Press "Enter" to skip to content

Posts tagged as “counting”

花花酱 LeetCode 982. Triples with Bitwise AND Equal To Zero

Given an array of integers A, find the number of triples of indices (i, j, k) such that:

  • 0 <= i < A.length
  • 0 <= j < A.length
  • 0 <= k < A.length
  • A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.

Example 1:

Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2

Note:

  1. 1 <= A.length <= 1000
  2. 0 <= A[i] < 2^16

Solution: Counting

Time complexity: O(n^2 + n * max(A))
Space complexity: O(max(A))

C++

花花酱 LeetCode 115. Distinct Subsequences

Problem

Given a string S and a string T, count the number of distinct subsequences of S which equals T.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).

Example 1:

Input: S = "rabbbit", T = "rabbit"
Output: 3
Explanation:  As shown below, there are 3 ways you can generate "rabbit" from S. (The caret symbol ^ means the chosen letters) 
rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^

Example 2:

Input: S = "babgbag", T = "bag"
Output: 5
Explanation:  As shown below, there are 5 ways you can generate "bag" from S. (The caret symbol ^ means the chosen letters)
babgbag
^^ ^
babgbag
^^ ^
babgbag
^ ^^
babgbag
^ ^^
babgbag
^^^

Solution: DP

Time complexity: O(|s| * |t|)

Space complexity: O(|s| * |t|)

C++

Related Problems:

花花酱 LeetCode 576. Out of Boundary Paths

Problem

There is an m by n grid with a ball. Given the start coordinate (i,j) of the ball, you can move the ball to adjacent cell or cross the grid boundary in four directions (up, down, left, right). However, you can at most move N times. Find out the number of paths to move the ball out of grid boundary. The answer may be very large, return it after mod 109 + 7.

Example 1:

Input:m = 2, n = 2, N = 2, i = 0, j = 0
Output: 6
Explanation:

Example 2:

Input:m = 1, n = 3, N = 3, i = 0, j = 1
Output: 12
Explanation:

Note:

  1. Once you move the ball out of boundary, you cannot move it back.
  2. The length and height of the grid is in range [1,50].
  3. N is in range [0,50].

Solution

Time complexity: O(m*n + N * m * n)

Space complexity: O(N*m*n) -> O(m*n)

C++

Recursion with memorization

no loops

 

Related Problems

花花酱 LeetCode 552. Student Attendance Record II

Problem

Given a positive integer n, return the number of all possible attendance records with length n, which will be regarded as rewardable. The answer may be very large, return it after mod 109 + 7.

A student attendance record is a string that only contains the following three characters:

  1. ‘A’ : Absent.
  2. ‘L’ : Late.
  3. ‘P’ : Present.

A record is regarded as rewardable if it doesn’t contain more than one ‘A’ (absent) or more than two continuous ‘L’ (late).

Example 1:

Input: n = 2
Output: 8 
Explanation:
There are 8 records with length 2 will be regarded as rewardable:
"PP" , "AP", "PA", "LP", "PL", "AL", "LA", "LL"
Only "AA" won't be regarded as rewardable owing to more than one absent times. 

Note: The value of n won’t exceed 100,000.

Solution

C++

DFS w/ memorization (stack overflow)

DP

Time complexity: O(n)

Space complexity: O(1)

花花酱 LeetCode 383. Ransom Note

题目大意:给你一个字符串,问能否用它其中的字符组成另外一个字符串,每个字符只能使用一次。

Problem:

Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.

Each letter in the magazine string can only be used once in your ransom note.

Note:
You may assume that both strings contain only lowercase letters.

canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true

Solution: HashTable

Time complexity: O(n + m)

Space complexity: O(128)

C++

 

Python3