# String | 450 DSA | Love Babbar

Posted on Jan 1, 0001

### Problem 1: Reverse a string

Write a function that reverses a string. The input string is given as an array of characters s.

#### Worst/ Better

Loop through the string from the back and store the elements in a new string.

Time Complexity: \$ O(n) \$

Auxiliary Space: \$ O(n) \$

#### Optimal

Make two pointers pointing towards the extreme ends and swap till the front < rear.

Time Complexity: \$ O(n) \$

Auxiliary Space: \$ O(1) \$

``````class Solution {
public:
void reverseString(vector<char>& s) {
char ch;
for(int i=0, j= s.size() - 1; i<j; i++, j--){
ch = s[j];
s[j] = s[i];
s[i] = ch;
}
}
};
``````

### Problem 2: Check for palindrome

Given a string S, check if it is palindrome or not.

#### Worst/ Better/Optimal

Make two pointers pointing towards the extreme ends and check if ch[front] == ch[rear] till the front < rear.

Time Complexity: \$ O(n) \$

Auxiliary Space: \$ O(1) \$

``````class Solution {
public:
void reverseString(vector<char>& s) {
char ch;
for(int i=0, j= s.size() - 1; i<j; i++, j--){
ch = s[j];
s[j] = s[i];
s[i] = ch;
}
}
};
``````

### Problem 3: Print duplicates

Print all the duplicates in the input string.

#### Worst/Better/Optimal

Either create a hash-array or a hash-map.

Time Complexity: \$ O(n) \$

Auxiliary Space: \$ O(k) \$ (number of distinct characters)

### Problem 4: Subjective

Why String is Immutable or Final in Java

### Problem 5: To check if strings are rotations of each other or not

Given a string s1 and a string s2, write a snippet to say whether s2 is a rotation of s1? Given s1 = ABCD and s2 = CDAB, return true, given s1 = ABCD, and s2 = ACBD , return false.

#### Using substr

``````bool solve(string s1,string s2){
int n = s1.size(), m = s2.size();
if(n != m) return false;
s1 += s1;
return (s1.find(s2) != string::npos);
}
``````

Time Complexity: \$ O(m*n) \$

Auxiliary Space: \$ O(1) \$

#### Using queue

``````bool solve(string s1, string s2)
{
int n = s1.size(), m = s2.size();
if(n != m) return false;

queue<char> q1;
for (int i = 0; i < n; i++)
q1.push(s[i]);

queue<char> q2;
for (int i = 0; i < m; i++)
q2.push(s2[i]);

while (m--) {
char ch = q2.front();
q2.pop();
q2.push(ch);
if (q2 == q1)
return true;
}
return false;
}
``````

Time Complexity: \$ O(m*n) \$

Auxiliary Space: \$ O(m+n) \$