You are given a 0-indexed integer array nums
. You are also given an integer key
, which is present in nums
.
For every unique integer target
in nums
, count the number of times target
immediately follows an occurrence of key
in nums
. In other words, count the number of indices i
such that:
-
0 <= i <= nums.length - 2
, -
nums[i] == key
and, -
nums[i + 1] == target
.
Return the target
with the maximum count. The test cases will be generated such that the target
with maximum count is unique.
Example 1:
Input: nums = [1,100,200,1,100], key = 1
Output: 100
Explanation: For target = 100, there are 2 occurrences at indices 1 and 4 which follow an occurrence of key.
No other integers follow an occurrence of key, so we return 100.
Example 2:
Input: nums = [2,2,2,2,3], key = 2
Output: 2
Explanation: For target = 2, there are 3 occurrences at indices 1, 2, and 3 which follow an occurrence of key.
For target = 3, there is only one occurrence at index 4 which follows an occurrence of key.
target = 2 has the maximum number of occurrences following an occurrence of key, so we return 2.
Constraints:
-
2 <= nums.length <= 1000
-
1 <= nums[i] <= 1000
- The test cases will be generated such that the answer is unique.
SOLUTION:
class Solution:
def mostFrequent(self, nums: List[int], key: int) -> int:
n = len(nums)
ctr = {}
for i in range(n):
if nums[i] == key:
if i < n - 1:
ctr[nums[i + 1]] = ctr.get(nums[i + 1], 0) + 1
return max((v, k) for k, v in ctr.items())[1]
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