# Bulk Shift Letters - Microsoft Top Interview Questions

### Problem Statement :

```You are given a lowercase alphabet string s and a list of integers shifts.

Each element shifts[i] means to shift the first i + 1 letters of s by shifts[i] positions.

Shifting a letter should wrap over "z" to "a". For example, shifting “z” by 2 results in “b”.

Return the resulting string after applying shifts to s.

Constraints

1 ≤ n ≤ 100,000 where n is the length of s and shifts

Example 1

Input

s = "afz"

shifts = [1, 2, 1]

Output

"eia"
Explanation

We shift the first 1 letter by 1 position to get: "bfz"

We shift the first 2 letters by 2 position to get: "dhz"

We shift the first 3 letters by 1 position to get: "eia"```

### Solution :

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

string solve(string s, vector<int>& shifts) {
int d = 0;
for (int i = s.size(); i-- > 0;) {
(d += shifts[i]) %= 26;
if (d) s[i] = (s[i] - 'a' + d) % 26 + 'a';
}
return s;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public String solve(String s, int[] shifts) {
for (int i = shifts.length - 2; i >= 0; i--) {
shifts[i] += shifts[i + 1];
shifts[i] %= 26;
}
StringBuilder ret = new StringBuilder();
for (int i = 0; i < s.length(); i++) {
ret.append(convert(s.charAt(i), shifts[i]));
}
return ret.toString();
}
//'a' = 97
//'z' = 122
public char convert(char c, int shift) {
int ret = (int) (c) + shift;
if (ret > 122) {
while (ret > 122) {
ret -= 26;
}
}
return (char) (ret);
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, s, shifts):
ns = ""
sum1 = []
count = 0
i = len(shifts) - 1
while i >= 0:
count += shifts[i]
sum1.append(count)
i -= 1
sum2 = sum1[::-1]
for i in range(len(shifts)):
ns = ns + chr(((ord(s[i]) + sum2[i] - ord("a")) % 26) + ord("a"))
return ns```
```

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