1658. Minimum Operations to Reduce X to Zero 🌟🌟

Prefix and Suffix Sum with map

• Simple - Just think.

Sliding Window

• In the problem its given that we can remove element from start or end to subtract its value form x so that it will become 0. Also we have to minimize the number of such operations.
• Instead of doing so we can simplify the process.
• ex.XXXOOOXX, here if we need to find sum of all X we can simply do it with total sum - sum of O, same idea can be used in this problem.
• We just need to find subarray with sum = reqSum, Where reqSum = totalSumOfArray - x
• And we need to Maximize the length of such subarray.
• The answer will be total length - max length of subarray
• Time Complexity: O(N)
• Space Complexity: O(1)

Code

class Solution {
public:
    int minOperations(vector<int>& nums, int x)
    {
        int n = nums.size();

        int reqSum = 0, windowSum = 0;
        reqSum = accumulate(nums.begin(), nums.end(), 0);
        reqSum -= x;
        // if required sum is 0, then total array length is answer.
        if (reqSum == 0) return n;

        int i = 0, maxLen = 0;
        // find sliding window whose sum=reqSum and length is maximum
        for (int j = 0; j < n; j++) {
            windowSum += nums[j];
            while (i < n && windowSum > reqSum) {
                // if window sum exceeds required sum, reduce window size from front
                windowSum -= nums[i];
                i++;
            }
            if (windowSum == reqSum) {
                // we found one of the subarray with required sum
                maxLen = max(maxLen, j - i + 1);
            }
        }

        // if don't find any subarray of required sum, return -1
        if (maxLen == 0) return -1;
        // else return n-maxLen
        return n - maxLen;
    }
};