Example Usage:

python
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# Example 1
print(containsDuplicate([1, 2, 3, 1])) # Output: True

# Example 2
print(containsDuplicate([1, 2, 3, 4])) # Output: False

# Example 3
print(containsDuplicate([1, 1, 1, 3, 3, 4, 3, 2, 4, 2])) # Output: True

Constraints:

 The function works efficiently with arrays of reasonable length. However, for
extremely large arrays, the performance will still be good due to the set's average time
complexity of O(1) for both insert and lookup operations.
 The function assumes that the input is always a valid list of integers as described.

You said:

i want in every solution in javajava

ChatGPT said:

ChatGPT
Memory updated

Here's how you can solve the problem in Java:

java
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import java.util.HashSet;

public class Solution {

public boolean containsDuplicate(int[] nums) {
HashSet<Integer> seen = new HashSet<>();
for (int num : nums) {
if (seen.contains(num)) {
return true;
}
seen.add(num);
}
return false;
}

public static void main(String[] args) {

Solution solution = new Solution();

// Example 1
int[] nums1 = {1, 2, 3, 1};
System.out.println(solution.containsDuplicate(nums1)); // Output:
true

// Example 2
int[] nums2 = {1, 2, 3, 4};
 System.out.println(solution.containsDuplicate(nums2)); // Output:
false

// Example 3
int[] nums3 = {1, 1, 1, 3, 3, 4, 3, 2, 4, 2};
System.out.println(solution.containsDuplicate(nums3)); // Output:
true
}
}

Explanation:

 HashSet: We use a HashSet because it doesn't allow duplicate elements and offers
average time complexity O(1) for add and contains operations.
 containsDuplicate Method:
o We loop through each element in the nums array.
o If the element is already in the HashSet, we return true, indicating a duplicate
exists.
o If not, we add the element to the HashSet.
o If no duplicates are found by the end of the loop, we return false.

Example Outputs:

 For the input [1, 2, 3, 1], the output is true.

 For the input [1, 2, 3, 4], the output is false.
 For the input [1, 1, 1, 3, 3, 4, 3, 2, 4, 2], the output is true.

1/2

You said:

Given an array of N integers where each value represents the number of chocolates in a
packet. Each packet can have a variable number of chocolates. There are m students, the task
is to distribute chocolate packets such that: Each student gets one packet. The difference
between the number of chocolates in the packet with maximum chocolates and the packet
with minimum chocolates given to the students is minimum. Examples: Input : arr[] = {7, 3,
2, 4, 9, 12, 56} , m = 3 Output: Minimum Difference is 2 Explanation: We have seven
packets of chocolates and we need to pick three packets for 3 students If we pick 2, 3 and 4,
we get the minimum difference between maximum and minimum packet sizes.
ChatGPT said:

ChatGPT

To solve this problem, you can follow these steps:

1. Sort the Array: First, sort the array of chocolates. This will allow us to easily find the
smallest possible difference between the largest and smallest numbers in any subarray
of size m.
2. Sliding Window Technique: Once the array is sorted, use a sliding window of size m
to traverse the array. Calculate the difference between the maximum and minimum
chocolates in each window. Track the minimum difference found during the traversal.
 3. Return the Minimum Difference: The minimum difference found during the
traversal is the answer.

Here’s how you can implement this in Java:

java
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import java.util.Arrays;

public class ChocolateDistribution {

public int findMinDiff(int[] arr, int n, int m) {

if (m == 0 || n == 0) {
return 0;
}

// Sort the given packets

Arrays.sort(arr);

// Number of students cannot be more than number of packets

if (n < m) {
return -1;
}

// Find the minimum difference

int minDiff = Integer.MAX_VALUE;

for (int i = 0; i + m - 1 < n; i++) {

int diff = arr[i + m - 1] - arr[i];
if (diff < minDiff) {
minDiff = diff;
}
}
return minDiff;
}

public static void main(String[] args) {

ChocolateDistribution distribution = new ChocolateDistribution();

int[] arr = {7, 3, 2, 4, 9, 12, 56};

int m = 3; // Number of students

int result = distribution.findMinDiff(arr, arr.length, m);

System.out.println("Minimum Difference is " + result); // Output:
2
}
}

Explanation:

 Sorting: The array is first sorted. This allows us to efficiently check subarrays of
length m to find the one with the smallest difference between the maximum and
minimum values.
 Sliding Window: We then slide a window of size m across the sorted array and
calculate the difference between the maximum and minimum elements in that
window.
  Minimum Difference: The smallest difference found during this process is returned
as the result.

Example:

Given arr[] = {7, 3, 2, 4, 9, 12, 56} and m = 3, the sorted array is [2, 3, 4, 7,
9, 12, 56]. The sliding window process evaluates differences:

 Difference between 2 and 4 (window [2, 3, 4]) = 2

 Difference between 3 and 7 (window [3, 4, 7]) = 4
 Difference between 4 and 9 (window [4, 7, 9]) = 5
 Difference between 7 and 12 (window [7, 9, 12]) = 5
 Difference between 9 and 56 (window [9, 12, 56]) = 47

The minimum difference is 2, which occurs between packets containing 2 and 4 chocolates.

You said:

Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k
(1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1],
nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated
at pivot index 3 and become [4,5,6,7,0,1,2]. Given the array nums after the possible rotation
and an integer target, return the index of target if it is in nums, or -1 if it is not in nums. You
must write an algorithm with O(log n) runtime complexity. Example 1: Input: nums =
[4,5,6,7,0,1,2], target = 0 Output: 4 Example 2: Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
ChatGPT said:

ChatGPT

To solve this problem with O(log⁡n)O(\log n)O(logn) runtime complexity, you can use a
modified version of binary search. The array is partially sorted, and the key is to identify
which part of the array to search in by comparing elements.

Here’s the plan:

1. Identify the sorted half: In each step of binary search, you determine whether the left
or right half of the array is properly sorted.
2. Narrow the search: Once you identify the sorted half, you can determine whether the
target lies within that half. If it does, continue searching within that half. Otherwise,
search the other half.

Here’s the implementation in Java:

java
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public class Solution {
public int search(int[] nums, int target) {
int left = 0, right = nums.length - 1;
 while (left <= right) {
int mid = left + (right - left) / 2;

// Check if mid is the target

if (nums[mid] == target) {
return mid;
}

// Determine which half is sorted

if (nums[left] <= nums[mid]) { // Left half is sorted
if (nums[left] <= target && target < nums[mid]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else { // Right half is sorted
if (nums[mid] < target && target <= nums[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
}

return -1; // Target not found

}

public static void main(String[] args) {

Solution solution = new Solution();

// Example 1
int[] nums1 = {4, 5, 6, 7, 0, 1, 2};
int target1 = 0;
System.out.println(solution.search(nums1, target1)); // Output: 4

// Example 2
int[] nums2 = {4, 5, 6, 7, 0, 1, 2};
int target2 = 3;
System.out.println(solution.search(nums2, target2)); // Output: -1
}
}

Explanation:

1. Binary Search: We start with the normal binary search approach by calculating the
mid-point of the array.
2. Check for Target: If the middle element is the target, return its index.
3. Identify Sorted Half:
o If the left half of the array (from left to mid) is sorted, check if the target lies
within this half. If it does, adjust the right pointer to search this half.
o Otherwise, search in the right half (from mid + 1 to right).
4. Repeat: Continue adjusting the search range until the target is found or the range is
exhausted.
5. Return Result: If the target is found, return its index. If the loop exits without finding
the target, return -1.

Example:
For the input nums = [4, 5, 6, 7, 0, 1, 2] and target = 0:

 The algorithm identifies the sorted portion and determines that 0 must be in the right
half, eventually finding it at index 4.

For target = 3, it will determine that 3 is not present and return -1.

You said:

A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and
removing all non-alphanumeric characters, it reads the same forward and backward.
Alphanumeric characters include letters and numbers. Given a string s, return true if it is a
palindrome, or false otherwise. Example 1: Input: s = "A man, a plan, a canal: Panama"
Output: true Explanation: "amanaplanacanalpanama" is a palindrome. Example 2: Input: s =
"race a car" Output: false Explanation: "raceacar" is not a palindrome.
ChatGPT said:

ChatGPT

To solve the problem of determining if a string is a palindrome after normalizing it (i.e.,

converting all uppercase letters to lowercase and removing non-alphanumeric characters),
you can follow these steps:

1. Normalize the String: Convert the string to lowercase and remove any characters
that are not alphanumeric.
2. Check for Palindrome: Compare the cleaned string with its reverse to determine if it
reads the same forwards and backwards.

Here’s how you can implement this in Java:

java
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public class Solution {
public boolean isPalindrome(String s) {
// Step 1: Normalize the string by removing non-alphanumeric
characters and converting to lowercase
StringBuilder cleaned = new StringBuilder();

for (char c : s.toCharArray()) {

if (Character.isLetterOrDigit(c)) {
cleaned.append(Character.toLowerCase(c));
}
}

// Step 2: Check if the cleaned string is a palindrome

String cleanedStr = cleaned.toString();
String reversedStr = cleaned.reverse().toString();

return cleanedStr.equals(reversedStr);
}

public static void main(String[] args) {

Solution solution = new Solution();
 // Example 1
String s1 = "A man, a plan, a canal: Panama";
System.out.println(solution.isPalindrome(s1)); // Output: true

// Example 2
String s2 = "race a car";
System.out.println(solution.isPalindrome(s2)); // Output: false
}
}

Explanation:

1. Normalization:
o We iterate over each character in the string s.
o We use Character.isLetterOrDigit(c) to check if a character is
alphanumeric.
o If it is, we convert it to lowercase using Character.toLowerCase(c) and
append it to a StringBuilder.
2. Palindrome Check:
o After constructing the cleaned string, we reverse it using cleaned.reverse()
and compare it to the original cleaned string.
o If both strings are equal, the original string is a palindrome, so return true.
Otherwise, return false.

Example:

 For the input "A man, a plan, a canal: Panama":

o The normalized string is "amanaplanacanalpanama", which is a palindrome,
so the output is true.
 For the input "race a car":
o The normalized string is "raceacar", which is not a palindrome, so the output
is false.

You said:

Given a string S, the task is to print all the duplicate characters with their occurrences in the
given string. Example: Input: S = “geeksforgeeks” Output: e, count = 4 g, count = 2 k, count
= 2 s, count = 2 Explanation: g occurs 2 times in the string, k occurs 2 times in the string, s
occurs 2 times in the string while e occurs 4 times in the string rest of the characters have
unique occurrences.
ChatGPT said:

ChatGPT

To solve this problem, you can use a HashMap to keep track of the occurrences of each
character in the string. Then, you can iterate over the HashMap to print the characters that
have more than one occurrence.

Here’s how you can implement this in Java:

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

public class DuplicateCharacters {

public static void printDuplicateCharacters(String s) {
// Step 1: Create a HashMap to store character frequencies
HashMap<Character, Integer> charCountMap = new HashMap<>();

// Step 2: Traverse the string and count occurrences of each

character
for (char c : s.toCharArray()) {
charCountMap.put(c, charCountMap.getOrDefault(c, 0) + 1);
}

// Step 3: Print characters that appear more than once

for (Map.Entry<Character, Integer> entry : charCountMap.entrySet())
{
if (entry.getValue() > 1) {
System.out.println(entry.getKey() + ", count = " +
entry.getValue());
}
}
}

public static void main(String[] args) {

String s = "geeksforgeeks";
printDuplicateCharacters(s);
}
}

Explanation:

1. Character Frequency Count:

o We use a HashMap where the keys are characters and the values are their
frequencies in the string.
o As we iterate through the string, for each character, we either add it to the map
(if it’s not already present) or increment its count (if it is already present).
2. Printing Duplicates:
o After populating the HashMap, we iterate over the map entries.
o For each entry, we check if the value (frequency) is greater than 1, which
indicates that the character is a duplicate.
o We then print the character and its count.

Example:

For the input "geeksforgeeks":

 The characters e, g, k, and s appear more than once.

 The output will be:

Copy code
e, count = 4
g, count = 2
k, count = 2
s, count = 2
This approach ensures that all characters with more than one occurrence are printed with their
respective counts.

You said:

Remove Duplicates from an Unsorted Linked List Last Updated : 02 Apr, 2024 Given an
unsorted Linked List, the task is to remove duplicates from the list. Examples: Input:
linked_list = 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21 Output: 12 -> 11 -> 21 -> 41 -> 43
Explanation: Second occurrence of 12 and 21 are removed. Input: linked_list = 12 -> 11 ->
12 -> 21 -> 41 -> 43 -> 21 Output: 12 -> 11 -> 21 -> 41 -> 43
ChatGPT said:

ChatGPT

To remove duplicates from an unsorted linked list, we can use a HashSet to keep track of the
values that have already been seen as we iterate through the linked list. If a value has been
seen before, we skip the node; otherwise, we add it to the HashSet and continue.

Steps to Implement:

1. Initialize a HashSet to store the values of the nodes.

2. Iterate through the linked list while keeping track of the previous node.
3. For each node:
o If the value is not in the HashSet, add it to the set and move to the next node.
o If the value is already in the HashSet, remove the node by adjusting the next
pointer of the previous node.
4. Continue until you reach the end of the list.

Implementation in Java:
java
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import java.util.HashSet;

class LinkedList {
Node head;

class Node {
int data;
Node next;

Node(int d) {
data = d;
next = null;
}
}

// Function to remove duplicates from an unsorted linked list

void removeDuplicates() {
HashSet<Integer> seen = new HashSet<>();
Node current = head;
Node prev = null;
 while (current != null) {
// If the current node's data is already in the HashSet, remove
it
if (seen.contains(current.data)) {
prev.next = current.next;
} else {
// If the current node's data is not in the HashSet, add it
to the HashSet
seen.add(current.data);
prev = current;
}
current = current.next;
}
}

// Function to add a new node at the end of the list

public void append(int new_data) {
Node new_node = new Node(new_data);

if (head == null) {
head = new Node(new_data);
return;
}

new_node.next = null;

Node last = head;

while (last.next != null) {
last = last.next;
}

last.next = new_node;
}

// Function to print the linked list

public void printList() {
Node n = head;
while (n != null) {
System.out.print(n.data + " -> ");
n = n.next;
}
System.out.println("null");
}

public static void main(String[] args) {

LinkedList list = new LinkedList();

list.append(12);
list.append(11);
list.append(12);
list.append(21);
list.append(41);
list.append(43);
list.append(21);

System.out.println("Linked List before removing duplicates:");

list.printList();

list.removeDuplicates();

System.out.println("Linked List after removing duplicates:");

 list.printList();
}
}

Explanation:

1. Node Class: Represents a single node in the linked list, holding the data and a
reference to the next node.
2. removeDuplicates Method:
o We use a HashSet to track the values we've already encountered.
o As we traverse the list, we check each node's data:
 If the data is already in the set, we skip the node by adjusting the next
pointer of the previous node.
 If it's not in the set, we add the data to the set and move to the next
node.
3. append Method: Adds new nodes to the end of the linked list.
4. printList Method: Prints the linked list from the head to the last node.

Example Output:

For the input 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21:

 Before removing duplicates:

rust
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12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21 -> null

 After removing duplicates:

rust
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12 -> 11 -> 21 -> 41 -> 43 -> null

This method effectively removes duplicate nodes while preserving the order of the first
occurrence of each value in the list.

You said:

Given the head of a singly linked list, reverse the list, and return the reversed list. Example 1:
Input: head = [1,2,3,4,5] Output: [5,4,3,2,1] Example 2: Input: head = [1,2] Output: [2,1]
Example 3: Input: head = [] Output: [] Constraints: The number of nodes in the list is the
range [0, 5000]. -5000 <= Node.val <= 5000
ChatGPT said:

You’re giving feedback on an experimental version of ChatGPT.

Which response do you prefer? Responses may take a moment to load.
ChatGPT
Response 1
To reverse a singly linked list, you can iteratively reverse the pointers of each node so that the
head node becomes the tail, and the tail node becomes the head.

Steps to Reverse the Linked List:

1. Initialize three pointers:

o prev (initially null): This will eventually become the new head of the
reversed list.
o current (initially head): This is used to traverse the list.
o next (initially null): This temporarily stores the next node during the
traversal.
2. Iterate through the list:
o Store the next node of current in next.
o Reverse the current node's next pointer to point to prev.
o Move prev to current, and current to next.
o Continue until current becomes null (i.e., you’ve reached the end of the
list).
3. Return prev as the new head of the reversed list.

Implementation in Java:
java
Copy code
class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}

public class Solution {

public ListNode reverseList(ListNode head) {
ListNode prev = null;
ListNode current = head;
ListNode next = null;

while (current != null) {

// Store next node
next = current.next;
// Reverse the current node's pointer
current.next = prev;
// Move pointers one position ahead
prev = current;
current = next;
}

// Prev will be the new head after the loop ends

return prev;
}

public static void main(String[] args) {

// Creating the linked list 1 -> 2 -> 3 -> 4 -> 5
ListNode head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(3);
head.next.next.next = new ListNode(4);
head.next.next.next.next = new ListNode(5);
 Solution solution = new Solution();
ListNode reversedHead = solution.reverseList(head);

// Printing the reversed linked list

while (reversedHead != null) {
System.out.print(reversedHead.val + " ");
reversedHead = reversedHead.next;
}
}
}

Explanation:

 Node Class (ListNode): Represents a single node in the linked list, with an integer
value (val) and a reference to the next node (next).
 reverseList Method:
o We initialize three pointers: prev, current, and next.
o We traverse the list using the current pointer, reversing the links as we go.
o After the loop, prev points to the new head of the reversed list, which we
return.

Example:

 For the input linked list [1, 2, 3, 4, 5], the output after reversal will be [5, 4,
3, 2, 1].
 For the input linked list [1, 2], the output will be [2, 1].
 For an empty list [], the output will be [] (i.e., null).

This algorithm runs in O(n)O(n)O(n) time complexity, where nnn is the number of nodes in
the linked list, and it uses O(1)O(1)O(1) space complexity since it only requires a few
pointers.

I prefer this response

ChatGPT
Response 2

To reverse a singly linked list, you need to change the direction of the pointers between the
nodes. Instead of pointing to the next node, each node will point to the previous node. This
can be accomplished by iterating through the list and adjusting the pointers step by step.

Steps to Reverse the Linked List:

1. Initialize Three Pointers:

o prev to keep track of the previous node (initially null).
o current to keep track of the current node (initially head).
o nextNode to keep track of the next node in the list.
2. Iterate through the List:
o For each node, store the reference to the next node.
o Reverse the next pointer of the current node to point to prev.
o Move prev and current one step forward.
 3. Return the New Head:
o After the loop, prev will point to the new head of the reversed list.

Implementation in Java:
java
Copy code
class ListNode {
int val;
ListNode next;

ListNode(int val) {
this.val = val;
this.next = null;
}
}

public class Solution {

public ListNode reverseList(ListNode head) {
ListNode prev = null;
ListNode current = head;

while (current != null) {

ListNode nextNode = current.next; // Store the next node
current.next = prev; // Reverse the link
prev = current; // Move prev to current
current = nextNode; // Move to the next node
}

return prev; // prev will be the new head of the reversed list
}

// Helper function to print the linked list

public static void printList(ListNode head) {
ListNode temp = head;
while (temp != null) {
System.out.print(temp.val + " ");
temp = temp.next;
}
System.out.println();
}

public static void main(String[] args) {

Solution solution = new Solution();

// Example 1: 1 -> 2 -> 3 -> 4 -> 5

ListNode head1 = new ListNode(1);
head1.next = new ListNode(2);
head1.next.next = new ListNode(3);
head1.next.next.next = new ListNode(4);
head1.next.next.next.next = new ListNode(5);

System.out.println("Original List:");
printList(head1);

ListNode reversed1 = solution.reverseList(head1);

System.out.println("Reversed List:");
printList(reversed1);

// Example 2: 1 -> 2
 ListNode head2 = new ListNode(1);
head2.next = new ListNode(2);

System.out.println("Original List:");
printList(head2);

ListNode reversed2 = solution.reverseList(head2);

System.out.println("Reversed List:");
printList(reversed2);

// Example 3: Empty list []

ListNode head3 = null;

System.out.println("Original List:");
printList(head3);

ListNode reversed3 = solution.reverseList(head3);

System.out.println("Reversed List:");
printList(reversed3);
}
}

Explanation:

1. reverseList Method:
o prev is initialized to null because the new tail (old head) will point to null.
o current starts at the head of the list.
o In the while loop, for each node:
 nextNode stores the next node to ensure we don’t lose the reference.
 The current.next is reversed to point to prev.
 prev and current are moved forward to the next nodes.
o Once the loop is done, prev is the new head of the reversed list.
2. printList Method:
o This helper method prints out the linked list from the head to the end.

Example Output:

 Input: 1 -> 2 -> 3 -> 4 -> 5

 Output: 5 -> 4 -> 3 -> 2 -> 1
 Input: 1 -> 2
 Output: 2 -> 1
 Input: []
 Output: []

This solution effectively reverses the linked list in O(n)O(n)O(n) time complexity, where nnn
is the number of nodes in the list, and uses O(1)O(1)O(1) additional space, making it efficient
for large linked lists.

I prefer this response