2190 - Most Frequent Number Following Key In an Array (Easy)
Problem Link
https://leetcode.com/problems/most-frequent-number-following-key-in-an-array/
Problem Statement
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 <= n - 2,nums[i] == keyand,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 <= 10001 <= nums[i] <= 1000- The test cases will be generated such that the answer is unique.
Approach 1: Counting
We iterate the input to find the possible and store its occurrence. Return the maximum one.
class Solution {
public:
int mostFrequent(vector<int>& nums, int key) {
int n = nums.size(), ans = 0;
// use hash map to store the occurrence of a possible target
unordered_map<int, int> m;
for (int i = 1; i < n; i++) {
// the previous one is key
// nums[i] is target
if (nums[i - 1] == key) {
// count occurrence
m[nums[i]]++;
}
}
int mx = 0;
for (auto x : m) {
// check if it is greater than the current max count
if (x.second > mx) {
// store the number
ans = x.first;
// update the max count
mx = x.second;
}
}
return ans;
}
};Once we get the idea, we can further optimise the above solution a bit. We only set answer if the current number is and its occurrence is greater than the current max count.
class Solution {
public:
int mostFrequent(vector<int>& nums, int key) {
int n = nums.size(), ans = 0;
unordered_map<int, int> m;
for (int i = 1; i < n; i++) {
if (nums[i - 1] == key && ++m[nums[i]] > m[ans]) {
ans = nums[i];
}
}
return ans;
}
};