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1658. Minimum Operations to Reduce X to Zero

Description

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

Solutions

  • class Solution {
    public int minOperations(int[] nums, int x) {
        x = -x;
        for (int v : nums) {
            x += v;
        }
        Map<Integer, Integer> vis = new HashMap<>();
        vis.put(0, -1);
        int n = nums.length;
        int ans = 1 << 30;
        for (int i = 0, s = 0; i < n; ++i) {
            s += nums[i];
            vis.putIfAbsent(s, i);
            if (vis.containsKey(s - x)) {
                int j = vis.get(s - x);
                ans = Math.min(ans, n - (i - j));
            }
        }
        return ans == 1 << 30 ? -1 : ans;
    }
}

  • class Solution {
public:
    int minOperations(vector<int>& nums, int x) {
        x = accumulate(nums.begin(), nums.end(), 0) - x;
        unordered_map<int, int> vis{ {0, -1} };
        int n = nums.size();
        int ans = 1 << 30;
        for (int i = 0, s = 0; i < n; ++i) {
            s += nums[i];
            if (!vis.count(s)) {
                vis[s] = i;
            }
            if (vis.count(s - x)) {
                int j = vis[s - x];
                ans = min(ans, n - (i - j));
            }
        }
        return ans == 1 << 30 ? -1 : ans;
    }
};

  • class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        x = sum(nums) - x
        vis = {0: -1}
        ans = inf
        s, n = 0, len(nums)
        for i, v in enumerate(nums):
            s += v
            if s not in vis:
                vis[s] = i
            if s - x in vis:
                j = vis[s - x]
                ans = min(ans, n - (i - j))
        return -1 if ans == inf else ans


  • func minOperations(nums []int, x int) int {
	x = -x
	for _, v := range nums {
		x += v
	}
	vis := map[int]int{0: -1}
	ans := 1 << 30
	s, n := 0, len(nums)
	for i, v := range nums {
		s += v
		if _, ok := vis[s]; !ok {
			vis[s] = i
		}
		if j, ok := vis[s-x]; ok {
			ans = min(ans, n-(i-j))
		}
	}
	if ans == 1<<30 {
		return -1
	}
	return ans
}

  • function minOperations(nums: number[], x: number): number {
    x = nums.reduce((a, b) => a + b, 0) - x;
    const vis = new Map();
    vis.set(0, -1);
    const n = nums.length;
    let ans = 1 << 30;
    for (let i = 0, s = 0; i < n; ++i) {
        s += nums[i];
        if (!vis.has(s)) {
            vis.set(s, i);
        }
        if (vis.has(s - x)) {
            const j = vis.get(s - x);
            ans = Math.min(ans, n - (i - j));
        }
    }
    return ans == 1 << 30 ? -1 : ans;
}


  • impl Solution {
    pub fn min_operations(nums: Vec<i32>, x: i32) -> i32 {
        let n = nums.len();
        let target = nums.iter().sum::<i32>() - x;
        if target < 0 {
            return -1;
        }
        let mut ans = i32::MAX;
        let mut sum = 0;
        let mut i = 0;
        for j in 0..n {
            sum += nums[j];
            while sum > target {
                sum -= nums[i];
                i += 1;
            }
            if sum == target {
                ans = ans.min((n - 1 - (j - i)) as i32);
            }
        }
        if ans == i32::MAX {
            return -1;
        }
        ans
    }
}

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