# Posts tagged as “subsequence”

Given an integer array arr and an integer difference, return the length of the longest subsequence in arr which is an arithmetic sequence such that the difference between adjacent elements in the subsequence equals difference.

Example 1:

Input: arr = [1,2,3,4], difference = 1
Output: 4
Explanation: The longest arithmetic subsequence is [1,2,3,4].

Example 2:

Input: arr = [1,3,5,7], difference = 1
Output: 1
Explanation: The longest arithmetic subsequence is any single element.


Example 3:

Input: arr = [1,5,7,8,5,3,4,2,1], difference = -2
Output: 4
Explanation: The longest arithmetic subsequence is [7,5,3,1].


Constraints:

• 1 <= arr.length <= 10^5
• -10^4 <= arr[i], difference <= 10^4

## Solution: DP

dp[i] := max length of sequence ends with x
dp[x] = max(0, dp[x – diff]) + 1

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

## C++

Given two strings str1 and str2, return the shortest string that has both str1 and str2 as subsequences.  If multiple answers exist, you may return any of them.

(A string S is a subsequence of string T if deleting some number of characters from T (possibly 0, and the characters are chosen anywherefrom T) results in the string S.)

Example 1:

Input: str1 = "abac", str2 = "cab"
Output: "cabac"
Explanation:
str1 = "abac" is a substring of "cabac" because we can delete the first "c".
str2 = "cab" is a substring of "cabac" because we can delete the last "ac".
The answer provided is the shortest such string that satisfies these properties.


Note:

1. 1 <= str1.length, str2.length <= 1000
2. str1 and str2 consist of lowercase English letters.

## Solution: LCS

Find the LCS (longest common sub-sequence) of two strings, and insert unmatched characters into the LCS.

Time complexity: O(mn)
Space complexity: O(mn)

## C++

We are given an array A of N lowercase letter strings, all of the same length.

Now, we may choose any set of deletion indices, and for each string, we delete all the characters in those indices.

For example, if we have an array A = ["babca","bbazb"] and deletion indices {0, 1, 4}, then the final array after deletions is ["bc","az"].

Suppose we chose a set of deletion indices D such that after deletions, the final array has every element (row) in lexicographic order.

For clarity, A is in lexicographic order (ie. A <= A <= ... <= A[A.length - 1]), A is in lexicographic order (ie. A <= A <= ... <= A[A.length - 1]), and so on.

Return the minimum possible value of D.length.

Example 1:

Input: ["babca","bbazb"]
Output: 3
Explanation: After deleting columns 0, 1, and 4, the final array is A = ["bc", "az"].
Both these rows are individually in lexicographic order (ie. A <= A and A <= A).
Note that A > A - the array A isn't necessarily in lexicographic order.


Example 2:

Input: ["edcba"]
Output: 4
Explanation: If we delete less than 4 columns, the only row won't be lexicographically sorted.


Example 3:

Input: ["ghi","def","abc"]
Output: 0
Explanation: All rows are already lexicographically sorted.


Note:

1. 1 <= A.length <= 100
2. 1 <= A[i].length <= 100

# Solution: DP

dp[i] := max length of increasing sub-sequence (of all strings) ends with i-th letter.
dp[i] = max(dp[j] + 1) if all A[*][j] <= A[*][i], j < i
Time complexity: (n*L^2)
Space complexity: O(L)

# Problem

Given a string S, count the number of distinct, non-empty subsequences of S .

Since the result may be large, return the answer modulo 10^9 + 7.

Example 1:

Input: "abc"
Output: 7
Explanation: The 7 distinct subsequences are "a", "b", "c", "ab", "ac", "bc", and "abc".


Example 2:

Input: "aba"
Output: 6
Explanation: The 6 distinct subsequences are "a", "b", "ab", "ba", "aa" and "aba".


Example 3:

Input: "aaa"
Output: 3
Explanation: The 3 distinct subsequences are "a", "aa" and "aaa".

Note:

1. S contains only lowercase letters.
2. 1 <= S.length <= 2000

# Solution: DP

counts[i][j] := # of distinct sub sequences of s[1->i] and ends with letter j. (‘a'<= j <= ‘z’)

Initialization:

counts[*][*] = 0

Transition:

counts[i][j] = sum(counts[i-1]) + 1 if s[i] == j  else counts[i-1][j]

ans = sum(counts[n])

e.g. S = “abc”

counts = {‘a’ : 1}
counts = {‘a’ : 1, ‘b’ : 1 + 1 = 2}
counts = {‘a’ : 1, ‘b’ : 2, ‘c’: 1 + 2 + 1 = 4}
ans = sum(counts) = 1 + 2 + 4 = 7

Time complexity: O(N*26)

Space complexity: O(N*26) -> O(26)

# Problem

Given an array of integers A, consider all non-empty subsequences of A.

For any sequence S, let the width of S be the difference between the maximum and minimum element of S.

Return the sum of the widths of all subsequences of A.

As the answer may be very large, return the answer modulo 10^9 + 7.

Example 1:

Input: [2,1,3]
Output: 6
Explanation:
Subsequences are , , , [2,1], [2,3], [1,3], [2,1,3].
The corresponding widths are 0, 0, 0, 1, 1, 2, 2.
The sum of these widths is 6.


Note:

• 1 <= A.length <= 20000
• 1 <= A[i] <= 20000 # Solution: Math

Sort the array, for A[i]:

• i numbers <= A[i]. A[i] is the upper bound of 2^i subsequences.
• n – i – 1 numbers >= A[i]. A[i] is the lower bound of 2^(n – i – 1) subsequences.
• A[i] contributes A[i] * 2^i – A[i] * 2^(n – i – 1) to the ans.

$$ans = \sum\limits_{i=0}^{n-1}A_{i}2^{i} – A_{i}2^{n – i – 1} =\sum\limits_{i=0}^{n-1}(A_i – A_{n-i-1})2^{i}$$

Time complexity: O(nlogn)

Space complexity: O(1)

Time complexity: O(n)

Space complexity: O(n)

Counting sort

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