HackerRank Zig Zag Sequence Java Solution Explained
Problem Overview
Original Problem: Zig Zag Sequence | HackerRank
Difficulty: Medium
Category: Arrays, Sorting
Description: Given an array of n distinct integers (where n is odd), transform it into a lexicographically smallest zig zag sequence. A zig zag sequence has the first k elements in increasing order and the last k elements in decreasing order, where k = (n+1)/2. Debug the provided function to correctly produce this sequence, modifying at most three lines of code.
This problem, part of HackerRank’s One Week Preparation Kit, is a great exercise for beginners to practice array manipulation, sorting, and debugging in Java, key skills for coding interviews.
Understanding the Problem
The Zig Zag Sequence problem requires transforming an array of n distinct integers into a sequence where the first k elements are in ascending order, and the remaining elements are in descending order, ensuring the result is the lexicographically smallest possible sequence. For example, for the array [1, 2, 3, 4, 5] (n = 5), the zig zag sequence is [1, 2, 5, 4, 3]: the first three elements [1, 2, 5] are increasing, and the last two [4, 3] are decreasing. The task is to debug the given code, which sorts the array and attempts to rearrange it, by modifying at most three lines to achieve the correct zig zag pattern.
Input and Output
Input:
First line: An integer t, the number of test cases.
For each test case:
First line: An integer n, the array size.
Second line: n space-separated distinct integers a[i].
Output:
For each test case, a single line of n space-separated integers representing the lexicographically smallest zig zag sequence.
Constraints:
1 ≤ t ≤ 20
1 ≤ n ≤ 10,000 (n is odd)
1 ≤ a[i] ≤ 10^9
All elements in the array are distinct.
Initial Reasoning
To create a zig zag sequence:
Sort the array to ensure we can construct the lexicographically smallest sequence.
Place the smallest k elements in the first k positions in ascending order.
Place the remaining elements in the last k positions in descending order, with the largest element at the peak (position k-1).
Debug the provided code, which sorts the array but incorrectly swaps elements, leading to an invalid zig zag sequence.
The challenge is to identify and fix the bug within three lines to achieve the correct arrangement efficiently.
Thought Process
Let’s analyze the problem and the provided code:
Zig Zag Sequence Structure:
For n = 5, k = (5+1)/2 = 3. The sequence should be [a, b, max, c, d], where a < b < max (first three elements increasing) and c > d (last two decreasing).
The lexicographically smallest sequence uses the smallest elements for the increasing part and arranges the remaining elements to peak at the middle with the largest value.
Debugging the Code:
The provided code sorts the array (Arrays.sort(a)), which is correct for lexicographical ordering.
It calculates mid = (n+1)/2, intending to place the largest element (a[n-1]) at the middle, but this is incorrect because mid = 3 for n = 5 places the largest element at index 3 (fourth position), not the peak of the increasing part (index 2).
The swap logic (while (st <= ed)) incorrectly reverses elements from mid+1 to n-1, causing the decreasing part to be in the wrong order.
Fixing the Bug:
Correct the mid calculation to mid = n/2 to place the largest element at the peak (index k-1).
Adjust the swap logic to reverse elements from mid+1 to n-2, placing the second-largest element at the end.
Ensure the output format remains unchanged.
Constraints Handling:
n ≤ 10,000 allows for O(n log n) sorting with Arrays.sort.
Elements up to 10^9 fit in int (no overflow in sorting).
Multiple test cases (t ≤ 20) are handled independently.
Odd n and distinct elements simplify the zig zag construction.
Edge Cases:
If n = 1, the sequence is a single element (already a zig zag).
For n = 3, the sequence should be [smallest, largest, middle] (e.g., [1, 3, 2]).
The constraints ensure valid inputs and odd n.
This approach requires modifying only two lines in the provided code to achieve the correct zig zag sequence, staying within the three-line limit.
Pseudocode
// findZigZagSequence method:
Sort array in ascending order
Calculate middle index for maxiumum value
Place largest element at middle
For each element between the midpoint and the end
Swap the element to reverse the order
Print the array
Solution Code
// Note: This is one of many possible solutions
// Note: Due to modification constraints, variable names are not updated in the code below to follow Java naming conventions.
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;
public class Main {
public static void main(String[] args) throws java.lang.Exception {
Scanner kb = new Scanner(System.in);
int test_cases = kb.nextInt();
for (int cs = 1; cs <= test_cases; cs++) {
int n = kb.nextInt();
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = kb.nextInt();
}
findZigZagSequence(a, n);
}
}
public static void findZigZagSequence(int[] a, int n) {
Arrays.sort(a);
int mid = n / 2; // Fix 1: Correct mid to n/2 to place largest element at peak (k-1)
int temp = a[mid];
a[mid] = a[n - 1];
a[n - 1] = temp;
int st = mid + 1;
int ed = n - 2; // Fix 2: Adjust swap loop to reverse from mid+1 to n-2
while (st <= ed) {
temp = a[st];
a[st] = a[ed];
a[ed] = temp;
st = st + 1;
ed = ed - 1; // Changed from ed + 1 to ed - 1 for proper reversal
}
for (int i = 0; i < n; i++) {
if (i > 0)
System.out.print(" ");
System.out.print(a[i]);
}
System.out.println();
}
}Step-by-Step Explanation
Let’s break down the debugged Java solution for beginners:
Input Handling:
The main method reads:
test_cases (t), the number of test cases, using kb.nextInt().
For each test case:
n, the array size.
n integers into array a using a loop with kb.nextInt().
Sort the Array:
In findZigZagSequence, sort the array a using Arrays.sort(a) to get the elements in ascending order (e.g., [1, 2, 3, 4, 5]).
This ensures the lexicographically smallest sequence by using the smallest elements first.
Fix the Middle Element:
Calculate mid = n / 2 (e.g., 2 for n = 5), which is the index of the peak (k-1, where k = (n+1)/2).
Swap a[mid] with a[n-1] to place the largest element at the peak (e.g., a[2] = 5).
Reverse the Decreasing Part:
Set st = mid + 1 (e.g., 3) and ed = n - 2 (e.g., 3) to reverse elements from mid+1 to n-2.
In the loop, swap a[st] and a[ed], increment st, and decrement ed until st > ed.
For n = 5, this swaps a[3] and a[3], placing the second-largest element (4) at the end, resulting in [1, 2, 5, 4, 3].
Output the Result:
Print the array with spaces between elements and a newline at the end using System.out.print
Common Mistakes to Avoid
Here are common pitfalls for beginners:
Incorrect Middle Index: Using (n+1)/2 for mid places the largest element too far right, breaking the increasing part’s length. Use n/2 to align with k-1.
Wrong Swap Logic: Reversing too many or too few elements (e.g., including n-1) disrupts the decreasing order. Only reverse from mid+1 to n-2.
Not Ensuring Lexicographical Order: Failing to sort the array first or misplacing elements can produce a valid zig zag sequence that isn’t the smallest lexicographically.
Real-World Application
The Zig Zag Sequence problem is relevant in data arrangement tasks, such as optimizing display patterns or signal processing, where elements need to follow a specific order (e.g., increasing then decreasing). In algorithmic trading, similar patterns might organize data for trend analysis. This problem teaches sorting, array manipulation, and debugging, skills used in software development and competitive programming.
FAQs
Q: Why is the lexicographically smallest sequence required?
A: Sorting the array first ensures the smallest elements are used in the increasing part, and the largest element is placed at the peak, minimizing the sequence lexicographically.
Q: Why modify only three lines?
A: The problem is designed as a debugging challenge, requiring minimal changes to fix the existing code, teaching efficient debugging skills within constraints.
Q: What if n = 1?
A: For n = 1, the array has one element, which is already a zig zag sequence (k = 1, single element is both increasing and decreasing). The code handles this correctly by skipping swaps.