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The given linked list is 1 -> 2 -> 3 -> 4-> NULL. Then the reverse linked list is 4 -> 3 -> 2 -> 1 -> NULL and the head of the reversed linked list will be 4.
Can you solve this problem in O(N) time and O(1) space complexity?



insert(word) - To insert a string "word" in Trie
search(word) - To check if string "word" is present in Trie or not.
startsWith(word) - To check if there is any string in the Trie that starts with the given prefix string "word".
Type 1: To insert a string "word" in Trie.
1 word
Type 2: To check if the string "word" is present in Trie or not.
2 word
Type 3: To check if there is any string in the Trie that starts with the given prefix string "word".
3 word



Consider 0-based indexing.
Consider the array 1, 2, 3, 4, 5, 6
We can Jump from index 0 to index 1
Then we jump from index 1 to index 2
Then finally make a jump of 3 to reach index N-1
There is also another path where
We can Jump from index 0 to index 1
Then we jump from index 1 to index 3
Then finally make a jump of 2 to reach index N-1
So multiple paths may exist but we need to return the minimum number of jumps in a path to end which here is 3.


Suppose given number is ‘13’ so we return ‘3’ as the answer as we can write 13 = 3 ^ 2 + 3 ^ 1 + 3 ^ 0. '3' is the smallest number 'B for which 13 base 'B' is ‘1 1 1’. So, the 'NINJA BASE' of 'N' = 13 is 3.
You are not required to print anything explicitly. It has already been taken care of. Just implement the function.

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What is recursion?