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.
class Solution:
def mostFrequent(self, nums: List[int], key: int) -> int:
cnt = Counter()
mx = ans = 0
for i, v in enumerate(nums[:-1]):
if v == key:
target = nums[i + 1]
cnt[target] += 1
if mx < cnt[target]:
mx = cnt[target]
ans = nums[i + 1]
return ans
class Solution {
public int mostFrequent(int[] nums, int key) {
int[] cnt = new int[1010];
int mx = 0, ans = 0;
for (int i = 0; i < nums.length - 1; ++i) {
if (nums[i] == key) {
int target = nums[i + 1];
++cnt[target];
if (mx < cnt[target]) {
mx = cnt[target];
ans = nums[i + 1];
}
}
}
return ans;
}
}
class Solution {
public:
int mostFrequent(vector<int>& nums, int key) {
vector<int> cnt(1010);
int mx = 0, ans = 0;
for (int i = 0; i < nums.size() - 1; ++i) {
if (nums[i] == key) {
int target = nums[i + 1];
++cnt[target];
if (mx < cnt[target]) {
mx = cnt[target];
ans = nums[i + 1];
}
}
}
return ans;
}
};
func mostFrequent(nums []int, key int) int {
cnt := make([]int, 1010)
mx, ans := 0, 0
for i, v := range nums[:len(nums)-1] {
if v == key {
target := nums[i+1]
cnt[target]++
if mx < cnt[target] {
mx = cnt[target]
ans = nums[i+1]
}
}
}
return ans
}