Question

1658. Minimum Operations to Reduce X to Zero

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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

We can transform the problem into finding the maximum length of a continuous subarray in the middle, such that the sum of the subarray is $x = sum(nums) - x$.

Define a hash table vis, where vis[s] represents the smallest index with a prefix sum of $s$.

Traverse the array nums. For each element $nums[i]$, we first add $nums[i]$ to the prefix sum $s$. If $s$ does not exist in the hash table, we add it to the hash table, and its value is the current index $i$. Then we check whether $s - x$ exists in the hash table. If it does, it means that there exists an index $j$ such that the sum of $nums[j + 1,..i]$ is $x$. At this time, we update the minimum value of the answer, that is, $ans = min(ans, n - (i - j))$.

At the end of the traversal, if no subarray meets the condition, return $-1$, otherwise return $ans$.

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

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

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;
  }
};

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
  }
}

#define min(a, b) (((a) < (b)) ? (a) : (b))

int minOperations(int* nums, int numsSize, int x) {
  int target = -x;
  for (int i = 0; i < numsSize; i++) {
    target += nums[i];
  }
  if (target < 0) {
    return -1;
  }
  int ans = INT_MAX;
  int sum = 0;
  int i = 0;
  for (int j = 0; j < numsSize; j++) {
    sum += nums[j];
    while (sum > target) {
      sum -= nums[i++];
    }
    if (sum == target) {
      ans = min(ans, numsSize - 1 - (j - i));
    }
  }
  if (ans == INT_MAX) {
    return -1;
  }
  return ans;
}

Solution 2: Two Pointers

Similar to Solution 1, we need to find a subarray such that the sum of the subarray is $x = sum(nums) - x$.

Define two pointers $j$ and $i$, initially $i = j = 0$. Then we move the pointer $i$ to the right, add $nums[i]$ to the prefix sum $s$. If $s > x$, then we move the pointer $j$ to the right in a loop, and subtract $nums[j]$ from the prefix sum $s$, until $s \le x$. If $s = x$, we can update the minimum value of the answer, that is, $ans = min(ans, n - (i - j + 1))$. Continue to move the pointer $i$ to the right, and repeat the above process.

Finally, if no subarray meets the condition, return $-1$, otherwise return $ans$.

The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array nums.

class Solution:
  def minOperations(self, nums: List[int], x: int) -> int:
    x = sum(nums) - x
    ans = inf
    n = len(nums)
    s = j = 0
    for i, v in enumerate(nums):
      s += v
      while j <= i and s > x:
        s -= nums[j]
        j += 1
      if s == x:
        ans = min(ans, n - (i - j + 1))
    return -1 if ans == inf else ans

class Solution {
  public int minOperations(int[] nums, int x) {
    x = -x;
    for (int v : nums) {
      x += v;
    }
    int n = nums.length;
    int ans = 1 << 30;
    for (int i = 0, j = 0, s = 0; i < n; ++i) {
      s += nums[i];
      while (j <= i && s > x) {
        s -= nums[j++];
      }
      if (s == x) {
        ans = Math.min(ans, n - (i - j + 1));
      }
    }
    return ans == 1 << 30 ? -1 : ans;
  }
}

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

func minOperations(nums []int, x int) int {
  x = -x
  for _, v := range nums {
    x += v
  }
  ans := 1 << 30
  s, n := 0, len(nums)
  j := 0
  for i, v := range nums {
    s += v
    for j <= i && s > x {
      s -= nums[j]
      j++
    }
    if s == x {
      ans = min(ans, n-(i-j+1))
    }
  }
  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 n = nums.length;
  let ans = 1 << 30;
  for (let i = 0, j = 0, s = 0; i < n; ++i) {
    s += nums[i];
    while (j <= i && s > x) {
      s -= nums[j++];
    }
    if (s == x) {
      ans = Math.min(ans, n - (i - j + 1));
    }
  }
  return ans == 1 << 30 ? -1 : ans;
}