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To make the problem simpler, we first calculate the total sum of the array, and then look for the longest subarray whose sum is equal to totalSum - x. This is because removing elements to get a sum of x is equivalent to finding a subarray with this complementary sum.
If we find this subarray, the number of operations is the total length of the array minus the length of this subarray. We use a sliding window to find the longest subarray with the target sum.
Time Complexity: O(n), where n is the length of the array. This is because each element is processed at most twice (once added and once removed).
Space Complexity: O(1), as we only use a few extra variables.
1import java.util.*;
2
3class Solution {
4 public int minOperations(int[] nums, int x) {
5 int totalSum = Arrays.stream(nums).sum();
6 int target = totalSum - x;
7 if (target < 0) return -1;
8
9 int left = 0, currSum = 0, maxLength = -1;
10 for (int right = 0; right < nums.length; ++right) {
11 currSum += nums[right];
12
13 while (currSum > target && left <= right) {
14 currSum -= nums[left];
15 ++left;
16 }
17
18 if (currSum == target) {
19 maxLength = Math.max(maxLength, right - left + 1);
20 }
21 }
22
23 return maxLength == -1 ? -1 : nums.length - maxLength;
24 }
25}
26This Java solution employs a sliding window technique similarly. The `stream.sum()` is used to simplify the calculation of the total array sum. It efficiently finds the longest subarray whose sum is the needed target, and thus the minimum operations.
This alternative method uses dynamic programming to explore all possible choices at any instance, storing results for optimal decisions. We define a DP array dp[i] that represents the minimum operations needed to reduce x using the first i elements from either side. We initialize by removing elements and updating DP states iteratively.
We optimize storage by using only needed previous state references, ensuring computations are rapid and effective. This dynamic programming approach is relatively complex due to combination subproblems hence less preferable but a dividing strategy worth studying.
Time Complexity: O(n*m), where n is the total number and m denotes max target value.
Space Complexity: O(m) in size for storage structures.
1from typing import List
2
3def minOperations(nums:
The Python solution leverages dynamic programming using a 1D array. As we iterate through nums, each element influences potential sums achievable by contributing to subsets, thus establishing a table of values dp representing the maximum number achieved at every potential sum. The solution emerges when final target accessibility is evaluated post iterations.