[LeetCode January Challange] Day 14 - 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’s 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 <= 10^5
• 1 <= nums[i] <= 10^4
• 1 <= x <= 10^9

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Solution

Two Pointer (Indirect)

Time complexity : O(n)
Space complexity : O(1)

class Solution {
public:
    int minOperations(vector<int>& nums, int x) {
        const int n = nums.size();
        int total = accumulate(nums.begin(), nums.end(), 0);
        int l = 0, max_len = -1, cur_sum = 0;
        for (int r=0; r<n; ++r) {
            cur_sum += nums[r];
            while (total-x < cur_sum && l <= r) 
                cur_sum -= nums[l++];
            if (cur_sum == total-x)
                max_len = max(max_len, r-l+1);
        }
        return -1 != max_len ? n-max_len : -1;
    }
};

反向思考：找出最長的 sub array，使得其 sub array 總和為 total - x。

Two Pointer (Direct)

Time complexity : O(n)
Space complexity : O(1)

class Solution {
public:
    int minOperations(vector<int>& nums, int x) {
        const int n = nums.size();
        int cur_sum = accumulate(nums.begin(), nums.end(), 0);
        int l = 0, min_len = INT_MAX;
        for (int r=0; r<n; ++r) {
            cur_sum -= nums[r];
            while (cur_sum < x && l <= r) 
                cur_sum += nums[l++];
            if (cur_sum == x)
                min_len = min(min_len, ((n-1)-r)+l);
        }
        return INT_MAX != min_len ? min_len : -1;
    }
};