1658. Minimum Operations to Reduce X to Zero

• We only allowed to delete from beginning and the end.
• we have to choose the shortest subarray from left and right which make sum = x.
• or we can rephrase it.
• we have to choose the LONGEST subarray from …….middle…… which makes sum = TOTAL_SUM - x
• thats where the sliding window jumps in.

int minOperations(vector<int>& nums, int x) {
    int ans = 0;
    int l = 0;
    int n = nums.size();
    int sum = 0;

    if (x == accumulate(nums.begin(), nums.end(), 0)) {
        return nums.size();
    }
    int target = accumulate(nums.begin(), nums.end(), 0) - x;

    for (int r = 0; r < n; r++) {
        sum += nums[r];
        while (l <= r && sum > target) {
            sum -= nums[l++];
        }
        if (sum == target) {
            ans = max(ans, r - l + 1);
        }
    }
    if (ans == 0)
        return -1;
    return n - ans;
}