Today I have done leetcode's October LeetCoding Challenge with cpp.

October LeetCoding Challenge 26

Description

Continuous Subarray Sum

Given an integer array nums and an integer k, return true if nums has a continuous subarray of size at least two whose elements sum up to a multiple of k, or false otherwise.

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

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  bool checkSubarraySum(vector<int>& nums, int k) {
    unordered_map<int, int> count{{0, -1}};
    int sum = 0;
    for(int i = 0; i < nums.size(); ++i) {
      sum += nums[i];
      sum %= k;
      if(count.count(sum)) {
        if(i - count[sum] > 1) return true;
      } else {
        count[sum] = i;
      }
    }
    return false;
  }
};

// Accepted
// 97/97 cases passed (352 ms)
// Your runtime beats 70.44 % of cpp submissions
// Your memory usage beats 88.57 % of cpp submissions (112.6 MB)