Arranging Flower Sticks in a Bouquet: Sorting Problem

The flower sticks bouquet problem splits an array at index K, sorts the left portion ascending and the right portion descending, then returns the combined result.

That description covers the entire problem. The name sounds like it might involve permutation counting or combinatorics, but no factorials are needed here. The word “arranging” in the title is a framing choice, not a hint about the algorithm. This is a two-pass sort on subranges of the same array.

What the Problem Asks

The function receives three inputs:

• num (integer): the total count of flower sticks, N
• random (integer): the split index K (the count to sort ascending)
• arr (list of integers): the lengths of the N sticks, in their original order

Return a rearranged list where:

• The first K elements (positions 0 to K-1) are in ascending order of length.
• The remaining N-K elements (positions K to N-1) are in descending order of length.

Constraints from the original problem: random < num and 0 < num < 1000000.

The split is positional. “First K flower sticks” means positions 0 through K-1 in the input array as given, not the globally K shortest or K tallest sticks. That distinction matters for getting the output right.

Worked Example

Starting values:

• num = 7, random = 3, arr = [8, 3, 6, 7, 2, 9, 5]

Step-by-step:

• Extract the first K=3 elements by position: [8, 3, 6]
• Sort that subarray ascending: [3, 6, 8]
• Extract the remaining N-K=4 elements by position: [7, 2, 9, 5]
• Sort that subarray descending: [9, 7, 5, 2]
• Concatenate both halves: [3, 6, 8, 9, 7, 5, 2]

Verification:

• Ascending check: 3 < 6 < 8 — correct.
• Descending check: 9 > 7 > 5 > 2 — correct.

One common mistake: reading the problem as “sort the K smallest values ascending, then the rest descending.” That would give a different output. The problem says first K by position, regardless of their values relative to the rest of the array.

Algorithm

Three steps, no data structure beyond the original array:

1. Slice positions 0 to K-1. Sort ascending.
2. Slice positions K to N-1. Sort descending.
3. Return the concatenation of both sorted slices.

The choice of language determines whether you sort in-place (C++) or create sorted copies (Python, Java). Both produce the same output.

Code Solutions

C++

The C++ standard library std::sort accepts iterator pairs, so you can sort subranges of the same vector without copying. The descending half passes greater<int>() as a comparator.

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

vector<int> arrangeFlowerSticks(int num, int k, vector<int>& arr) {
    // Sort first K elements ascending
    sort(arr.begin(), arr.begin() + k);
    // Sort remaining N-K elements descending
    sort(arr.begin() + k, arr.end(), greater<int>());
    return arr;
}

int main() {
    vector<int> arr = {8, 3, 6, 7, 2, 9, 5};
    int num = 7, k = 3;
    vector<int> result = arrangeFlowerSticks(num, k, arr);
    for (int val : result) cout << val << " ";
    // Output: 3 6 8 9 7 5 2
    return 0;
}

This modifies arr in-place. The two sort calls each operate on a contiguous subrange of the same backing memory: no copies are created, just O(log N) stack space from introsort’s recursion.

Python

The Python sorted() built-in creates a new sorted list from any iterable. reverse=True handles the descending half.

def arrange_flower_sticks(num, k, arr):
    first_part = sorted(arr[:k])                  # ascending
    second_part = sorted(arr[k:], reverse=True)   # descending
    return first_part + second_part

# Example
arr = [8, 3, 6, 7, 2, 9, 5]
print(arrange_flower_sticks(7, 3, arr))
# Output: [3, 6, 8, 9, 7, 5, 2]

arr[:k] and arr[k:] create new list objects (slice copies), so the original arr is not modified. Space cost is O(N) for the two output lists.

Java

Collections.sort() and List.sort(Collections.reverseOrder()) work on ArrayList views. Copy each subrange first to avoid issues with fixed-size list implementations.

import java.util.*;

public class FlowerBouquet {
    public static List<Integer> arrangeFlowerSticks(int num, int k, List<Integer> arr) {
        // Copy subranges to new ArrayLists before sorting
        List<Integer> firstPart = new ArrayList<>(arr.subList(0, k));
        List<Integer> secondPart = new ArrayList<>(arr.subList(k, num));
        Collections.sort(firstPart);
        secondPart.sort(Collections.reverseOrder());
        firstPart.addAll(secondPart);
        return firstPart;
    }

    public static void main(String[] args) {
        List<Integer> arr = Arrays.asList(8, 3, 6, 7, 2, 9, 5);
        System.out.println(arrangeFlowerSticks(7, 3, arr));
        // Output: [3, 6, 8, 9, 7, 5, 2]
    }
}

The copy step (new ArrayList<>(arr.subList(...))) matters: Arrays.asList() returns a fixed-size list, and calling addAll() on a sublist view of a fixed-size list throws UnsupportedOperationException. The copy produces a regular, resizable ArrayList.

Complexity Analysis

Sorting each half separately:

• Time: Sorting the first K elements costs O(K log K). Sorting the remaining N-K elements costs O((N-K) log(N-K)). Combined: O(K log K + (N-K) log(N-K)). Worst case is when K = N/2, giving O(N log N).
• Space (C++): Sorts in-place. Introsort uses O(log N) stack frames from recursion, treated as O(1) auxiliary heap space in most analyses.
• Space (Python): Creates two new lists totalling N elements. O(N) auxiliary space.
• Space (Java): Copies two sublists totalling N elements. O(N) auxiliary space.
• Scale note: For N = 1,000,000, O(N log N) is roughly 20 million comparisons; that is well within a two-second time limit on a standard online judge.

You cannot do better than O(N log N) in the general comparison model regardless of the split point K, because each half still requires at least O(M log M) comparisons to sort M unsorted elements.

Edge Cases and Common Mistakes

Input boundary cases:

• K = 0: The ascending segment is empty. The entire array is sorted descending. The constraint 0 < num < 1000000 says num is positive but does not prevent K from being 0 in practice; handle it explicitly if the calling code can pass K = 0.
• K = 1: Single-element ascending segment (trivially sorted). Remaining N-1 elements sorted descending.
• K = N-1: First N-1 elements sorted ascending. Last element forms a one-element descending segment and stays in place.
• Duplicate values: Python’s Timsort is stable, so duplicates preserve relative input order within each sorted half. C++ std::sort is not guaranteed stable; use std::stable_sort if the relative order of equal elements matters. Java’s Collections.sort is a stable merge sort.
• All elements equal: Both sorts produce the same sequence. Output equals input.

Algorithm-reading mistakes to avoid:

• Value-based split vs. positional split: “Sort the first K flower sticks by length” means positions 0 to K-1 of the input array, sorted. It does not mean “take the K shortest sticks and sort them.” The split index K partitions the array by position.
• Off-by-one on the split: arr.begin() + k in C++ is the start of the descending segment. The two sort calls share this iterator as the exclusive end of the first range and the start of the second, with no overlap and no gap.

For more array-traversal patterns that appear alongside this type of question, the data structures interview guide covers the 20 most common patterns across sorting, searching, and tree traversal. The smallest and largest element problem covers single-pass array traversal with similar index-boundary discipline.

The decompose-sort-merge structure here is the same skeleton as a retrieval pipeline: split the input, process each segment with different logic, merge the results. TinkerLLM starts with that parallel as its first exercise: ₹299 entry, and you will have a working two-pass LLM pipeline by session two, built on the same decompose-process-merge instinct this problem exercises.