Short Palindrome
Problem Statement :
Consider a string, , of lowercase English letters where each character, (, denotes the letter at index in . We define an palindromic tuple of to be a sequence of indices in satisfying the following criteria: , meaning the characters located at indices and are the same. , meaning the characters located at indices and are the same. , meaning that , , , and are ascending in value and are valid indices within string . Given , find and print the number of tuples satisfying the above conditions. As this value can be quite large, print it modulo . unction Description Complete the function shortPalindrome in the editor below. shortPalindrome has the following paramter(s): - string s: a string Returns - int: the number of tuples, modulo Input Format A single string, . Constraints It is guaranteed that only contains lowercase English letters.
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
char s[1000000] = "hello";
int n = 0;
long long* map[26];
int maxLen = 0;
long long* work;
long long* psum;
long long* ssum;
long long* rsum;
const long long modul = 1000000007;
long long calc(int len);
long long* getArr(char c);
void solve(long long* c1, long long* c2);
long long solveOne(long long l);
int main(int argc, char* argv[]) {
char c = ' ';
for (int i = 0; i <= 1000005; ++i) {
c = getchar();
if (('\n' == c) || (EOF == c)) {
s[i] = '\0';
n = i;
break;
} else {
s[i] = c - 'a';
}
}
for (char c = 0; c < 26; ++c) {
map[c] = getArr(c);
//printf("%lld\n", map[c][0]);
if (map[c][0] > maxLen)
maxLen = map[c][0];
}
work = malloc((maxLen + 1) * sizeof(long long));
psum = malloc((maxLen + 1) * sizeof(long long));
ssum = malloc((maxLen + 1) * sizeof(long long));
rsum = malloc((maxLen + 1) * sizeof(long long));
long long ans = 0;
for (char c1 = 0; c1 < 26; ++c1) {
for (char c2 = 0; c2 < 26; ++c2) {
if (c1 == c2) {
if (map[c1][0] > 0) {
//printf("%d\n", map[c1][0]);
long long t = solveOne(map[c1][0]);
ans = (ans + t) % modul;
}
} else {
if ((map[c1][0] > 0) && (map[c2][0] > 0)) {
solve(map[c1], map[c2]);
//for (int i = 0; i < map[c1][0] + 1; ++i) printf("%d ", work[i]);
//printf("\n");
ans += calc(map[c1][0] + 1);
ans = ans % modul;
}
}
}
}
printf("%lld\n", ans);
return 0;
}
long long solveOne(long long l) {
long long res = 0;
for (long long i = 1; i <= (l - 3); ++i) {
res += (i * (l - 2 - i) * (l - 1 - i) / 2) % modul;
res = res % modul;
}
res = res % modul;
return res;
}
long long calc(int len) {
psum[0] = work[0];
//printf("psum: %d ", psum[0]);
for (int i = 1; i < len; ++i) {
psum[i] = psum[i - 1] + work[i];
//printf("%d ", psum[i]);
}
ssum[0] = psum[0];
//printf("\nssum: %d ", ssum[0]);
for (int i = 1; i < len; ++i) {
ssum[i] = (ssum[i - 1] + psum[i]) % modul;
//printf("%d ", ssum[i]);
}
rsum[len - 1] = work[len - 1];
for (int i = len - 2; i >= 0; --i) {
rsum[i] = rsum[i + 1] + work[i];
}
/*(printf("\nrsum: ");
for (int i = 0; i < len; ++i) {
printf("%d ", rsum[i]);
}
printf("\n");*/
long long res = 0;
for (int i = 2; i < len; ++i) {
res += (rsum[i] * ssum[i - 2]) % modul;
}
res = res % modul;
return res;
}
void solve(long long* c1, long long* c2) {
long long l1 = 1;
long long l2 = 1;
long long r = 0;
long long count = 0;
for (;;) {
if ((l1 > c1[0]) && (l2 > c2[0])) {
work[r] = count;
++r;
break;
} else if (l1 > c1[0]) {
++l2;
++count;
} else if (l2 > c2[0]) {
++l1;
work[r] = count;
count = 0;
++r;
} else if (c1[l1] > c2[l2]) {
++l2;
++count;
} else if (c1[l1] < c2[l2]) {
++l1;
work[r] = count;
count = 0;
++r;
}
}
}
long long* getArr(char c) {
long long count = 0;
for (long long i = 0; i < n; ++i)
if (c == s[i])
++count;
long long *res = malloc(sizeof(long long) * (count + 1));
long long cur = 1;
res[0] = count;
for (long long i = 0; i < n; ++i)
if (c == s[i])
res[cur++] = i;
return res;
}
Solution in C++ :
In C++ :
#include<bits/stdc++.h>
#include<iostream>
using namespace std;
#define fre freopen("0.in","r",stdin);freopen("0.out","w",stdout)
#define abs(x) ((x)>0?(x):-(x))
#define MOD 1000000007
#define LL signed long long int
#define scan(x) scanf("%d",&x)
#define print(x) printf("%d\n",x)
#define scanll(x) scanf("%lld",&x)
#define printll(x) printf("%lld\n",x)
#define rep(i,from,to) for(int i=(from);i <= (to); ++i)
#define pii pair<int,int>
vector<int> G[2*100000+5];
LL suf[654];
LL pre[654];
LL dp[29][29];
int main(){
//fre;
string S;
cin>>S;
int N = S.size();
LL ans = 0;
S = " " + S;
for(int i=1;i<=N;++i)suf[S[i]-'a'+1]++;
for(int i=1;i<=N;++i){
int x = S[i]-'a'+1;
suf[x]--;
for(int j=1;j<=26;++j){
ans += dp[j][x] * suf[j];
ans %= MOD;
}
for(int j=1;j<=26;++j){
dp[j][x] += pre[j];
}
pre[x]++;
}
cout<<ans;
}
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) throws Exception{
BufferedReader inR = new BufferedReader(new InputStreamReader(System.in));
String s = inR.readLine().trim();
long MOD = 1000000000L+7;
long[] cnt = new long[26];
long[][] sumK = new long[26][26];
long[][] delta = new long[26][26];
long ans = 0;
for(int i=0; i<s.length(); i++) {
int c = s.charAt(i) - 'a';
for(int j=0; j<26; j++) {
ans = (ans + delta[c][j]) % MOD;
}
for(int j=0; j<26; j++) {
delta[j][c] = (delta[j][c] + sumK[j][c]) % MOD;
sumK[j][c] = (sumK[j][c] + cnt[j]) % MOD;
}
cnt[c] += 1;
}
System.out.println(ans % MOD);
inR.close();
}
}
Solution in Python :
In Python3 :
#!/bin/python3
import math
import os
import random
import re
import sys
import collections
# Complete the shortPalindrome function below.
def shortPalindrome(s):
arr1 = [0]*26
arr2 = [[0]*26 for i in range(26)]
arr3 = [0]*26
ans = 0
for i in range(len(s)):
idx = ord(s[i]) - ord('a')
ans += arr3[idx]
for j in range(26):
arr3[j] += arr2[j][idx]
for j in range(26):
arr2[j][idx] += arr1[j]
arr1[idx] += 1
return ans % (10**9+7)
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
s = input()
result = shortPalindrome(s)
fptr.write(str(result) + '\n')
fptr.close()
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