There is a problem using the erase function.
when I input aaabbbbcc, only aaa is output.
But I want to print aaacc.
What is the problem with this code?
int main() {
int i = 3;
string a, temp_a;
cin >> a;
a.erase(i, i + 3);
printf("%s", a.c_str());
}
Related
I want to remove K characters from a string such that the occurrence of each character is at a minimum.
For example:
String: abcdefghijkllllll
K: 5
Answer: 12 (abcdeghijkl)
String: ababac
K: 4
Answer: 3 (aac)
String: aaaab
K: 4
Answer: 1(b)
I want to remove 5 characters. Those characters would be 5 l's
What I've done so far is count the occurence of each character using a map
But I'm stuck as to what to do next.
#include <bits/stdc++.h>
using namespace std;
string s;
int l, k;
map<char, int> m;
int main() {
getline(cin, s);
scanf("%d %d", &l, &k);
for(int i=0; i<s.length(); i++) {
m[s[i]]++;
}
for(auto &x : m) {
cout << x.second << "\n";
}
return 0;
}
The expected result is the minimum length of a string after removing the characters of any given string (can be sorted or unsorted).
You can remove any character in the String
Update:
#include <bits/stdc++.h>
using namespace std;
string s;
int l, k;
map<char, int> m;
int main() {
getline(cin, s);
cin >> l >> k;
for(int i = 0; i < s.length(); i++) {
m[s[i]]++;
}
for(auto it = m.end(); it != m.begin(); it--) {
// cout << it->second << "\n";
}
vector<pair<int, int>> pairs;
for (auto itr = m.begin(); itr != m.end(); itr++) {
pairs.push_back(*itr);
}
sort(pairs.begin(), pairs.end(), [=](pair<int, int>& a, pair<int, int>& b) { return a.second < b.second; } );
for(auto it = m.end(); it != m.begin(); it--) {
if(it->second - k >= 1) {
it->second-=k;
k -= it->second;
}
}
int sum = 0;
for(auto it = m.end(); it != m.begin(); it--) {
sum += it->second;
// cout << it->second << "\n";
}
cout << sum << "\n";
return 0;
}
The current problem with this is that it doesn't read all the characters and map them correctly to the map.
I'm uncertain from your description and test cases what you're looking for. Your answer is returning the number of characters remaining in the string, and your updated function returns the sum variable. If that's the case why not just return the length of the string minus k?
Your second test case is:
String: ababac
K: 4
Answer: 3 (aac)
Removing 4 characters from "ababac" (length 6) would give it a length of 2, not 3. How does this work?
Can the characters be removed in any order? For the third test case you have:
String: aaaab
K: 4
Answer: 1(b)
Given the description: I want to remove K characters from a string such that the occurrence of each character is at a minimum. Removing 3 characters gives the result "ab". Removing the 4th could result in either "a" or "b". What do you do in this case?
There's a lot of ambiguity in this question, and the test cases are a bit confusing. For example, given "aabbccdddd" k=3, what would be the accepted answer? "abcdddd" or "aabbccd"? "abcdddd" would increase the number of characters that are at a minimum whereas "aabbccd" would reduce the number of the most frequently occurring character.
I've put together an answer using a max priority queue / max-heap (in Java) with the later example from above. This assumes that all your input is good.
import java.util.*;
public class SO {
//Helper class to put in the priority queue.
class CharInt {
char c;
int count;
public CharInt(char c) {
this.c = c;
this.count = 1;
}
void increment() {
this.count++;
}
void decrement() {
this.count--;
}
}
public int minChar(String s, int k) {
Map<Character, CharInt> map = new HashMap<Character, CharInt>();
for (Character c : s.toCharArray()) {
if (map.get(c) == null) {
map.put(c, new CharInt(c));
}else {
map.get(c).increment();
}
}
//Makes a Max-Heap from a PriorityQueue object. The comparator makes sure the top of the PriorityQueue is the character with the highest count.
PriorityQueue<CharInt> maxHeap = new PriorityQueue<CharInt>(new Comparator<CharInt>() {
#Override
public int compare(CharInt o1, CharInt o2) {
return - Integer.compare(o1.count, o2.count);
}
});
//Add all values to the heap.
for (CharInt c : map.values()) {
maxHeap.add(c);
}
//Take the top value off, decrement its count, add it back to the heap. Do this k times.
while (k-- > 0) {
CharInt c = maxHeap.poll();
c.decrement();
maxHeap.add(c);
}
StringBuilder builder = new StringBuilder(); // Used to make output string. Can be left out.
int sum = 0;
//Remove every element from the heap and get its count value.
while(!maxHeap.isEmpty()) {
CharInt c = maxHeap.poll();
for (int i = 0; i < c.count; i++) {
sum += c.count;
builder.append(c.c); // Used to make output string. Can be left out.
}
}
char[] chars = builder.toString().toCharArray(); // Used to make output string. Can be left out.
Arrays.sort(chars); // Used to make output string. Can be left out.
System.out.println(chars); // Used to make output string. Can be left out.
return sum;
}
public static void main(String...bannaa) {
int s = new SO().minChar("abcdefghijkllllll", 5);
int s2 = new SO().minChar("ababac", 4);
int s3 = new SO().minChar("aaaab", 4);
int s4 = new SO().minChar("abbbccc", 4);
System.out.println(s + " " + s2 + " " + s3 + " " + s4);
}
}
Output:
abcdefghijkl
ac
a
abc
12 2 1 3
For problem statement, please see this.
(I have omitted libraries.)
My code works fine (when online judge tests the code) except for this case:
aaaaaaaaaaaaaaaaaaaaaaaaaaaaa
bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb
I would love to know why output it expects to be 104 when intersection is zero?
std::string string_intersection;
int number_needed(string a, string b) {
int y = a.length();
int z = b.length();
sort(a.begin(), a.end());
sort(b.begin(), b.end());
std::set_intersection(a.begin(), a.end(), b.begin(), b.end(), std::back_inserter(string_intersection));
int num, x = string_intersection.length();
if(x==0)
num = 0;
else
num = y + z - 2*x;
return num;
}
int main(){
string a;
cin >> a;
string b;
cin >> b;
cout << number_needed(a, b) << endl;
return 0;
}
Just change
if(x==0)
num = y+z; //as everything needs to be deleted
else
num = y + z - 2*x;
Shil has a string S , consisting of N lowercase English letters. In one operation, he can delete any pair of adjacent letters with same value. For example, string "aabcc" would become either "aab" or "bcc" after operation.
Shil wants to reduce S as much as possible. To do this, he will repeat the above operation as many times as it can be performed. Help Shil out by finding and printing 's non-reducible form!
If the final string is empty, print Empty String; otherwise, print the final non-reducible string.
Sample Input 0
aaabccddd
Sample Output 0
abd
Sample Input 1
baab
Sample Output 1
Empty String
Sample Input 2
aa
Sample Output 2
Empty String
Explanation
Sample Case 0: Shil can perform the following sequence of operations to get the final string:
Thus, we print .
Sample Case 1: Shil can perform the following sequence of operations to get the final string: aaabccddd -> abccddd
abccddd -> abddd
abddd -> abd
Thus we print abd
Sample case 1: baab -> bb
bb -> Empty String.
in my code in the while loop when i assign s[i] to str[i].the value of s[i] is not getting assigned to str[i].the str[i] has a garbage value.
my code :
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
string s;
cin>>s;
int len = s.length();
int len1 = 0;
string str;
for(int i = 0;i < len-1;i++){
if(s[i]!= '*'){
for(int j=i+1;j < len;j++){
if(s[j] != '*'){
if(s[i] == s[j]){
s[i] = s[j] = '*';
}
}
}
}
}
int i = 0;
while(i<len){
if(s[i] != '*'){
str[len1] = s[i];
len1++;
}
i++;
}
if(len1 != 0){
cout<<str;
}
else{
cout<<"Empty String";
}
return 0;
}
#include <iostream>
using namespace std;
int main(){
string s,tempS;
bool condition = false;
cin >> s;
tempS = s;
while(condition==false){
for(int i=1; i<s.size(); i++){
if(s[i]==s[i-1]){
s.erase(s.begin()+i-1, s.begin()+i+1);
}
}
if(tempS == s){
condition = true;
} else{
tempS = s;
}
}
if(s.size()==0){
cout << "Empty String" ;
} else{
cout << s;
}
return 0;
}
the first while loop keeps on modifying the string until and unless it becomes equal to temp (which is equal to the string pre-modification)
It is comparing the adjacent elements if they are equal or not and then deleting them both.
As soon as string becomes equal to temp after modifications, string has reached it's most reduced state !
I know this is extremely basic but I'm very new to C++ and can't seem to find an answer. I'm simply trying to convert a few integers to strings. This method works:
int a = 10;
stringstream ss;
ss << a;
string str = ss.str();
but when I need to convert the second and third ones like this:
int b = 13;
stringstream ss;
ss << a;
string str2 = ss.str();
int c = 15;
stringstream ss;
ss << b;
string str3 = ss.str();
I get this error:
'std::stringstream ss' previously declared here
Do I need to somehow close stringstream? I've noticed that if I put them far away from each other in my code the compiler doesn't mind but that doesn't seem like something I should do. Does anyone have suggestions?
You're trying to redeclare the same stringstream with the same name. You can modify your code like this in order to work :
int b = 13;
stringstream ss2;
ss2 << a;
string str2 = ss2.str();
Or like this if you don't want to redeclare it :
int b = 13;
ss.str(""); // empty the stringstream
ss.clear();
ss << a;
string str2 = ss.str();
You can also use , which is much quicker :
int c = 42;
std::string s = std::to_string(c);
The stringstream is a variable just like any other. To make the idiom work twice without change, put it in braces:
int a = 13;
string str2;
{
stringstream ss;
ss << a;
str2 = ss.str();
} // ss ceases to exist at the closing brace
int b = 15;
string str3;
{
stringstream ss; // now we can make a new ss
ss << b;
str3 = ss.str();
}
Since C++11, you can do
std::string str4 = std::to_string( b );
or (with c of arbitrary type)
std::string str5 = ( std::stringstream() << c ).str();
There are other solutions, of course.
The particular error you are getting is telling you that you can't declare two variables with the same name (ss in your case) within the same scope.
If you wanted to create a new stringstream you could call it something else.
But there are other ways to convert an int to a string.. you could use std::to_string().
try
#include <string>
int a = 10;
std::string s = std::to_string(a);
Make sure you declare:
using namespace std;
Then insert the code:
int a = 10;
ostringstream ssa;
ssa << a;
string str = ssa.str();
int b = 13;
ostringstream ssb;
ssb << b;
string str2 = ssb.str();
int c = 15;
ostringstream ssc;
ssc << c;
string str3 = ssc.str();
One of the simplest way i can think of doing it is:
string str = string(itoa(num));
How can I add an integer variable to a string and char* variable? for example:
int a = 5;
string St1 = "Book", St2;
char *Ch1 = "Note", Ch2;
St2 = St1 + a --> Book5
Ch2 = Ch1 + a --> Note5
Thanks
The C++ way of doing this is:
std::stringstream temp;
temp << St1 << a;
std::string St2 = temp.str();
You can also do the same thing with Ch1:
std::stringstream temp;
temp << Ch1 << a;
char* Ch2 = new char[temp.str().length() + 1];
strcpy(Ch2, temp.str().c_str());
for char* you need to create another variable that is long enough for both, for instance. You can 'fix' the length of the output string to remove the chance of overrunning the end of the string. If you do that, be careful to make this large enough to hold the whole number, otherwise you might find that book+50 and book+502 both come out as book+50 (truncation).
Here's how to manually calculate the amount of memory required. This is most efficient but error-prone.
int a = 5;
char* ch1 = "Book";
int intVarSize = 11; // assumes 32-bit integer, in decimal, with possible leading -
int newStringLen = strlen(ch1) + intVarSize + 1; // 1 for the null terminator
char* ch2 = malloc(newStringLen);
if (ch2 == 0) { exit 1; }
snprintf(ch2, intVarSize, "%s%i", ch1, a);
ch2 now contains the combined text.
Alternatively, and slightly less tricky and also prettier (but less efficient) you can also do a 'trial run' of printf to get the required length:
int a = 5;
char* ch1 = "Book";
// do a trial run of snprintf with max length set to zero - this returns the number of bytes printed, but does not include the one byte null terminator (so add 1)
int newStringLen = 1 + snprintf(0, 0, "%s%i", ch1, a);
char* ch2 = malloc(newStringLen);
if (ch2 == 0) { exit 1; }
// do the actual printf with real parameters.
snprintf(ch2, newStringLen, "%s%i", ch1, a);
if your platform includes asprintf, then this is a lot easier, since asprintf automatically allocates the correct amount of memory for your new string.
int a = 5;
char* ch1 = "Book";
char* ch2;
asprintf(ch2, "%s%i", ch1, a);
ch2 now contains the combined text.
c++ is much less fiddly, but I'll leave that to others to describe.
You need to create another string large enough to hold the original string followed by the number (i.e. append the character corresponding to each digit of the number to this new string).
Try this out:
char *tmp = new char [ stelen(original) ];
itoa(integer,intString,10);
output = strcat(tmp,intString);
//use output string
delete [] tmp;