package com.fishercoder.solutions;

import com.fishercoder.common.classes.ListNode;

import java.util.HashSet;
import java.util.Set;

/**
 * 160. Intersection of Two Linked Lists
 Write a program to find the node at which the intersection of two singly linked lists begins.


 For example, the following two linked lists:

 A:          a1 → a2
                     ↘
                     c1 → c2 → c3
                     ↗
 B:     b1 → b2 → b3
 begin to intersect at node c1.

 Example 1:
 A:             4 → 1
                     ↘
                     8 → 4 → 5
                     ↗
 B:          5 → 0 → 1


 Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
 Output: Reference of the node with value = 8
 Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect).
 From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5].
 There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.


 Example 2:
 A:         0 -> 9 → 1
                     ↘
                     2 → 4
                     ↗
 B:                 3

 Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
 Output: Reference of the node with value = 2
 Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect).
 From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4].
 There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.


 Example 3:
 A:          2 → 6 -> 4

 B:          1 -> 5

 Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
 Output: null
 Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5].
 Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
 Explanation: The two lists do not intersect, so return null.


 Notes:

 If the two linked lists have no intersection at all, return null.
 The linked lists must retain their original structure after the function returns.
 You may assume there are no cycles anywhere in the entire linked structure.
 Your code should preferably run in O(n) time and use only O(1) memory.
 */

public class _160 {

    public static class Solution1 {

        public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
            int lenA = findLen(headA);
            int lenB = findLen(headB);
            /**align headA and headB to the same starting point and then move together until we find the intersection point*/
            while (lenA < lenB) {
                headB = headB.next;
                lenB--;
            }

            while (lenB < lenA) {
                headA = headA.next;
                lenA--;
            }

            while (headA != headB) {
                headA = headA.next;
                headB = headB.next;
            }

            return headA;
        }

        private int findLen(ListNode head) {
            int len = 0;
            while (head != null) {
                head = head.next;
                len++;
            }
            return len;
        }

    }

    public static class Solution2 {
        /**
         * Most optimal solution:
         *
         * O(m+n) time
         * O(1) space
         * credit: https://discuss.leetcode.com/topic/28067/java-solution-without-knowing-the-difference-in-len*/
        public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
            if (headA == null || headB == null) {
                return null;
            }

            ListNode a = headA;
            ListNode b = headB;

            /**if a and b have different lengths, then it will stop the loop after second iteration*/
            while (a != b) {
                /**for the first iteration, it'll just reset the pointer to the head of another linkedlist*/
                a = a == null ? headB : a.next;
                b = b == null ? headA : b.next;
            }
            return a;
        }
    }

    public static class Solution3 {
        /**
         * O(m+n) time
         * O(Math.max(m, n)) space
         * */
        public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
            Set<ListNode> set = new HashSet<>();
            while (headA != null) {
                set.add(headA);
                headA = headA.next;
            }

            while (headB != null) {
                if (set.contains(headB)) {
                    return headB;
                }
                headB = headB.next;
            }
            return null;
        }
    }
}

@@ -131,2244 +131,8 @@
 t;%0A%0A
-/**%0A * 160. Intersection of Two Linked Lists%0A Write a program to find the node at which the intersection of two singly linked lists begins.%0A%0A%0A For example, the following two linked lists:%0A%0A A:          a1 %E2%86%92 a2%0A                     %E2%86%98%0A                     c1 %E2%86%92 c2 %E2%86%92 c3%0A                     %E2%86%97%0A B:     b1 %E2%86%92 b2 %E2%86%92 b3%0A begin to intersect at node c1.%0A%0A Example 1:%0A A:             4 %E2%86%92 1%0A                     %E2%86%98%0A                     8 %E2%86%92 4 %E2%86%92 5%0A                     %E2%86%97%0A B:          5 %E2%86%92 0 %E2%86%92 1%0A%0A%0A Input: intersectVal = 8, listA = %5B4,1,8,4,5%5D, listB = %5B5,0,1,8,4,5%5D, skipA = 2, skipB = 3%0A Output: Reference of the node with value = 8%0A Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect).%0A From the head of A, it reads as %5B4,1,8,4,5%5D. From the head of B, it reads as %5B5,0,1,8,4,5%5D.%0A There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.%0A%0A%0A Example 2:%0A A:         0 -%3E 9 %E2%86%92 1%0A                     %E2%86%98%0A                     2 %E2%86%92 4%0A                     %E2%86%97%0A B:                 3%0A%0A Input: intersectVal = 2, listA = %5B0,9,1,2,4%5D, listB = %5B3,2,4%5D, skipA = 3, skipB = 1%0A Output: Reference of the node with value = 2%0A Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect).%0A From the head of A, it reads as %5B0,9,1,2,4%5D. From the head of B, it reads as %5B3,2,4%5D.%0A There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.%0A%0A%0A Example 3:%0A A:          2 %E2%86%92 6 -%3E 4%0A%0A B:          1 -%3E 5%0A%0A Input: intersectVal = 0, listA = %5B2,6,4%5D, listB = %5B1,5%5D, skipA = 3, skipB = 2%0A Output: null%0A Input Explanation: From the head of A, it reads as %5B2,6,4%5D. From the head of B, it reads as %5B1,5%5D.%0A Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.%0A Explanation: The two lists do not intersect, so return null.%0A%0A%0A Notes:%0A%0A If the two linked lists have no intersection at all, return null.%0A The linked lists must retain their original structure after the function returns.%0A You may assume there are no cycles anywhere in the entire linked structure.%0A Your code should preferably run in O(n) time and use only O(1) memory.%0A */%0A%0A
 publ
@@ -1156,16 +1156,20 @@
        *
+ %3Cp%3E
 %0A       
@@ -1317,16 +1317,26 @@
 e-in-len
+%0A         
 */%0A     
@@ -2062,24 +2062,24 @@
 ) space%0A
+
         
  * */%0A  
@@ -2070,18 +2070,16 @@
         
- *
  */%0A    

refactor 160