Reverse Shuffle Merge
Problem Statement :
Given a string, A, we define some operations on the string as follows: a. denotes the string obtained by reversing string . Example: b. denotes any string that's a permutation of string . Example: c. denotes any string that's obtained by interspersing the two strings & , maintaining the order of characters in both. For example, & , one possible result of could be , another could be , another could be and so on. Given a string such that for some string , find the lexicographically smallest . For example, s = abab. We can split it into two strings of ab. The reverse is ba and we need to find a string to shuffle in to get abab . The middle two characters match our reverse string, leaving the a and b at the ends. Our shuffle string needs to be ab . Lexicographically ab < ba , so our answer is ab. Function Description Complete the reverseShuffleMerge function in the editor below. It must return the lexicographically smallest string fitting the criteria. reverseShuffleMerge has the following parameter(s): s: a string Input Format A single line containing the string s. Constraints s contains only lower-case English letters, ascii[a-z] 1 <= | s | <= 10000 Output Format Find and return the string which is the lexicographically smallest valid A.
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
char s[10000],c[5000];
int a[26],b[26],i=0,len,pos,limit,j,index;
scanf("%s",s);
len=strlen(s);
pos=len-1;
limit=len>>1;
while(s[i])
a[s[i++]-97]++;
for(i=0;i<26;i++)
b[i]=a[i]/2;
for(i=0;i<limit;i++)
{
char best;
int x=0;
for(j=pos;j>=0;j--)
{
if((!x||s[j]<best)&&b[s[j]-97])
{
x=1;
best=s[j];
index=j;
}
a[s[j]-97]--;
if(a[s[j]-97]<b[s[j]-97])
break;
}
for(; j < index; ++j)
{
++a[s[j]-97];
}
c[i]=best;
b[best-97]--;
pos=index-1;
}
printf("%s",c);
return 0;
}
Solution in C++ :
In C++ :
#include <algorithm>
#include <cassert>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <iostream>
#include <iterator>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <utility>
#include <vector>
using namespace std;
const int MAXN = 10000;
int cnt[26][MAXN+1];
int nxt[MAXN][26];
int shuffcnt[26];
int Acnt[26];
int totals[26];
int main() {
string S;
cin >> S;
int n = S.size();
assert(n%2 == 0);
reverse(S.begin(), S.end());
for (int i=0; i<n; ++i) {
for (int j=0; j<26; ++j) {
cnt[j][i+1] = cnt[j][i];
}
++cnt[S[i]-'a'][i+1];
}
for (int j=0; j<26; ++j) {
nxt[n-1][j] = -1;
}
nxt[n-1][S[n-1]-'a'] = n-1;
for (int i=n-2; i>=0; --i) {
for (int j=0; j<26; ++j) {
nxt[i][j] = nxt[i+1][j];
}
nxt[i][S[i]-'a'] = i;
}
for (int c=0; c<26; ++c) {
assert(cnt[c][n]%2 == 0);
totals[c] = cnt[c][n]/2;
}
string sol;
int start = 0;
while ((int)sol.size() < n/2) {
assert(start < n);
for (int c=0; c<26; ++c) {
if (Acnt[c] == totals[c]) continue;
int p = nxt[start][c];
if (p == -1) continue;
bool ok = true;
for (int j=0; j<26; ++j) {
if (shuffcnt[j]+(cnt[j][p]-cnt[j][start]) > totals[j]) {
ok = false;
break;
}
}
if (ok) {
sol += char(c + 'a');
for (int j=0; j<26; ++j) {
shuffcnt[j] += cnt[j][p] - cnt[j][start];
}
++Acnt[c];
start = p + 1;
break;
}
}
}
assert(int(sol.size()) == n/2);
cout << sol << '\n';
vector<int> tst(26);
for (int i=0; i<(int)sol.size(); ++i) {
++tst[sol[i]-'a'];
}
for (int j=0; j<26; ++j) {
assert(tst[j] == totals[j]);
}
return 0;
}
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
import java.math.*;
public class Solution implements Runnable {
static BufferedReader in;
static PrintWriter out;
static StringTokenizer st;
static Random rnd;
private void solve() throws IOException {
int tests = 1;
for (int test = 0; test < tests; test++)
solveOne();
}
private void solveOne() throws IOException {
String s = nextToken();
s = reverseString(s);
final int alphaSize = 26;
int[] count = new int[alphaSize];
for (int i = 0; i < s.length(); i++)
++count[s.charAt(i) - 'a'];
int needLength = 0;
for (int i = 0; i < alphaSize; i++) {
if (count[i] % 2 != 0)
throw new AssertionError();
count[i] /= 2;
needLength += count[i];
}
StringBuilder result = new StringBuilder();
int[][] counts = new int[s.length()][alphaSize];
for (int i = s.length() - 1; i >= 0; i--) {
for (int j = 0; j < alphaSize; j++)
counts[i][j] = (i + 1 == s.length() ? 0 : counts[i + 1][j]);
counts[i][s.charAt(i) - 'a']++;
}
int leftPointer = 0;
for (int it = 0; it < needLength; it++) {
int resultIndex = -1;
for (int i = leftPointer; i < s.length(); i++) {
// out.println(it + " " + i + " " + resultIndex);
if (count[s.charAt(i) - 'a'] > 0) {
if (resultIndex == -1
|| s.charAt(i) < s.charAt(resultIndex)) {
if (isOk(count, counts[i]))
resultIndex = i;
}
}
}
result.append(s.charAt(resultIndex));
--count[s.charAt(resultIndex) - 'a'];
leftPointer = resultIndex + 1;
// out.println(resultIndex + " " + result);
// out.flush();
}
out.println(result);
}
private boolean isOk(int[] a, int[] b) {
for (int i = 0; i < a.length; i++)
if (a[i] > b[i])
return false;
return true;
}
private String reverseString(String s) {
return new StringBuilder(s).reverse().toString();
}
public static void main(String[] args) {
new Solution().run();
}
public void run() {
try {
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
rnd = new Random();
solve();
out.close();
} catch (IOException e) {
e.printStackTrace();
System.exit(1);
}
}
private String nextToken() throws IOException {
while (st == null || !st.hasMoreTokens()) {
String line = in.readLine();
if (line == null)
return null;
st = new StringTokenizer(line);
}
return st.nextToken();
}
private int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
private long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
private double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
}
Solution in Python :
In Python3 :
from collections import defaultdict
S = input()
S = S[::-1]
count = defaultdict(int)
for c in S:
count[c] += 1
need = {}
for c in count:
need[c] = count[c] / 2
solution = []
i = 0
while len(solution) < len(S) / 2:
min_char_at = -1
while True:
c = S[i]
if need[c] > 0 and (min_char_at < 0 or c < S[min_char_at]):
min_char_at = i
count[c] -= 1
if count[c] < need[c]:
break
i += 1
for j in range(min_char_at+1, i+1):
count[S[j]] += 1
need[S[min_char_at]] -= 1
solution.append(S[min_char_at])
i = min_char_at + 1
print(''.join(solution))
View More Similar Problems
Self-Driving Bus
Treeland is a country with n cities and n - 1 roads. There is exactly one path between any two cities. The ruler of Treeland wants to implement a self-driving bus system and asks tree-loving Alex to plan the bus routes. Alex decides that each route must contain a subset of connected cities; a subset of cities is connected if the following two conditions are true: There is a path between ever
View Solution →Unique Colors
You are given an unrooted tree of n nodes numbered from 1 to n . Each node i has a color, ci. Let d( i , j ) be the number of different colors in the path between node i and node j. For each node i, calculate the value of sum, defined as follows: Your task is to print the value of sumi for each node 1 <= i <= n. Input Format The first line contains a single integer, n, denoti
View Solution →Fibonacci Numbers Tree
Shashank loves trees and math. He has a rooted tree, T , consisting of N nodes uniquely labeled with integers in the inclusive range [1 , N ]. The node labeled as 1 is the root node of tree , and each node in is associated with some positive integer value (all values are initially ). Let's define Fk as the Kth Fibonacci number. Shashank wants to perform 22 types of operations over his tree, T
View Solution →Pair Sums
Given an array, we define its value to be the value obtained by following these instructions: Write down all pairs of numbers from this array. Compute the product of each pair. Find the sum of all the products. For example, for a given array, for a given array [7,2 ,-1 ,2 ] Note that ( 7 , 2 ) is listed twice, one for each occurrence of 2. Given an array of integers, find the largest v
View Solution →Lazy White Falcon
White Falcon just solved the data structure problem below using heavy-light decomposition. Can you help her find a new solution that doesn't require implementing any fancy techniques? There are 2 types of query operations that can be performed on a tree: 1 u x: Assign x as the value of node u. 2 u v: Print the sum of the node values in the unique path from node u to node v. Given a tree wi
View Solution →Ticket to Ride
Simon received the board game Ticket to Ride as a birthday present. After playing it with his friends, he decides to come up with a strategy for the game. There are n cities on the map and n - 1 road plans. Each road plan consists of the following: Two cities which can be directly connected by a road. The length of the proposed road. The entire road plan is designed in such a way that if o
View Solution →