Merge pull request #58 from Samyak137/main

linked list important questions

Author: gautamankoji

Diff for: interview_questions/linked-list/merge_list.cpp

@@ -0,0 +1,71 @@
+// Title: Merge Two Sorted Linked Lists
+
+// Problem: Given two sorted linked lists, merge them into a single sorted linked list.
+// Approach: Use a two-pointer approach to compare nodes and merge them into a new list.
+
+
+#include <iostream>
+using namespace std;
+
+struct ListNode {
+    int val;
+    ListNode* next;
+    ListNode(int x) : val(x), next(nullptr) {}
+};
+
+ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
+    if (!l1) return l2;
+    if (!l2) return l1;
+
+    if (l1->val < l2->val) {
+        l1->next = mergeTwoLists(l1->next, l2);
+        return l1;
+    } else {
+        l2->next = mergeTwoLists(l1, l2->next);
+        return l2;
+    }
+}
+
+void printList(ListNode* head) {
+    while (head) {
+        cout << head->val << " ";
+        head = head->next;
+    }
+    cout << endl;
+}
+
+ListNode* createList(int n) {
+    int value;
+    ListNode* head = nullptr;
+    ListNode* tail = nullptr;
+    cout << "Enter values in sorted order: ";
+    for (int i = 0; i < n; ++i) {
+        cin >> value;
+        ListNode* newNode = new ListNode(value);
+        if (!head) {
+            head = newNode;
+            tail = newNode;
+        } else {
+            tail->next = newNode;
+            tail = newNode;
+        }
+    }
+    return head;
+}
+
+int main() {
+    int n1, n2;
+    cout << "Enter number of nodes in first sorted list: ";
+    cin >> n1;
+    ListNode* l1 = createList(n1);
+
+    cout << "Enter number of nodes in second sorted list: ";
+    cin >> n2;
+    ListNode* l2 = createList(n2);
+
+    ListNode* mergedHead = mergeTwoLists(l1, l2);
+    cout << "Merged list: ";
+    printList(mergedHead);
+
+    return 0;
+}

Diff for: interview_questions/linked-list/middle_of_linked_list.cpp

@@ -0,0 +1,51 @@
+// Title: Find the Middle of a Linked List
+
+// Problem: Find the middle node of a linked list. If there are two middle nodes, return the second one.
+// Approach: Use two pointers, one moving one step at a time (slow) and the other moving two steps at a time (fast).
+
+
+#include <iostream>
+using namespace std;
+
+struct ListNode {
+    int val;
+    ListNode* next;
+    ListNode(int x) : val(x), next(nullptr) {}
+};
+
+ListNode* findMiddle(ListNode* head) {
+    ListNode* slow = head;
+    ListNode* fast = head;
+    while (fast && fast->next) {
+        slow = slow->next;
+        fast = fast->next->next;
+    }
+    return slow;
+}
+
+int main() {
+    int n, value;
+    cout << "Enter the number of nodes in the list: ";
+    cin >> n;
+
+    ListNode* head = nullptr;
+    ListNode* tail = nullptr;
+
+    cout << "Enter values for each node: ";
+    for (int i = 0; i < n; ++i) {
+        cin >> value;
+        ListNode* newNode = new ListNode(value);
+        if (!head) {
+            head = newNode;
+            tail = newNode;
+        } else {
+            tail->next = newNode;
+            tail = newNode;
+        }
+    }
+
+    ListNode* middleNode = findMiddle(head);
+    cout << "The middle node's value is: " << middleNode->val << endl;
+
+    return 0;
+}

Diff for: interview_questions/linked-list/remove_nth_node_from_end.cpp

@@ -0,0 +1,74 @@
+// Title: Remove N-th Node from End of List
+// Problem: Given the head of a linked list, remove the N-th node from the end of the list.
+
+
+#include <iostream>
+using namespace std;
+
+struct ListNode {
+    int val;
+    ListNode* next;
+    ListNode(int x) : val(x), next(nullptr) {}
+};
+
+ListNode* removeNthFromEnd(ListNode* head, int n) {
+    ListNode* dummy = new ListNode(0);
+    dummy->next = head;
+    ListNode* first = dummy;
+    ListNode* second = dummy;
+
+    // Move first n+1 steps ahead
+    for (int i = 0; i <= n; ++i) {
+        first = first->next;
+    }
+
+    // Move both pointers until first reaches the end
+    while (first) {
+        first = first->next;
+        second = second->next;
+    }
+
+    // Skip the desired node
+    second->next = second->next->next;
+
+    return dummy->next;
+}
+
+void printList(ListNode* head) {
+    while (head) {
+        cout << head->val << " ";
+        head = head->next;
+    }
+    cout << endl;
+}
+
+int main() {
+    int n, nodes, value;
+    cout << "Enter the number of nodes: ";
+    cin >> nodes;
+
+    ListNode* head = nullptr;
+    ListNode* tail = nullptr;
+
+    cout << "Enter values for each node: ";
+    for (int i = 0; i < nodes; ++i) {
+        cin >> value;
+        ListNode* newNode = new ListNode(value);
+        if (!head) {
+            head = newNode;
+            tail = newNode;
+        } else {
+            tail->next = newNode;
+            tail = newNode;
+        }
+    }
+
+    cout << "Enter the value of n (from end): ";
+    cin >> n;
+
+    head = removeNthFromEnd(head, n);
+    cout << "List after removing " << n << "-th node from end: ";
+    printList(head);
+
+    return 0;
+}

Diff for: interview_questions/linked-list/reverse_list.cpp

@@ -0,0 +1,65 @@
+// Title: Reverse List
+// Problem: Given list have to print in reverse order
+
+
+#include <iostream>
+using namespace std;
+
+struct ListNode {
+    int val;
+    ListNode* next;
+    ListNode(int x) : val(x), next(nullptr) {}
+};
+
+ListNode* reverseList(ListNode* head) {
+    ListNode* prev = nullptr;
+    ListNode* curr = head;
+    while (curr) {
+        ListNode* nextTemp = curr->next;
+        curr->next = prev;
+        prev = curr;
+        curr = nextTemp;
+    }
+    return prev;
+}
+
+void printList(ListNode* head) {
+    while (head) {
+        cout << head->val << " ";
+        head = head->next;
+    }
+    cout << endl;
+}
+
+int main() {
+    int n, value;
+    cout << "Enter the number of nodes: ";
+    cin >> n;
+
+    ListNode* head = nullptr;
+    ListNode* tail = nullptr;
+
+    cout << "Enter the values: ";
+    for (int i = 0; i < n; ++i) {
+        cin >> value;
+        ListNode* newNode = new ListNode(value);
+        if (!head) {
+            head = newNode;
+            tail = newNode;
+        } else {
+            tail->next = newNode;
+            tail = newNode;
+        }
+    }
+
+    cout << "Original list: ";
+    printList(head);
+
+    ListNode* reversedHead = reverseList(head);
+
+
+    cout << "Reversed list: ";
+    printList(reversedHead);
+
+    return 0;
+}

Diff for: interview_questions/linked-list/reverse_node_in_kGroup.cpp

@@ -0,0 +1,74 @@
+// Title: Reverse Nodes in k-Group
+
+// Problem: Given a linked list, reverse nodes in groups of k.
+// Approach: Reverse every k nodes iteratively, then move to the next k nodes.
+
+#include <iostream>
+using namespace std;
+
+struct ListNode {
+    int val;
+    ListNode* next;
+    ListNode(int x) : val(x), next(nullptr) {}
+};
+
+ListNode* reverseKGroup(ListNode* head, int k) {
+    ListNode* curr = head;
+    int count = 0;
+    while (curr && count < k) {
+        curr = curr->next;
+        count++;
+    }
+    if (count < k) return head;
+
+    curr = head;
+    ListNode* prev = nullptr;
+    ListNode* next = nullptr;
+    for (int i = 0; i < k; ++i) {
+        next = curr->next;
+        curr->next = prev;
+        prev = curr;
+        curr = next;
+    }
+    head->next = reverseKGroup(curr, k);
+    return prev;
+}
+
+void printList(ListNode* head) {
+    while (head) {
+        cout << head->val << " ";
+        head = head->next;
+    }
+    cout << endl;
+}
+
+int main() {
+    int n, k, value;
+    cout << "Enter the number of nodes: ";
+    cin >> n;
+
+    ListNode* head = nullptr;
+    ListNode* tail = nullptr;
+
+    cout << "Enter values for each node: ";
+    for (int i = 0; i < n; ++i) {
+        cin >> value;
+        ListNode* newNode = new ListNode(value);
+        if (!head) {
+            head = newNode;
+            tail = newNode;
+        } else {
+            tail->next = newNode;
+            tail = newNode;
+        }
+    }
+
+    cout << "Enter value of k: ";
+    cin >> k;
+
+    ListNode* newHead = reverseKGroup(head, k);
+    cout << "List after reversing every k group: ";
+    printList(newHead);
+
+    return 0;
+}