2845. Count of Interesting Subarrays

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 $nums$.

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}