Total numbers having frequency k in a given range

Question

How to find total numbers having frequency=k in a particular range(l,r) in a given array. There are total 10^5 queries of format l,r and each query is built on the basis of previous query's answer. In particular, after each query we increment l by the result of the query, swapping l and r if l > r. Note that 0<=a[i]<=10^9. Total elements in array is n=10^5.

My Attempt:

n,k,q = map(int,input().split())
a = list(map(int,input().split()))
ans = 0
for _ in range(q):
    l,r = map(int,input().split())
    l+=ans
    l%=n
    r+=ans
    r%=n
    if l>r:
        l,r = r,l
    d = {}
    for i in a[l:r+1]:
        try:
            d[i]+=1
        except:
            d[i] = 1
    curr_ans = 0
    for i in d.keys():
        if d[i]==k:
            curr_ans+=1
    ans = curr_ans
    print(ans)

Sample Input:
5 2 3
7 6 6 5 5
0 4
3 0
4 1

Sample Output:
2
1
1

Answer 1

If the number of different values in the array is not too large, you may consider storing arrays as long as the input array, one per unique value, counting the number of appearances of the value until each point. Then you just need to subtract the end values from the beginning values to find how many frequency matches are there:

def range_freq_queries(seq, k, queries):
    n = len(seq)
    c = freq_counts(seq)
    result = [0] * len(queries)
    offset = 0
    for i, (l, r) in enumerate(queries):
        result[i] = range_freq_matches(c, offset, l, r, k, n)
        offset = result[i]
    return result

def freq_counts(seq):
    s = {v: i for i, v in enumerate(set(seq))}
    counts = [None] * (len(seq) + 1)
    counts[0] = [0] * len(s)
    for i, v in enumerate(seq, 1):
        counts[i] = list(counts[i - 1])
        j = s[v]
        counts[i][j] += 1
    return counts

def range_freq_matches(counts, offset, start, end, k, n):
    start, end = sorted(((start + offset) % n, (end + offset) % n))
    num = 0
    return sum(1 for cs, ce in zip(counts[start], counts[end + 1]) if ce - cs == k)

seq = [7, 6, 6, 5, 5]
k = 2
queries = [(0, 4), (3, 0), (4, 1)]
print(range_freq_queries(seq, k, queries))
# [2, 1, 1]

You can do it faster with NumPy, too. Since each result depends on the previous one, you will have to loop in any case, but you can use Numba to really accelerate things up:

import numpy as np
import numba as nb

def range_freq_queries_np(seq, k, queries):
    seq = np.asarray(seq)
    c = freq_counts_np(seq)
    return _range_freq_queries_np_nb(seq, k, queries, c)

@nb.njit  # This is not necessary but will make things faster
def _range_freq_queries_np_nb(seq, k, queries, c):
    n = len(seq)
    offset = np.int32(0)
    out = np.empty(len(queries), dtype=np.int32)
    for i, (l, r) in enumerate(queries):
        l = (l + offset) % n
        r = (r + offset) % n
        l, r = min(l, r), max(l, r)
        out[i] = np.sum(c[r + 1] - c[l] == k)
        offset = out[i]
    return out

def freq_counts_np(seq):
    uniq = np.unique(seq)
    seq_pad = np.concatenate([[uniq.max() + 1], seq])
    comp = seq_pad[:, np.newaxis] == uniq
    return np.cumsum(comp, axis=0)

seq = np.array([7, 6, 6, 5, 5])
k = 2
queries = [(0, 4), (3, 0), (4, 1)]
print(range_freq_queries_np(seq, k, queries))
# [2 1 2]

Let's compare it with the original algorithm:

from collections import Counter

def range_freq_queries_orig(seq, k, queries):
    n = len(seq)
    ans = 0
    counter = Counter()
    out = [0] * len(queries)
    for i, (l, r) in enumerate(queries):
        l += ans
        l %= n
        r += ans
        r %= n
        if l > r:
            l, r = r, l
        counter.clear()
        counter.update(seq[l:r+1])
        ans = sum(1 for v in counter.values() if v == k)
        out[i] = ans
    return out

Here is a quick test and timing:

import random
import numpy

# Make random input
random.seed(0)
seq = random.choices(range(1000), k=5000)
queries = [(random.choice(range(len(seq))), random.choice(range(len(seq))))
           for _ in range(20000)]
k = 20
# Input as array for NumPy version
seq_arr = np.asarray(seq)
# Check all functions return the same result
res1 = range_freq_queries_orig(seq, k, queries)
res2 = range_freq_queries(seq, k, queries)
print(all(r1 == r2 for r1, r2 in zip(res1, res2)))
# True
res3 = range_freq_queries_np(seq_arr, k, queries)
print(all(r1 == r3 for r1, r3 in zip(res1, res3)))
# True

# Timings
%timeit range_freq_queries_orig(seq, k, queries)
# 3.07 s ± 1.11 s per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit range_freq_queries(seq, k, queries)
# 1.1 s ± 307 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit range_freq_queries_np(seq_arr, k, queries)
# 265 ms ± 726 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)

Obviously the effectiveness of this depends on the characteristics of the data. In particular, if there are fewer repeated values the time and memory cost to construct the counts table will approach O(n²).

Answer 2

Let's say the input array is A, |A|=n. I'm going to assume that the number of distinct elements in A is much smaller than n.

We can divide A into sqrt(n) segments each of size sqrt(n). For each of these segments, we can calculate a map from element to count. Building these maps takes O(n) time.

With that preprocessing done, we can answer each query by adding together all the maps wholly contained in (l,r), of which there are at most sqrt(n), then adding any extra elements (or going one segment over and subtracting), also sqrt(n).

If there are k distinct elements, this takes O(sqrt(n) * k) so in the worst case O(n) if in fact every element of A is distinct.

You can keep track of the elements that have the desired count while combining the hashes and extra elements.