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Home Code Solutions Hackerrank Algorithms

String Function Calculation – HackerRank Solution

String Function Calculation - HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

bhautik bhalala by bhautik bhalala
May 24, 2022
Reading Time: 2 mins read
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Table of Contents

  • String Function Calculation – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • Solutions of Algorithms Data Structures Hard HackerRank:
    • Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts
  • C++ String Function Calculation HackerRank Solution
  • Java String Function Calculation HackerRank Solution
  • Python 3 String Function Calculation HackerRank Solution
  • Python 2 String Function Calculation HackerRank Solution
  • C String Function Calculation HackerRank Solution
      • Related posts:

String Function Calculation – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Solutions of Algorithms Data Structures Hard HackerRank:

Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts

C++ String Function Calculation HackerRank Solution


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#include <cstdio>   
#include <cstdlib>   
#include <string>
#include <cstring>
#include <vector>
#include <algorithm>
#include <stack>
using namespace std;

const int N = 201000;   
int wa[N], wb[N], ws[N*2], wv[N];   
int Rank[N], sa[N], height[N], r[N];   
char s[N];
   
int cmp( int* r, int a, int b, int L ){   
    return r[a]== r[b] && r[a+ L]== r[b+ L];   
}

long long mul(long long x,long long y) {
	return x * y;
}   
   
void da( int* r, int* sa, int n, int m ){   
    int i, j, p, *x= wa, *y= wb, *t;   
    for( i= 0; i< m; ++i ) ws[i]= 0;   
    for( i= 0; i< n; ++i ) ws[ x[i]= r[i] ]++;   
    for( i= 1; i< m; ++i ) ws[i]+= ws[i-1];   
    for( i= n- 1; i>= 0; i-- ) sa[ --ws[ x[i] ] ]= i;   
   
    for( j= 1, p= 1; p< n; j*= 2, m= p ){   
        for( p= 0, i= n- j; i< n; ++i ) y[p++]= i;   
        for( i= 0; i< n; ++i )   
            if( sa[i]>= j ) y[p++]= sa[i]- j;   
   
        for( i= 0; i< n; ++i ) wv[i]= x[y[i]];   
        for( i= 0; i< m; ++i ) ws[i]= 0;   
        for( i= 0; i< n; ++i ) ws[ wv[i] ]++;   
        for( i= 1; i< m; ++i ) ws[i]+= ws[i-1];   
        for( i= n- 1; i>= 0; i-- ) sa[ --ws[ wv[i] ] ]= y[i];   
   
        t= x, x= y, y= t, p= 1; x[ sa[0] ]= 0;   
        for( i= 1; i< n; ++i )   
            x[ sa[i] ]= cmp( y, sa[i-1], sa[i], j )? p- 1: p++;   
    }   
}


long long largestRectangleArea(vector<int> &height) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int n = height.size();
	long long result = 0;
        stack<int> s;
        for (int i = 0; i < n; ++i) {
	   //printf("%d\n",height[i]);
            while ((!s.empty()) && (height[s.top()] > height[i])) {
                int h = height[s.top()];
                s.pop();
                result = max(result, mul((i  - (s.empty()?(-1):s.top())) , h));
                
            }
            s.push(i);
        }
        while (!s.empty()) {
            int h = height[s.top()];
            s.pop();
	    //printf("h = %d\n",h);
            result = max(result, mul((n  - (s.empty()?(-1):s.top())) , h));
        }
        return result;
        
    }   
   
void callheight( int* r, int*sa, int n ){   
    int i, j, k= 0;   
    for( i= 1; i<= n; ++i ) Rank[ sa[i] ]= i;   
   
    for( i= 0; i< n; height[ Rank[i++] ]= k )   
        for( k?k--:0, j= sa[ Rank[i]- 1]; r[i+k]== r[j+k]; k++ );   
   
}   
   
   
int main(){   
    scanf("%s",s );
    int n = strlen(s); 
    for(int i= 0; i < n; ++i ){   
        r[i] = s[i] - 'a' + 1; 
    }   
    r[n]= 0;   
    da( r, sa, n + 1, 27);   
    callheight( r, sa, n );  
    vector<int> a; 
    for (int i = 0; i <= n; ++i) {
	//printf("%d\n",height[i]);
	a.push_back(height[i]);
    }
    printf("%lld\n", max((long long) n, largestRectangleArea(a)));
    return 0;
}
	

 

Java String Function Calculation HackerRank Solution


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import java.io.*;
import java.util.ArrayList;
import java.util.List;

public class Solution {

    static class SuffixAutomata {

        static class Vertex {
            Vertex suffixLink = null;
            Vertex[] edges;
            int log = 0;

            int terminals;
            boolean visited;

            public Vertex(Vertex o, int log) {
                edges = o.edges.clone();
                this.log = log;
            }

            public Vertex(int log) {
                edges = new Vertex[26];
                this.log = log;
            }

            long dp() {
                if (visited) {
                    return 0;
                }
                visited = true;
                long r = 0;
                for (Vertex v : edges) {
                    if (v != null) {
                        r = Math.max(r, v.dp());
                        terminals += v.terminals;
                    }
                }
                return Math.max(r, 1L * log * terminals);
            }
        }

        Vertex root, last;

        public SuffixAutomata(String str) {
            last = root = new Vertex(0);
            for (int i = 0; i < str.length(); i++) {
                addChar(str.charAt(i));
            }
            addTerm();
        }

        private void addChar(char c) {
            Vertex cur = last;
            last = new Vertex(cur.log + 1);
            while (cur != null && cur.edges[c - 'a'] == null) {
                cur.edges[c - 'a'] = last;
                cur = cur.suffixLink;
            }
            if (cur != null) {
                Vertex q = cur.edges[c - 'a'];
                if (q.log == cur.log + 1) {
                    last.suffixLink = q;
                } else {
                    Vertex r = new Vertex(q, cur.log + 1);
                    r.suffixLink = q.suffixLink;
                    q.suffixLink = r;
                    last.suffixLink = r;
                    while (cur != null) {
                        if (cur.edges[c - 'a'] == q) {
                            cur.edges[c - 'a'] = r;
                        } else {
                            break;
                        }
                        cur = cur.suffixLink;
                    }
                }
            } else {
                last.suffixLink = root;
            }
        }

        private void addTerm() {
            Vertex cur = last;
            while (cur != null) {
                cur.terminals++;
                cur = cur.suffixLink;
            }
        }
    }

    public static void solve(Input in, PrintWriter out) throws IOException {
        String s = in.next();
        SuffixAutomata a = new SuffixAutomata(s);
        out.println(a.root.dp());
    }

    public static void main(String[] args) throws IOException {
        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 (" \n\r\t".indexOf(c) == -1) {
                    sb.append((char)c);
                    break;
                }
            }
            while (true) {
                int c = in.read();
                if (c == -1 || " \n\r\t".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());
        }
    }
}

Python 3 String Function Calculation HackerRank Solution


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from itertools import groupby
from operator import itemgetter
def SuffixArray(tx, _step=16):
    size = len(tx); step = min(max(_step, 1), size)
    sa = sorted(range(size), key=lambda i: tx[i:i + step])
    grpstart = [False]*size+[True]
    rsa = [None]*size
    stgrp, igrp = '', 0
    for i, pos in enumerate(sa):
        st = tx[pos:pos + step]
        if st != stgrp:
            grpstart[igrp] = (igrp < i - 1); stgrp = st; igrp = i
        rsa[pos] = igrp; sa[i] = pos
    grpstart[igrp] = (igrp < size - 1 or size == 0)
    while grpstart.index(True) < size:
        nextgr = grpstart.index(True)
        while nextgr < size:
            igrp = nextgr; nextgr = grpstart.index(True, igrp + 1); glist = []
            for ig in range(igrp, nextgr):
                pos = sa[ig]
                if rsa[pos] != igrp: break
                newgr = rsa[pos + step] if pos + step < size else -1
                glist.append((newgr, pos))
            glist.sort()
            for ig, g in groupby(glist, key=itemgetter(0)):
                g = [x[1] for x in g]; sa[igrp:igrp + len(g)] = g
                grpstart[igrp] = (len(g) > 1)
                for pos in g: rsa[pos] = igrp
                igrp += len(g)
        step *= 2
    return sa, rsa

def LCPArray(tx, sa = None, rsa = None):
    if sa == None: sa, rsa = SuffixArray(tx)
    size = len(tx)
    lcp = [None]*size; h = 0
    for i in range(size):
        if rsa[i] > 0:
            j = sa[rsa[i] - 1]
            while i+h<size and j+h<size and tx[i+h] == tx[j+h]: h += 1
            lcp[rsa[i]] = h
            if h: h -= 1
    if size: lcp[0] = 0
    return sa, lcp

def LRIHP1(h):
    n = len(h); h.append(0)
    stack = []; ans = len(s); i = 0
    while i < n+1:
        if not stack or h[stack[-1]] <= h[i]:
            stack.append(i); i+= 1; continue
        top = stack.pop()
        left = i-stack[-1] if stack else i+1
        ans = max(ans, h[top]*left)
    h.pop()
    return ans

s = input()
sa, lcp = LCPArray(s)
print(LRIHP1(lcp))

 

Python 2 String Function Calculation HackerRank Solution


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#!/usr/bin/python

def suffix_array(line):
    isa, sa, lb = [0] * len(line), [0] * len(line), [0] * len(line)
    for i in xrange(len(line)): sa[i] = ord(line[i])

    size = 1
    while size <= len(line):
        for i in xrange(len(lb)):
            if i + size < len(line): lb[i] = ((sa[i], sa[i + size]), i)
            else: lb[i] = ((sa[i], -1), i)
        lb.sort()
        sa[lb[0][1]] = 0
        for i in xrange(1, len(lb)):
            cls, idx = lb[i]
            if cls == lb[i - 1][0]: sa[idx] = sa[lb[i - 1][1]]
            else: sa[idx] = sa[lb[i - 1][1]] + 1
        size *= 2

    for i, p in enumerate(sa): isa[p] = i
    return isa, sa

def lcp(line, sa, rank):
    lcp = [0] * len(sa)
    h = 0
    for i in xrange(len(line)):
        if rank[i] == 0:
            continue
        j = sa[rank[i] - 1]
        while line[i + h] == line[j + h]: h += 1
        lcp[rank[i]] = h
        if h > 0: h -= 1
    return lcp

def solve1(line):
    line = line + chr(0)
    sa, rank = suffix_array(line)
    lp = lcp(line, sa, rank)
    lp.append(0)

    ans, st = len(line) - 1, []
    suffix, length, count = 0, len(line) - 1, 1
    for i, l in enumerate(lp):
        pos = i
        while st and st[-1][1] > l:
            j, h = st.pop()
            pos = j
            if (i - j + 1) * h > ans:
                ans = (i - j + 1) * h
                suffix, length, count = sa[j - 1], h, i - j + 1
        if not st or st[-1][1] < l:
            st.append((pos, l))

    #print sa
    #print lp

    #print 'Sub[{}:{}] count={}'.format(suffix, suffix + length, count)
    #print line[suffix:suffix + length]
    return ans

def solve2(line):
    class Substr:
         def __init__(self, line, i, j):
             self.line = line
             self.b = i
             self.e = j
             self._hash = hash(line[i:j])
         def __hash__(self):
             return self._hash
         def __eq__(self, other):
             if hash(self) != hash(other):
                 return False
             return self.line[self.b:self.e] == other.line[other.b:other.e]
         def __ne__(self, other):
             if hash(self) != hash(other):
                 return True
             return self.line[self.b:self.e] != other.line[other.b:other.e]
         def __len__(self):
             return self.e - self.b
    subs = {}
    ans = len(line)
    suffix, length, count = 0, len(line), 1
    for i in xrange(len(line)):
        for j in xrange(i + 1, len(line) + 1):
            sub = Substr(line, i, j)
            occ = subs.get(sub, 0) + 1
            subs[sub] = occ
            if occ * len(sub) > ans:
                ans = occ * len(sub)
                suffix, length, count = i, j - i, occ
    print 'Sub[{}:{}] count={}'.format(suffix, suffix + length, count)
    print line[suffix:suffix + length]
    return ans

def main():
    line = raw_input()
    print solve1(line)
    #assert solve1(line) == solve2(line)

main()

 

C String Function Calculation HackerRank Solution


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/*
aaaaaa
12

abcabcddd
9
*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define NN 100001
typedef struct _node{
  int u;
  int w;
} node;
void heap_insert(node *heap,node *elem,int *size,int *heap_list);
void heap_update(node *heap,node *elem,int size,int *heap_list);
int init( int n ,int *N,int *tree);
void range_increment( int i, int j, int val ,int N,int *tree);
int query( int i ,int N,int *tree);
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);
void merge(int*a,int*left,int*right,int left_size, int right_size);
void sort_a(int*a,int size);
void suffixSort(int n);
void LCP(int n);
char str[NN]; //input
int rank[NN], pos[NN], lcp[NN]; //output
int cnt[NN], next[NN]; //internal
int bh[NN], b2h[NN];

int main(){
  scanf("%s",str);
  int len,i,c=0,heap_size=0,group,N,diff,tmax;
  int *index,*p,*u,*v,*ans,*tree;
  node *heap,temp;
  long long max=-1;
  len=strlen(str);
  suffixSort(len);
  LCP(len);
  index=(int*)malloc(len*sizeof(int));
  p=(int*)malloc(len*sizeof(int));
  u=(int*)malloc(len*sizeof(int));
  v=(int*)malloc(len*sizeof(int));
  ans=(int*)malloc(len*sizeof(int));
  tree=(int*)malloc(3*len*sizeof(int));
  heap=(node*)malloc(2*len*sizeof(node));
  for(i=0;i<len;i++)
    index[i]=i;
  sort_a2(lcp,index,len);
  u[0]=0;
  v[0]=len-1;
  temp.u=0;
  temp.w=len;
  c++;
  init(len,&N,tree);
  heap_insert(heap,&temp,&heap_size,p);
  for(i=0;i<len;i++){
    if(i && lcp[i]!=lcp[i-1])
      ans[lcp[i-1]]=heap[1].w;
    group=query(index[i]+1,N,tree);
    temp.u=group;
    temp.w=v[group]-index[i];
    u[c]=u[group];
    u[group]=index[i]+1;
    heap_update(heap,&temp,heap_size,p);
    if(index[i]!=u[c]){
      v[c]=index[i]-1;
      temp.u=c;
      temp.w=index[i]-u[c];
      heap_insert(heap,&temp,&heap_size,p);
      diff=c-group;
      range_increment(u[c]+1,v[c]+1,diff,N,tree);
      c++;
    }
  }
  ans[lcp[i-1]]=heap[1].w;
  tmax=len-1;
  for(i=0;i<len;i++)
    if(ans[i]==-1)
      ans[i]=tmax;
    else
      tmax=ans[i];
  for(i=0;i<len;i++)
    if(max==-1 || (ans[i]+1)*(long long)(i+1)>max)
      max=(ans[i]+1)*(long long)(i+1);
  printf("%lld",max);
  return 0;
}
void heap_insert(node *heap,node *elem,int *size,int *heap_list){
  (*size)++;
  int index=(*size);
  while(index>1 && elem->w>heap[index/2].w){
    heap[index]=heap[index/2];
    heap_list[heap[index].u]=index;
    index/=2;
  }
  heap[index]=(*elem);
  heap_list[elem->u]=index;
  return;
}
void heap_update(node *heap,node *elem,int size,int *heap_list){
  int index=heap_list[elem->u];
  int max,maxi;
  while(1){
    if(2*index<=size){
      max=heap[2*index].w;
      maxi=2*index;
    }
    else
      break;
    if(2*index+1<=size && heap[2*index+1].w>max){
      max=heap[2*index+1].w;
      maxi=2*index+1;
    }
    if(elem->w<max){
      heap[index]=heap[maxi];
      heap_list[heap[index].u]=index;
      index=maxi;
    }
    else
      break;
  }
  heap[index]=(*elem);
  heap_list[elem->u]=index;
  return;
}
// initialises a tree for n data entries
int init( int n ,int *N,int *tree)
{
  (*N) = 1;
  while( (*N) < n ) (*N) *= 2;
  int i;
  for( i = 1; i < (*N) + n; i++ ) tree[i] = 0;
  return 0;
}

// increments a[k] by val for all k between i and j, inclusive
void range_increment( int i, int j, int val ,int N,int *tree)
{
  for( i += N, j += N; i <= j; i = ( i + 1 ) / 2, j = ( j - 1 ) / 2 )
  {
    if( i % 2 == 1 ) tree[i] += val;
    if( j % 2 == 0 ) tree[j] += val;
  }
}

// compute a[i]
int query( int i ,int N,int *tree)
{
  int ans = 0,j;
  for( j = i + N; j; j /= 2 ) ans += tree[j];
  return ans;
}
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;
}
void merge(int*a,int*left,int*right,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[j];
            j++;
        } else if (j == right_size) {
            a[i+j] = left[i];
            i++;
        } else if (str[left[i]] <= str[right[j]]) {
            a[i+j] = left[i];
            i++;                
        } else {
            a[i+j] = right[j];
            j++;
        }
    }
    return;
}
void sort_a(int*a,int size){
  if (size < 2)
    return;
  int m = (size+1)/2,i;
  int *left,*right;
  left=(int*)malloc(m*sizeof(int));
  right=(int*)malloc((size-m)*sizeof(int));
  for(i=0;i<m;i++)
    left[i]=a[i];
  for(i=0;i<size-m;i++)
    right[i]=a[i+m];
  sort_a(left,m);
  sort_a(right,size-m);
  merge(a,left,right,m,size-m);
  free(left);
  free(right);
  return;
}
 
void suffixSort(int n){
  //sort suffixes according to their first characters
  int h,i,j,k;
  for (i=0; i<n; ++i){
    pos[i] = i;
  }
  sort_a(pos, n);
  //{pos contains the list of suffixes sorted by their first character}
 
  for (i=0; i<n; ++i){
    bh[i] = i == 0 || str[pos[i]] != str[pos[i-1]];
    b2h[i] = 0;
  }
 
  for (h = 1; h < n; h <<= 1){
    //{bh[i] == false if the first h characters of pos[i-1] == the first h characters of pos[i]}
    int buckets = 0;
    for (i=0; i < n; i = j){
      j = i + 1;
      while (j < n && !bh[j]) j++;
      next[i] = j;
      buckets++;
    }
    if (buckets == n) break; // We are done! Lucky bastards!
    //{suffixes are separted in buckets containing strings starting with the same h characters}
 
    for (i = 0; i < n; i = next[i]){
      cnt[i] = 0;
      for (j = i; j < next[i]; ++j){
        rank[pos[j]] = i;
      }
    }
 
    cnt[rank[n - h]]++;
    b2h[rank[n - h]] = 1;
    for (i = 0; i < n; i = next[i]){
      for (j = i; j < next[i]; ++j){
        int s = pos[j] - h;
        if (s >= 0){
          int head = rank[s];
          rank[s] = head + cnt[head]++;
          b2h[rank[s]] = 1;
        }
      }
      for (j = i; j < next[i]; ++j){
        int s = pos[j] - h;
        if (s >= 0 && b2h[rank[s]]){
          for (k = rank[s]+1; !bh[k] && b2h[k]; k++) b2h[k] = 0;
        }
      }
    }
    for (i=0; i<n; ++i){
      pos[rank[i]] = i;
      bh[i] |= b2h[i];
    }
  }
  for (i=0; i<n; ++i){
    rank[pos[i]] = i;
  }
}
// End of suffix array algorithm

void LCP(int n){
  int l=0,i,j,k;
  for(i=0;i<n;i++){
    k=rank[i];
    if(!k)
      continue;
    j=pos[k-1];
    while(str[i+l]==str[j+l])
      l++;
    lcp[k]=l;
    if(l>0)
      l--;
  }
  lcp[0]=0;
  return;
}

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