One97 Interview Experience | Set 3 (Backend/Node js Developer)

Swap nodes in a linked list without swapping data

Last Updated : 20 Sep, 2024

Given a singly linked list with two values x and y, the task is to swap two nodes having values x and y without swapping data.

Examples:

Input: 1->2->3->4->5, x = 2, y = 4
Output: 1->4->3->2->5

Input: 10->15->12->13->20->14, x = 10, y = 20
Output: 20->15->12->13->10->14

Possible cases to handle:

When swapping 2 nodes, x and y, in a linked list, there are a few cases that can arise:

x and y are adjacent to each other, and
- One of x or y is the head node of the list.
- None of x or y is either a head node or a tail node.
x and y are not adjacent to each other, and:
- One of x or y is the head node of the list.
- None of x or y is either a head node or a tail node.
x or y don’t even exist in the linked list.

Therefore, our solution must handle all these cases.

[Naive Approach] By Keeping Track of Previous Node – O(n) Time and O(1) Space:

The idea is to first search x and y nodes in the given linked list. If any of them is not present, then return. While searching for x and y, keep track of current and previous pointers. First change next of previous pointers, then change next of current pointers.

Below is the diagram of the case where either Node X or Node Y is the head node of the list.

Swap-nodes-in-a-linked-list-without-swapping-data_1

Swapping when Node X is the head of the List.

Below is the diagram of the case where Nodes X and Node Y are both not the head node of the list.

Swap-nodes-in-a-linked-list-without-swapping-data-2

Swapping when neither Node X or Node Y is the head of the list.

Below is the implementation of the above approach:

C++

// C++ program to swap the nodes of a linked list rather
// than swapping the data from the nodes.

#include <iostream>
using namespace std;

class Node {
public:
    int data;
    Node* next;
    Node(int x) {
        data = x;
        next = nullptr;
    }
};

// Function to swap nodes x and y in linked list by
// changing links and return the updated head 
Node* swapNodes(Node* head, int x, int y) {
  
    // Nothing to do if x and y are the same
    if (x == y) return head;

    Node *prevX = nullptr, *currX = nullptr;
    Node *prevY = nullptr, *currY = nullptr;
    Node *curr = head;

    // First loop to find x
    while (curr != nullptr) {
        if (curr->data == x) {
            currX = curr;
            break;
        }
        prevX = curr;
        curr = curr->next;
    }

    curr = head;

    // Second loop to find y
    while (curr != nullptr) {
        if (curr->data == y) {
            currY = curr;
            break;
        }
        prevY = curr;
        curr = curr->next;
    }

    // If either x or y is not present, nothing to do
    if (currX == nullptr || currY == nullptr) return head;

    // If x is not head of the linked list
    if (prevX != nullptr) {
        prevX->next = currY;
    } else { 
        head = currY;
    }

    // If y is not head of the linked list
    if (prevY != nullptr) {
        prevY->next = currX;
    } else {
        head = currX;
    }

    // Swap next pointers
    Node* temp = currY->next;
    currY->next = currX->next;
    currX->next = temp;

    return head;
}

void print(Node* curr) {
    while (curr != nullptr) {
        cout << curr->data << " ";
        curr = curr->next;
    }
    cout << endl;
}

int main() {
  
    // Constructed linked list:
    // 1->2->3->4->5
    Node* head = new Node(1);
    head->next = new Node(2);
    head->next->next = new Node(3);
    head->next->next->next = new Node(4);
    head->next->next->next->next = new Node(5);

    head = swapNodes(head, 4, 3);
    print(head);

    return 0;
}

C

// C program to swap the nodes of a linked list rather
// than swapping the data from the nodes.
#include <stdio.h>
#include <stdlib.h>

struct Node {
    int data;
    struct Node* next;
};

// Function to swap nodes x and y in linked list by
// changing links and return the updated head 
struct Node* swapNodes(struct Node* head, int x, int y) {
  
    // Nothing to do if x and y are the same
    if (x == y)
        return head;

    struct Node *prevX = NULL, *currX = NULL;
    struct Node *prevY = NULL, *currY = NULL;
    struct Node *curr = head;

    // First loop to find x
    while (curr != NULL) {
        if (curr->data == x) {
            currX = curr;
            break;
        }
        prevX = curr;
        curr = curr->next;
    }

    curr = head; 

    // Second loop to find y
    while (curr != NULL) {
        if (curr->data == y) {
            currY = curr;
            break;
        }
        prevY = curr;
        curr = curr->next;
    }

    // If either x or y is not present, nothing to do
    if (currX == NULL || currY == NULL)
        return head;

    // If x is not head of the linked list
    if (prevX != NULL)
        prevX->next = currY;
    else
        head = currY;

    // If y is not head of the linked list
    if (prevY != NULL)
        prevY->next = currX;
    else
        head = currX; 

    // Swap next pointers
    struct Node* temp = currY->next;
    currY->next = currX->next;
    currX->next = temp;

    return head;
}

void print(struct Node* curr) {
    while (curr != NULL) {
        printf("%d ", curr->data);
        curr = curr->next;
    }
    printf("\n");
}

struct Node* createNode(int x) {
    struct Node* node = 
      (struct Node*)malloc(sizeof(struct Node));
    node->data = x;
    node->next = NULL;
    return node;
}

int main() {
  
    // Constructed linked list: 1->2->3->4->5
    struct Node* head = createNode(1);
    head->next = createNode(2);
    head->next->next = createNode(3);
    head->next->next->next = createNode(4);
    head->next->next->next->next = createNode(5);

    head = swapNodes(head, 4, 3);
    print(head);

    return 0;
}

Java

// Java program to swap the nodes of a linked list rather
// than swapping the data from the nodes.
class Node {
    int data;
    Node next;

    Node(int data) {
        this.data = data;
        this.next = null;
    }
}

class GfG {

    // Function to swap nodes x and y in linked list by changing links
    static Node swapNodes(Node head, int x, int y) {
      
        // Nothing to do if x and y are the same
        if (x == y) {
            return head;
        }

        Node prevX = null, currX = null;
        Node prevY = null, currY = null;
        Node curr = head;

        // First loop to find x
        while (curr != null) {
            if (curr.data == x) {
                currX = curr;
                break;
            }
            prevX = curr;
            curr = curr.next;
        }

        curr = head; 

        // Second loop to find y
        while (curr != null) {
            if (curr.data == y) {
                currY = curr;
                break;
            }
            prevY = curr;
            curr = curr.next;
        }

        // If either x or y is not present, nothing to do
        if (currX == null || currY == null) {
            return head;
        }

        // If x is not head of the linked list
        if (prevX != null) {
            prevX.next = currY;
        } else {
            head = currY; 
        }

        // If y is not head of the linked list
        if (prevY != null) {
            prevY.next = currX;
        } else {
            head = currX;
        }

        // Swap next pointers
        Node temp = currY.next;
        currY.next = currX.next;
        currX.next = temp;

        return head;
    }

    static void printList(Node curr) {
        while (curr != null) {
            System.out.print(curr.data + " ");
            curr = curr.next;
        }
        System.out.println();
    }

    public static void main(String[] args) {
      
        // Constructed linked list: 1->2->3->4->5
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);
        head.next.next.next.next = new Node(5);

        head = swapNodes(head, 4, 3);
        printList(head);
    }
}

Python

# Python program to swap the nodes of a linked list rather
# than swapping the data from the nodes.
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

# Function to swap nodes x and y in linked
# list by changing links
def swapNodes(head, x, y):
  
    # Nothing to do if x and y are the same
    if x == y:
        return head

    prevX = None
    currX = head
    while currX and currX.data != x:
        prevX = currX
        currX = currX.next

    prevY = None
    currY = head
    while currY and currY.data != y:
        prevY = currY
        currY = currY.next

    # If either x or y is not present, nothing to do
    if currX is None or currY is None:
        return head

    # If x is not head of the linked list
    if prevX:
        prevX.next = currY
    else:
        head = currY 

    # If y is not head of the linked list
    if prevY:
        prevY.next = currX
    else:
        head = currX 

    # Swap next pointers
    currX.next, currY.next = currY.next, currX.next

    return head

def printList(curr):
    while curr:
        print(curr.data, end=" ")
        curr = curr.next
    print()

if __name__ == "__main__":
  
    # Constructed linked list: 1->2->3->4->5
    head = Node(1)
    head.next = Node(2)
    head.next.next = Node(3)
    head.next.next.next = Node(4)
    head.next.next.next.next = Node(5)

    head = swapNodes(head, 4, 3)
    printList(head)

C#

// C# program to swap the nodes of a linked list rather
// than swapping the data from the nodes.
using System;

class Node {
    public int data;
    public Node next;

    public Node(int data) {
        this.data = data;
        this.next = null;
    }
}

class GfG {

    // Function to swap nodes x and y in linked list by changing links
    static Node swapNodes(Node head, int x, int y) {

        // Nothing to do if x and y are the same
        if (x == y)
            return head;

        Node prevX = null;
        Node currX = head;

        // Find currX and prevX
        while (currX != null && currX.data != x) {
            prevX = currX;
            currX = currX.next;
        }

        Node prevY = null;
        Node currY = head;

        // Find currY and prevY
        while (currY != null && currY.data != y) {
            prevY = currY;
            currY = currY.next;
        }

        // If either x or y is not present, nothing to do
        if (currX == null || currY == null)
            return head;

        // If x is not head of the linked list
        if (prevX != null)
            prevX.next = currY;
        else
            head = currY;

        // If y is not head of the linked list
        if (prevY != null)
            prevY.next = currX;
        else
            head = currX;

        // Swap next pointers
        Node temp = currY.next;
        currY.next = currX.next;
        currX.next = temp;

        return head;
    }

    static void printList(Node curr) {
        while (curr != null) {
            Console.Write(curr.data + " ");
            curr = curr.next;
        }
        Console.WriteLine();
    }

    static void Main(string[] args) {
      
        // Constructed linked list: 1->2->3->4->5
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);
        head.next.next.next.next = new Node(5);

        head = swapNodes(head, 4, 3);
        printList(head); 
    }
}

JavaScript

// Javascript program to swap the nodes of a linked list rather
// than swapping the data from the nodes.
class Node {
    constructor(data) {
        this.data = data;
        this.next = null;
    }
}

// Function to swap nodes x and y in linked 
// list by changing links
function swapNodes(head, x, y) {

    // Nothing to do if x and y are the same
    if (x === y) return head;

    let prevX = null, currX = head;

    // Find currX and prevX
    while (currX && currX.data !== x) {
        prevX = currX;
        currX = currX.next;
    }

    let prevY = null, currY = head;

    // Find currY and prevY
    while (currY && currY.data !== y) {
        prevY = currY;
        currY = currY.next;
    }

    // If either x or y is not present, nothing to do
    if (!currX || !currY) return head;

    // If x is not head of the linked list
    if (prevX) prevX.next = currY;
    else head = currY;

    // If y is not head of the linked list
    if (prevY) prevY.next = currX;
    else head = currX; 

    // Swap next pointers
    let temp = currY.next;
    currY.next = currX.next;
    currX.next = temp;

    return head;
}

function printList(curr) {
    while (curr) {
        console.log(curr.data + " ");
        curr = curr.next;
    }
    console.log();
}

// Constructed linked list:
// 1->2->3->4->5
let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);

head = swapNodes(head, 4, 3);
printList(head);

Output

1 2 4 3 5

Time Complexity: O(n), where n is the number of nodes in linked list.
Auxiliary Space: O(1)

[Efficient Approach] Using Single traversal – O(n) Time and O(1) Space:

The above code can be optimized to search x and y in a single traversal. Two loops are used to keep the program simple.

Follow the steps below to solve the problem:

If x and y are the same, return as no need to swap.
Start traversing the List to search for x and y nodes by keeping track of the previous nodes prevX and prevY.
If either x or y is not found in the list, return without making any changes.
else, update connections:
- If x is not the head, set prevX->next to currY. Otherwise, set currY as the new head.
- If y is not the head, set prevY->next to currX. Otherwise, set currX as the new head.
Swap the next pointers of currX and currY to complete the swap.

Below is the implementation of the above approach:

C++

// C++ program to swaps the nodes of linked list rather
// than swapping the data from the nodes.

#include <iostream>
using namespace std;

class Node {
public:
    int data;
    Node* next;
  	Node(int x) {
      	data = x;
      	next = nullptr;
    }
};

// Function to swap nodes x and y in linked list by
// changing links and return the updated head 
Node* swapNodes(Node* head, int x, int y) {
  
    // Nothing to do if x and y are same
    if (x == y)
        return head;

    Node *prevX = nullptr, *currX = nullptr;
    Node *prevY = nullptr, *currY = nullptr;
    Node *prev = nullptr, *curr = head;

    // Single loop to find both x and y
    while (curr != nullptr) {
        if (curr->data == x) {
            prevX = prev;
            currX = curr;
        } else if (curr->data == y) {
            prevY = prev;
            currY = curr;
        }
        prev = curr;
        curr = curr->next;
    }

    // If either x or y is not present, nothing to do
    if (currX == nullptr || currY == nullptr)
        return head;

    // If x is not head of the linked list
    if (prevX != nullptr)
        prevX->next = currY;
    else // Else make y the new head
        head = currY;

    // If y is not head of the linked list
    if (prevY != nullptr)
        prevY->next = currX;
    else 
      	// Else make x the new head
        head = currX;

    // Swap next pointers
    Node* temp = currY->next;
    currY->next = currX->next;
    currX->next = temp;

    return head;
}

void print(Node* curr) {
  
    while (curr != nullptr) {
        cout << curr->data << " ";
        curr = curr->next;
    }
}

int main() {
  
    // Constructed linked list:
    // 1->2->3->4->5
    Node* head = new Node(1);
    head->next = new Node(2);
    head->next->next = new Node(3);
    head->next->next->next = new Node(4);
    head->next->next->next->next = new Node(5);

    head = swapNodes(head, 4, 3);
    print(head);

    return 0;
}

C

// C program to swaps the nodes of linked list rather
// than swapping the data from the nodes.

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

struct Node {
    int data;
    struct Node* next;
};

// Function to swap nodes x and y in linked list by
// changing links and return the updated head 
struct Node* swapNodes(struct Node* head, int x, int y) {
  
    // Nothing to do if x and y are same
    if (x == y)
        return head;

    struct Node *prevX = NULL, *currX = NULL;
    struct Node *prevY = NULL, *currY = NULL;
    struct Node *prev = NULL, *curr = head;

    // Single loop to find both x and y
    while (curr != NULL) {
        if (curr->data == x) {
            prevX = prev;
            currX = curr;
        } else if (curr->data == y) {
            prevY = prev;
            currY = curr;
        }
        prev = curr;
        curr = curr->next;
    }

    // If either x or y is not present, nothing to do
    if (currX == NULL || currY == NULL)
        return head;

    // If x is not head of the linked list
    if (prevX != NULL)
        prevX->next = currY;
    else
        head = currY;

    // If y is not head of the linked list
    if (prevY != NULL)
        prevY->next = currX;
    else 
        head = currX;

    // Swap next pointers
    struct Node* temp = currY->next;
    currY->next = currX->next;
    currX->next = temp;

    return head;
}

void print(struct Node* curr) {
  
    while (curr != NULL) {
        printf("%d ", curr->data);
        curr = curr->next;
    }
}

struct Node* createNode(int x) {
    struct Node* node = 
      (struct Node*)malloc(sizeof(struct Node));
    node->data = x;
    node->next = NULL;
    return node;
}

int main() {
  
    // Constructed linked list:
    // 1->2->3->4->5
    struct Node* head = createNode(1);
    head->next = createNode(2);
    head->next->next = createNode(3);
    head->next->next->next = createNode(4);
    head->next->next->next->next = createNode(5);

    head = swapNodes(head, 4, 3);
    print(head);

    return 0;
}

Java

// Java program to swaps the nodes of linked list rather
// than swapping the data from the nodes.

class Node {
    int data;
    Node next;

    Node(int data) {
        this.data = data;
        this.next = null;
    }
}

class GfG {
  
    // Function to swap nodes x and y in linked list by
    // changing links and return the updated head
    static Node swapNodes(Node head, int x, int y) {
      
        // Nothing to do if x and y are same
        if (x == y)
            return head;

        Node prevX = null, currX = null;
        Node prevY = null, currY = null;
        Node prev = null, curr = head;

        // Single loop to find both x and y
        while (curr != null) {
            if (curr.data == x) {
                prevX = prev;
                currX = curr;
            } else if (curr.data == y) {
                prevY = prev;
                currY = curr;
            }
            prev = curr;
            curr = curr.next;
        }

        // If either x or y is not present, nothing to do
        if (currX == null || currY == null)
            return head;

        // If x is not head of the linked list
        if (prevX != null)
            prevX.next = currY;
        else
            // Else make y the new head
            head = currY;

        // If y is not head of the linked list
        if (prevY != null)
            prevY.next = currX;
        else
            // Else make x the new head
            head = currX;

        // Swap next pointers
        Node temp = currY.next;
        currY.next = currX.next;
        currX.next = temp;

        return head;
    }

    static void printList(Node curr) {
        while (curr != null) {
            System.out.print(curr.data + " ");
            curr = curr.next;
        }
    }

    public static void main(String[] args) {

        // Constructed linked list:
        // 1->2->3->4->5
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);
        head.next.next.next.next = new Node(5);

        head = swapNodes(head, 4, 3);
        printList(head);
    }
}

Python

# Python program to swaps the nodes of linked list rather
# han swapping the data from the nodes.

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

# Function to swap nodes x and y in linked list by
# changing links and return the updated head 
def swapNodes(head, x, y):
  
    # Nothing to do if x and y are same
    if x == y:
        return head

    prevX = None
    currX = None
    prevY = None
    currY = None
    prev = None
    curr = head

    # Single loop to find both x and y
    while curr:
        if curr.data == x:
            prevX = prev
            currX = curr
        elif curr.data == y:
            prevY = prev
            currY = curr
        prev = curr
        curr = curr.next

    # If either x or y is not present, nothing to do
    if currX is None or currY is None:
        return head

    # If x is not head of the linked list
    if prevX is not None:
        prevX.next = currY
    else: 
        head = currY

    # If y is not head of the linked list
    if prevY is not None:
        prevY.next = currX
    else: 
      
        # Else make x the new head
        head = currX

    # Swap next pointers
    temp = currY.next
    currY.next = currX.next
    currX.next = temp

    return head

def printList(curr):
    while curr:
        print(curr.data, end=" ")
        curr = curr.next

if __name__ == "__main__":
  
    # Constructed linked list:
    # 1->2->3->4->5
    head = Node(1)
    head.next = Node(2)
    head.next.next = Node(3)
    head.next.next.next = Node(4)
    head.next.next.next.next = Node(5)

    head = swapNodes(head, 4, 3)
    printList(head)

C#

// C# program to swaps the nodes of linked list rather
// than swapping the data from the nodes.

using System;

class Node {
    public int data;
    public Node next;

    public Node(int data) {
        this.data = data;
        this.next = null;
    }
}

class GfG {
  
    // Function to swap nodes x and y in linked list by
    // changing links and return the updated head
    static Node swapNodes(Node head, int x, int y) {
      
        // Nothing to do if x and y are same
        if (x == y)
            return head;

        Node prevX = null, currX = null;
        Node prevY = null, currY = null;
        Node prev = null, curr = head;

        // Single loop to find both x and y
        while (curr != null) {
            if (curr.data == x) {
                prevX = prev;
                currX = curr;
            } else if (curr.data == y) {
                prevY = prev;
                currY = curr;
            }
            prev = curr;
            curr = curr.next;
        }

        // If either x or y is not present, nothing to do
        if (currX == null || currY == null)
            return head;

        // If x is not head of the linked list
        if (prevX != null)
            prevX.next = currY;
        else
            // Else make y the new head
            head = currY;

        // If y is not head of the linked list
        if (prevY != null)
            prevY.next = currX;
        else
            // Else make x the new head
            head = currX;

        // Swap next pointers
        Node temp = currY.next;
        currY.next = currX.next;
        currX.next = temp;

        return head;
    }

    static void printList(Node curr) {
        while (curr != null) {
            Console.Write(curr.data + " ");
            curr = curr.next;
        }
    }

    static void Main(string[] args) {
      
        // Constructed linked list:
        // 1->2->3->4->5
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);
        head.next.next.next.next = new Node(5);

        head = swapNodes(head, 4, 3);
        printList(head);
    }
}

JavaScript

// Javascript program to swaps the nodes of linked list rather
// than swapping the data from the nodes.

class Node {
    constructor(data) {
        this.data = data;
        this.next = null;
    }
}

// Function to swap nodes x and y in linked list by
// changing links and return the updated head 
function swapNodes(head, x, y) {

    // Nothing to do if x and y are same
    if (x === y) return head;

    let prevX = null, currX = null;
    let prevY = null, currY = null;
    let prev = null, curr = head;

    // Single loop to find both x and y
    while (curr) {
        if (curr.data === x) {
            prevX = prev;
            currX = curr;
        } else if (curr.data === y) {
            prevY = prev;
            currY = curr;
        }
        prev = curr;
        curr = curr.next;
    }

    // If either x or y is not present, nothing to do
    if (!currX || !currY) return head;

    // If x is not head of the linked list
    if (prevX) prevX.next = currY;
    else head = currY;

    // If y is not head of the linked list
    if (prevY) prevY.next = currX;
    else head = currX;

    // Swap next pointers
    let temp = currY.next;
    currY.next = currX.next;
    currX.next = temp;

    return head;
}

function printList(curr) {
    while (curr) {
        console.log(curr.data + " ");
        curr = curr.next;
    }
}

// Constructed linked list:
// 1->2->3->4->5
let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);

head = swapNodes(head, 4, 3);
printList(head);

Output

1 2 4 3 5

Time Complexity: O(n), where n is the number of nodes in linked list.
Auxiliary Space: O(1)

One97 Interview Experience | Set 3 (Backend/Node js Developer)

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