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))
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