1658. Minimum Operations to Reduce X to Zero 

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class Solution {
    public int minOperations(int[] nums, int x) {
        int n = nums.length;

        int[] doubleNums = new int[n + n];
        for (int i = 0; i < n; i++) {
            doubleNums[i] = doubleNums[i + n] = nums[i];
        }

        int minOperations = Integer.MAX_VALUE / 2;

        int windowSum = 0;
        int i = 0;
        for (int j = 0; j < doubleNums.length; j++) {
            windowSum += doubleNums[j];

            while (windowSum > x) {
                windowSum -= doubleNums[i++];
            }

            // 注意需要排除掉非边界区间的元素和
            if (windowSum == x && !(i >= 1 && j <= n - 2) && !(i >= n + 1 && j <= 2 * n - 2)) {
                minOperations = Math.min(minOperations, j - i + 1);
            }
        }

        return minOperations == Integer.MAX_VALUE / 2 || minOperations > nums.length ? -1 : minOperations;
    }
}

References

1658. Minimum Operations to Reduce X to Zero