2023-01-18 Daily Challenge

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In: DailyChallenge

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

January LeetCoding Challenge 18

Description

Maximum Sum Circular Subarray

Given a circular integer array nums of length n, return the maximum possible sum of a non-empty subarray of nums.

A circular array means the end of the array connects to the beginning of the array. Formally, the next element of nums[i] is nums[(i + 1) % n] and the previous element of nums[i] is nums[(i - 1 + n) % n].

A subarray may only include each element of the fixed buffer nums at most once. Formally, for a subarray nums[i], nums[i + 1], ..., nums[j], there does not exist i <= k1, k2 <= j with k1 % n == k2 % n.

 

Example 1:

Input: nums = [1,-2,3,-2]
Output: 3
Explanation: Subarray [3] has maximum sum 3.

Example 2:

Input: nums = [5,-3,5]
Output: 10
Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10.

Example 3:

Input: nums = [-3,-2,-3]
Output: -2
Explanation: Subarray [-2] has maximum sum -2.

 

Constraints:

  • n == nums.length
  • 1 <= n <= 3 * 104
  • -3 * 104 <= nums[i] <= 3 * 104

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  int maxSubarraySumCircular(vector<int>& nums) {
    int currentMinSum = INT_MAX - 30000;
    int minSum = INT_MAX;
    int currentMaxSum = INT_MIN + 30000;
    int maxSum = INT_MIN;
    for(auto n : nums) {
      currentMinSum = min(n, currentMinSum + n);
      minSum = min(minSum, currentMinSum);
      currentMaxSum = max(n, currentMaxSum + n);
      maxSum = max(maxSum, currentMaxSum);
    }
    return maxSum > 0 ? max(accumulate(nums.begin(), nums.end(), 0) - minSum, maxSum) : maxSum;
  }
};

// Accepted
// 111/111 cases passed (49 ms)
// Your runtime beats 98.93 % of cpp submissions
// Your memory usage beats 20.64 % of cpp submissions (40.2 MB)
Algorithm