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

Solution 1: Hash Table + Prefix Sum

According to the problem description, we need to remove elements from both ends of the array $nums$ so that the sum of the removed elements equals $x$, and the number of removed elements is minimized. We can transform the problem into: find the longest consecutive subarray in the array $nums$ such that the sum of the subarray $s = \sum_{i=0}^{n} nums[i] - x$. In this way, we can transform the problem into finding the length $mx$ of the longest consecutive subarray in the array $nums$ with a sum of $s$, and the answer is $n - mx$.

We initialize $mx = -1$, and then use a hash table $vis$ to store the prefix sum, where the key is the prefix sum and the value is the index corresponding to the prefix sum.

Traverse the array $nums$, for the current element $nums[i]$, calculate the prefix sum $t$, if $t$ is not in the hash table, add $t$ to the hash table; if $t - s$ is in the hash table, update $mx = \max(mx, i - vis[t - s])$.

Finally, if $mx = -1$, return $-1$, otherwise return $n - mx$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.

Python3

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

Java

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

C++

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

Go

func minOperations(nums []int, x int) int {
	s := -x
	for _, v := range nums {
		s += v
	}
	vis := map[int]int{0: -1}
	mx, t := -1, 0
	for i, v := range nums {
		t += v
		if _, ok := vis[t]; !ok {
			vis[t] = i
		}
		if j, ok := vis[t-s]; ok {
			mx = max(mx, i-j)
		}
	}
	if mx == -1 {
		return -1
	}
	return len(nums) - mx
}

TypeScript

function minOperations(nums: number[], x: number): number {
    const s = nums.reduce((acc, cur) => acc + cur, -x);
    const vis: Map<number, number> = new Map([[0, -1]]);
    let [mx, t] = [-1, 0];
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        t += nums[i];
        if (!vis.has(t)) {
            vis.set(t, i);
        }
        if (vis.has(t - s)) {
            mx = Math.max(mx, i - vis.get(t - s)!);
        }
    }
    return ~mx ? n - mx : -1;
}

Rust

use std::collections::HashMap;

impl Solution {
    pub fn min_operations(nums: Vec<i32>, x: i32) -> i32 {
        let s = nums.iter().sum::<i32>() - x;
        let mut vis: HashMap<i32, i32> = HashMap::new();
        vis.insert(0, -1);
        let mut mx = -1;
        let mut t = 0;
        for (i, v) in nums.iter().enumerate() {
            t += v;
            if !vis.contains_key(&t) {
                vis.insert(t, i as i32);
            }
            if let Some(&j) = vis.get(&(t - s)) {
                mx = mx.max((i as i32) - j);
            }
        }
        if mx == -1 {
            -1
        } else {
            (nums.len() as i32) - mx
        }
    }
}

Solution 2: Two Pointers

Based on the analysis of Solution 1, we need to find the length $mx$ of the longest consecutive subarray in the array $nums$ with a sum of $s$. Since all elements in the array $nums$ are positive integers, the prefix sum of the array will only increase monotonically, so we can use two pointers to solve this problem.

We initialize pointer $j = 0$, prefix sum $t = 0$, and the length of the longest consecutive subarray $mx = -1$.

Traverse the array $nums$, for the current element $nums[i]$, calculate the prefix sum $t += nums[i]$. If $t > s$, then move the pointer $j$ until $t \leq s$. If $t = s$, then update $mx = \max(mx, i - j + 1)$.