HackerRank String Function Calculation Solution

Hello Programmers, In this post, you will learn 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 straight away look for the solutions, first try to solve the problems by yourself. If you find any difficulty after trying several times, then look for the solutions. We are going to solve the HackerRank Algorithms problems using C, CPP, JAVA, PYTHON, JavaScript & SCALA Programming Languages.

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HackerRank String Function Calculation Solution
HackerRank String Function Calculation Solution

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HackerRank String Function Calculation

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 Solution

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 Chase2learn. This tutorial is only for Educational and Learning purposes.

FAQ:

1. How do you solve the first question in HackerRank?

If you want to solve the first question of Hackerrank then you have to decide which programing language you want to practice i.e C programming, Cpp Programing, or Java programming then you have to start with the first program HELLO WORLD.

2. How do I find my HackerRank ID?

You will receive an email from HackerRank to confirm your access to the ID. Once you have confirmed your email, the entry will show up as verified on the settings page. You will also have an option to “Make primary”. Click on that option. Read more

3. Does HackerRank detect cheating?

yes, HackerRank uses a powerful tool to detect plagiarism in the candidates’ submitted code. The Test report of a candidate highlights any plagiarized portions in the submitted code and helps evaluators to verify the integrity of answers provided in the Test.

4. Does HackerRank use camera?

No for coding practice Hackerrank does not use camera but for companies’ interviews code submission time Hackerrank uses the camera.

5. Should I put HackerRank certificate on resume?

These certificates are useless, and you should not put them on your resume. The experience you gained from getting them is not useless. Use it to build a portfolio, and link to it on your resume. 

6. Can I retake HackerRank test?

The company which sent you the HackerRank Test invite owns your Test submissions and results. It’s their discretion to permit a reattempt for a particular Test. If you wish to retake the test, we recommend that you contact the concerned recruiter who invited you to the Test and request a re-invite. 

7. What is HackerRank?

HackerRank is a tech company that focuses on competitive programming challenges for both consumers and businesses. Developers compete by writing programs according to provided specifications. Wikipedi


Finally, we are now, in the end, I just want to conclude some important message for you

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