the subsequence by decrementing the count of subsequences ending in `x-1` and incrementing the count of subsequences ending in `x`. If `x` cannot extend an existing subsequence, it checks if a new subsequence of length 3 can be started, by checking for the presence of `x+1` and `x+2`. If so, it decrements their frequencies and adds a new subsequence ending at `x+2`. If neither of these is possible, the array cannot be split as required, and the function returns `false`. If the entire array is processed, and no issues are encountered, the function returns `true`.

Step-by-Step Algorithm

Step 1: Initialize two frequency counters (hash maps), `freq` and `tails`. `freq` stores the remaining frequency of each number in `nums`. `tails` stores the number of subsequences ending with a particular number.
Step 2: Iterate through the input array `nums`.
Step 3: For each number `x`, check if its frequency in `freq` is greater than 0. If not, skip to the next number.
Step 4: Decrement the frequency of `x` in `freq`.
Step 5: Check if we can extend an existing subsequence (i.e., a subsequence ending in `x-1`). If `tails[x-1] > 0`, then decrement `tails[x-1]` and increment `tails[x]`.
Step 6: If we cannot extend an existing subsequence, check if we can start a new subsequence of length 3, by checking if `freq[x+1] > 0` and `freq[x+2] > 0`. If so, decrement `freq[x+1]`, decrement `freq[x+2]`, and increment `tails[x+2]`.
Step 7: If neither extending an existing subsequence nor starting a new subsequence of length 3 is possible, return `false`.
Step 8: After iterating through all numbers in `nums`, return `true`.

Key Insights

Insight 1: The core idea is to greedily extend existing subsequences whenever possible to minimize the creation of new ones, as subsequences must have a length of at least 3.
Insight 2: Using two hash maps (or frequency counters) is crucial. One (`freq`) tracks the remaining frequency of each number in the input array. The other (`tails`) tracks the number of subsequences ending with a particular number.
Insight 3: The problem relies on the array being sorted. This allows us to process the numbers in order and greedily extend existing subsequences without needing to backtrack or consider all possible combinations.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Greedy, Heap (Priority Queue).

Companies

Asked at: PhonePe.

Split Array into Consecutive Subsequences | LeetCode 659

Medium

Array Hash Table Greedy Heap (Priority Queue)

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public boolean isPossible(int[] nums) {
        Map<Integer, Integer> freq = new HashMap<>();
        Map<Integer, Integer> tails = new HashMap<>();

        for (int x : nums) {
            freq.put(x, freq.getOrDefault(x, 0) + 1);
        }

        for (int x : nums) {
            if (freq.getOrDefault(x, 0) == 0) {
                continue;
            }

            freq.put(x, freq.get(x) - 1);

            if (tails.getOrDefault(x - 1, 0) > 0) {
                tails.put(x - 1, tails.get(x - 1) - 1);
                tails.put(x, tails.getOrDefault(x, 0) + 1);
            } else if (freq.getOrDefault(x + 1, 0) > 0 && freq.getOrDefault(x + 2, 0) > 0) {
                freq.put(x + 1, freq.get(x + 1) - 1);
                freq.put(x + 2, freq.get(x + 2) - 1);
                tails.put(x + 2, tails.getOrDefault(x + 2, 0) + 1);
            } else {
                return false;
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 659: Split Array into Consecutive Subsequences - Detailed explanation and solution approaches.

Topics

Array Hash Table Greedy Heap (Priority Queue)

Companies

PhonePe

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:659

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize two frequency counters (hash maps), `freq` and `tails`. `freq` stores the remaining frequency of each number in `nums`. `tails` stores the number of subsequences ending with a particular number.
2
Step 2: Iterate through the input array `nums`.
3
Step 3: For each number `x`, check if its frequency in `freq` is greater than 0. If not, skip to the next number.
4
Step 4: Decrement the frequency of `x` in `freq`.
5
Step 5: Check if we can extend an existing subsequence (i.e., a subsequence ending in `x-1`). If `tails[x-1] > 0`, then decrement `tails[x-1]` and increment `tails[x]`.
6
Step 6: If we cannot extend an existing subsequence, check if we can start a new subsequence of length 3, by checking if `freq[x+1] > 0` and `freq[x+2] > 0`. If so, decrement `freq[x+1]`, decrement `freq[x+2]`, and increment `tails[x+2]`.
7
Step 7: If neither extending an existing subsequence nor starting a new subsequence of length 3 is possible, return `false`.
8
Step 8: After iterating through all numbers in `nums`, return `true`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because we iterate through the `nums` array once. Each operation within the loop (hash map lookups, updates, and comparisons) takes O(1) time on average. The space complexity is linear because in the worst-case scenario (e.g., when all elements are distinct), both the `freq` and `tails` hash maps will store n elements.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Priority Queue Approach

Maintain a priority queue (min-heap) for each number encountered, storing the length of the subsequence ending at that number. When a number is encountered, either extend the shortest subsequence ending with the previous number, or create a new subsequence of length 1. At the end, check if the length of each subsequence is at least 3.

Time: O(n log n)Space: O(n)

Trade-offs: While conceptually similar, the priority queue approach introduces a logarithmic factor due to heap operations, making it less efficient than the greedy approach in terms of time complexity. The greedy approach is preferred due to its simplicity and better performance.

✨

Key Insights

1Insight 1: The core idea is to greedily extend existing subsequences whenever possible to minimize the creation of new ones, as subsequences must have a length of at least 3.
2Insight 2: Using two hash maps (or frequency counters) is crucial. One (`freq`) tracks the remaining frequency of each number in the input array. The other (`tails`) tracks the number of subsequences ending with a particular number.
3Insight 3: The problem relies on the array being sorted. This allows us to process the numbers in order and greedily extend existing subsequences without needing to backtrack or consider all possible combinations.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: Empty input array. The provided code implicitly handles this as the loops will not execute, and it will return `true`, which is arguably the correct behaviour as an empty array trivially satisfies the condition.
Edge case 2: Input array with repeated numbers. The frequency counters `freq` and `tails` correctly track the counts of each number, ensuring that the algorithm considers the number of subsequences and remaining numbers accurately.
Edge case 3: Input array where it is impossible to form any subsequence of length 3 or more, such as [1,2,4]. The algorithm correctly identifies this and returns `false`.
Edge case 4: A very long increasing consecutive sequence can be handled correctly by the greedy approach because the algorithm always tries to extend previous sequences.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the frequencies correctly. If the frequency of a number is not tracked accurately, it may lead to incorrect decisions about whether to extend an existing subsequence or start a new one.

✗Common mistake 2: Not checking the length of the subsequences at the end. It is crucial to ensure that all formed subsequences have a length of at least 3. The tails approach elegantly incorporates this requirement within the greedy iteration.

✗Common mistake 3: Not considering the sorted nature of the array. Some solutions might attempt to use dynamic programming or other approaches that do not leverage the sorted property, leading to increased complexity.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation - Imagine allocating tasks to workers, where tasks need to be executed in sequence, and each worker needs to work on at least 3 consecutive tasks. This problem mirrors the subsequence problem, where tasks are numbers, and workers are subsequences.

🏦Real-world scenario 2: Stock market analysis - Identifying patterns of consecutive increasing stock prices for a certain period. Subsequences of increasing prices of at least a specific length could indicate a potential bullish trend.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 28, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Split Array into Consecutive Subsequences"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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