Posts tagged as “prefix”

Given a string s. An awesome substring is a non-empty substring of s such that we can make any number of swaps in order to make it palindrome.

Return the length of the maximum length awesome substring of s.

Example 1:

Input: s = "3242415"
Output: 5
Explanation: "24241" is the longest awesome substring, we can form the palindrome "24142" with some swaps.


Example 2:

Input: s = "12345678"
Output: 1


Example 3:

Input: s = "213123"
Output: 6
Explanation: "213123" is the longest awesome substring, we can form the palindrome "231132" with some swaps.


Example 4:

Input: s = "00"
Output: 2


Constraints:

• 1 <= s.length <= 10^5
• s consists only of digits.

For a palindrome all digits must occurred even times expect one. We can use a 10 bit mask to track the occurrence of each digit for prefix s[0~i]. 0 is even, 1 is odd.

We use a hashtable to track the first index of each prefix state.
If s[0~i] and s[0~j] have the same state which means every digits in s[i+1~j] occurred even times (zero is also even) and it’s an awesome string. Then (j – (i+1) + 1) = j – i is the length of the palindrome. So far so good.

But we still need to consider the case when there is a digit with odd occurrence. We can enumerate all possible ones from 0 to 9, and temporarily flip the bit of the digit and see whether that state happened before.

fisrt_index[0] = -1, first_index[*] = inf
ans = max(ans, j – first_index[mask])

Time complexity: O(n)
Space complexity: O(2^10) = O(1)

Python3

Given a sentence that consists of some words separated by a single space, and a searchWord.

You have to check if searchWord is a prefix of any word in sentence.

Return the index of the word in sentence where searchWord is a prefix of this word (1-indexed).

If searchWord is a prefix of more than one word, return the index of the first word (minimum index). If there is no such word return -1.

prefix of a string S is any leading contiguous substring of S.

Example 1:

Input: sentence = "i love eating burger", searchWord = "burg"
Output: 4
Explanation: "burg" is prefix of "burger" which is the 4th word in the sentence.


Example 2:

Input: sentence = "this problem is an easy problem", searchWord = "pro"
Output: 2
Explanation: "pro" is prefix of "problem" which is the 2nd and the 6th word in the sentence, but we return 2 as it's the minimal index.


Example 3:

Input: sentence = "i am tired", searchWord = "you"
Output: -1
Explanation: "you" is not a prefix of any word in the sentence.


Example 4:

Input: sentence = "i use triple pillow", searchWord = "pill"
Output: 4


Example 5:

Input: sentence = "hello from the other side", searchWord = "they"
Output: -1


Constraints:

• 1 <= sentence.length <= 100
• 1 <= searchWord.length <= 10
• sentence consists of lowercase English letters and spaces.
• searchWord consists of lowercase English letters.

Solution 1: Brute Force

Time complexity: O(n)
Space complexity: O(1)

C++

Given an array of integers arr.

We want to select three indices ij and k where (0 <= i < j <= k < arr.length).

Let’s define a and b as follows:

• a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]
• b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]

Note that ^ denotes the bitwise-xor operation.

Return the number of triplets (ij and k) Where a == b.

Example 1:

Input: arr = [2,3,1,6,7]
Output: 4
Explanation: The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4)


Example 2:

Input: arr = [1,1,1,1,1]
Output: 10


Example 3:

Input: arr = [2,3]
Output: 0


Example 4:

Input: arr = [1,3,5,7,9]
Output: 3


Example 5:

Input: arr = [7,11,12,9,5,2,7,17,22]
Output: 8


Constraints:

• 1 <= arr.length <= 300
• 1 <= arr[i] <= 10^8

Solution 1: Brute Force (TLE)

Time complexity: O(n^4)
Space complexity: O(1)

Solution 2: Prefix XORs

Let xors[i] = arr[0] ^ arr[1] ^ … ^ arr[i-1]
arr[i] ^ arr[i + 1] ^ … ^ arr[j – 1] = (arr[0] ^ … ^ arr[j – 1]) ^ (arr[0] ^ … ^ arr[i-1]) = xors[j] ^ xors[i]

We then can compute a and b in O(1) time.

Time complexity: O(n^3)
Space complexity: O(n)

Solution 3: Prefix XORs II

a = arr[i] ^ arr[i + 1] ^ … ^ arr[j – 1]
b = arr[j] ^ arr[j + 1] ^ … ^ arr[k]
a == b => a ^ b == 0
XORs(i ~ k) == 0
XORS(0 ~ k) ^ XORs(0 ~ i – 1) = 0

Problem => find all pairs of (i – 1, k) such that xors[k+1] == xors[i]
For each pair (i – 1, k), there are k – i positions we can insert j.

Time complexity: O(n^2)
Space complexity: O(1)

Solution 3: HashTable

Similar to target sum, use a hashtable to store the frequency of each prefix xors.

Time complexity: O(n)
Space complexity: O(n)

C++

KMP Algorithm, KMP 字符串搜索算法

Time complexity: O(m+n)
Space complexity: O(m)

Applications

LeetCode 28. strStr()

C++

LeetCode 459. Repeated Substring Pattern

C++

1392. Longest Happy Prefix

C++

There are n soldiers standing in a line. Each soldier is assigned a unique rating value.

You have to form a team of 3 soldiers amongst them under the following rules:

• Choose 3 soldiers with index (ijk) with rating (rating[i]rating[j]rating[k]).
• A team is valid if:  (rating[i] < rating[j] < rating[k]) or (rating[i] > rating[j] > rating[k]) where (0 <= i < j < k < n).

Return the number of teams you can form given the conditions. (soldiers can be part of multiple teams).

Example 1:

Input: rating = [2,5,3,4,1]
Output: 3
Explanation: We can form three teams given the conditions. (2,3,4), (5,4,1), (5,3,1).


Example 2:

Input: rating = [2,1,3]
Output: 0
Explanation: We can't form any team given the conditions.


Example 3:

Input: rating = [1,2,3,4]
Output: 4


Constraints:

• n == rating.length
• 1 <= n <= 200
• 1 <= rating[i] <= 10^5

Solution 1: Brute Force

Time complexity: O(n^3)
Space complexity: O(1)

Solution 2: Math

For each soldier j, count how many soldiers on his left has smaller ratings as left[j], count how many soldiers his right side has larger ratings as right[j]

ans = sum(left[j] * right[j] + (j – left[j]) * (n – j – 1 * right[j])

Time complexity: O(n^2)
Space complexity: O(1)

C++

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