Have you heard of YANDEX? | Fixed Size Sliding Window Problem - Permutation in String - Leetcode 567

dynamic programming, leetcode, coding interview question, data structures, data structures and algorithms, faang

[GregHogg]

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[sanjeevsinghrajput]

Yandex have a pretty good reverse image search

[timuchin]

It is a russian company. Basically, russian Google

[dial-upking]

Only difference I had was using a map instead of just a basic array for the frequency graphs. If I had to guess, the array is probably better in both compute time and memory compared to a map with it's overhead, but it only works assuming the strings only contain the 26 English lowercase alphabet characters. Here would be my Java submission:

private static final BiFunction<Integer, Integer, Integer> ADD = (a, b) -> {
int s = a + b;
return s == 0 ? null : s;
};

private static boolean perm(String s1, String s2) {
HashMap<Character, Integer> f1 = new HashMap<>(), f2 = new HashMap<>();
for (char c : s1.toCharArray()) {
f1.merge(c, 1, ADD);
}
char[] s2a = s2.toCharArray();
int l1 = s1.length(), l2 = s2a.length, n = l1;
for (byte i = 0; i < l1; ++i) {
f2.merge(s2a[i], 1, ADD);
}
do {
if (f1.equals(f2)) {
return true;
}
f2.merge(s2a[n - l1], -1, ADD);
f2.merge(s2a[n], 1, ADD);
} while (++n < l2);
return false;
}

[MichaelOnayan]

love your content it's fantastic and full of valuable lessons

[SKOBRY]

Isn’t "d", "c" and "" also permutations of "dc"?

[Daimakaicho]

and nobody said it cannot contain capital letters, numbers, chars in that string not forgetting some countriers wierd characters basically unicode ...

[dohunkim]

how efficient is it to use list instead of hashmap in this case? ive heard that using map is better when it comes to alphabets but I dont get why

[ivanmorales]

"...well neither have I" xD!

[michaelmatveev]

lol have you heard of Yandex? ?? That’s gold.

I’ll say that no one should know them

[micah]

Can't u just do something like this:

1. If s1 is longer then s2 return false
2. For each char x in s1
-> if s2 contains x
-> -> remove x from s2
-> else return false
Return true