1658. Minimum Operations to Reduce X to Zero

Problem Statement

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 <= 105

• 1 <= nums[i] <= 104

• 1 <= x <= 109

Intuition

Approach:

We want elements from end
If we take middle window target and subtract, that might give us answer

Eg ->  1 1 4 2 3  X=5
Take total(arrsum)-X = 6
Find 6 using Sliding Window

(1 1 4) 2 3
subtarct window size

Links

https://leetcode.com/problems/minimum-operations-to-reduce-x-to-zero/description/?envType=daily-question&envId=2023-09-20

Video Links

https://www.youtube.com/watch?v=xumn16n7njs&ab_channel=NeetCodeIO

Approach 1:

C++

class Solution {
public:
    int minOperations(vector<int>& nums, int x) {
        int n = nums.size();
        int sum = accumulate(nums.begin(), nums.end(), 0);
        int low=0, high=0, window_sum=0, ans = INT_MAX, target=sum-x;
        if(target < 0)
            return -1;

        while(high<n){
            window_sum += nums[high];
            while(window_sum > target)
                window_sum -= nums[low++];

            if(window_sum == target)
                ans = min(ans, n-(high-low+1));

            high++;
        }

        return ans == INT_MAX ? -1 : ans;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++

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