LeetCode 1658. Minimum Operations to Reduce X to Zero

原题链接在这里：https://leetcode.com/problems/minimum-operations-to-reduce-x-to-zero/

题目：

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

题解：

After minimum operation to substract x, the remaining subarray with remaining sum = totalSum - x.

Thus this question is the same as maximum size subarray sum equals to totalSum - x.

Return nums.length - maxSize.

Note: when totalSum - x = 0. Return nums.length. Since we can't find subarray sum = 0. But the correct answer is to remove all the elements.

Time Complexity: O(n). n = nums.length.

Space: O(n).

AC Java:

class Solution {
    public int minOperations(int[] nums, int x) {
        int target = 0 - x;
        for(int num : nums){
            target += num;
        }
        
        if(target == 0){
            return nums.length;
        }
        
        int sum = 0;
        HashMap<Integer, Integer> sumToInd = new HashMap<>();
        sumToInd.put(0, -1);
        int subArrLen = -1;
        for(int i = 0; i < nums.length; i++){
            sum += nums[i];
            if(sumToInd.containsKey(sum - target)){
                subArrLen = Math.max(subArrLen, i - sumToInd.get(sum - target));
            }
            
            sumToInd.putIfAbsent(sum, i);
        }
        
        return subArrLen == -1 ? -1 : nums.length - subArrLen;
    }
}

类似Maximum Size Subarray Sum Equals k.