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Medium
Array
Hash Table
Math
Prefix Sum

523. Continuous Subarray Sum

中文文档

Description

Given an integer array nums and an integer k, return true if nums has a good subarray or false otherwise.

A good subarray is a subarray where:

  • its length is at least two, and
  • the sum of the elements of the subarray is a multiple of k.

Note that:

  • A subarray is a contiguous part of the array.
  • An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k.

 

Example 1:

Input: nums = [23,2,4,6,7], k = 6
Output: true
Explanation: [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.

Example 2:

Input: nums = [23,2,6,4,7], k = 6
Output: true
Explanation: [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.

Example 3:

Input: nums = [23,2,6,4,7], k = 13
Output: false

 

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 109
  • 0 <= sum(nums[i]) <= 231 - 1
  • 1 <= k <= 231 - 1

Solutions

Solution 1: Prefix Sum + Hash Table

According to the problem description, if there exist two positions $i$ and $j$ ($j &lt; i$) where the remainders of the prefix sums modulo $k$ are the same, then the sum of the subarray $\text{nums}[j+1..i]$ is a multiple of $k$.

Therefore, we can use a hash table to store the first occurrence of each remainder of the prefix sum modulo $k$. Initially, we store a key-value pair $(0, -1)$ in the hash table, indicating that the remainder $0$ of the prefix sum $0$ appears at position $-1$.

As we iterate through the array, we calculate the current prefix sum's remainder modulo $k$. If the current prefix sum's remainder modulo $k$ has not appeared in the hash table, we store the current prefix sum's remainder modulo $k$ and its corresponding position in the hash table. Otherwise, if the current prefix sum's remainder modulo $k$ has already appeared in the hash table at position $j$, then we have found a subarray $\text{nums}[j+1..i]$ that meets the conditions, and thus return $\text{True}$.

After completing the iteration, if no subarray meeting the conditions is found, we return $\text{False}$.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $\text{nums}$.

Python3

class Solution:
    def checkSubarraySum(self, nums: List[int], k: int) -> bool:
        d = {0: -1}
        s = 0
        for i, x in enumerate(nums):
            s = (s + x) % k
            if s not in d:
                d[s] = i
            elif i - d[s] > 1:
                return True
        return False

Java

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        Map<Integer, Integer> d = new HashMap<>();
        d.put(0, -1);
        int s = 0;
        for (int i = 0; i < nums.length; ++i) {
            s = (s + nums[i]) % k;
            if (!d.containsKey(s)) {
                d.put(s, i);
            } else if (i - d.get(s) > 1) {
                return true;
            }
        }
        return false;
    }
}

C++

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        unordered_map<int, int> d{{0, -1}};
        int s = 0;
        for (int i = 0; i < nums.size(); ++i) {
            s = (s + nums[i]) % k;
            if (!d.contains(s)) {
                d[s] = i;
            } else if (i - d[s] > 1) {
                return true;
            }
        }
        return false;
    }
};

Go

func checkSubarraySum(nums []int, k int) bool {
	d := map[int]int{0: -1}
	s := 0
	for i, x := range nums {
		s = (s + x) % k
		if _, ok := d[s]; !ok {
			d[s] = i
		} else if i-d[s] > 1 {
			return true
		}
	}
	return false
}

TypeScript

function checkSubarraySum(nums: number[], k: number): boolean {
    const d: Record<number, number> = { 0: -1 };
    let s = 0;
    for (let i = 0; i < nums.length; ++i) {
        s = (s + nums[i]) % k;
        if (!d.hasOwnProperty(s)) {
            d[s] = i;
        } else if (i - d[s] > 1) {
            return true;
        }
    }
    return false;
}