2023-09-20 Daily Challenge

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Posted modified:
In: DailyChallenge

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

September LeetCoding Challenge 20

Description

Minimum Operations to Reduce X to Zero

You are given an integer array nums and an integer x. In one operation, you can either remove the leftmost or the rightmost element from the array nums and subtract its value from x. Note that this modifies the array for future operations.

Return the minimum number of operations to reduce x to exactly 0 if it is possible, otherwise, return -1.

 

Example 1:

Input: nums = [1,1,4,2,3], x = 5
Output: 2
Explanation: The optimal solution is to remove the last two elements to reduce x to zero.

Example 2:

Input: nums = [5,6,7,8,9], x = 4
Output: -1

Example 3:

Input: nums = [3,2,20,1,1,3], x = 10
Output: 5
Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 104
  • 1 <= x <= 109

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  int minOperations(vector<int>& nums, int x) {
    unordered_map<int, int> sumLeft, sumRight;
    int len = nums.size();
    int sum = 0;
    for(int i = 0; i < len; ++i) {
      sum += nums[i];
      sumLeft[sum] = i+1;
    }
    sumLeft[0] = 0;
    sum = 0;
    for(int i = 1; i <= len; ++i) {
      sum += nums[len-i];
      sumRight[sum] = i;
    }
    sumRight[0] = 0;
    
    int answer = INT_MAX;
    for(auto [sum, cnt] : sumLeft) {
      if(sumRight.count(x-sum) && cnt+sumRight[x-sum] <= len) answer = min(answer, cnt+sumRight[x-sum]);
    }
    return answer == INT_MAX ? -1 : answer;
  }
};

// Accepted
// 94/94 cases passed (577 ms)
// Your runtime beats 5.05 % of cpp submissions
// Your memory usage beats 5.04 % of cpp submissions (238.1 MB)
Algorithm