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.
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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:
- Add a character to the end of S for A dollars.
- 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 space–separated integers, N, A, and B, respectively.
The second contains S (the string Greg wishes to build).
Constraints
- 1 <= T <= 3
- 1 <= N <= 3 x 104
- 1 <= A, B <= 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.
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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
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4. Does HackerRank use camera?
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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.
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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
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