Question

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

class Solution:
  def checkSubarraySum(self, nums: List[int], k: int) -> bool:
    s = 0
    mp = {0: -1}
    for i, v in enumerate(nums):
      s += v
      r = s % k
      if r in mp and i - mp[r] >= 2:
        return True
      if r not in mp:
        mp[r] = i
    return False

class Solution {
  public boolean checkSubarraySum(int[] nums, int k) {
    Map<Integer, Integer> mp = new HashMap<>();
    mp.put(0, -1);
    int s = 0;
    for (int i = 0; i < nums.length; ++i) {
      s += nums[i];
      int r = s % k;
      if (mp.containsKey(r) && i - mp.get(r) >= 2) {
        return true;
      }
      if (!mp.containsKey(r)) {
        mp.put(r, i);
      }
    }
    return false;
  }
}

class Solution {
public:
  bool checkSubarraySum(vector<int>& nums, int k) {
    unordered_map<int, int> mp;
    mp[0] = -1;
    int s = 0;
    for (int i = 0; i < nums.size(); ++i) {
      s += nums[i];
      int r = s % k;
      if (mp.count(r) && i - mp[r] >= 2) return true;
      if (!mp.count(r)) mp[r] = i;
    }
    return false;
  }
};

func checkSubarraySum(nums []int, k int) bool {
  mp := map[int]int{0: -1}
  s := 0
  for i, v := range nums {
    s += v
    r := s % k
    if j, ok := mp[r]; ok && i-j >= 2 {
      return true
    }
    if _, ok := mp[r]; !ok {
      mp[r] = i
    }
  }
  return false
}