1658. Minimum Operations to Reduce X to Zero

Problem

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 <= 10^5

• 1 <= nums[i] <= 10^4

• 1 <= x <= 10^9

Solution (Java)

class Solution {
    public int minOperations(int[] nums, int x) {
        int totalArraySum = 0;
        for (int each : nums) {
            totalArraySum += each;
        }
        if (totalArraySum == x) {
            return nums.length;
        }
        int target = totalArraySum - x;
        // as we need to find value equal to x so that x-x=0,
        // and we need to search the longest sub array with sum equal t0 total array sum -x;
        int sum = 0;
        int result = -1;
        int start = 0;
        for (int end = 0; end < nums.length; end++) {
            sum += nums[end];
            while (sum > target && start < nums.length) {
                sum -= nums[start];
                start++;
            }
            if (sum == target) {
                result = Math.max(result, end + 1 - start);
            }
        }
        if (result == -1) {
            return result;
        } else {
            return nums.length - result;
        }
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).