Hello Programmers, In this post, you will Know how to solve HackerRank String Function Calculation Solution. This problem is a part of the HackerRank Algorithms Series.
One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.
HackerRank String Function Calculation Solution
Task
Jane loves strings more than anything. She has a string t with her, and value of string s over function f can be calculated as given below:
f(s) = |s| x Number of times s occurs in t
Jane wants to know the maximum value of f(s) among all the substrings (s) of string t. Can you help her?
Input Format
A single line containing string t.
Output Format
Print the maximum value of f(s) among all the substrings (s) of string t.
Constraints
- 1 <= |t| <= 105
- The string consists of lowercase English alphabets.
Sample Input 0
aaaaaa
Sample Output 0
12
Explanation 0
f(‘a’) = 6
f(‘aa’) = 10
f(‘aaa’) = 12
f(‘aaaa’) = 12
f(‘aaaaa‘) = 10
f(‘aaaaaa’) = 6
Sample Input 1
abcabcddd
Sample Output 1
9
Explanation 1
f values of few of the substrings are shown below:
f(“a”) = 2
f(“b”) = 2
f(“c”) = 2
f(“ab”) = 4
f(“bc”) = 4
f(“ddd“) = 3
f(“abc”) = 6
f(“abcabcddd”) = 9
Among the function values 9 is the maximum one.
HackerRank String Function Calculation Solutions
String Function Calculation Solution in C
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define MAXN 100000+2
char str[MAXN];
int sa[MAXN];
int rank[MAXN];
int cnt[MAXN];
int wb[MAXN];
int wv[MAXN];
int height[MAXN];
int stack[MAXN];
inline
int max(int a, int b) {
return a > b? a : b;
}
int cmp(int *r, int a, int b, int k) {
return r[a] == r[b] && r[a+k] == r[b+k];
}
void gen_sa(char *str, int n, int *sa, int *rank) {
int m = 128, p;
int i, j, k;
int *x, *y, *t;
x = rank; y = wb;
memset(cnt, 0, sizeof(int) * m);
for (i = 0; i < n; ++ i) ++ cnt[x[i] = str[i]];
for (i = 1; i < m; ++ i) cnt[i] += cnt[i-1];
for (i = n-1; i >= 0; -- i) sa[--cnt[x[i]]] = i;
for (k = 1; k <= n; k = k << 1) {
for (p = 0, i = n-k; i < n; ++ i) y[p++] = i;
for (i = 0; i < n; ++ i) if (sa[i] >= k) y[p++] = sa[i] - k;
memset(cnt, 0, sizeof(int) * m);
for (i = 0; i < n; ++ i) {
wv[i] = x[y[i]];
++ cnt[wv[i]];
}
for (i = 1; i < m; ++ i) cnt[i] += cnt[i-1];
for (i = n-1; i >= 0; -- i) sa[--cnt[wv[i]]] = y[i];
t = x; x = y; y = t;
x[sa[0]] = 0;
for (p = 1, i = 0; i < n; ++ i) {
x[sa[i]] = cmp(y, sa[i], sa[i-1], k) ? p-1: p++;
}
m = p;
}
if (x != rank) memcpy(rank, x, sizeof(int)*n);
}
void gen_height(char *str, int n, int *sa, int *rank, int *height) {
int i, j, k;
height[0] = 0;
k = 0;
for (i = 0; i < n-1; ++ i) {
if (k) -- k;
j = rank[i]-2;
if (j == -1) continue;
for (j = sa[j]; str[i+k] == str[j+k]; ) {
++ k;
}
height[rank[i]-1] = k;
}
}
int max_rectangle(int *height, int n) {
int i, j, left, right, cur, top = -1;
int result = 0;
height[n] = 0;
stack[++top] = 0;
for (i = 0; i <= n; ++ i) {
while (top > -1 && height[i] < height[stack[top]]) {
cur = stack[top--];
left = (top > -1? cur-stack[top]: cur+1) * height[cur];
right = (i - cur - 1) * height[cur];
result = max(result, left+right+height[cur]);
}
stack[++top] = i;
}
return max(result, n-1);
}
//void test(int n) {
// int i;
//
// printf("suffix array:\n");
// for (i = 0; i < n; ++ i) printf("%s\n", str + sa[i]);
//
// printf("rank array:\n");
// for (i = 0; i < n; ++ i) printf("%s, %d\n", str+i, rank[i]);
//
// printf("height array:\n");
// for (i = 0; i < n; ++ i) printf("%d\n", height[i]);
//}
int main() {
// freopen("input.txt", "r", stdin);
int n, result;
scanf("%s", str);
n = strlen(str);
gen_sa(str, n+1, sa, rank);
gen_height(str, n+1, sa, rank, height);
result = max_rectangle(height, n+1);
printf("%d\n", result);
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
return 0;
}String Function Calculation Solution in Cpp
#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;
}
String Function Calculation Solution in Java
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());
}
}
}
String Function Calculation Solution in Python
#!/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()String Function Calculation Solution using JavaScript
'use strict';
const fs = require('fs');
process.stdin.resume();
process.stdin.setEncoding('utf-8');
let inputString = '';
let currentLine = 0;
process.stdin.on('data', inputStdin => {
inputString += inputStdin;
});
process.stdin.on('end', _ => {
inputString = inputString.replace(/\s*$/, '')
.split('\n')
.map(str => str.replace(/\s*$/, ''));
main();
});
function readLine() {
return inputString[currentLine++];
}
// Complete the maxValue function below.
// nlogn algorithm for building suffix array
function buildSuffixArray (txt) {
const len = txt.length
const suffixes = new Array(len)
const aCode = 'a'.charCodeAt(0)
// sort based on first 2 characters
for (let i = 0; i !== len; i++) {
const nextIndex = i + 1
suffixes[i] = {
index: i,
rank: [
txt.charCodeAt(i) - aCode,
nextIndex < len ? txt.charCodeAt(nextIndex) - aCode : -1
]
}
}
suffixes.sort(compare)
// console.log(JSON.stringify(suffixes))
// sort based on first 4 characters and so on
const ind = new Array(len) // get the index in suffixes[] from origin index
for (let k = 4; k < 2 * len; k*=2) {
// assign rank to suffixes[0]
let rank = 0
let prevRank = suffixes[0].rank[0]
suffixes[0].rank[0] = rank
ind[suffixes[0].index] = 0
// assign rank from suffixes[1] to suffixes[len -1]
for (let i = 1; i < len; i++) {
if (suffixes[i].rank[0] === prevRank && suffixes[i].rank[1] === suffixes[i - 1].rank[1]) {
prevRank = suffixes[i].rank[0]
suffixes[i].rank[0] = rank
} else {
prevRank = suffixes[i].rank[0]
suffixes[i].rank[0] = ++rank
}
ind[suffixes[i].index] = i
}
// assign next rank from suffixes[0] to suffixes[len -1]
for (let i = 0; i < len; i++) {
let nextIndex = suffixes[i].index + Math.floor(k / 2) // origin index
suffixes[i].rank[1] = nextIndex < len ? suffixes[ind[nextIndex]].rank[0]: -1
}
// sort the suffixes according to first k characters
suffixes.sort(compare)
}
// console.log(JSON.stringify(suffixes))
// build suffix array
const suffixArray = suffixes.map(suffix => suffix.index)
// for (let i = 0; i < len; i++) {
// suffixArray[i] = suffixes[i].index
// }
return suffixArray
}
function compare (a, b) {
if (a.rank[0] === b.rank[0]) {
if (a.rank[1] < b.rank[1]) {
return -1
} else if (a.rank[1] > b.rank[1]) {
return 1
} else {
return 0
}
} else if (a.rank[0] < b.rank[0]) {
return -1
} else {
return 1
}
}
// kasai algorithm for building lcp array
function buildLcpKasai (suffixArr, txt) {
const len = suffixArr.length
const lcp = new Array(len)
// An auxiliary array to store inverse of suffix array
// elements. For example if suffixArr[0] is 5, the
// invSuff[5] would store 0.
// In fact invSuff[i], i present index in original text,
// also present the suffix string,
// and invSuff[i] present index in suffixArr.
// You can take invSuff as a map between origin text index(suffix string) and suffixArr index.
const invSuff = new Array(len)
// init
for (let i = 0; i !== len; i++) {
lcp[i] = 0
invSuff[suffixArr[i]] = i
}
// build lcp
let nextLcp = 0
for (let i = 0; i !== len; i++) {
// i is the index of origin text, so in fact we process
// all suffix in origin text one by one
// remember invSuff[i] is index in suffixArr.
// lcp[len - 1] is zero
if (invSuff[i] === len - 1) {
nextLcp = 0
continue
}
const nextSuffixIndex = suffixArr[invSuff[i] + 1] // index in origin text
while (i + nextLcp < len
&& nextSuffixIndex + nextLcp < len
&& txt[i + nextLcp] === txt[nextSuffixIndex + nextLcp]) {
nextLcp++
}
lcp[invSuff[i]] = nextLcp
// because lcp of next suffix in text will be at least ${nextLcp - 1}
nextLcp > 0 && (nextLcp--)
}
// return lcp
return lcp
}
function stringFuctionCalculation (txt) {
const suffixArr = buildSuffixArray(txt)
const lcp = buildLcpKasai(suffixArr, txt)
const len = txt.length
let result = len
for (let i = 0; i < len; i++) {
// because it's common prefix, means at least there are two of the common prefix
let count = 2
for (let j = i - 1; j >= 0; j--) {
if (lcp[j] >= lcp[i]) {
count++
} else {
break
}
}
for (let j = i + 1; j < len; j++) {
if (lcp[j] >= lcp[i]) {
count++
} else {
break
}
}
result = Math.max(result, count * lcp[i])
}
return result
}
function main() {
const ws = fs.createWriteStream(process.env.OUTPUT_PATH);
const t = readLine();
let result = stringFuctionCalculation(t);
ws.write(result + "\n");
ws.end();
}String Function Calculation Solution in Scala
object Solution {
import io.StdIn._
import collection.mutable.Stack
def visitLen(a:Array[Int]) = {
var h = Stack[(Int, Int)]((-1, -1))
Array.tabulate(a.size){i => {
while (h.top._2 >= a(i)) h.pop
val r = i - h.top._1
h.push(i -> a(i)); r
}}
}
def main(args:Array[String]) {
val sa = new SuffixArray(readLine.view.map{_.toLong}.toArray)
val a = Array.tabulate(sa.sa.size - 1) {
i => sa.lcp(sa.sa(i), sa.sa(i + 1))
}
var m = a.size
val aL = visitLen(a)
val aR = visitLen(a.reverse).reverse
for (i <- 0 until a.size) {
val k = (aL(i) + aR(i)) * a(i)
if (k > m) m = k
}
println(m)
}
class SuffixArray(a:Array[Long]) {
val n = a.size
val m = Math.getExponent(n) + 1
val b = Array.fill(m, n + 1){0L}
b(0) = a :+ 0L
def cityHash(x1:Long, x2:Long) = {
val kMul = 0x9ddfea08eb382d69L
var a = x1 * kMul
a ^= a >>> 47
var b = (a ^ x2) * kMul
b ^ (b >>> 47)
}
for (i <- 1 until m; j <- 0 until n) b(i)(j) = {
val j0 = j + (1 << i - 1)
cityHash(b(i - 1)(j), if (j0 <= n) b(i - 1)(j0) else 0L)
}
def lcp(n1:Int, n2:Int) = {
var k = 0
for (i <- Range(m - 1, -1, -1)) {
if (b(i)(n1 + k) == b(i)(n2 + k)) k += 1 << i
}; k
}
def less(n1:Int, n2:Int):Boolean = {
if (b(0)(n1) < b(0)(n2)) return true
if (b(0)(n1) > b(0)(n2)) return false
val k = lcp(n1, n2)
b(0)(n1 + k) < b(0)(n2 + k)
}
lazy val sa = Array.range(0, n + 1).sortWith{less}
}
}String Function Calculation Solution in Pascal
program j01; const maxn=100086; var t:array[0..5*maxn]of record son:array['a'..'z']of longint;dis,fa:longint; end; s:ansistring; len,cnt,root,last:longint; ans:int64; id,sum:array[0..5*maxn]of longint; num:array[0..5*maxn]of int64; function max(a,b:int64):int64;inline;begin if a>b then exit(a) else exit(b); end; function newnode(d:longint):longint;inline;begin inc(cnt);t[cnt].dis:=d;exit(cnt); end; procedure ins(ch:char); var p,np,q,nq:longint; begin p:=last;np:=newnode(t[p].dis+1);last:=np; num[np]:=1; while(p>0)and(t[p].son[ch]=0)do begin t[p].son[ch]:=np;p:=t[p].fa; end; if p=0 then t[np].fa:=root else begin q:=t[p].son[ch]; if t[q].dis=t[p].dis+1 then t[np].fa:=q else begin nq:=newnode(t[p].dis+1); t[nq].son:=t[q].son; t[nq].fa:=t[q].fa;t[q].fa:=nq;t[np].fa:=nq; while t[p].son[ch]=q do begin t[p].son[ch]:=nq;p:=t[p].fa; end; end; end; end; procedure build; var i:longint; begin root:=1;cnt:=1;last:=1;len:=length(s); for i:=1 to len do ins(s[i]); fillchar(sum,sizeof(sum),0); for i:=1 to cnt do inc(sum[t[i].dis]); for i:=1 to len do inc(sum[i],sum[i-1]); for i:=cnt downto 1 do begin id[sum[t[i].dis]]:=i;dec(sum[t[i].dis]); end; end; procedure solve; var i,p:longint; begin ans:=0; for i:=cnt downto 1 do begin p:=id[i];ans:=max(ans,num[p]*t[p].dis); //writeln(num[p],' ',t[p].dis); inc(num[t[p].fa],num[p]); end; end; begin readln(s); build; solve; writeln(ans); end.
Disclaimer: This problem (String Function Calculation) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.