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])
View More Similar Problems
Array Manipulation
Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu
View Solution →Print the Elements of a Linked List
This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode
View Solution →Insert a Node at the Tail of a Linked List
You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink
View Solution →Insert a Node at the head of a Linked List
Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below
View Solution →Insert a node at a specific position in a linked list
Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e
View Solution →Delete a Node
Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo
View Solution →