# Posts tagged as “hashtable”

You are given a 0-indexed array of positive integers nums. Find the number of triplets (i, j, k) that meet the following conditions:

• 0 <= i < j < k < nums.length
• nums[i]nums[j], and nums[k] are pairwise distinct.
• In other words, nums[i] != nums[j]nums[i] != nums[k], and nums[j] != nums[k].

Return the number of triplets that meet the conditions.

Example 1:

Input: nums = [4,4,2,4,3]
Output: 3
Explanation: The following triplets meet the conditions:
- (0, 2, 4) because 4 != 2 != 3
- (1, 2, 4) because 4 != 2 != 3
- (2, 3, 4) because 2 != 4 != 3
Since there are 3 triplets, we return 3.
Note that (2, 0, 4) is not a valid triplet because 2 > 0.


Example 2:

Input: nums = [1,1,1,1,1]
Output: 0
Explanation: No triplets meet the conditions so we return 0.


Constraints:

• 3 <= nums.length <= 100
• 1 <= nums[i] <= 1000

## Solution 1: Brute Force

Enumerate i, j, k.

Time complexity: O(n3)
Space complexity: O(1)

## C++

Given an integer array nums that does not contain any zeros, find the largest positive integer k such that -k also exists in the array.

Return the positive integer k. If there is no such integer, return -1.

Example 1:

Input: nums = [-1,2,-3,3]
Output: 3
Explanation: 3 is the only valid k we can find in the array.


Example 2:

Input: nums = [-1,10,6,7,-7,1]
Output: 7
Explanation: Both 1 and 7 have their corresponding negative values in the array. 7 has a larger value.


Example 3:

Input: nums = [-10,8,6,7,-2,-3]
Output: -1
Explanation: There is no a single valid k, we return -1.


Constraints:

• 1 <= nums.length <= 1000
• -1000 <= nums[i] <= 1000
• nums[i] != 0

## Solution 1: Hashtable

We can do in one pass by checking whether -x in the hashtable and update ans with abs(x) if so.

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

## Solution 2: Sorting

Sort the array by abs(x) in descending order.

[-1,10,6,7,-7,1] becomes = [-1, 1, 6, -7, 7, 10]

Check whether arr[i] = -arr[i-1].

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

## Solution 3: Two Pointers

Sort the array.

Let sum = nums[i] + nums[j], sum == 0, we find one pair, if sum < 0, ++i else –j.

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

## C++

You are given a 0-indexed string s consisting of only lowercase English letters, where each letter in s appears exactly twice. You are also given a 0-indexed integer array distance of length 26.

Each letter in the alphabet is numbered from 0 to 25 (i.e. 'a' -> 0'b' -> 1'c' -> 2, … , 'z' -> 25).

In a well-spaced string, the number of letters between the two occurrences of the ith letter is distance[i]. If the ith letter does not appear in s, then distance[i] can be ignored.

Return true if s is a well-spaced string, otherwise return false.

Example 1:

Input: s = "abaccb", distance = [1,3,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: true
Explanation:
- 'a' appears at indices 0 and 2 so it satisfies distance[0] = 1.
- 'b' appears at indices 1 and 5 so it satisfies distance[1] = 3.
- 'c' appears at indices 3 and 4 so it satisfies distance[2] = 0.
Note that distance[3] = 5, but since 'd' does not appear in s, it can be ignored.
Return true because s is a well-spaced string.


Example 2:

Input: s = "aa", distance = [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: false
Explanation:
- 'a' appears at indices 0 and 1 so there are zero letters between them.
Because distance[0] = 1, s is not a well-spaced string.


Constraints:

• 2 <= s.length <= 52
• s consists only of lowercase English letters.
• Each letter appears in s exactly twice.
• distance.length == 26
• 0 <= distance[i] <= 50

## Solution: Hashtable

Use a hastable to store the index of first occurrence of each letter.

Time complexity: O(n)
Space complexity: O(26)

## C++

Given a string s, partition the string into one or more substrings such that the characters in each substring are unique. That is, no letter appears in a single substring more than once.

Return the minimum number of substrings in such a partition.

Note that each character should belong to exactly one substring in a partition.

Example 1:

Input: s = "abacaba"
Output: 4
Explanation:
Two possible partitions are ("a","ba","cab","a") and ("ab","a","ca","ba").
It can be shown that 4 is the minimum number of substrings needed.


Example 2:

Input: s = "ssssss"
Output: 6
Explanation:
The only valid partition is ("s","s","s","s","s","s").


Constraints:

• 1 <= s.length <= 105
• s consists of only English lowercase letters.

## Solution: Greedy

Extend the cur string as long as possible unless a duplicate character occurs.

You can use hashtable / array or bitmask to mark whether a character has been seen so far.

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

## C++

Given an integer array nums, return the most frequent even element.

If there is a tie, return the smallest one. If there is no such element, return -1.

Example 1:

Input: nums = [0,1,2,2,4,4,1]
Output: 2
Explanation:
The even elements are 0, 2, and 4. Of these, 2 and 4 appear the most.
We return the smallest one, which is 2.

Example 2:

Input: nums = [4,4,4,9,2,4]
Output: 4
Explanation: 4 is the even element appears the most.


Example 3:

Input: nums = [29,47,21,41,13,37,25,7]
Output: -1
Explanation: There is no even element.


Constraints:

• 1 <= nums.length <= 2000
• 0 <= nums[i] <= 105

## Solution: Hashtable

Use a hashtable to store the frequency of even numbers.

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