added intersection of two linked list #1581

Linked_list/Intersection_of_Two_Linked_List/program.c

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+#include <stdio.h>
+#include <stdlib.h>
+
+struct ListNode {
+    int val;
+    struct ListNode *next;
+};
+
+// Helper function to create a new node
+struct ListNode* createNode(int value) {
+    struct ListNode* newNode = (struct ListNode*)malloc(sizeof(struct ListNode));
+    newNode->val = value;
+    newNode->next = NULL;
+    return newNode;
+}
+
+// Function to find the intersection node of two linked lists
+struct ListNode* getIntersectionNode(struct ListNode* headA, struct ListNode* headB) {
+    if (!headA || !headB) return NULL;
+
+    struct ListNode* ptrA = headA;
+    struct ListNode* ptrB = headB;
+
+    // Traverse both lists; when one pointer reaches the end, start from the beginning of the other list
+    while (ptrA != ptrB) {
+        ptrA = (ptrA == NULL) ? headB : ptrA->next;
+        ptrB = (ptrB == NULL) ? headA : ptrB->next;
+    }
+
+    // Either both pointers meet at the intersection node, or they meet at NULL
+    return ptrA;
+}
+
+// Helper function to print the linked list
+void printList(struct ListNode* head) {
+    while (head != NULL) {
+        printf("%d -> ", head->val);
+        head = head->next;
+    }
+    printf("NULL\n");
+}
+
+int main() {
+    struct ListNode* listA = createNode(4);
+    listA->next = createNode(1);
+    struct ListNode* intersectNode = createNode(9);
+    listA->next->next = intersectNode;
+    intersectNode->next = createNode(4);
+    intersectNode->next->next = createNode(5);
+
+    struct ListNode* listB = createNode(5);
+    listB->next = createNode(6);
+    listB->next->next = createNode(1);
+    listB->next->next->next = intersectNode;
+
+    printf("List A: ");
+    printList(listA);
+    printf("List B: ");
+    printList(listB);
+
+    struct ListNode* intersection = getIntersectionNode(listA, listB);
+
+    if (intersection) {
+        printf("Intersected at '%d'\n", intersection->val);
+    } else {
+        printf("No intersection\n");
+    }
+
+    return 0;
+}

Linked_list/Intersection_of_Two_Linked_List/readme.md

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+# Intersection of Two Linked Lists
+
+## Problem Statement
+
+Given the heads of two singly linked-lists, `headA` and `headB`, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return `NULL`.
+
+### Example
+
+**Input**:
+- `listA = [4,1,8,4,5]`
+- `listB = [5,6,1,8,4,5]`
+
+**Output**:
+- `Intersected at '8'`
+
+**Explanation**:
+- Both lists intersect at the node with value `8`.
+
+### Constraints
+- Each linked list is non-cyclical.
+- The intersection, if it exists, is a reference to a node, not a new node with the same value.
+
+## Approach
+
+### Two-Pointer Technique
+
+1. **Set up two pointers,** `ptrA` starting at the head of `listA` and `ptrB` starting at the head of `listB`.
+2. **Traverse the lists**:
+   - If a pointer reaches the end of one list, redirect it to the head of the other list.
+   - Continue moving both pointers forward one node at a time until they meet.
+3. **Why This Works**:
+   - When the lists intersect, both pointers will meet at the intersection node after traversing the combined length of both lists.
+   - If they don’t intersect, both pointers will eventually reach `NULL` at the same time.
+
+This method is efficient with a time complexity of `O(m + n)`, where `m` and `n` are the lengths of `listA` and `listB`. It also has `O(1)` space complexity as it modifies the list in place without additional data structures.