Posts tagged as “subsequence”

花花酱 LeetCode 891. Sum of Subsequence Widths

By zxi on August 20, 2018

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 [1], [2], [3], [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)

// Author: Huahua

// Running time: 56 ms

class Solution {

public:

int sumSubseqWidths(vector<int>& A) {

constexpr long kMod = 1e9 + 7;

const int n = A.size();

sort(begin(A), end(A));

long ans = 0;

long p = 1;

for (int i = 0; i < n; ++i) {

ans = (ans + (A[i] - A[n - i - 1]) * p) % kMod;

p = (p << 1) % kMod;

}

return (ans + kMod) % kMod;

}

};

Time complexity: O(n)

Space complexity: O(n)

Counting sort

// Author: Huahua

// Running time: 16 / 44 ms (beats 100%)

static auto x = [](){std::ios::sync_with_stdio(false);cin.tie(nullptr);return nullptr;}();

class Solution {

public:

int sumSubseqWidths(vector<int>& A) {

constexpr long kMod = 1e9 + 7;

const int n = A.size();

countingSort(A);

long ans = 0;

long p = 1;

for (int i = 0; i < n; ++i) {

ans = (ans + (A[i] - A[n - i - 1]) * p) % kMod;

p = (p << 1) % kMod;

}

return (ans + kMod) % kMod;

}

private:

void countingSort(vector<int>& A) {

int max_v = INT_MIN;

int min_v = INT_MAX;

for (int a : A) {

if (a > max_v) max_v = a;

if (a < min_v) min_v = a;

}

vector<int> c(max_v - min_v + 1);

for (int a : A)

++c[a - min_v];

int i = 0;

int j = 0;

while (i < c.size()) {

if (--c[i] >= 0)

A[j++] = i + min_v;

else

++i;

}

};

花花酱 LeetCode 115. Distinct Subsequences

By zxi on July 19, 2018

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++

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int numDistinct(string s, string t) {

int ls = s.length();

int lt = t.length();

vector<vector<int>> dp(lt + 1, vector<int>(ls + 1));

fill(begin(dp[0]), end(dp[0]), 1);

for (int i = 1; i <= lt; ++i)

for (int j = 1; j <= ls; ++j)

dp[i][j] = dp[i][j - 1] + (t[i - 1] == s[j - 1] ? dp[i - 1][j - 1] : 0);

return dp[lt][ls];

}

};

Solution: DFS

Time complexity: O(2^n)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 140 ms (<99.88%)

class Solution {

public:

vector<vector<int>> findSubsequences(vector<int>& nums) {

vector<vector<int>> ans;

vector<int> cur;

dfs(nums, 0, cur, ans);

return ans;

}

private:

void dfs(const vector<int>& nums, int s, vector<int>& cur, vector<vector<int>>& ans) {

unordered_set<int> seen;

for (int i = s; i < nums.size(); ++i) {

if (!cur.empty() && nums[i] < cur.back()) continue;

// each number can be used only once at the same depth.

if (seen.count(nums[i])) continue;

seen.insert(nums[i]);

cur.push_back(nums[i]);

if (cur.size() > 1)

ans.push_back(cur);

dfs(nums, i + 1, cur, ans);

cur.pop_back();

}

};

花花酱 LeetCode 334. Increasing Triplet Subsequence

By zxi on May 28, 2018

Problem

题目大意：判断一个数组中是否存在长度为3的单调递增子序列。

Formally the function should:

Return true if there exists i, j, k
such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.

Your algorithm should run in O(n) time complexity and O(1) space complexity.

Examples:
Given [1, 2, 3, 4, 5],
return true.

Given [5, 4, 3, 2, 1],
return false.

Credits:
Special thanks to @DjangoUnchained for adding this problem and creating all test cases.

Solution: Greedy

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 7 ms

class Solution {

public:

bool increasingTriplet(vector<int>& nums) {

int min1 = INT_MAX;

int min2 = INT_MAX;

for (int num : nums) {

if (num > min2) return true;

else if (num < min1) {

min1 = num;

} else if (num > min1 && num < min2) {

min2 = num;

}

return false;

}

};

花花酱 LeetCode 516. Longest Palindromic Subsequence

By zxi on May 28, 2018

Problem

题目大意：找出最长回文子序列的长度。

https://leetcode.com/problems/longest-palindromic-subsequence/description/

Given a string s, find the longest palindromic subsequence’s length in s. You may assume that the maximum length of s is 1000.

Example 1:
Input:

"bbbab"

Output:

One possible longest palindromic subsequence is “bbbb”.

Example 2:
Input:

"cbbd"

Output:

One possible longest palindromic subsequence is “bb”.

Solution: DP

Time complexity: O(n^2)

Space complexity: O(n^2)

C++

// Author: Huahua

// Running time: 102 ms

class Solution {

public:

int longestPalindromeSubseq(string s) {

const int n = s.length();

vector<vector<int>> dp(n, vector<int>(n, 0));

for (int l = 1; l <= n; ++l)

for (int i = 0; i <= n - l; ++i) {

int j = i + l - 1;

if (i == j) {

dp[i][j] = 1;

continue;

}

dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);

if (s[i] == s[j])

dp[i][j] = dp[i + 1][j - 1] + 2;

}

return dp[0][n - 1];

}

};

Time complexity: O(n^2)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 28 ms

class Solution {

public:

int longestPalindromeSubseq(string s) {

const int n = s.length();

vector<int> dp0(n); // sols of len = l

vector<int> dp1(n); // sols of len = l - 1

vector<int> dp2(n); // sols of len = l - 2

for (int l = 1; l <= n; ++l) {

for (int i = 0; i <= n - l; ++i) {

int j = i + l - 1;

if (i == j) {

dp0[i] = 1;

continue;

}

if (s[i] == s[j])

dp0[i] = dp2[i + 1] + 2;

else

dp0[i] = max(dp1[i + 1], dp1[i]);

}

dp0.swap(dp1);

dp2.swap(dp0);

}

return dp1[0];

}

};

Python3

"""

Author: Huahua

Running time: 1276 ms

"""

class Solution:

def longestPalindromeSubseq(self, s):

n = len(s)

dp0 = [0] * n

dp1 = [1] * n

dp2 = [0] * n

for l in range(2, n + 1):

for i in range(0, n - l + 1):

j = i + l - 1

if s[i] == s[j]:

dp0[i] = dp2[i + 1] + 2

else:

dp0[i] = max(dp1[i + 1], dp1[i])

dp0, dp1, dp2 = dp2, dp0, dp1

return dp1[0]

// Author: Huahua

// Running time: 116 ms (beats 100%)

public class Solution {

public int LongestPalindromeSubseq(string s) {

var n = s.Length;

var dp0 = new int[n];

var dp1 = new int[n];

var dp2 = new int[n];

for (var l = 1; l <= n; ++l) {

for (var i = 0; i <= n - l; ++i) {

int j = i + l - 1;

if (i == j) {

dp0[i] = 1;

continue;

}

if (s[i] == s[j])

dp0[i] = dp2[i + 1] + 2;

else

dp0[i] = Math.Max(dp1[i], dp1[i + 1]);

}

var tmp = dp2;

dp2 = dp1;

dp1 = dp0;

dp0 = tmp;

}

return dp1[0];

}

Posts tagged as “subsequence”

花花酱 LeetCode 891. Sum of Subsequence Widths

Problem

Solution: Math

花花酱 LeetCode 115. Distinct Subsequences

Problem

Solution: DP

Related Problems:

花花酱 LeetCode 491. Increasing Subsequences

Problem

Solution: DFS

花花酱 LeetCode 334. Increasing Triplet Subsequence

Problem

Solution: Greedy

花花酱 LeetCode 516. Longest Palindromic Subsequence

Problem

Solution: DP