**Shortest Window Substring in Order π - 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 β€ n β€ 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) {
vector<vector<int>> adj(26);
for (int i = 0; i < s.length(); i++) {
adj[s[i] - 'a'].push_back(i);
}
int ans = 1e7;
for (auto& r : adj) sort(r.begin(), r.end());
for (int i = 0; i < adj[0].size(); i++) {
int s = adj[0][i], curr = 1, next = adj[0][i];
while (curr != 26) {
int temp_next =
upper_bound(adj[curr].begin(), adj[curr].end(), next) - adj[curr].begin();
if (temp_next == adj[curr].size()) break;
next = adj[curr][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[26], dp1 = new int[26]; // 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[25] != -1) {
res = Math.min(res, dp1[25]);
}
}
}
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[0]}
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
```

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