How to Solve Most Frequent Element in an Array
Most Frequent Element in an Array Introduction
Most Frequent Element in an Array Introduction
The Most Frequent Element in an Array problem asks us to element that appears the most in a given array. This problem can be solved simply by using a map to increment the frequent count of each element as we come across them and return the integer with the highest count at the end.
Most Frequent Element in an Array Problem
Most Frequent Element in an Array Problem
Given an array of integers, find the most frequent element in the array. Write a method that takes an array of integers and returns an integer. If there is a tie, you can just return any.
Example Inputs and Outputs
Example 1
Input: arr = [1, 2, 3, 3]
Output: 3
Most Frequent Element in an Array Solutions
Most Frequent Element in an Array Solutions
To solve this problem, we can use a map or a map-like data structure to iterate through the given array and keep track of the count of each integer. We start by initializing an empty count map and two variables: mostFrequent to store the current most frequent integer, and maxCount to keep track of its count. Next, we iterate through the array element by element. For each integer we find, we check if it exists in the count map. If it does, we increment its count; otherwise, we add it to the count map with an initial count of 1.
During the iteration, we also update the mostFrequent and maxCount variables whenever we find an integer with a count higher than the current maximum count. This way, by the end of the iteration, we'll have identified the integer with the highest count. This integer represents the most frequent integer in the array.
def findMostFrequent(nums):
countMap = {}
maxCount = 0
mostFrequent = None
for num in nums:
if num in countMap:
countMap[num] += 1
else:
countMap[num] = 1
if countMap[num] > maxCount:
maxCount = countMap[num]
mostFrequent = num
return mostFrequent
# Test case
arr = [1, 2, 3, 3]
result = findMostFrequent(arr)
print(result) # Output: 31def findMostFrequent(nums):
2 countMap = {}
3 maxCount = 0
4 mostFrequent = None
5
6 for num in nums:
7 if num in countMap:
8 countMap[num] += 1
9 else:
10 countMap[num] = 1
11
12 if countMap[num] > maxCount:
13 maxCount = countMap[num]
14 mostFrequent = num
15
16 return mostFrequent
17
18# Test case
19arr = [1, 2, 3, 3]
20result = findMostFrequent(arr)
21print(result) # Output: 3Time/Space Complexity Analysis
- Time Complexity: O(N), where N is the size of the input array.
- Space Complexity: O(K), where k is the number of unique integers in the input array.
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