Divide Array Into Increasing Sequences - Practice Coding Problems

Divide Array Into Increasing Sequences - Problem

Array Counting Hard

Given a sorted integer array nums in non-decreasing order and an integer k, your task is to determine if this array can be divided into disjoint increasing subsequences where each subsequence has at least k elements.

An increasing subsequence means each element must be strictly greater than the previous element in the same subsequence. The subsequences must be disjoint, meaning each element from the original array can only be used in exactly one subsequence.

Example: For array [1,2,2,3,3,4,4] and k=3, we can form two increasing subsequences: [1,2,3,4] and [2,3,4], so return true.

Input & Output

example_1.py — Basic Valid Case

$ Input: nums = [1,2,2,3,3,4,4], k = 3

› Output: true

💡 Note: We can form two valid increasing subsequences: [1,2,3,4] and [2,3,4]. Both have at least 3 elements and are strictly increasing.

example_2.py — Insufficient Elements

$ Input: nums = [5,6,6,7,8], k = 3

› Output: false

💡 Note: We cannot form valid subsequences. We have two 6's but need them in separate increasing subsequences, which would require at least two 7's and two 8's, but we only have one of each.

example_3.py — Edge Case k=1

$ Input: nums = [1,2,3,4], k = 1

› Output: true

💡 Note: When k=1, we can always divide the array since each element can form its own subsequence of length 1.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ k ≤ nums.length
-10⁵ ≤ nums[i] ≤ 10⁵
nums is sorted in non-decreasing order

Visualization

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Understanding the Visualization

Count Players by Skill

First, count how many players we have at each skill level

Process Skills in Order

Handle players from lowest to highest skill level

Assign to Existing Teams

First, assign players to teams that are already waiting for this skill level

Form New Teams

Use remaining players to start new teams, each needing k-1 more players

Key Takeaway

🎯 Key Insight: We can solve this optimally by counting frequencies and using a greedy approach - always satisfy existing subsequences first, then start new ones with remaining elements.

Asked in

G Google 35 a Amazon 28 ⊞ Microsoft 22 f Meta 18

The optimal solution uses a frequency map with greedy assignment approach. We count the frequency of each element, then process elements in ascending order. For each element, we first satisfy existing subsequences that need this element, then use any remaining occurrences to start new subsequences. The key insight is that we need exactly enough elements at each level to support all active subsequences requiring them.

Common Approaches

Approach	Time	Space	Notes
✓ Frequency Map with Greedy Assignment (Optimal)	O(n)	O(n)	Count frequencies and use greedy approach to assign elements to subsequences
Brute Force (Generate All Partitions)	O(2^n × n)	O(n)	Generate all possible ways to partition the array and check each one

Frequency Map with Greedy Assignment (Optimal) — Algorithm Steps

Count frequency of each unique element
Iterate through unique elements in ascending order
For each element, determine how many new subsequences to start
Update the count of active subsequences needing each future element
Check if we can satisfy all requirements

Visualization

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Step-by-Step Walkthrough

Count frequencies

Build frequency map of all elements

Process in order

Handle elements in ascending order

Satisfy existing needs

First satisfy existing subsequences

Start new subsequences

Use remaining elements to start new subsequences

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

// Simplified C implementation
// Would need hash table implementation for full solution
bool canDivideArray(int* nums, int numsSize, int k) {
    if (k <= 1) return true;
    
    // This would require implementing hash tables in C
    // For brevity, showing structure only
    return false; // Placeholder
}

int main() {
    // Input parsing and function call
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array to count frequencies, then one pass through unique elements

✓ Linear Growth

Space Complexity

O(n)

Hash map to store frequencies and active subsequence counts

⚡ Linearithmic Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Frequency Map with Greedy Assignment (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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