Introduction to Hashing

Hashing is a technique in Competitive Programming which is mostly used with string. In many cases, using hash can relieve you from doing other complex data structures or complicated dynamic programming optimization.

A drawback of hashing is that: you are not completely sure that your solution will work on all cases. If you are really really unlucky, your solution may fail one test-case and receive Wrong Answer. But the chance of failing is very low, and you can even lower that chance by using multiple hashes. So, in the scope of a programming contest, hashing is acceptable and can give you generous advantages. When applicable, hashing often helps you solve problems with a shorter time compared to the official solution.

It’s about 90% when we use Hash that we only care about lower-case English letters (not upper-case letters, digits nor other symbols). So in this post, I will assume that the string we hash contains only lower-case letters.

Hash function

There are several useful hash-functions. I will introduce in this post the most common one. It is common because it is simple and versatile. Let’s see:

Given a string s contains only lower-case English letters. As the length of s is s.length, we have:

s = s[0..(s.length-1)]

What we will do is to calculate an array, let’s name it ‘hash’, which has the same size as s, and hash[i] is a number that represents the substring s[0..i]. What I mean by “represents” is: when we compare 2 strings (or 2 substrings), we don’t need to loop through every character of the 2 strings (or substrings) to check if they are equal. Instead, we can compare their hash values. If their hash values are different, we are 100% sure that the 2 strings (or substrings) are different. If their hash values are equal, then the 2 strings or (substrings) are probably equal. 2 identical strings will have the same hash value. 2 different strings usually have different hash values, but there is very little chance that they can have the same hash value.

Back to the case of string s. Because s contains only lower-case letters (from a-z), so it can contain only 26 different characters. Let’s encode a equals to 1, b equals to 2, etc , z equals to 26. We will obtain array hash as follow:

hash[0] = s[0]

hash[1] = hash[0] + s[1]*26

hash[2] = hash[1] + s[2]*26^2

hash[3] = hash[2] + s[3]*26^3

hash[s.length – 1] = hash[s.length – 2] + s[s.length – 1]*26^{s.length - 1}

For example, let s = “apple”. We have the array hash as:

hash[0] = 1 (as ‘a’ = 1)

hash[1] = 1 + 16*26 = 417

hash[2] = 417 + 16*26^2 = 11233

hash[3] = 11233 + 12*26^3 = 222145

hash[4] = 222145 + 5*26^4 = 2507025

Finally, hash = [1, 417, 11233, 222145, 2597025].

We can model this process on a table:

Encode char to int11616125
Multiplier at index12626^226^326^4

As we said above, hash[i] is a representation of s[0..i]. So for example, the hash value of ‘app’ is 11233, the hash value of ‘appl’ is 222145, etc.
But what is the hash value of ‘ppl’ (‘ppl’ is s[1..3])? It turns out that we can easily compute it using the hash array.

Let’s recall, normally, we will compute the hash value of ‘ppl’ as:
Hash value of ‘ppl’ = 16 + 16*26 + 12*26^2 = 8544. Or:

Encode char to int161612
Multiplier at index12626^2

But wait, we already had the hash array of string ‘apple’. And the table of ‘apple’ and ‘ppl’ have something very similar. This is your exercise: let’s compare the 2 tables and try to obtain the hash value of ‘ppl’ using the table of ‘apple’!

And here is the result: On the table of ‘apple’, first we will remove the index 4 because the character ‘e’ has nothing to do with ‘ppl’. Then, we also remove the index 0, because ‘a’ is also not in ‘ppl’. Then the table becomes:

Encode char to int161612
Multiplier at index12626^226^326^4

Ok, we have just the string ‘ppl’ on the table now. But we have to do one more thing: shift the string ‘ppl’ to the left. Because as in the table of ‘ppl’, its characters start from index position 0.

By shifting to the left 1 unit, we just have to divide the hash value for 26. That is divide 222144 by 26, and we will get 8544, which is the hash value of ‘ppl’.

Formally, we have:

hash value of s[i..j]  = \frac{hash[j] - hash[i-1]}{26^i}

This computation is done in just O(1), given we pre-calculated array hash and 26^i for all i from 0 to s.length – 1.

MOD value

As you might have noticed, when we do as above, if the length of string s is longer, values in our array hash will be overflowed, even if we use long long type. That’s why we need a MOD value. That is, in each computation above, we always take the result mod the MOD value.

hash[i] =(hash[i-1] + s[i]*26*i) \mod MOD, for 1 \leq i \leq s.length -1.
hash value of s[i..j]  =
((hash[j] - hash[i-1]) *modular-inverse(26^i)) \mod MOD.
(If you don’t know about modular-inverse, just search for “Fermat’s Little Theorem and Modular Inverse” on the internet.)

This will prevent our values from overflowing.

My code for Hash function is below.

struct HashFunction 
    string s;
    ll *hash;
    ll *modular_inverse;
    ll *pexp;
    ll p, HMOD;
    ll mul_mod(ll x, ll y, ll MOD) {
        if (MOD < (1<<31)) {
            if (x > MOD)
                x %= MOD;
            if (y > MOD)
                y %= MOD;
            return (x * y) % MOD;
        else {
            ll res = 0, tmp = x%MOD;
            while (y > 0) {
                if (y&1)
                    res = (res + tmp) % MOD;
                y >>= 1;
                tmp = (tmp + tmp) % MOD;
            return res;
    ll pow_mod(ll x, ll exp, ll MOD) {
        if (exp == 0)
            return 1;
        if (x > MOD)
            x %= MOD;
        ll half = pow_mod(x, exp >> 1, MOD);
        if (exp & 1) {
            return mul_mod(mul_mod(half, half, MOD), x, MOD);
        else {
            return mul_mod(half, half, MOD);
    HashFunction(string s_, ll p_ = 26, ll HMOD_ = 1000000007) : s(s_), p(p_), HMOD(HMOD_) {
        // compute hashing array and modular-inverse
        hash = new ll[s.size()];
        modular_inverse = new ll[s.size()];
        ll seed = 1;
        hash[0] = s[0] - '`';
        modular_inverse[0] = 1;
        pexp[0] = 1;
        for (int i = 1; i < s.size(); ++i) {
            seed = (seed * p) % HMOD;
            hash[i] =  (hash[i-1] + (s[i] - '`')*seed) % HMOD;
            modular_inverse[i] = pow_mod(seed, HMOD-2, HMOD);
            pexp[i] = pexp[i-1] * p % MOD;
    ll getHash(int i, int j) { // get hash of s[i..j]
        if (i == 0)
            return hash[j];
            return ((hash[j] - hash[i-1] + HMOD) * modular_inverse[i]) % HMOD;
    ll isEqualSubstring(int i, int j, int u, int v) { // compare s[i..j] and s[u..v]
        if (i > u) {
            swap(i, u);
            swap(j, v);
        ll hashFirst, hashSecond;
        if (i == 0)
            hashFirst = hash[j];
            hashFirst = ((hash[j] - hash[i-1] + HMOD) % HMOD);
        if (u == 0)
            hashSecond = hash[v];
            hashSecond = (hash[v] - hash[u-1] + HMOD) % HMOD;
        return (hashFirst * pexp[u-i] % HMOD) == hashSecond ;

// Example:
// string s = "onetwothreetwoone";
// HashFunction hash(s);
// cout << hash.isEqualSubstring(0, 2, 14, 16) << endl;
// cout << hash.isEqualSubstring(0, 2, 13, 15) << endl;

Practice problems

Codeforces 126B

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