Morgan and a String
Problem Statement :
Jack and Daniel are friends. Both of them like letters, especially uppercase ones. They are cutting uppercase letters from newspapers, and each one of them has his collection of letters stored in a stack. One beautiful day, Morgan visited Jack and Daniel. He saw their collections. He wondered what is the lexicographically minimal string made of those two collections. He can take a letter from a collection only when it is on the top of the stack. Morgan wants to use all of the letters in their collections. Note the choice when there was a tie at CA and CF. Function Description Complete the morganAndString function in the editor below. morganAndString has the following parameter(s): string a: Jack's letters, top at index 0 string b: Daniel's letters, top at index 0 Returns - string: the completed string Input Format The first line contains the an integer t, the number of test cases. The next t pairs of lines are as follows: - The first line contains string a - The second line contains string b. Constraints 1 <= T <= 5 1 <= | a |, | b | <= 10^5 a and b contain upper-case letters only, ascii[A-Z].
Solution :
Solution in C :
In C++ :
// macros {{{
#include <bits/stdc++.h>
using namespace std;
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())
#define BIT(n) ((1LL) << (long long)(n))
#define FOR(i,c) for (auto i=(c).begin(); i != (c).end(); ++i)
#define REP(i,n) for (int i = 0; i < (int)(n); ++i)
#define REP1(i,a,b) for (int i=(int)(a); i <= (int)(b); ++i)
#define MP make_pair
#define PB push_back
#define Fst first
#define Snd second
#ifdef WIN32
#define LLD "%I64d"
#else
#define LLD "%lld"
#endif
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> PII;
typedef vector<int> VI;
#define runtime() ((double)clock() / CLOCKS_PER_SEC)
const double eps = 1e-7;
// }}}
#define MAX 100005
#define BASE 27
#define MOD 1000000007
char sa[MAX];
char sb[MAX];
ll ta[MAX];
ll tb[MAX];
ll pw[MAX];
int n, m;
ll getHash(int l, int r, ll table[])
{
ll tr = l > 0 ? table[l-1] : 1;
return (table[r] - pw[r - l + 1] * tr % MOD + MOD) % MOD;
}
bool isSame(int a, int b, int l, int r)
{
return getHash(a, b, ta) == getHash(l, r, tb);
}
bool check(int i, int j)
{
int l = 0, r = min(n - i - 1, m - j - 1);
int ans = r + 1;
while (l <= r)
{
int mid = (l + r) / 2;
if (isSame(i, i+mid, j, j+mid))
l = mid + 1;
else
r = mid - 1, ans = mid;
}
if (not sa[i + ans]) return false;
if (not sb[j + ans]) return true;
return sa[i + ans] <= sb[j + ans];
}
int main()
{
int T;
scanf("%d", &T);
pw[0] = 1;
for (int i = 1; i < MAX; ++i)
pw[i] = pw[i-1] * BASE % MOD;
while (T--)
{
scanf("%s %s", sa, sb);
n = strlen(sa);
m = strlen(sb);
ll hash;
hash = 1;
for (int i = 0; sa[i]; ++i)
{
hash = (hash * BASE + sa[i] - 'A') % MOD;
ta[i] = hash;
}
hash = 1;
for (int i = 0; sb[i]; ++i)
{
hash = (hash * BASE + sb[i] - 'A') % MOD;
tb[i] = hash;
}
string ans;
int i = 0, j = 0;
while (sa[i] and sb[j])
{
if (check(i, j)) // i <= j
ans.push_back(sa[i++]);
else
ans.push_back(sb[j++]);
}
while (sa[i])
ans.push_back(sa[i++]);
while (sb[j])
ans.push_back(sb[j++]);
printf("%s\n", ans.c_str());
}
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
private static final int MAXSIZE = 200100;
private static final int ALPHABET = 128;
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int testcase = Integer.parseInt(in.nextLine());
for (int i = 0; i < testcase; i++) {
String str1 = in.nextLine();
String str2 = in.nextLine();
System.out.println(solution(str1 + "a", str2 + "b"));
}
}
private static List<Integer> buildSuffixArray(String str) {
int[] p = new int[MAXSIZE];
int[] c = new int[MAXSIZE];
int[] cnt = new int[MAXSIZE];
int[] pn = new int[MAXSIZE];
int[] cn = new int[MAXSIZE];
Arrays.fill(cnt, 0);
int n = str.length();
for (int i = 0; i < n; i++) {
++cnt[str.charAt(i)];
}
for (int i = 1; i < ALPHABET; i++) {
cnt[i] += cnt[i - 1];
}
for (int i = 0; i < n; i++) {
p[--cnt[str.charAt(i)]] = i;
}
int count = 1;
c[p[0]] = count - 1;
for (int i = 1; i < n; i++) {
if (str.charAt(p[i]) != str.charAt(p[i - 1])) {
++count;
}
c[p[i]] = count - 1;
}
for (int h = 0; (1 << h) < n; ++h) {
for (int i =0; i < n; i++) {
pn[i] = p[i] - (1 << h);
if (pn[i] < 0) {
pn[i] += n;
}
}
Arrays.fill(cnt, 0);
for (int i = 0; i < n; i++) {
++cnt[c[i]];
}
for (int i = 1; i < count; i++) {
cnt[i] += cnt[i - 1];
}
for (int i = n - 1; i >= 0; i--) {
p[--cnt[c[pn[i]]]] = pn[i];
}
count = 1;
cn[p[0]] = count - 1;
for (int i = 1; i < n; i++) {
int pos1 = (p[i] + (1 << h)) % n;
int pos2 = (p[i - 1] + (1 << h)) % n;
if (c[p[i]] != c[p[i - 1]] || c[pos1] != c[pos2]) {
++count;
}
cn[p[i]] = count - 1;
}
for (int i = 0; i < n; i++) {
c[i] = cn[i];
}
}
List<Integer> res = new ArrayList<Integer>(n);
for (int i = 0; i < n; i++) {
res.add(c[i]);
}
return res;
}
private static String solution(String str1, String str2) {
StringBuilder sb = new StringBuilder(str1).append(str2);
List<Integer> suffix = buildSuffixArray(sb.toString());
StringBuilder rst = new StringBuilder();
int start1 = 0;
int start2 = 0;
while (start1 < str1.length() - 1 || start2 < str2.length() - 1) {
if (start1 >= str1.length() - 1) {
rst.append(str2.charAt(start2++));
continue;
}
if (start2 >= str2.length() - 1) {
rst.append(str1.charAt(start1++));
continue;
}
if (suffix.get(start1) < suffix.get(str1.length() + start2)) {
rst.append(str1.charAt(start1++));
} else {
rst.append(str2.charAt(start2++));
}
}
return rst.toString();
}
}
In C :
#include <stdio.h>
#include <string.h>
int is_a(const char* a, const char* b) {
while (*a && *b) {
if (*a == *b) {
a++;
b++;
} else {
return *a < *b;
}
}
return *a != 0;
}
void merge(const char* a, const char* b, char* c) {
int is_a_preferred = is_a(a, b);
while (*a && *b) {
if (*a == *b) {
*c = (is_a_preferred)? *(a++) : *(b++);
} else {
*c = (*a < *b)? *(a++) : *(b++);
is_a_preferred = is_a(a, b);
}
++c;
}
while (*a) {
*c++ = *a++;
}
while (*b) {
*c++ = *b++;
}
*c = '\0';
}
int main() {
int t;
char a[100001];
char b[100001];
char c[200001];
scanf("%d", &t);
while (t--) {
scanf("%s\n%s", a, b);
merge(a, b, c);
printf("%s\n", c);
}
}
In Python3 :
from collections import deque
from builtins import range
from builtins import input
for _ in range(int(input())):
a = input() + '['
b = input() + '['
string = deque()
while len(a) > 1 and len(b) > 1:
if a < b:
string.append(a[0])
a = a[1:]
else:
string.append(b[0])
b = b[1:]
print(''.join(string) + a[:-1] + b[:-1])
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