# Partition String - Amazon Top Interview Questions

### Problem Statement :

```Given a lowercase alphabet string s, partition s into as many pieces as possible such that each letter appears in at most one piece and return the sizes of the partitions as a list.

Constraints

0 ≤ n ≤ 100,000 where n is the length of s

Example 1

Input

s = "cocoplaydae"

Output

[4, 1, 1, 4, 1]

Explanation

The string is split to["coco", "p", "l", "ayda", "e"].```

### Solution :

```                        ```Solution in C++ :

vector<int> solve(string s) {
// store the last index of all the chars
vector<int> m(26);
for (int i = 0; i < s.size(); i++) {
m[s[i] - 'a'] = i;
}
// integer which stores the index where we have to go(at least) for creating a piece
int curr = 0;
// interger which stores the last index, from where we had started
int last = 0;
vector<int> res;
for (int i = 0; i < s.size(); i++) {
// take the max index where we have to go for creating a piece
curr = max(curr, m[s[i] - 'a']);
if (curr == i) {
// if the last index of current substring is the current index itself, create a piece
res.push_back(curr - last + 1);
// update the last index to next index
last = curr + 1;
}
}
return res;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int[] solve(String s) {
int[] arr = new int;
List<Integer> list = new ArrayList<>();
int end;

for (int i = 0; i < s.length(); i++) {
arr[s.charAt(i) - 'a'] = i;
}

for (int i = 0; i < s.length(); i++) {
end = arr[s.charAt(i) - 'a'];

for (int j = i; j < end; j++) {
end = Math.max(end, arr[s.charAt(j) - 'a']);
}

i = end;
}

int[] res = new int[list.size()];

for (int i = 0; i < res.length; i++) res[i] = list.get(i);

return res;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, s):
last = {c: i for i, c in enumerate(s)}
lo = r = 0
res = []

for hi, c in enumerate(s):
r = max(r, last[c])
if hi == r:
res.append(hi - lo + 1)
lo = r = hi + 1

return res```
```

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