**Look and Say - Google Top Interview Questions**

### Problem Statement :

The "look and say" sequence is defined as follows: beginning with the term 1, each subsequent term visually describes the digits appearing in the previous term. The first few terms are as follows: 1 11 <- 1 one 21 <- 2 ones 1211 <- 1 two, 1 one 111221 <- 1 one, 1 two, 2 ones 312211 <- 3 ones, 2 twos, 1 one Given an integer n, return the nth term of this sequence as a string. Constraints 1 ≤ n ≤ 40 Example 1 Input n = 3 Output "21" Example 2 Input n = 4 Output "1211" Example 3 Input n = 5 Output "111221" Solved 2,691 Attempted 2,822

### Solution :

` ````
Solution in C++ :
string solve(int n) {
string s = "1", t;
for (int i = 1; i < n; i++) {
for (int j = 0, k = 0; j < s.size();) {
char c = s[k];
k++;
while (k < s.size() && s[k] == s[k - 1]) k++;
t.append(to_string(k - j));
t.push_back(c);
j = k;
}
t.swap(s);
t.clear();
}
return s;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public String solve(int n) {
// base case
String s = "1";
// for every k after 1, we need to apply the logic
for (int k = 1; k < n; k++) {
StringBuilder next = new StringBuilder();
// count starts at 1, as we compare previous with current
int count = 1;
for (int i = 1; i <= s.length(); i++) {
// if we reached the end of the string, or the char has changed,
// we append the count and the previous char. restart the count to 1
if (i == s.length() || s.charAt(i) != s.charAt(i - 1)) {
next.append(count).append(s.charAt(i - 1));
count = 1;
} else {
// same char, just increase the count
count++;
}
}
s = next.toString();
}
return s;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, n):
s = "1"
for _ in range(n - 1):
temp = ""
for x, y in groupby(s):
temp += str(len(list(y))) + x
s = temp
return s
```

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