HackerRank Build a String Solution

Hello Programmers, In this post, you will learn how to solve HackerRank Build a String 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.

You can practice and submit all HackerRank problem solutions in one place. Find a solution for other domains and Sub-domain. I.e. Hacker Rank solution for HackerRank C ProgrammingHackerRank C++ ProgrammingHackerRank Java Programming, HackerRank Python ProgrammingHackerRank Linux ShellHackerRank SQL Programming, and HackerRank 10 days of Javascript.

HackerRank Build a String Solution
HackerRank Build a String Solution

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HackerRank Build a String

Task

Greg wants to build a string, S of length N. Starting with an empty string, he can perform 2 operations:

  1. Add a character to the end of S for A dollars.
  2. Copy any substring of S, and then add it to the end of S for B dollars.

Calculate minimum amount of money Greg needs to build S.

Input Format

The first line contains number of testcases T.

The 2 x T subsequent lines each describe a test case over 2 lines:
The first contains 3 spaceseparated integers, NA, and B, respectively.
The second contains S (the string Greg wishes to build).

Constraints

  • 1 <= T <= 3
  • 1 <= N <= 3 x 104
  • 1 <= AB <= 10000
  • S is composed of lowercase letters only.

Output Format

On a single line for each test case, print the minimum cost (as an integer) to build S.

Sample Input

2
9 4 5
aabaacaba
9 8 9
bacbacacb

Sample Output

26
42

Explanation

Test Case 0:
Sinitial = “”; Sfinal “aabaacaba
Append “a“; S = “a; cost is 4
Append “a“; S = “aa“; cost is 4
Append “b“; S = “aab“; cost is 4
Copy and append “aa“; S = “aabaa“; cost is 5
Append “c“; S = “aabaac“; cost is 4
Copy and append “aba“; S = “aabaacaba“; cost is 5

Summing each cost, we get 4 + 4 + 4 + 5 + 4 + 5 = 26, so our output for Test Case 1 is 26.

Test Case 1:
Sinitial = “”; Sfinal “bacbacacb
Append “b“; S = “b“; cost is $8
Append “a“; S = “ba“; cost is $8
Append “c; S = “bac“; cost is $8
Copy and append “bac“; S = “bacbac“; cost is $9
Copy and append “acb“; S = “bacbacacb“; cost is $9

Summing each cost, we get 8 + 8 + 8 + 9 + 9 = 42, so our output for Test Case 2 is 42.

HackerRank Build a String Solution

Build a String Solution in C

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

char s[30001];
int sublens[30001] = { 0 };

void longestsubstr(int pos) {
    int i, max = 0;
    for (i = 0; i < pos; i++) {
        if (s[i] != s[pos]) sublens[i] = 0;
        else {
            sublens[i] = sublens[i+1] + 1;
            if (i + sublens[i] > pos) sublens[i] = pos - i;
            if (sublens[i] > max) max = sublens[i];
        }
    }
    sublens[pos] = max;
}

int main() {
    int t, t1;
    scanf("%d", &t);
    for (t1 = 0; t1 < t; t1++) {
        int n, a, b, sublen, i, j, temp;
        scanf("%d %d %d", &n, &a, &b);
        scanf("%s", s);
        int ar[30001];
        for (i = 0; i < n; i++) {
            ar[i] = 0x7FFFFFFF;
            sublens[i] = 0;
        }
        for (i = n - 1; i >= 1; i--) longestsubstr(i);
        ar[0] = a;
        for (i = 1; i < n; i++) {
            if (ar[i-1] + a < ar[i]) ar[i] = ar[i-1] + a;
            sublen = sublens[i];
            temp = ar[i-1] + b;
            for (j = 0; j < sublen; j++) if (temp < ar[i+j]) ar[i+j] = temp;
        }
        printf("%d\n", ar[n-1]);
    }
    return 0;
}

Build a String Solution in Cpp

#include <bits/stdc++.h>
using namespace std;

const int MaxLen = 200001;
const int MaxLog = 21;
int lg[MaxLen], tmp[MaxLen];
int S[MaxLen];

struct SuffixArray
{
	int rank [MaxLen], SA[MaxLen], h[MaxLen], D[MaxLen];
	int n, dep , count_rank [MaxLen], f[MaxLog][MaxLen];
	void Build ()
	{
		for(int len = 1; len < n; len <<= 1)
		{
			fill ( count_rank , count_rank + 1 + n, 0);
			for (int i=1;i <=n;++ i)
			++ count_rank [ rank [SA[i]+ len ]];
			for (int i=1;i <=n;++ i)
			count_rank [i]+= count_rank [i -1];
			for (int i=n;i >0;--i)
			D[ count_rank [ rank [SA[i]+ len ]]--] = SA[i];
			fill ( count_rank , count_rank + 1 + n, 0);
			for (int i=1;i <=n;++ i)
			++ count_rank [ rank [SA[i ]]];
			for (int i=1;i <=n;++ i)
			count_rank [i]+= count_rank [i -1];
			for (int i=n;i >0;--i)
			SA[ count_rank [ rank [D[i]]] --] = D[i];
			copy (rank , rank + 1 + n, D);
			rank [SA [1]]=1;
			for (int i=2;i <=n;++ i)
			if(D[SA[i]] != D[SA[i -1]] ||
			D[SA[i]+ len] != D[SA[i -1] + len ])
			rank [SA[i ]]= rank [SA[i -1]]+1;
			else
			rank [SA[i ]]= rank [SA[i -1]];
			if( rank [SA[n]] == n) break ;
		}
	}
	int strsuf (int *p, int *q)
	{
		int ret =0;
		for (; *p == *q; ++p, ++q, ++ ret );
		return ret;
	}
	void CalcHeight ()
	{
		for (int i=1;i <=n;++ i)
		{
			if( rank [i] == 1)
			h[i] = 0;
			else
			if(i == 1 || h[i -1] <= 1)
			h[i]= strsuf (S+i, S+SA[ rank [i] -1]);
			else
			h[i]= strsuf (S+i+h[i -1] -1 ,
			S+SA[ rank [i] -1]+h[i -1] -1)+ h[i -1] -1;
			f [0][ rank [i]]=h[i];
		}
		dep =1;
		for (int len =1; len *2 <=n;len <<=1 , dep ++)
		for(int i=1; i+len *2 -1 <=n;++ i)
		f[dep ][i]= min(f[dep -1][ i],f[dep -1][ i+len ]);
	}
	void init ( int _n) // String Stored in (S +1)
	{
		lg[1] = 0;
		for(int i = 2; i <= _n; i++)
			lg[i] = lg[i/2] + 1;
		n = _n;
		memset (tmp ,0, sizeof ( tmp ));
		for (int i=1;i <=n;++ i)
		++ tmp[S[i ]];
		for (int i=1;i <MaxLen;++ i)tmp [i]+= tmp [i -1];
		for (int i=n;i >0;--i)
		SA[ tmp[S[i]] --]=i;
		rank [SA [1]]=1;
		for (int i=2;i <=n;++ i)
		if(S[SA[i]] != S[SA[i -1]])
		rank [SA[i]] = rank [SA[i -1]]+1;
		else
		rank [SA[i]] = rank [SA[i -1]];
		Build ();
		CalcHeight ();
	}
	inline int lcp( int a, int b)
	{ // lcp of S[a] and S[b]
		if(a == b) return n - a + 1;
		a = rank [a], b = rank [b];
		if(a > b) swap (a, b);
		int d = lg[b - a];
		if ((1 << d) == (b - a)) return f[d][a +1];
		else return min(f[d][a+1] , f[d][b -(1<<d )+1]);
	}
}mySA;

/*  Note
	1. Set MaxLen, MaxLog: length of string, log of it
	2. S[i] > 0 (Can't be zero!)
*/

/*  Eaxmple
	S[1] = 'a';
	S[2] = 'b';
	S[3] = 'a';
	mySA.init(3);
	cout << mySA.lcp(1,3) << endl;
*/

int n;
string s = "xxabab";
int myRank[30001];
int rankToIndex[30001];
int minimal[30001][21];
int INF = 1000000001;
int maxJump[30001];

vector <int> dpIndex;
vector <int> dpValue;
int dp[30001];

int rmq(int L, int R)
{
	int t = 0;
	while((1<<(t+1)) <= R-L+1) t ++;
	return min(minimal[L][t], minimal[R - (1<<t) + 1][t]);
}

bool check(int to, int front)
{

	if(to == front) return true;
	int r = myRank[to];
	//cout << "myRank = " << r << endl;
	{
		int L = 0, R = r, M;
		while(R - L > 1)
		{
			M = (L + R) / 2;
			if(rmq(M, r) <= front)
				L = M;
			else
				R = M;
		}
		//cout << "get : " << L << endl;
		if(L > 0)
		{
			int index = rankToIndex[L];
			//cout << index << endl;
			if(mySA.lcp(n+1-index, n+1-to) >= (to - front)) return true;
		}
	}
	{
		int L = r, R = n+1, M;
		while(R - L > 1)
		{
			M = (L + R) / 2;
			if(rmq(r, M) <= front)
				R = M;
			else
				L = M;
		}
		//cout << "get : " << R <<  endl;
		if(R <= n)
		{
			int index = rankToIndex[R];
			//cout << index << endl;
			if(mySA.lcp(n+1-index, n+1-to) >= (to - front)) return true;
		}
	}
	return false;
}

int MAIN()
{
	int T;
	cin >> T;
	while(T--)
	{
		int A, B, len;
		cin >> len >> A >> B >> s;
		n = s.length();
		for(int i = 0; i < n; i++)
			S[i+1] = s[n-1-i];
		mySA.init(n);
		for(int i = 1; i <= n; i++)
		{
			rankToIndex[mySA.rank[i]] = n+1-i;
			myRank[n+1-i] = mySA.rank[i];
		}
		/*for(int i = 1; i <= n; i++)
			cout << myRank[i] << " ";
		cout << endl;*/

		for(int i = 1; i <= n; i++)
		{
			minimal[i][0] = rankToIndex[i];
		}
		for(int k = 1; k <= 20; k++)
			for(int i = 1; i <= n; i++)
			{
				minimal[i][k] = INF;
				int t = i + (1<<(k-1));
				if(t <= n)
					minimal[i][k] = min(minimal[i][k-1], minimal[t][k-1]);
			}
		for(int i = 1; i <= n; i++)
		{
			int L = 0, R = i, M;
			while(R - L > 1)
			{
				M = (L + R) / 2;
				if(check(i, i - M))
					L = M;
				else
					R = M;
			}
			maxJump[i] = L;
			//cout << i << " : " << L << endl;
		}
		dpValue.clear();
		dpIndex.clear();
		dp[1] = A;
		dpValue.push_back(A);
		dpIndex.push_back(1);
		for(int i = 2; i <= n; i++)
		{
			dp[i] = dp[i-1] + A;
			if(maxJump[i] > 0)
			{
				int L = -1, R = dpValue.size(), M;
				while(R - L > 1)
				{
					M = (L + R);
					if(dpIndex[M] >= i - maxJump[i])
						R = M;
					else
						L = M;
				}
				if(R < dpValue.size())
				{
					dp[i] = min(dp[i], dpValue[R] + B);
				}
			}
			while(dpValue.size() > 0 && dp[i] < dpValue[dpValue.size()-1])
			{
				dpValue.pop_back();
				dpIndex.pop_back();
			}
			dpValue.push_back(dp[i]);
			dpIndex.push_back(i);
		}
		cout << dp[n] << endl;
	}
	
	return 0;
}

int main()
{
	int start = clock();
	#ifdef LOCAL_TEST
		freopen("in.txt", "r", stdin);
		freopen("out.txt", "w", stdout);
	#endif
	ios :: sync_with_stdio(false);
	cout << fixed << setprecision(16);
	int ret = MAIN();
	#ifdef LOCAL_TEST
		cout << "[Finished in " << clock() - start << " ms]" << endl;
	#endif
	return ret;
}

Build a String Solution in Java

import java.awt.*;
import java.awt.event.*;
import java.awt.geom.*;
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*; 
import java.util.*;
import java.util.regex.*;
/*
	  br = new BufferedReader(new FileReader("input.txt"));
	  pw = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
	  br = new BufferedReader(new InputStreamReader(System.in));
	  pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
 */


public class Solution {
	private static BufferedReader br;
	private static StringTokenizer st;
	private static PrintWriter pw;

	public static void main(String[] args) throws Exception {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		//int qq = 1;
		//int qq = Integer.MAX_VALUE;
		int qq = readInt();
		for(int casenum = 1; casenum <= qq; casenum++)	{
			int n = readInt();
			int a = readInt();
			int b = readInt();
			String s = nextToken();
			int[] list = new int[n];
			for(int i = 0; i < n; i++) {
				list[i] = s.charAt(i) - 'a';
			}

			int[] dp = new int[n+1];
			Arrays.fill(dp, 1 << 30);
			dp[0] = 0;

			ArrayList<int[]> edges = new ArrayList<int[]>();
			ArrayList<Integer> link = new ArrayList<Integer>();
			ArrayList<Integer> length = new ArrayList<Integer>();
			edges.add(empty(26));
			link.add(-1);
			length.add(0);
			int last = 0;
			for(int i = 0; i < n; i++) {
				
				dp[i+1] = Math.min(dp[i+1], dp[i] + a);
				int len = 0;
				int currSuffixLoc = 0;
				while(currSuffixLoc < edges.size() && i + len < list.length) {
					if(edges.get(currSuffixLoc)[list[i+len]] == -1) {
						break;
					}
					currSuffixLoc = edges.get(currSuffixLoc)[list[i+len]];
					len++;
				}
				
				dp[i+len] = Math.min(dp[i+len], dp[i] + b);
				
				// construct r
				edges.add(empty(26));
				length.add(i+1);
				link.add(0);
				int r = edges.size() - 1;
				int p = last;
				while(p >= 0 && edges.get(p)[list[i]] == -1) {
					edges.get(p)[list[i]] = r;
					p = link.get(p);
				}
				if(p != -1) {
					int q = edges.get(p)[list[i]];
					if(length.get(p) + 1 == length.get(q)) {
						link.set(r, q);
					} 
					else {
						// we have to split, add q'
						edges.add(deepCopy(edges.get(q))); // copy edges of q
						length.add(length.get(p) + 1);
						link.add(link.get(q).intValue()); // copy parent of q
						int qqq = edges.size()-1;
						// add qq as the new parent of q and r
						link.set(q, qqq);
						link.set(r, qqq);
						// move short classes pointing to q to point to q'
						while(p >= 0 && edges.get(p)[list[i]] == q) {
							edges.get(p)[list[i]] = qqq;
							p = link.get(p);
						}
					}
				}
				last = r;
			}
			pw.println(dp[n]);
		}
		exitImmediately();
	}

	public static int[] deepCopy(int[] list) {
		int[] ret = new int[list.length];
		for(int i = 0; i < ret.length; i++) {
			ret[i] = list[i];
		}
		return ret;
	}
	
	public static int[] empty(int len) {
		int[] ret = new int[len];
		Arrays.fill(ret, -1);
		return ret;
	}

	private static void exitImmediately() {
		pw.close();
		System.exit(0);
	}

	private static long readLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	private static double readDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	private static int readInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	private static String nextLine() throws IOException  {
		if(!br.ready()) {
			exitImmediately();
		}
		st = null;
		return br.readLine();
	}

	private static String nextToken() throws IOException  {
		while(st == null || !st.hasMoreTokens())  {
			if(!br.ready()) {
				exitImmediately();
			}
			st = new StringTokenizer(br.readLine().trim());
		}
		return st.nextToken();
	}
}

Build a String Solution in Python

# Enter your code here. Read input from STDIN. Print output to STDOUT
def sol():
    N, A, B = map(int, raw_input().strip().split())
    S = raw_input().strip()
    res = [0]*N
    res[0] = A
    maxl = 0
    for i in range(1,N):
        minv = res[i-1] + A
        cp, idx, newl = False, i, 0
        for k in range(maxl,-1,-1):
            if S[i-k:i+1] in S[0:i-k]:
                cp, idx, newl = True, i-k, k+1
                break
        if cp: minv = min(minv, res[idx-1]+B)
        maxl = newl
        res[i] = minv
    print res[-1]
T = int(raw_input().strip())
for x in range(T):
    sol()

Build a String Solution using JavaScript

// given a string, and the remainder of the string that we are looking for, find the length of the biggest substring we could possibly add
function findBiggestSubstring(string, remainder, min) {
    for (var i=min; i <= remainder.length; i++) {
        if (string.indexOf(remainder.substring(0, i)) == -1) {
            return i - 1;
        }
    }
    return i - 1;
}

// given a string, the cost to add and the cost to copy, determine minimum cost
function calculate(string, costAdd, costCopy) {
    var costPerState = [];
    costPerState[string.length-1] = 0;
    biggestPerState = [];
    var biggest = 0;
    for (var i=0; i < string.length; i++) {
        var substring = string.substring(0, i+1);
        var remainder = string.substring(i+1);
        biggest = findBiggestSubstring(substring, remainder, biggest);        
        biggestPerState[i] = biggest;
    }
    for (var i=string.length-2; i >= 0; i--) {
        var minCost = costAdd + costPerState[i+1];
        for (var j=1; j <= biggestPerState[i]; j++) {
            var cost = costCopy + costPerState[i+j];
            minCost = Math.min(minCost, cost);
        }
        costPerState[i] = minCost;
    }
    return costAdd + costPerState[0];
}

function processData(input) {
    var lines = input.split("\n");
    var cases = lines[0];
    for (var t=0; t < cases; t++) {
        var ints = lines[t*2+1].split(' ');
        var string = lines[t*2+2];
        var result = calculate(string, parseInt(ints[1]), parseInt(ints[2]));
        console.log(result);
    }
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});

Build a String Solution in Scala

import scala.reflect.ClassTag

class SuffixArray (_s: String) {
  val s = _s :+ 0.toChar
  val n = _s.length
  val alphabet_size = (_s.fold (0.toChar) ( (x, y) => x.max (y))).toInt + 1

  private def counting_sort (m: Int, c: Array[Int], p: Seq[Int], o: Array[Int]): Unit = {
    val cnt:Array[Int] = Array.ofDim (m)
    p.foreach (pi => cnt(c(pi)) = cnt(c(pi)) + 1)
    (1 until m).foreach (i => cnt(i) = cnt(i) + cnt(i-1))
    p.reverseIterator.foreach (pi => {
      cnt(c(pi)) = cnt(c(pi)) - 1
      o(cnt(c(pi))) = pi
    })
  }
  private def build ():Array[Int] = {
    val l = s.length
    var p: Array[Int] = Array.ofDim (l)
    var c = s.map (c => c.toInt).toArray
    var q: Array[Int] = Array.ofDim (l)
    counting_sort (alphabet_size, c, (0 until l), p)
    c(p(0)) = 0
    var m = 0
    for (i <- 1 until l) {
      if (s(p(i)) != s(p(i-1))) {
        m = m + 1
      }
      c(p(i)) = m
    }
    m = m + 1
    var step = 1
    while (step < l) {
      counting_sort (m, c, p.map (v => (v + l - step) % l), q)
      val t1 = p; p = q; q = t1
      q(p(0)) = 0
      m = 0
      for (i <- 1 until l) {
        if (c(p(i)) != c(p(i-1)) || c((p(i) + step) % l) != c((p(i-1) + step) % l)) {
          m = m + 1
        }
        q(p(i)) = m
      }
      m = m + 1
      val t2 = c; c = q; q = t2
      step = 2 * step
    }
    p.slice (1, l)
  }
  val o = build
  private def lcp_build () = {
    var q: Array[Int] = Array.ofDim (n)
    val r: Array[Int] = Array.ofDim (n)
    (0 until n).foreach (i => r(o(i)) = i)
    var l = 0
    for (j <- 0 until n) {
      l = 0.max (l - 1)
      val i = r(j)
      if (i > 0) {
        val k = o(i - 1)
        while ((j + l < n) && (k + l < n) && s(j + l) == s(k + l)) {
          l = l + 1
        }
      } else {
        l = 0
      }
      q(i) = l
    }
    (q, r)
  }
  val (lcp, r) = lcp_build
}

class SegmentTree[T : ClassTag] (_a: Array[T], op: (T, T) => T) {
  val a = _a
  val n = a.length
  val t = Array.ofDim (4 * n)
  private def build (v: Int, l: Int, r: Int): Unit = {
    if (l == r) {
      t(v) = a(l)
    } else {
      val m = (l + r) >> 1
      build (v << 1, l, m)
      build ((v << 1) + 1, m + 1, r)
      t(v) = op (t(v << 1), t((v << 1) + 1))
    }
  }
  build (1, 0, n - 1)
  private def reduce (v: Int, l: Int, r: Int, a: Int, b: Int): T =  {
    if (a == l && b == r) {
      t(v)
    } else {
      val m = (l + r) >> 1
      val x = b.min (m)
      val y = a.max (m + 1)
      val w = 2 * v
      if (a <= x) {
        if (y <= b) {
          op (reduce (w, l, m, a, x), reduce (w + 1, m + 1, r, y, b))
        } else {
          reduce (w, l, m, a, x)
        }
      } else {
        reduce (w + 1, m + 1, r, y, b)
      }
    }
  }
  def reduce (a: Int, b: Int): T = reduce (1, 0, n - 1, a, b)
  def update (i: Int, new_value: T) = {
    var l = 0
    var r = n - 1
    var v = 1
    while (l < r) {
      val m = (l + r) >> 1
      v <<= 1
      if (i <= m) {
        r = m
      } else {
        v += 1
        l = m + 1
      }
    }
    t(v) = new_value
    while (v > 1) {
      v &= ~1
      t(v >> 1) = op (t(v), t(v + 1))
      v >>= 1
    }
  }
}

class ArraySet (_a: Array[Int]) {
  val a = _a
  val n = a.length
  def union (that: ArraySet): ArraySet = {
    val b = Array.ofDim[Int] (n + that.n);
    var i = 0
    var j = 0
    var k = 0
    while (i < n && j < that.n) {
      if (a(i) < that.a(j)) {
        b(k) = a(i)
        i += 1
        k += 1
      } else if (a(i) > that.a(j)) {
        b(k) = that.a(j)
        j += 1
        k += 1
      } else {
        b(k) = a(i)
        i += 1
        j += 1
        k += 1
      }
    }
    while (i < n) {
      b(k) = a(i)
      i += 1
      k += 1
    }
    while (j < that.n) {
      b(k) = that.a(j)
      j += 1
      k += 1
    }
    new ArraySet (b.slice (0, k))
  }
  def binsearch (v: Int) = {
    var l = -1
    var r = n
    while (r - l > 1) {
      val m = (l + r) >> 1
      if (a(m) <= v) l = m else r = m
    }
    l
  }
  def upper (v: Int): Int = {
    val l = binsearch (v)
    if (l < 0) a(0)
    else if (l >= n) Int.MaxValue
    else if (a(l) > v) a(l)
    else if (l + 1 < n) a(l + 1)
    else Int.MaxValue
  }
  def lower (v: Int): Int = {
    val l = binsearch (v)
    if (l < 0) Int.MinValue
    else if (l >= n) a(n - 1)
    else if (a(l) < v) a(l)
    else if (l > 0) a (l - 1)
    else Int.MinValue
  }
}


class SegmentTree2D[T : ClassTag] (_a: Array[T], build_op: (T, T) => T) extends SegmentTree[T] (_a, build_op) {
  private def reduce2d[U] (v: Int, l: Int, r: Int, a: Int, b: Int, extract_op: (T) => U, reduce_op: (U, U) => U): U =  {
    if (a == l && b == r) {
      extract_op (t(v))
    } else {
      val m = (l + r) >> 1
      val x = b.min (m)
      val y = a.max (m + 1)
      val w = 2 * v
      if (a <= x) {
        if (y <= b) {
          reduce_op (reduce2d (w, l, m, a, x, extract_op, reduce_op), reduce2d (w + 1, m + 1, r, y, b, extract_op, reduce_op))
        } else {
          reduce2d (w, l, m, a, x, extract_op, reduce_op)
        }
      } else {
        reduce2d (w + 1, m + 1, r, y, b, extract_op, reduce_op)
      }
    }
  }
  def reduce2d[U] (a: Int, b: Int, extract_op: (T) => U, reduce_op: (U, U) => U): U = reduce2d (1, 0, n - 1, a, b, extract_op, reduce_op)
}


object Solution {
  import scala.io._
  //import scala.collection.immutable.TreeSet
  var lineit:Iterator[String] = Source.stdin.getLines.filterNot(_.isEmpty).flatMap(i => i.split(" "))
  def rs() : String = lineit.next
  def ri() : Int = rs.toInt
  def compute_prefix_function (s: String): Array[Int] = {
    val n = s.length 
    val p = Array.ofDim[Int] (n + 1)
    p(0) = 0
    var k = -1
    for (q <- 1 until n) {
      while (k > 0 && s(k+1) != s(q)) {
        k = p(k)
      }
      if (s(k+1) == s(q)) {
        k += 1
      }
      p(q) = k
    }
    p
  }
  def solve (s: String, n: Int, a: Int, b: Int) = {
    val f = Array.fill (n) (Int.MaxValue)
    f(0) = a
    val t = new SegmentTree[Int] (f, (x, y) => x.min (y))
    val sa = new SuffixArray (s)
    val t2 = new SegmentTree2D[ArraySet] (Array.tabulate (n) (i => new ArraySet (Array (sa.r(i)))), (a, b) => a.union (b))
    val tl = new SegmentTree[Int] (sa.lcp, (x, y) => x.min (y))
    for (i <- 1 until n) {
      val o = sa.o(i)
      //val r = sa.r(o)
      def check (l: Int) = {
        val start = i + 1 - l
        val r = sa.r(start)
        def next (t: ArraySet) = t.upper (r)     // t.from (r + 1).headOption.getOrElse (Int.MaxValue)
        def prev (t: ArraySet) = t.lower (r)    // t.to (r - 1).lastOption.getOrElse (Int.MinValue)
        val v = t2.reduce2d[Int] (0, start - l, (t => next (t)), (a, b) => a.min (b))
        if (v < Int.MaxValue && tl.reduce (r + 1, v) >= l) true        
        else {
          val u = t2.reduce2d[Int] (0, start - l, (t => prev (t)), (a, b) => a.max (b))
          (u > Int.MinValue && tl.reduce (u + 1, r) >= l)
        }
      }
      def binsearch (a: Int, b: Int): Int = {
        if (a >= b) a
        else {
          val m = (a + b + 1) >> 1
          if (check (m)) binsearch (m, b) else binsearch (a, m - 1)
        }
      }
      f(i) = f(i - 1) + a
      val l = binsearch (0, (i + 1) >> 1)
      if (l > 0) {
        f(i) = f(i).min (t.reduce (i - l, i - 1) + b)
      }
      t.update (i, f(i))
    }

    f(n-1)
  }
  def main(args: Array[String]) {
    val nt = ri
    for (t <- 1 to nt) {
      val n = ri
      val a = ri
      val b = ri
      val s = rs
      println (solve (s, n, a, b))
    }
  }
}

Build a String Solution in Pascal

var f:array [0..30000] of longint;

procedure process;
        var i,n,m,k,a,b,l,r,rs:longint; s,t:ansistring;
        begin
                readln(n,a,b);
                readln(s);
                for i:=1 to n do f[i]:=a*i;
                for i:=1 to n do
                        begin
                        l:=1; r:=i-1; rs:=0;
                        while l<=r do
                                begin
                                m:=(l+r) div 2;
                                t:=copy(s,i-m+1,m);
                                k:=pos(t,s);
                                if k+m-1<i-m+1 then
                                        begin
                                        if m>rs then rs:=m;
                                        l:=m+1;
                                        end
                                else r:=m-1;
                                end;
                        f[i]:=f[i-1]+a;
                        if rs>0 then
                                if f[i-rs]+b<f[i] then f[i]:=f[i-rs]+b;
                        end;                           
                writeln(f[n]);
        end;

procedure dosth;
        var t:longint;
        begin
                readln(t);
                while t>0 do
                        begin
                        dec(t);
                        process;
                        end;
        end;


begin
        dosth;
end.

Disclaimer: This problem (Build a String) 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

Note:- I compile all programs, if there is any case program is not working and showing an error please let me know in the comment section. If you are using adblocker, please disable adblocker because some functions of the site may not work correctly.

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