Step through the algorithm visually — use Play or the step buttons (inspired by AlgoMaster / visualgo).
Oracle interview context: Permutation in String is a Medium Sliding Window problem — Maintain a window [L, R] and update counts as you expand or contract.
Use the animation above to step through each move before writing code.
Pattern: Sliding Window
Read from stdin, write to stdout. Classic interview problem #567.
Permutation in String — Oracle interview prep · Sliding Window
Classic interview problem #567.
Input (stdin)
Line 1: s1\nLine 2: s2
Output (stdout)
true if permutation of s1 in s2
Your program must read from stdin and write the answer to stdout (no extra debug text).
ab eidbaooo
true
Toolliyo Coach
Progressive help: Nudge → Guide → Approach. Full solution stays behind the Solution tab.
| Test | Status | Details |
|---|
Ready — edit the code above and click Run or Submit.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
class Program
{
static int[] Ria(string line = null)
{
line ??= Console.ReadLine();
if (string.IsNullOrWhiteSpace(line)) return Array.Empty<int>();
return line.Trim().Split(new[] { ' ', ',', '\t' }, StringSplitOptions.RemoveEmptyEntries)
.Select(int.Parse).ToArray();
}
static string[] Rsa()
{
int n = int.Parse(Console.ReadLine());
var arr = new string[n];
for (int i = 0; i < n; i++) arr[i] = Console.ReadLine();
return arr;
}
static void W(params object[] parts) => Console.WriteLine(string.Join(" ", parts));
static void Wb(bool v) => Console.WriteLine(v ? "true" : "false");
static void Wi(int v) => Console.WriteLine(v);
static void Ws(string v) => Console.WriteLine(v);
static void Main()
{
string s1 = Console.ReadLine();
string s2 = Console.ReadLine();
var need = new int[26], have = new int[26];
foreach (var c in s1) need[c - 'a']++;
int matches = 0;
for (int i = 0; i < 26; i++) if (need[i] == 0) matches++;
bool found = false;
for (int i = 0; i < s2.Length; i++) {
int idx = s2[i] - 'a';
have[idx]++;
if (have[idx] == need[idx]) matches++;
else if (have[idx] == need[idx] + 1) matches--;
if (i >= s1.Length) {
int outIdx = s2[i - s1.Length] - 'a';
have[outIdx]--;
if (have[outIdx] == need[outIdx]) matches++;
else if (have[outIdx] == need[outIdx] - 1) matches--;
}
if (matches == 26) { found = true; break; }
}
Wb(found);
}
}
Try solving on your own first, then reveal the official answer.