# Shortest Window Substring in Order 🏂 - Google Top Interview Questions

### Problem Statement :

```Given a lowercase alphabet string s, return the length of the shortest substring containing all alphabet characters in order from "a" to "z". If there's no solution, return -1.

Constraints

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

Example 1

Input

s = "aaaaabcbbdefghijklmnopqrstuvwxyzzz"

Output

28

Explanation

The shortest such substring is "abcbbdefghijklmnopqrstuvwxyz". The two additional "b"s contribute to
the 2 extra characters.

Example 2

Input

s = "zyxwvutsrqponmlkjihgfedcba"

Output

-1

Explanation

Even though this string has all the characters in the alphabet, it's not in order from "a" to "z".```

### Solution :

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

int solve(string s) {
for (int i = 0; i < s.length(); i++) {
}
int ans = 1e7;
for (auto& r : adj) sort(r.begin(), r.end());
for (int i = 0; i < adj.size(); i++) {
while (curr != 26) {
int temp_next =
curr += 1;
}
if (curr == 26) ans = min(ans, next - s + 1);
}
return ans == 1e7 ? -1 : ans;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(String s) {
int[] dp0 = new int, dp1 = new int; // position, length
Arrays.fill(dp0, -1);
int res = Integer.MAX_VALUE;
for (int j = 0; j != s.length(); j++) {
final int ch = s.charAt(j) - 'a';
if (ch == 0) {
dp0[ch] = j;
dp1[ch] = 1;
} else {
if (dp0[ch - 1] != -1) {
dp0[ch] = j;
dp1[ch] = j - dp0[ch - 1] + dp1[ch - 1];
}
if (dp0 != -1) {
res = Math.min(res, dp1);
}
}
}
return res == Integer.MAX_VALUE ? -1 : res;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, S):
inds = [[] for _ in range(26)]
for i, c in enumerate(S):
inds[ord(c) - ord("a")].append(i)

roots = {i: i for i in inds}
for row1, row2 in zip(inds, inds[1:]):
roots2 = {}
i = 0
for jx in row2:
while i < len(row1) and jx > row1[i]:
if row1[i] in roots:
roots2[jx] = roots[row1[i]]
i += 1

roots = roots2

return min(k - v + 1 for k, v in roots.items()) if roots else -1```
```

## Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

## Left Rotation

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d

## Sparse Arrays

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun

## Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

## Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

## Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink