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523. Continuous Subarray Sum
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 ofk
if there exists an integern
such thatx = n * k
.0
is always a multiple ofk
.
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
-
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; } };
-
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
-
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 }
-
function checkSubarraySum(nums: number[], k: number): boolean { const n = nums.length; let prefixSum = 0; const map = new Map([[0, 0]]); for (let i = 0; i < n; i++) { prefixSum += nums[i]; const remainder = prefixSum % k; if (!map.has(remainder)) { map.set(remainder, i + 1); } else if (i - map.get(remainder)! > 0) return true; } return false; }