Question

2845. Count of Interesting Subarrays

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Description

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

Solutions

Solution 1: Hash Table + Prefix Sum

The problem requires the number of indices $i$ in an interval that satisfy $nums[i] \bmod modulo = k$. We can transform the array $nums$ into a $0-1$ array $arr$, where $arr[i] = 1$ indicates $nums[i] \bmod modulo = k$, otherwise $arr[i] = 0$.

For an interval $[l, r]$, we can calculate the number of $1$s in $arr[l..r]$ through the prefix sum array $s$, i.e., $s[r] - s[l - 1]$, where $s[0] = 0$.

We use a hash table $cnt$ to record the number of occurrences of the prefix sum $s \bmod modulo$, initially $cnt[0]=1$.

Next, we traverse the array $arr$, calculate the prefix sum $s$, add the number of occurrences of $(s-k) \bmod modulo$ to the answer, and then add $1$ to the number of occurrences of $s \bmod modulo$.

After the traversal ends, return the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

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

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;
  }
}

class Solution {
public:
  long long countInterestingSubarrays(vector<int>& nums, int modulo, int k) {
    int n = nums.size();
    vector<int> arr(n);
    for (int i = 0; i < n; ++i) {
      arr[i] = int(nums[i] % modulo == k);
    }
    unordered_map<int, int> cnt;
    cnt[0] = 1;
    long long ans = 0;
    int s = 0;
    for (int x : arr) {
      s += x;
      ans += cnt[(s - k + modulo) % modulo];
      cnt[s % modulo]++;
    }
    return ans;
  }
};

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
}

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;
}