Circular Palindromes


Problem Statement :


A palindrome is a string that reads the same from left to right as it does from right to left.

Given a string, , S of N lowercase English letters, we define a k-length rotation as cutting the first  kcharacters from the beginning of S and appending them to the end of S. For each S, there are N possible k-length rotations (where 0  <=  K  <  N ). 

See the Explanation section for examples.

Given N and S, find all N  k-length rotations of S; for each rotated string, Sk, print the maximum possible length of any palindromic substring of Sk on a new line.


Input Format

The first line contains an integer, N (the length of S ).
The second line contains a single string, S.


Constraints


1   <=   N   <=  5* 10^5
0  <=    K   <  N
S is  comprised of lower case English letters.

Output Format

There should be N  lines of output, where each line k contains an integer denoting the maximum length of any palindromic substring of rotation Sk.



Solution :



title-img


                            Solution in C :

In     C++  :







#include <algorithm>
#include <iostream>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <set>

using namespace std;

#define type(x) __typeof((x).begin())
#define foreach(i, x) for(type(x) i = (x).begin(); i != (x).end(); i++)
#define hash ___hash

typedef long long ll;
typedef pair < int, int > ii;

const int inf = 1e9 + 333;
const ll linf = 1e18 + 333;

const int N = 2e6 + 5;

int n;
char s[N];
bool h[N];
int go_odd[N], go_even[N], tmp[N << 1], rad[N << 1], ans[2][N];

void manacher() {
    memset(tmp, 0, sizeof(tmp));
    memset(rad, 0, sizeof(rad));
    int m = n * 2 + 1;
    for(int i = 0; i < m; i++)
        tmp[i] = '#';
    for(int i = 0; i < n; i++)
        tmp[i * 2 + 1] = s[i + 1];
    int i = 0, j = 0;
    while(i < m) {
        while(i - j >= 0 and i + j < m and tmp[i - j] == tmp[i + j])
            j++;
        rad[i] = j;
        int k = 1;
        while(rad[i - k] < rad[i] - k) {
            rad[i + k] = rad[i - k];
            k++;
        }
        i += k;
        j = max(0, j - k);
    }
    for(int i = 1; i <= n; i++)
        go_odd[i] = rad[(i - 1) * 2 + 1] / 2;//abcba --> go_odd[3] = 3
    for(int i = 1; i <= n; i++)
        go_even[i - 1] = rad[(i - 1) * 2] / 2;//abccba --> go_even[3] = 3
}

void solveOdd(bool w) {
    int oth = 0;
    set < ii > go;
    for(int i = 1; i <= (n + 1) / 2; i++) {
        go.insert({go_odd[i], i});
    }
    for(int i = 1; i <= n; i++) {
        while(go.size()) {
            int x = go.rbegin() -> first;
            int id = go.rbegin() -> second;
            if(id < i) {
                go.erase(*go.rbegin());
                continue;
            }
            int mx = (id - i + 1) * 2 - 1;
            if(x > mx) {
                oth = max(oth, id);
                go.erase(*go.rbegin());
                continue;
            }
            break;
        }
        if(go.size())
            ans[w][i] = max(ans[w][i], go.rbegin() -> first);
        if(oth >= i)
            ans[w][i] = max(ans[w][i], (oth - i + 1) * 2 - 1);
        //printf("ans[%d] = %d oth = %d\n", i, ans[w][i]);
        int add = (n + 1) / 2 + i;
        go.insert({go_odd[add], add});
    }
}

int get(int x, int y) {
    y -= x - 1;
    return min(y, n - y) * 2;
}

void solveEven(bool w) {
    int oth = 0;
    set < ii > go;
    for(int i = 1; i <= (n + 1) / 2; i++) {
        go.insert({go_even[i], i});
    }
    for(int i = 1; i <= n; i++) {
        while(go.size()) {
            int x = go.rbegin() -> first;
            int id = go.rbegin() -> second;
            if(id < i) {
                go.erase(*go.rbegin());
                continue;
            }
            int mx = get(i, id);
            if(x > mx) {
                oth = max(oth, id);
                go.erase(*go.rbegin());
                continue;
            }
            break;
        }
        if(go.size())
            ans[w][i] = max(ans[w][i], go.rbegin() -> first);
        if(oth >= i)
            ans[w][i] = max(ans[w][i], get(i, oth));
        //printf("ans[%d] = %d\n", i, ans[w][i]);
        int add = (n + 1) / 2 + i;
        go.insert({go_even[add], add});
    }
}


int main () {
    
    scanf("%d %s", &n, s + 1);
    
    for(int i = 1; i <= n; i++)
        s[i + n] = s[i];
    
    ////////////////////////////////////
    
    n *= 2;
    manacher();
    n /= 2;
    
    for(int i = 1; i <= n + n; i++) {
        go_odd[i] *= 2;
        go_odd[i] -= 1;
        go_even[i] *= 2;
    }
    
    solveOdd(0);
    solveEven(0);
    
    ///////////////////////////////////
    
    reverse(s + 1, s + n * 2 + 1);
    
    n *= 2;
    manacher();
    n /= 2;
    
    for(int i = 1; i <= n + n; i++) {
        go_odd[i] *= 2;
        go_odd[i] -= 1;
        go_even[i] *= 2;
    }
    
    solveOdd(1);
    solveEven(1);
    
    ///////////////////////////////////
    
    printf("%d\n", max(ans[0][1], ans[1][1]));
    
    for(int i = 2; i <= n; i++) {
        printf("%d\n", max(ans[0][i], ans[1][n + 2 - i]));
    }
    
    return 0;
    
}








In    Java  :







import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class E2 {
    InputStream is;
    PrintWriter out;
    String INPUT = "";
    
    void solve()
    {
        int n = ni();
        char[] s = ns(n);
        char[] s2 = new char[2*n];
        for(int i = 0;i < n;i++){
            s2[i] = s2[i+n] = s[i];
        }
        int[] pal = palindrome(s2);
//        tr(pal, pal.length, n);
        long[] es = new long[16*n];
        int p = 0;
        for(int i = 0;i < 4*n;i+=2){
            pal[i] = Math.min(pal[i], n-((n&1)^1));
            es[p++] = (long)(i/2)<<32|i;
            es[p++] = (long)(i/2+pal[i]/2)<<32|i;
            es[p++] = (long)(i/2+n-pal[i]/2-1)<<32|i;
            es[p++] = (long)(i/2+n)<<32|i;
        }
        for(int i = 1;i < 4*n;i+=2){
            pal[i] = Math.min(pal[i], n-((n&1)));
            es[p++] = (long)(i/2)<<32|i;
            es[p++] = (long)(i/2+pal[i]/2)<<32|i;
            es[p++] = (long)(i/2+n-pal[i]/2)<<32|i;
            es[p++] = (long)(i/2+n)<<32|i;
        }
        
        Arrays.sort(es, 0, p);
        MaxHeap inc = new MaxHeap(4*n+1);
        MaxHeap dec = new MaxHeap(4*n+1);
        MaxHeap flat = new MaxHeap(4*n+1);
        
        int[] st = new int[4*n];
        int q = 0;
        for(int i = 0;i < 2*n-1;i++){
            while(q < p && es[q]>>>32 <= i){
                int ind = (int)es[q];
                if(st[ind] == 0){
                    inc.add(ind, (pal[ind]&1)-2*i);
                }else if(st[ind] == 1){
                    inc.remove(ind);
                    flat.add(ind, pal[ind]);
                }else if(st[ind] == 2){
                    flat.remove(ind);
                    dec.add(ind, pal[ind]+2*i);
                }else if(st[ind] == 3){
                    dec.remove(ind);
                }
                st[ind]++;
                q++;
            }
            if(i >= n-1){
//                tr("i", i);
                int max = 0;
                if(inc.size() > 0)max = Math.max(inc.max()+2*i, max);
//                tr(max);
                if(dec.size() > 0)max = Math.max(dec.max()-2*i, max);
//                tr(max);
                max = Math.max(flat.max(), max);
//                tr(max);
                out.println(max);
            }
        }
    }
    public static class MaxHeap {
        public int[] a;
        public int[] map;
        public int[] imap;
        public int n;
        public int pos;
        public static int INF = Integer.MIN_VALUE;
        
        public MaxHeap(int m)
        {
            n = m+2;
            a = new int[n];
            map = new int[n];
            imap = new int[n];
            Arrays.fill(a, INF);
            Arrays.fill(map, -1);
            Arrays.fill(imap, -1);
            pos = 1;
        }
        
        public int add(int ind, int x)
        {
            int ret = imap[ind];
            if(imap[ind] < 0){
                a[pos] = x; map[pos] = ind; imap[ind] = pos;
                pos++;
                up(pos-1);
            }
            return ret != -1 ? a[ret] : x;
        }
        
        public int update(int ind, int x)
        {
            int ret = imap[ind];
            if(imap[ind] < 0){
                a[pos] = x; map[pos] = ind; imap[ind] = pos;
                pos++;
                up(pos-1);
            }else{
                int o = a[ret];
                a[ret] = x;
                up(ret);
                down(ret);
//                if(a[ret] < o){
//                    up(ret);
//                }else{
//                    down(ret);
//                }
            }
            return x;
        }
        
        public int remove(int ind)
        {
            if(pos == 1)return INF;
            if(imap[ind] == -1)return INF;
            
            pos--;
            int rem = imap[ind];
            int ret = a[rem];
            map[rem] = map[pos];
            imap[map[pos]] = rem;
            imap[ind] = -1;
            a[rem] = a[pos];
            a[pos] = INF;
            map[pos] = -1;
            
            up(rem);
            down(rem);
            return ret;
        }
        
        public int max() { return a[1]; }
        public int argmax() { return map[1]; }
        public int size() {    return pos-1; }
        
        private void up(int cur)
        {
            for(int c = cur, p = c>>>1;p >= 1 && a[p] < a[c];c>>>=1, p>>>=1){
                int d = a[p]; a[p] = a[c]; a[c] = d;
                int e = imap[map[p]]; imap[map[p]] = imap[map[c]]; imap[map[c]] = e;
                e = map[p]; map[p] = map[c]; map[c] = e;
            }
        }
        
        private void down(int cur)
        {
            for(int c = cur;2*c < pos;){
                int b = a[2*c] > a[2*c+1] ? 2*c : 2*c+1;
                if(a[b] > a[c]){
                    int d = a[c]; a[c] = a[b]; a[b] = d;
                    int e = imap[map[c]]; imap[map[c]] = imap[map[b]]; imap[map[b]] = e;
                    e = map[c]; map[c] = map[b]; map[b] = e;
                    c = b;
                }else{
                    break;
                }
            }
        }
    }
    
    public static int[] palindrome(char[] str)
    {
        int n = str.length;
        int[] r = new int[2*n];
        int k = 0;
        for(int i = 0, j = 0;i < 2*n;i += k, j = Math.max(j-k, 0)){
            // normally
            while(i-j >= 0 && i+j+1 < 2*n && str[(i-j)/2] == str[(i+j+1)/2])j++;
            r[i] = j;
            
            // skip based on the theorem
            for(k = 1;i-k >= 0 && r[i]-k >= 0 && r[i-k] != r[i]-k;k++){
                r[i+k] = Math.min(r[i-k], r[i]-k);
            }
        }
        return r;
    }

    
    void run() throws Exception
    {

is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
        out = new PrintWriter(System.out);
        
        long s = System.currentTimeMillis();
        solve();
        out.flush();
        if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
    }
    
    public static void main(String[] args) throws Exception { new E2().run(); }
    
    private byte[] inbuf = new byte[1024];
    private int lenbuf = 0, ptrbuf = 0;
    
    private int readByte()
    {
        if(lenbuf == -1)throw new InputMismatchException();
        if(ptrbuf >= lenbuf){
            ptrbuf = 0;
            try { lenbuf = is.read(inbuf); }
 catch (IOException e) { throw new InputMismatchException(); }
            if(lenbuf <= 0)return -1;
        }
        return inbuf[ptrbuf++];
    }
    
    private boolean isSpaceChar(int c) 
{ return !(c >= 33 && c <= 126); }
    private int skip()
{ int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
    
    private double nd() { return Double.parseDouble(ns()); }
    private char nc() { return (char)skip(); }
    
    private String ns()
    {
        int b = skip();
        StringBuilder sb = new StringBuilder();
        while(!(isSpaceChar(b)))
{ // when nextLine, (isSpaceChar(b) && b != ' ')
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    
    private char[] ns(int n)
    {
        char[] buf = new char[n];
        int b = skip(), p = 0;
        while(p < n && !(isSpaceChar(b))){
            buf[p++] = (char)b;
            b = readByte();
        }
        return n == p ? buf : Arrays.copyOf(buf, p);
    }
    
    private char[][] nm(int n, int m)
    {
        char[][] map = new char[n][];
        for(int i = 0;i < n;i++)map[i] = ns(m);
        return map;
    }
    
    private int[] na(int n)
    {
        int[] a = new int[n];
        for(int i = 0;i < n;i++)a[i] = ni();
        return a;
    }
    
    private int ni()
    {
        int num = 0, b;
        boolean minus = false;
        while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if(b == '-'){
            minus = true;
            b = readByte();
        }
        
        while(true){
            if(b >= '0' && b <= '9'){
                num = num * 10 + (b - '0');
            }else{
                return minus ? -num : num;
            }
            b = readByte();
        }
    }
    
    private long nl()
    {
        long num = 0;
        int b;
        boolean minus = false;

 while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if(b == '-'){
            minus = true;
            b = readByte();
        }
        
        while(true){
            if(b >= '0' && b <= '9'){
                num = num * 10 + (b - '0');
            }else{
                return minus ? -num : num;
            }
            b = readByte();
        }
    }
    
    private static void tr(Object... o) 
{ System.out.println(Arrays.deepToString(o)); }
}









In   C  :






#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void solve(char *str,int *a);
void init( int n );
void range_increment( int i, int j, int val );
int query( int i );
int max(int x,int y);
void update(int x,int y,int z);
void sort_a2(int*a,int*b,int size);
void merge2(int*a,int*left_a,int*right_a,int*b,int*left_b,int*right_b,int left_size, int right_size);
char str[1000001]={0};
int N,NN,a[2000004],tree[2000000],ans[500000],b[500000],c[500000];

int main(){
  int i,j;
  scanf("%d%s",&NN,str);
  strncpy(str+NN,str,NN);
  solve(str,a);
  init(NN);
  for(i=0;i<4*NN;i++)
    if(a[i])
      if(i%2)
        update(i/2-a[i]/2,i/2+a[i]/2,a[i]);
      else
        update(i/2-a[i]/2,i/2+a[i]/2-1,a[i]);
  for(i=0;i<NN;i++){
    ans[i]=query(i);
    b[i]=ans[i];
    c[i]=i;
  }
  sort_a2(b,c,NN);
  for(i=NN;i>=0;i--){
    for(j=c[i];1;j=(j-1+NN)%NN)
      if(ans[j]-ans[(j-1+NN)%NN]>2)
        ans[(j-1+NN)%NN]=ans[j]-2;
      else
        break;
    for(j=c[i];1;j=(j+1)%NN)
      if(ans[j]-ans[(j+1)%NN]>2)
        ans[(j+1)%NN]=ans[j]-2;
      else
        break;
  }
  for(i=0;i<NN;i++)
    printf("%d\n",ans[i]);
  return 0;
}
void solve(char *str,int *a){
  char *p;
  int len,R,Ri,i,j,mi;
  len=strlen(str);
  p=(char*)malloc(2*(len+1)*sizeof(char));
  for(i=0;i<len;i++){
    p[2*i]='#';
    p[2*i+1]=str[i];
  }
  p[2*i]='#';
  p[2*i+1]=0;
  a[0]=R=Ri=0;
  for(i=1;i<=len*2;i++)
    if(i>=R){
      if(p[i]!='#')
        a[i]=1;
      else
        a[i]=0;
      for(j=i+1;1;j++)
        if(j<=2*len && 2*i-j>=0 && p[j]==p[2*i-j]){
          if(p[j]!='#')
            a[i]+=2;
        }
        else{
          Ri=i;
          R=j-1;
          break;
        }
    }
    else{
      mi=2*Ri-i;
      if(i+a[mi]>=R || mi==a[mi]){
        a[i]=R-i;
        for(j=R+1;1;j++)
          if(j<=2*len && 2*i-j>=0 && p[j]==p[2*i-j]){
            if(p[j]!='#')
              a[i]+=2;
          }
          else{
            Ri=i;
            R=j-1;
            break;
          }
      }
      else
        a[i]=a[mi];
    }
  free(p);
  return;
}
void init( int n ){
  N = 1;
  while( N < n ) N *= 2;
  int i;
  for( i = 1; i < N + n; i++ ) tree[i] = 0;
}
void range_increment( int i, int j, int val ){
  for( i += N, j += N; i <= j; i = ( i + 1 ) / 2, j = ( j - 1 ) / 2 )
  {
    if( i % 2 == 1 ) tree[i] = max(tree[i],val);
    if( j % 2 == 0 ) tree[j] = max(tree[j],val);
  }
}
int query( int i ){
  int ans = 0,j;
  for( j = i + N; j; j /= 2 ) ans = max(ans,tree[j]);
  return ans;
}
int max(int x,int y){
  return (x>y)?x:y;
}
void update(int x,int y,int z){
  if(z>NN){
    int m=x+z/2;
    if(z%2)
      if(NN%2)
        update(m-NN/2,m+NN/2,NN);
      else
        update(m-NN/2+1,m+NN/2-1,NN-1);
    else
      if(NN%2)
        update(m-NN/2,m+NN/2-1,NN-1);
      else
        update(m-NN/2,m+NN/2-1,NN);
  }
  if(y<NN){
    range_increment(0,x,z);
    range_increment(y+1,NN-1,z);
  }
  else
    range_increment(y-NN+1,x,z);
  return;
}
void sort_a2(int*a,int*b,int size){
  if (size < 2)
    return;
  int m = (size+1)/2,i;
  int*left_a,*left_b,*right_a,*right_b;
  left_a=(int*)malloc(m*sizeof(int));
  right_a=(int*)malloc((size-m)*sizeof(int));
  left_b=(int*)malloc(m*sizeof(int));
  right_b=(int*)malloc((size-m)*sizeof(int));
  for(i=0;i<m;i++){
    left_a[i]=a[i];
    left_b[i]=b[i];
  }
  for(i=0;i<size-m;i++){
    right_a[i]=a[i+m];
    right_b[i]=b[i+m];
  }
  sort_a2(left_a,left_b,m);
  sort_a2(right_a,right_b,size-m);
  merge2(a,left_a,right_a,b,left_b,right_b,m,size-m);
  free(left_a);
  free(right_a);
  free(left_b);
  free(right_b);
  return;
}
void merge2(int*a,int*left_a,int*right_a,int*b,int*left_b,int*right_b,int left_size, int right_size){
  int i = 0, j = 0;
  while (i < left_size|| j < right_size) {
    if (i == left_size) {
      a[i+j] = right_a[j];
      b[i+j] = right_b[j];
      j++;
    } else if (j == right_size) {
      a[i+j] = left_a[i];
      b[i+j] = left_b[i];
      i++;
    } else if (left_a[i] <= right_a[j]) {
      a[i+j] = left_a[i];
      b[i+j] = left_b[i];
      i++;
    } else {
      a[i+j] = right_a[j];
      b[i+j] = right_b[j];
      j++;
    }
  }
  return;
}
                        








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This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

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Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

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