LeetCode 2121 Intervals Between Identical Elements

LeetCode

You are given a 0-indexed array of n integers arr.

The interval between two elements in arr is defined as the absolute difference between their indices. More formally, the interval between arr[i] and arr[j] is |i - j|.

Return an array intervals of length n where intervals[i] is the sum of intervals between arr[i] and each element in arr with the same value as arr[i].

Note: |x| is the absolute value of x.

Example 1:

Input: arr = [2,1,3,1,2,3,3]
Output: [4,2,7,2,4,4,5]
Explanation:
- Index 0: Another 2 is found at index 4. |0 - 4| = 4
- Index 1: Another 1 is found at index 3. |1 - 3| = 2
- Index 2: Two more 3s are found at indices 5 and 6. |2 - 5| + |2 - 6| = 7
- Index 3: Another 1 is found at index 1. |3 - 1| = 2
- Index 4: Another 2 is found at index 0. |4 - 0| = 4
- Index 5: Two more 3s are found at indices 2 and 6. |5 - 2| + |5 - 6| = 4
- Index 6: Two more 3s are found at indices 2 and 5. |6 - 2| + |6 - 5| = 5

Example 2:

Input: arr = [10,5,10,10]
Output: [5,0,3,4]
Explanation:
- Index 0: Two more 10s are found at indices 2 and 3. |0 - 2| + |0 - 3| = 5
- Index 1: There is only one 5 in the array, so its sum of intervals to identical elements is 0.
- Index 2: Two more 10s are found at indices 0 and 3. |2 - 0| + |2 - 3| = 3
- Index 3: Two more 10s are found at indices 0 and 2. |3 - 0| + |3 - 2| = 4

Constraints:

n == arr.length 1 <= n <= 10<sup>5</sup>
* 1 <= arr[i] <= 10<sup>5</sup>

题目大意：

求相同元素的下标差的和

解题思路：

暴力法是TLE，由于存在重复计算。用prefix和suffix来优化。
prefix为从某个元素为起点到前继节点下标差之和
suffix为从某个元素为起点到后继节点下标差之和
结果 = prefix[i] + suffix[i]

解题步骤：

1. 用Map来存储相同元素的下标
2. 用prefix和suffix来计算每个元素的结果

注意事项：

1. 一开始用presum表示从起点到其他下标的差之和。这不能用于某个元素的前继元素下标差之和。所以要改成从某个元素为起点到前继节点下标差之和

Python代码：

def getDistances(self, arr: List[int]) -> List[int]:
	val_to_index = collections.defaultdict(list)
	res = [0] * len(arr)
	for i in range(len(arr)):
		val_to_index[arr[i]].append(i)
	for val, indices in val_to_index.items():
		prefix, suffix = [0], [0]
		for i in range(1, len(indices)):
			prefix.append(prefix[-1] + i * abs(indices[i] - indices[i - 1]))
		for i in range(len(indices) - 2, -1, -1):
			suffix.append(suffix[-1] + (len(indices) - i - 1) * abs(indices[i] - indices[i + 1]))
		for i in range(len(indices)):
			res[indices[i]] = prefix[i] + suffix[len(indices) - i - 1]
	return res

算法分析：

时间复杂度为O(n)，空间复杂度O(n)