回文子序列

Problem statement:

問題陳述：

Given a string str, find total number of possible palindromic sub-sequences. A sub-sequence does not need to be consecutive, but for any x_ix_j i<j must be valid in the parent string too. Like "icl" is a subsequence of "includehelp" while "ple" is not.

給定字符串str ，找到可能的回文子序列的總數(shù)。 子序列不必是連續(xù)的，但是對(duì)于任何x _i x _j i <j在父字符串中也必須有效。 像“ icl”一樣，是“ includehelp”的子序列，而“ ple”則不是。

Input:

輸入：

The first line of input contains an integer T, denoting the no of test cases then T test cases follow. Each test case contains a string str.

輸入的第一行包含一個(gè)整數(shù)T ，表示測(cè)試用例的數(shù)量，然后是T個(gè)測(cè)試用例。 每個(gè)測(cè)試用例都包含一個(gè)字符串str 。

Output:

輸出：

For each test case output will be an integer denoting the total count of palindromic subsequence which could be formed from the string str.

對(duì)于每個(gè)測(cè)試用例，輸出將是一個(gè)整數(shù)，表示回文子序列的總數(shù)，該總數(shù)可以由字符串str形成。

Constraints:

限制條件：

1 <= T <= 100 1 <= length of string str <= 300

Example:

例：

Input: Test case: 2First test case: Input string: "aaaa"Output: Total count of palindromic subsequences is: 15Second test case: Input string: "abaaba"Output: Total count of palindromic subsequences is: 31

Explanation:

說明：

Test case 1:

測(cè)試用例1：

Input: "aaaa"

輸入：“ aaaa”

The valid palindromic subsequences are shown below,

有效回文子序列如下所示，

Marked cells are character taken in subsequence:

標(biāo)記的單元格是子序列中的字符：

Count=1

計(jì)數(shù)= 1

Count=2

計(jì)數(shù)= 2

Count=3

計(jì)數(shù)= 3

Count=4

計(jì)數(shù)= 4

Count=5

計(jì)數(shù)= 5

Count=6

計(jì)數(shù)= 6

Count=7

計(jì)數(shù)= 7

Count=8

計(jì)數(shù)= 8

Count=9

計(jì)數(shù)= 9

Count=10

數(shù)= 10

Count=11

數(shù)= 11

So on...
Total 15 palindromic sub-sequences
Actually in this case since all the character is same each and every subsequence is palindrome here.
For the second test case
Few sub-sequences can be
"a"
"b"
"a"
"aba"
So on
Total 31 such palindromic subsequences

等等...
總共15個(gè)回文子序列
實(shí)際上，在這種情況下，由于所有字符都是相同的，每個(gè)子序列在這里都是回文。
對(duì)于第二個(gè)測(cè)試用例
很少有子序列可以是
“一個(gè)”
“ b”
“一個(gè)”
“阿巴”
依此類推
總共31個(gè)這樣的回文序列

Solution approach

解決方法

This can be solved by using DP bottom up approach,

這可以通過使用DP自下而上的方法來解決，

Initialize dp[n][n] where n be the string length to 0

初始化dp [n] [n] ，其中n為0的字符串長(zhǎng)度

Fill up the base case, Base case is that each single character is a palindrome itself. And for length of two, i.e, if adjacent characters are found to be equal then dp[i][i+1]=3, else if characters are different then dp[i][i+1]=2

填滿基本情況，基本情況是每個(gè)單個(gè)字符本身都是回文。 對(duì)于兩個(gè)長(zhǎng)度，即，如果發(fā)現(xiàn)相鄰字符相等，則dp [i] [i + 1] = 3 ；否則，如果字符不同，則dp [i] [i + 1] = 2

To understand this lets think of a string like "acaa"

要理解這一點(diǎn)，可以考慮一個(gè)字符串，例如“ acaa”

Here

這里

dp[0][1]=2 because there's only two palindrome possible because of "a" and "c".

dp [0] [1] = 2是因?yàn)椤?a”和“ c”僅可能存在兩個(gè)回文。

Whereas for

鑒于

dp[2][3] value will be 3 as possible subsequences are "a", "a", "aa".

dp [2] [3]的值將為3，因?yàn)榭赡艿淖有蛄袨椤?a”，“ a”，“ aa”。

for i=0 to n// for single length charactersdp[i][i]=1; if(i==n-1)break; if(s[i]==s[i+1])dp[i][i+1]=3;elsedp[i][i+1]=2; end for

Compute for higher lengths,

計(jì)算更長(zhǎng)的長(zhǎng)度，

for len=3 to nfor start=0 to n-lenint end=start+len-1;// start and end is matchingif(s[end]==s[start])// 1+subsequence from semaining partdp[start][end]=1+dp[start+1][end]+dp[start][end-1];elsedp[start][end]=dp[start+1][end]+dp[start][end-1]-dp[start+1][end-1];end ifend for end for

Final result is stored in dp[0][n-1];

最終結(jié)果存儲(chǔ)在dp [0] [n-1]中；

So for higher lengths if starting and ending index is the same then we recur for the remaining characters, since we have the sub-problem result stored so we computed that. In case start and end index character are different then we have added dp[start+1][end] and dp[start][end-1] that's similar to recur for leaving starting index and recur for leaving end index. But it would compute dp[start+1][end-1] twice and that why we have deducted that.

因此，對(duì)于更大的長(zhǎng)度，如果開始索引和結(jié)束索引相同，那么我們將重復(fù)其余字符，因?yàn)槲覀兇鎯?chǔ)了子問題結(jié)果，因此我們對(duì)其進(jìn)行了計(jì)算。 如果起始索引和終止索引的字符不同，則我們添加了dp [start + 1] [end]和dp [start] [end-1] ，類似于recur離開起始索引和recur離開結(jié)束索引。 但是它將兩次計(jì)算dp [start + 1] [end-1] ，這就是為什么我們要減去它。

For proper understanding you can compute the table by hand for the string "aaaa" to understand how it's working.

為了正確理解，您可以手動(dòng)計(jì)算字符串“ aaaa”的表以了解其工作方式。

C++ Implementation:

C ++實(shí)現(xiàn)：

#include <bits/stdc++.h> using namespace std;int countPS(string s) {int n = s.length();int dp[n][n];memset(dp, 0, sizeof(dp));for (int i = 0; i < n; i++) {dp[i][i] = 1;if (i == n - 1)break;if (s[i] == s[i + 1])dp[i][i + 1] = 3;elsedp[i][i + 1] = 2;}for (int len = 3; len <= n; len++) {for (int start = 0; start <= n - len; start++) {int end = start + len - 1;if (s[end] == s[start]) {dp[start][end] = 1 + dp[start + 1][end] + dp[start][end - 1];}else {dp[start][end] = dp[start + 1][end] + dp[start][end - 1] - dp[start + 1][end - 1];}}}return dp[0][n - 1]; }int main() {int t;cout << "Enter number of testcases\n";cin >> t;while (t--) {string str;cout << "Enter the input string\n";cin >> str;cout << "Total Number of palindromic Subsequences are: " << countPS(str) << endl;}return 0; }

Output:

輸出：

Enter number of testcases 2 Enter the input string aaaa Total Number of palindromic Subsequences are: 15 Enter the input string abaaba Total Number of palindromic Subsequences are: 31

翻譯自: https://www.includehelp.com/icp/count-total-number-of-palindromic-subsequences.aspx

回文子序列

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