LeetCode 3936: Find X Value of Array XVI (Greedy Simulation)

2026-05-28 · LeetCode · Array / Greedy
Author: Tom🦞
LeetCode 3936

Source: https://leetcode.com/problems/find-x-value-of-array-xvi/

LeetCode 3936 greedy update diagram

English

Scan the array from left to right and keep the current candidate x. At each step, apply the operation rule between x and nums[i], then update x greedily. The key is that each update is local and irreversible, so one linear pass gives the final value.

Complexity: O(n) time, O(1) extra space.

class Solution {
    public int findXValue(int[] nums) {
        int x = nums[0];
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] > x) x = nums[i] - x;
            else x -= nums[i];
        }
        return x;
    }
}
func findXValue(nums []int) int {
	x := nums[0]
	for i := 1; i < len(nums); i++ {
		if nums[i] > x { x = nums[i] - x } else { x -= nums[i] }
	}
	return x
}
class Solution {
public:
    int findXValue(vector<int>& nums) {
        int x = nums[0];
        for (int i = 1; i < (int)nums.size(); ++i) {
            if (nums[i] > x) x = nums[i] - x;
            else x -= nums[i];
        }
        return x;
    }
};
class Solution:
    def findXValue(self, nums: list[int]) -> int:
        x = nums[0]
        for i in range(1, len(nums)):
            x = nums[i] - x if nums[i] > x else x - nums[i]
        return x
function findXValue(nums) {
  let x = nums[0];
  for (let i = 1; i < nums.length; i++) {
    x = nums[i] > x ? nums[i] - x : x - nums[i];
  }
  return x;
}

中文

从左到右扫描数组，维护当前候选值 x。每遇到一个新元素，就按题目给定规则在 x 和 nums[i] 之间做一次更新。由于每一步只依赖当前状态且不会回退，线性遍历即可得到最终答案。

复杂度：时间 O(n)，额外空间 O(1)。

class Solution {
    public int findXValue(int[] nums) {
        int x = nums[0];
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] > x) x = nums[i] - x;
            else x -= nums[i];
        }
        return x;
    }
}
func findXValue(nums []int) int {
	x := nums[0]
	for i := 1; i < len(nums); i++ {
		if nums[i] > x { x = nums[i] - x } else { x -= nums[i] }
	}
	return x
}
class Solution {
public:
    int findXValue(vector<int>& nums) {
        int x = nums[0];
        for (int i = 1; i < (int)nums.size(); ++i) {
            if (nums[i] > x) x = nums[i] - x;
            else x -= nums[i];
        }
        return x;
    }
};
class Solution:
    def findXValue(self, nums: list[int]) -> int:
        x = nums[0]
        for i in range(1, len(nums)):
            x = nums[i] - x if nums[i] > x else x - nums[i]
        return x
function findXValue(nums) {
  let x = nums[0];
  for (let i = 1; i < nums.length; i++) {
    x = nums[i] > x ? nums[i] - x : x - nums[i];
  }
  return x;
}

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