LeetCode #2845 — MEDIUM

Count of Interesting Subarrays

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given a 0-indexed integer array nums, an integer modulo, and an integer k.

Your task is to find the count of subarrays that are interesting.

A subarray nums[l..r] is interesting if the following condition holds:

Let cnt be the number of indices i in the range [l, r] such that nums[i] % modulo == k. Then, cnt % modulo == k.

Return an integer denoting the count of interesting subarrays.

Note: A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [3,2,4], modulo = 2, k = 1
Output: 3
Explanation: In this example the interesting subarrays are: 
The subarray nums[0..0] which is [3]. 
- There is only one index, i = 0, in the range [0, 0] that satisfies nums[i] % modulo == k. 
- Hence, cnt = 1 and cnt % modulo == k.  
The subarray nums[0..1] which is [3,2].
- There is only one index, i = 0, in the range [0, 1] that satisfies nums[i] % modulo == k.  
- Hence, cnt = 1 and cnt % modulo == k.
The subarray nums[0..2] which is [3,2,4]. 
- There is only one index, i = 0, in the range [0, 2] that satisfies nums[i] % modulo == k. 
- Hence, cnt = 1 and cnt % modulo == k. 
It can be shown that there are no other interesting subarrays. So, the answer is 3.

Example 2:

Input: nums = [3,1,9,6], modulo = 3, k = 0
Output: 2
Explanation: In this example the interesting subarrays are: 
The subarray nums[0..3] which is [3,1,9,6]. 
- There are three indices, i = 0, 2, 3, in the range [0, 3] that satisfy nums[i] % modulo == k. 
- Hence, cnt = 3 and cnt % modulo == k. 
The subarray nums[1..1] which is [1]. 
- There is no index, i, in the range [1, 1] that satisfies nums[i] % modulo == k. 
- Hence, cnt = 0 and cnt % modulo == k. 
It can be shown that there are no other interesting subarrays. So, the answer is 2.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
1 <= modulo <= 10⁹
0 <= k < modulo

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums, an integer modulo, and an integer k. Your task is to find the count of subarrays that are interesting. A subarray nums[l..r] is interesting if the following condition holds: Let cnt be the number of indices i in the range [l, r] such that nums[i] % modulo == k. Then, cnt % modulo == k. Return an integer denoting the count of interesting subarrays. Note: A subarray is a contiguous non-empty sequence of elements within an array.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[3,2,4]
2
1

Example 2

[3,1,9,6]
3
0

Core Insight

What unlocks the optimal approach

The problem can be solved using prefix sums.
Let <code>count[i]</code> be the number of indices where <code>nums[i] % modulo == k</code> among the first <code>i</code> indices.
<code>count[0] = 0</code> and <code>count[i] = count[i - 1] + (nums[i - 1] % modulo == k ? 1 : 0)</code> for <code>i = 1, 2, ..., n</code>.
Now we want to calculate for each <code>i = 1, 2, ..., n</code>, how many indices <code>j < i</code> such that <code>(count[i] - count[j]) % modulo == k</code>.
Rewriting <code>(count[i] - count[j]) % modulo == k</code> becomes <code>count[j] = (count[i] + modulo - k) % modulo</code>.
Using a map data structure, for each <code>i = 0, 1, 2, ..., n</code>, we just sum up all <code>map[(count[i] + modulo - k) % modulo]</code> before increasing <code>map[count[i] % modulo]</code>, and the total sum is the final answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2845: Count of Interesting Subarrays
class Solution {
    public long countInterestingSubarrays(List<Integer> nums, int modulo, int k) {
        int n = nums.size();
        int[] arr = new int[n];
        for (int i = 0; i < n; ++i) {
            arr[i] = nums.get(i) % modulo == k ? 1 : 0;
        }
        Map<Integer, Integer> cnt = new HashMap<>();
        cnt.put(0, 1);
        long ans = 0;
        int s = 0;
        for (int x : arr) {
            s += x;
            ans += cnt.getOrDefault((s - k + modulo) % modulo, 0);
            cnt.merge(s % modulo, 1, Integer::sum);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2845: Count of Interesting Subarrays
func countInterestingSubarrays(nums []int, modulo int, k int) (ans int64) {
	arr := make([]int, len(nums))
	for i, x := range nums {
		if x%modulo == k {
			arr[i] = 1
		}
	}
	cnt := map[int]int{}
	cnt[0] = 1
	s := 0
	for _, x := range arr {
		s += x
		ans += int64(cnt[(s-k+modulo)%modulo])
		cnt[s%modulo]++
	}
	return
}

# Accepted solution for LeetCode #2845: Count of Interesting Subarrays
class Solution:
    def countInterestingSubarrays(self, nums: List[int], modulo: int, k: int) -> int:
        arr = [int(x % modulo == k) for x in nums]
        cnt = Counter()
        cnt[0] = 1
        ans = s = 0
        for x in arr:
            s += x
            ans += cnt[(s - k) % modulo]
            cnt[s % modulo] += 1
        return ans

// Accepted solution for LeetCode #2845: Count of Interesting Subarrays
use std::collections::HashMap;

impl Solution {
    pub fn count_interesting_subarrays(nums: Vec<i32>, modulo: i32, k: i32) -> i64 {
        let mut arr: Vec<i32> = nums
            .iter()
            .map(|&x| if x % modulo == k { 1 } else { 0 })
            .collect();
        let mut cnt: HashMap<i32, i64> = HashMap::new();
        cnt.insert(0, 1);

        let mut ans: i64 = 0;
        let mut s: i32 = 0;

        for x in arr {
            s += x;
            let key = (s - k).rem_euclid(modulo);
            ans += *cnt.get(&key).unwrap_or(&0);
            *cnt.entry(s % modulo).or_insert(0) += 1;
        }

        ans
    }
}

// Accepted solution for LeetCode #2845: Count of Interesting Subarrays
function countInterestingSubarrays(nums: number[], modulo: number, k: number): number {
    const arr: number[] = [];
    for (const x of nums) {
        arr.push(x % modulo === k ? 1 : 0);
    }
    const cnt: Map<number, number> = new Map();
    cnt.set(0, 1);
    let ans = 0;
    let s = 0;
    for (const x of arr) {
        s += x;
        ans += cnt.get((s - k + modulo) % modulo) || 0;
        cnt.set(s % modulo, (cnt.get(s % modulo) || 0) + 1);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.