Given an array of integers, return the maximum value and its number of occurences.
Example:
Input: nums = [2, 7, 11, 8, 11, 8, 3, 11]
Output: [11, 3]
Explanation: The maximum value is 11 and it appears 3 times
Note:
Your algorithm should run in O(n) time and use O(1) space.
Follow up:
Could you do this in one pass (e.g. looping over the array only once)?
Let's start by thinking about an easier version of the problem. Suppose we only want to return the maximum value in the array and we don't care how many times it occurs. Have we solved a similar easier problem in the past? Can we adapt that solution here?
That similar problem is "Minimum Value Of Three". There we tried out each of the three numbers and used a variable
minVal to store the smallest one found so far throught the algorithm:
minVal = A
if B < minVal:
minVal = B
if C < minVal:
minVal = C
return minVal
How can we adapt this approach for an array of numbers instead of just three?
The principle is the same. We will iterate through each value in the array
nums and update the maximum accordingly.
The variable is now maxVal and stores the greatest element found so far:
maxVal = nums[0]
for val in nums:
if val > maxVal:
maxVal= val
return maxVal
Now how can we get the number of occurences of this value in nums?
One idea would be that after we found
maxVal to traverse the array again and count the occurences in another variable count:
maxVal = nums[0]
for val in nums:
if val > maxVal:
maxVal = val
count = 0
for val in nums:
if val == maxVal:
count += 1
return [maxVal, count]
This is good, but the follow up asks us to solve the question in one pass. We can also compute count in the same for loop where we compute maxVal. Take a small example and try to figure out how to update both maxVal and count for each val when necessary.
For each element
val of nums, there are two scenarios that are important:
val > maxVal=> We found a new maximum value, so we should updatemaxValtovaland resetcountto1as this is the first occurence ever of this value.val == maxVal=> We found a new occurence of the current solution, so we should incrementcount.
maxVal, count = nums[0], 0
for val in nums:
if val > maxVal:
maxVal = val
count = 1
else if val == maxVal:
count += 1
return [maxVal, count]
Walkthroughs of the solution. Try the steps first, then watch to compare approaches.
Solution #1 · O(n)
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