2367. 等差三元组的数目 

题目描述

给你一个下标从 0 开始、严格递增 的整数数组 nums 和一个正整数 diff 。如果满足下述全部条件，则三元组 (i, j, k) 就是一个 等差三元组 ：

i < j < k ，
nums[j] - nums[i] == diff 且
nums[k] - nums[j] == diff

返回不同 等差三元组 的数目。

示例 1：

输入：nums = [0,1,4,6,7,10], diff = 3
输出：2
解释：
(1, 2, 4) 是等差三元组：7 - 4 == 3 且 4 - 1 == 3 。
(2, 4, 5) 是等差三元组：10 - 7 == 3 且 7 - 4 == 3 。

示例 2：

输入：nums = [4,5,6,7,8,9], diff = 2
输出：2
解释：
(0, 2, 4) 是等差三元组：8 - 6 == 2 且 6 - 4 == 2 。
(1, 3, 5) 是等差三元组：9 - 7 == 2 且 7 - 5 == 2 。

提示：

3 <= nums.length <= 200
0 <= nums[i] <= 200
1 <= diff <= 50
nums 严格递增

解法

方法一：暴力枚举

我们注意到，数组 $nums$ 的长度只有不超过 $200$，因此可以直接暴力枚举 $i$, $j$, $k$，判断是否满足条件，若满足，累加三元组数目。

时间复杂度 $O(n^3)$，其中 $n$ 为数组 $nums$ 的长度。空间复杂度 $O(1)$。

Python3

class Solution:
    def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
        return sum(b - a == diff and c - b == diff for a, b, c in combinations(nums, 3))

Java

class Solution {
    public int arithmeticTriplets(int[] nums, int diff) {
        int ans = 0;
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                for (int k = j + 1; k < n; ++k) {
                    if (nums[j] - nums[i] == diff && nums[k] - nums[j] == diff) {
                        ++ans;
                    }
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int arithmeticTriplets(vector<int>& nums, int diff) {
        int ans = 0;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                for (int k = j + 1; k < n; ++k) {
                    if (nums[j] - nums[i] == diff && nums[k] - nums[j] == diff) {
                        ++ans;
                    }
                }
            }
        }
        return ans;
    }
};

Go

func arithmeticTriplets(nums []int, diff int) (ans int) {
	n := len(nums)
	for i := 0; i < n; i++ {
		for j := i + 1; j < n; j++ {
			for k := j + 1; k < n; k++ {
				if nums[j]-nums[i] == diff && nums[k]-nums[j] == diff {
					ans++
				}
			}
		}
	}
	return
}

TypeScript

function arithmeticTriplets(nums: number[], diff: number): number {
    const n = nums.length;
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        for (let j = i + 1; j < n; ++j) {
            for (let k = j + 1; k < n; ++k) {
                if (nums[j] - nums[i] === diff && nums[k] - nums[j] === diff) {
                    ++ans;
                }
            }
        }
    }
    return ans;
}

方法二：数组或哈希表

我们可以先将 $nums$ 中的元素存入哈希表或数组 $vis$ 中，然后枚举 $nums$ 中的每个元素 $x$，判断 $x+diff$, $x+diff+diff$ 是否也在 $vis$ 中，若是，累加三元组数目。

枚举结束后，返回答案。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 的长度。

Python3

class Solution:
    def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
        vis = set(nums)
        return sum(x + diff in vis and x + diff * 2 in vis for x in nums)

Java

class Solution {
    public int arithmeticTriplets(int[] nums, int diff) {
        boolean[] vis = new boolean[301];
        for (int x : nums) {
            vis[x] = true;
        }
        int ans = 0;
        for (int x : nums) {
            if (vis[x + diff] && vis[x + diff + diff]) {
                ++ans;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int arithmeticTriplets(vector<int>& nums, int diff) {
        bitset<301> vis;
        for (int x : nums) {
            vis[x] = 1;
        }
        int ans = 0;
        for (int x : nums) {
            ans += vis[x + diff] && vis[x + diff + diff];
        }
        return ans;
    }
};

Go

func arithmeticTriplets(nums []int, diff int) (ans int) {
	vis := [301]bool{}
	for _, x := range nums {
		vis[x] = true
	}
	for _, x := range nums {
		if vis[x+diff] && vis[x+diff+diff] {
			ans++
		}
	}
	return
}

TypeScript

function arithmeticTriplets(nums: number[], diff: number): number {
    const vis: boolean[] = new Array(301).fill(false);
    for (const x of nums) {
        vis[x] = true;
    }
    let ans = 0;
    for (const x of nums) {
        if (vis[x + diff] && vis[x + diff + diff]) {
            ++ans;
        }
    }
    return ans;
}