Shortest Window Substring in Order πŸ‚ - 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.


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

Example 1


s = "aaaaabcbbdefghijklmnopqrstuvwxyzzz"




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

Example 2


s = "zyxwvutsrqponmlkjihgfedcba"




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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