In this HackerRank Palindromic Border problem solution, we have given a string s, consisting only of the first 8 lowercase letters of the English alphabet. we need to find the sum of all the non-empty substrings of the string.
Problem solution in Python.
def is_palin(s):
head, tail = 0, len(s) - 1
while head < tail:
if s[head] != s[tail]:
return False
head += 1
tail -= 1
return True
#key is a palin, value is the times it appears
def calc_palin_borders(palin_dict):
#print('palin_dict= ', palin_dict)
output = 0
for palin, times in palin_dict.items():
output += times * (times - 1) // 2
return output
def mono_str(s):
cc = s[0]
for c in s:
if c != cc:
return False
return True
def mono_str_result(s):
output = 0
for i in range(2, len(s) + 1):
output += i * (i - 1) // 2
output %= 1000000007
return output
def pb(s):
if mono_str(s):
return mono_str_result(s)
output = 0
#palin tuple for substring of length 1
odd = [[], {}, 1]
for c in s:
if c not in odd[1]:
odd[1][c] = 0
odd[1][c] += 1
for i in range(len(s)):
odd[0].append(i)
output += calc_palin_borders(odd[1])
#print('odd = ', odd)
#palin tuple for substring of length 2
even = [[], {}, 1]
for i in range(len(s) - 1):
if s[i] == s[i + 1]:
even[0].append(i)
ss = s[i:i + 2]
if ss not in even[1]:
even[1][ss] = 0
even[1][ss] += 1
output += calc_palin_borders(even[1])
#print('even = ', even)
for l in range(3, len(s)):
#print('l = ', l)
#working tuple
if l % 2 == 0:
wt = even
else:
wt = odd
new_tuple = [[], {}, l]
for idx in wt[0]:
if idx - 1 >= 0 and idx + l - 2 < len(s) and s[idx - 1] == s[idx + l - 2]:
new_tuple[0].append(idx - 1)
ss = s[idx - 1:idx - 1 + l]
if ss not in new_tuple[1]:
new_tuple[1][ss] = 0
new_tuple[1][ss] += 1
#print('new_tuple= ', new_tuple)
output += calc_palin_borders(new_tuple[1])
output %= 1000000007
if l % 2 == 0:
even = new_tuple
else:
odd = new_tuple
return output
if __name__ == '__main__':
print(pb(input()))
Problem solution in Java.
import java.io.*;
import java.util.Arrays;
public class timus2040 {
static int[][] es;
static int[] slink, len, cnt;
static int free;
static int newNode(int l) {
len[free] = l;
return free++;
}
static int get(int i, char c) {
return es[c - 'a'][i];
}
static void set(int i, char c, int n) {
es[c - 'a'][i] = n;
}
public static void solve(Input in, PrintWriter out) throws IOException {
char[] s = in.next().toCharArray();
int n = s.length;
es = new int[8][n + 2];
for (int[] ar : es) {
Arrays.fill(ar, -1);
}
len = new int[n + 2];
slink = new int[n + 2];
cnt = new int[n + 2];
int root0 = newNode(0);
int rootm1 = newNode(-1);
slink[root0] = slink[rootm1] = rootm1;
int cur = root0;
for (int i = 0; i < n; ++i) {
while (i - len[cur] == 0 || s[i] != s[i - len[cur] - 1]) {
cur = slink[cur];
}
if (get(cur, s[i]) == -1) {
set(cur, s[i], newNode(len[cur] + 2));
if (cur == rootm1) {
slink[get(cur, s[i])] = root0;
} else {
int cur1 = slink[cur];
while (s[i] != s[i - len[cur1] - 1]) {
cur1 = slink[cur1];
}
slink[get(cur, s[i])] = get(cur1, s[i]);
}
}
cur = get(cur, s[i]);
cnt[cur]++;
}
long ans = 0;
for (int i = free - 1; i >= 0; --i) {
cnt[slink[i]] += cnt[i];
if (len[i] > 0) {
ans = (ans + 1L * cnt[i] * (cnt[i] - 1) / 2) % 1000000007;
}
}
out.println(ans);
}
public static void main(String[] args) throws IOException {
// FileWriter out = new FileWriter("output.txt");
// solve(new FileReader("input.txt"), out);
PrintWriter out = new PrintWriter(System.out);
solve(new Input(new BufferedReader(new InputStreamReader(System.in))), out);
out.close();
}
static class Input {
BufferedReader in;
StringBuilder sb = new StringBuilder();
public Input(BufferedReader in) {
this.in = in;
}
public Input(String s) {
this.in = new BufferedReader(new StringReader(s));
}
public String next() throws IOException {
sb.setLength(0);
while (true) {
int c = in.read();
if (c == -1) {
return null;
}
if (" nrt".indexOf(c) == -1) {
sb.append((char)c);
break;
}
}
while (true) {
int c = in.read();
if (c == -1 || " nrt".indexOf(c) != -1) {
break;
}
sb.append((char)c);
}
return sb.toString();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
}
}
Problem solution in C++.
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstring>
using namespace std;
#define ll long long
#define mod 1000000007
#define L 5000011
int sa[L];
int sai[L];
int lcp[L];
int v[L];
char s[L];
ll ts[L];
int p[L<<1];
char t[L<<1];
int m, n;
set<ll> found;
bool scomp(int i, int j) {
return s[i] < s[j];
}
bool tscomp(int i, int j) {
return ts[i] < ts[j];
}
void get_suffix_array() {
for (int i = 0; i < n; i++) v[i] = i;
sort(v, v + n, scomp);
sai[v[0]] = 1;
for (int i = 1; i < n; i++) {
if (s[v[i]] == s[v[i - 1]]) {
sai[v[i]] = sai[v[i - 1]];
} else {
sai[v[i]] = i+1;
}
}
for (int p = 1; p <= n; p <<= 1) {
for (int i = 0; i < n-p; i++) ts[i] = sai[i] * (n+1LL) + sai[i+p];
for (int i = n-p; i < n; i++) ts[i] = sai[i] * (n+1LL);
sort(v, v + n, tscomp);
sai[v[0]] = 1;
for (int i = 1; i < n; i++) {
if (ts[v[i]] == ts[v[i - 1]]) {
sai[v[i]] = sai[v[i - 1]];
} else {
sai[v[i]] = i+1;
}
}
}
for (int i = 0; i < n; i++) sai[i]--;
for (int i = 0; i < n; i++) sa[sai[i]] = i;
}
void get_lcp() {
for (int i = 0; i < n; i++) lcp[i] = 0;
int l = 0;
for (int i = 0; i < n-1; i++) {
int k = sai[i];
int j = k ? sa[k-1] : sa[n-1];
while (j + l < n and s[i + l] == s[j + l]) {
l++;
}
lcp[k] = l;
if (l > 0) {
l--;
}
}
}
void manacher() {
// from wikipedia
// t has been processed
int center = 0, end = 0, left = 0, right = 0;
for (int i = 1; i < m; i++) {
if (i > end) {
p[i] = 0;
left = i - 1;
right = i + 1;
} else {
int j = 2*center - i; // index on the other side
if (p[j] < end - i) { // whole palindrome is inside
p[i] = p[j];
left = -1; // so we don't enter the loop below
} else {
p[i] = end - i;
right = end + 1;
left = 2*i - right;
}
}
while (left >= 0 and right < m and t[left] == t[right]) {
p[i]++;
left--;
right++;
}
if (i + p[i] > end) {
center = i;
end = i + p[i];
}
}
}
struct Node {
int i, j, v;
Node *p, *l, *r;
Node(int i, int j, Node *p = NULL): i(i), j(j), p(p) {
if (j - i == 1) {
l = r = NULL;
v = lcp[i];
} else {
int k = i + j >> 1;
l = new Node(i, k, this);
r = new Node(k, j, this);
v = min(l->v, r->v);
}
}
};
int node_minocc(Node *node, int v, int i) {
// find maximum j, 0 <= j <= i such that a[j] < v
while (node->l) {
if (i < node->l->j) {
node = node->l;
} else {
node = node->r;
}
}
// now node->i = i < node->j = i+1
if (node->v < v) {
return node->i;
}
while (true) {
while (node->p and node->p->l == node) {
node = node->p;
}
if (!node->p) {
return 0;
}
node = node->p;
if (node->l->v < v) {
node = node->l;
break;
}
}
while (node->l) {
if (node->r->v < v) {
node = node->r;
} else {
node = node->l;
}
}
return node->i;
}
int node_maxocc(Node *node, int v, int i) {
// find maximum j, i <= j <= n such that a[j] >= v
if (i == n) {
return n;
}
while (node->l) {
if (i < node->l->j) {
node = node->l;
} else {
node = node->r;
}
}
// now node->i = i < node->j = i+1
if (node->v < v) {
return node->i;
}
while (true) {
while (node->p and node->p->r == node) {
node = node->p;
}
if (!node->p) {
return n;
}
node = node->p;
if (node->r->v < v) {
node = node->r;
break;
}
}
while (node->l) {
if (node->l->v < v) {
node = node->l;
} else {
node = node->r;
}
}
return node->i;
}
int main() {
scanf("%s", s);
n = strlen(s);
// suffix tree
get_suffix_array();
get_lcp();
Node *root = new Node(0, n);
m = 0;
for (int i = 0; i < n; i++) {
t[m++] = '#';
t[m++] = s[i];
}
t[m++] = '$';
t[0] = '^';
manacher();
ll ans = 0;
for (int i = 1; i < m-1; i++) {
int k = p[i];
if (t[i-k] == '#') k--;
for(; k >= 0; k -= 2) {
int start = sai[i-k>>1];
int mino = node_minocc(root,k+1,start);
ll hsh = k*(n+1LL)+mino;
if (found.find(hsh) != found.end()) {
break;
}
found.insert(hsh);
int maxo = node_maxocc(root,k+1,start+1);
ll c = maxo - mino;
ans += c*(c-1);
ans %= mod;
}
}
ans = ans * (mod+1>>1) % mod;
printf("%lldn", ans);
}