LeetCode #1248 — MEDIUM

Count Number of Nice Subarrays

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

Solve on LeetCode

The Problem

Problem Statement

Given an array of integers nums and an integer k. A continuous subarray is called nice if there are k odd numbers on it.

Return the number of nice sub-arrays.

Example 1:

Input: nums = [1,1,2,1,1], k = 3
Output: 2
Explanation: The only sub-arrays with 3 odd numbers are [1,1,2,1] and [1,2,1,1].

Example 2:

Input: nums = [2,4,6], k = 1
Output: 0
Explanation: There are no odd numbers in the array.

Example 3:

Input: nums = [2,2,2,1,2,2,1,2,2,2], k = 2
Output: 16

Constraints:

1 <= nums.length <= 50000
1 <= nums[i] <= 10^5
1 <= k <= nums.length

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: Given an array of integers nums and an integer k. A continuous subarray is called nice if there are k odd numbers on it. Return the number of nice sub-arrays.

Baseline thinking

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

Pattern signal: Array · Hash Map · Math · Sliding Window

Example 1

[1,1,2,1,1]
3

Example 2

[2,4,6]
1

Example 3

[2,2,2,1,2,2,1,2,2,2]
2

Related Problems

K Divisible Elements Subarrays (k-divisible-elements-subarrays)
Count Subarrays With Fixed Bounds (count-subarrays-with-fixed-bounds)
Ways to Split Array Into Good Subarrays (ways-to-split-array-into-good-subarrays)
Count of Interesting Subarrays (count-of-interesting-subarrays)

Step 02

Core Insight

What unlocks the optimal approach

After replacing each even by zero and every odd by one can we use prefix sum to find answer ?
Can we use two pointers to count number of sub-arrays ?
Can we store the indices of odd numbers and for each k indices count the number of sub-arrays that contains them ?

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 #1248: Count Number of Nice Subarrays
class Solution {
    public int numberOfSubarrays(int[] nums, int k) {
        int n = nums.length;
        int[] cnt = new int[n + 1];
        cnt[0] = 1;
        int ans = 0, t = 0;
        for (int v : nums) {
            t += v & 1;
            if (t - k >= 0) {
                ans += cnt[t - k];
            }
            cnt[t]++;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1248: Count Number of Nice Subarrays
func numberOfSubarrays(nums []int, k int) (ans int) {
	n := len(nums)
	cnt := make([]int, n+1)
	cnt[0] = 1
	t := 0
	for _, v := range nums {
		t += v & 1
		if t >= k {
			ans += cnt[t-k]
		}
		cnt[t]++
	}
	return
}

# Accepted solution for LeetCode #1248: Count Number of Nice Subarrays
class Solution:
    def numberOfSubarrays(self, nums: List[int], k: int) -> int:
        cnt = Counter({0: 1})
        ans = t = 0
        for v in nums:
            t += v & 1
            ans += cnt[t - k]
            cnt[t] += 1
        return ans

// Accepted solution for LeetCode #1248: Count Number of Nice Subarrays
struct Solution;

impl Solution {
    fn number_of_subarrays(nums: Vec<i32>, k: i32) -> i32 {
        let n = nums.len();
        let mut count = vec![0; n + 1];
        let k = k as usize;
        count[0] = 1;
        let mut prev = 0;
        let mut res = 0;
        for i in 0..n {
            if nums[i] % 2 == 1 {
                prev += 1;
            }
            count[prev] += 1;
            if prev >= k {
                res += count[prev - k];
            }
        }
        res as i32
    }
}

#[test]
fn test() {
    let nums = vec![1, 1, 2, 1, 1];
    let k = 3;
    let res = 2;
    assert_eq!(Solution::number_of_subarrays(nums, k), res);
    let nums = vec![2, 4, 6];
    let k = 1;
    let res = 0;
    assert_eq!(Solution::number_of_subarrays(nums, k), res);
    let nums = vec![2, 2, 2, 1, 2, 2, 1, 2, 2, 2];
    let k = 2;
    let res = 16;
    assert_eq!(Solution::number_of_subarrays(nums, k), res);
}

// Accepted solution for LeetCode #1248: Count Number of Nice Subarrays
function numberOfSubarrays(nums: number[], k: number): number {
    const n = nums.length;
    const cnt = Array(n + 1).fill(0);
    cnt[0] = 1;
    let [t, ans] = [0, 0];
    for (const v of nums) {
        t += v & 1;
        ans += cnt[t - k] ?? 0;
        cnt[t] += 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(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.