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LESSON 182 OF 409 · FREE CODING INTERVIEW QUESTIONS

Max Val and Number of Occurrences in O(n) Time Using C++

Practice O(n) Read the full explanation

Given an array of integers, return the maximum value and its number of occurrences.

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 is less than minVal: minVal = B
if C is less than 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 = the first element of nums
for each val in nums:
    if val is greater than maxVal: maxVal = val
return maxVal
Now how can we get the number of occurrences of this value in nums?

One idea would be that after we found maxVal to traverse the array again and count the occurrences in another variable count: 
maxVal = the first element of nums
for each val in nums:
    if val is greater than maxVal: maxVal = val

count = 0
for each val in nums:
    if val is equal to maxVal: count = count + 1

return maxVal and 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 update maxVal to val and reset count to 1 as this is the first occurrence ever of this value.
  • val equals maxVal => We found a new occurrence of the current solution, so we should increment count.
This is how the code would look like: 
maxVal = the first element of nums, count = 0

for each val in nums:
    if val is greater than maxVal:
        maxVal = val
        count = 1
    otherwise, if val is equal to maxVal:
        count = count + 1

return maxVal and count


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    C++ · FREE CODING INTERVIEW QUESTIONS

    Max Val And Number Of Occurrences in C++, explained

    Problem Definition

    We need to find the maximum value in an array of integers and count how many times this maximum value occurs. The solution should be efficient, running in O(n) time complexity and using O(1) space complexity.

    Input:

    An array of integers nums.

    Output:

    A list containing two elements: the maximum value and its number of occurrences.

    Constraints:

    • The array can contain both positive and negative integers.
    • The array is non-empty.

    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

    Understanding the Problem

    The core challenge is to find the maximum value in the array and count its occurrences efficiently. This problem is significant in various applications such as statistical analysis, data processing, and more. A common pitfall is to use a two-pass solution, which is not optimal for this problem.

    Approach

    To solve this problem efficiently, we can use a single-pass algorithm. The idea is to iterate through the array once, keeping track of the maximum value and its count simultaneously.

    Naive Solution

    A naive solution would involve two passes over the array: one to find the maximum value and another to count its occurrences. This approach is not optimal as it requires O(n) time complexity but involves two passes.

    Optimized Solution

    We can optimize the solution by combining the two passes into one. During a single iteration, we can update the maximum value and its count as follows:

    • If the current element is greater than the current maximum, update the maximum and reset the count to 1.
    • If the current element is equal to the current maximum, increment the count.

    Algorithm

    Here is a step-by-step breakdown of the algorithm:

    1. Initialize maxVal to the first element of the array and count to 0.
    2. Iterate through each element in the array.
    3. If the current element is greater than maxVal, update maxVal and reset count to 1.
    4. If the current element is equal to maxVal, increment count.
    5. Return [maxVal, count].

    Code Implementation

    
#include <iostream>
#include <vector>
using namespace std;

pair findMaxAndCount(const vector& nums) {
    // Initialize maxVal to the first element and count to 0
    int maxVal = nums[0];
    int count = 0;

    // Iterate through the array
    for (int val : nums) {
        if (val > maxVal) {
            // Found a new maximum value
            maxVal = val;
            count = 1; // Reset count to 1
        } else if (val == maxVal) {
            // Found another occurrence of the current maximum value
            count++;
        }
    }

    // Return the maximum value and its count
    return {maxVal, count};
}

int main() {
    vector nums = {2, 7, 11, 8, 11, 8, 3, 11};
    pair result = findMaxAndCount(nums);
    cout << "Max Value: " << result.first << ", Count: " << result.second << endl;
    return 0;
}

    Complexity Analysis

    The time complexity of this approach is O(n) because we only iterate through the array once. The space complexity is O(1) as we are using a constant amount of extra space.

    Edge Cases

    Consider the following edge cases:

    • Array with all identical elements: [5, 5, 5, 5]
    • Array with negative and positive integers: [-1, -2, 3, 4, -1, 4]
    • Array with a single element: [10]

    Each of these cases should be handled correctly by the algorithm.

    Testing

    To test the solution comprehensively, consider a variety of test cases:

    • Simple cases with small arrays.
    • Edge cases as mentioned above.
    • Large arrays to test performance.

    Using a testing framework like Google Test can help automate and manage these tests effectively.

    Thinking and Problem-Solving Tips

    When approaching such problems, consider the following tips:

    • Break down the problem into smaller, manageable parts.
    • Think about the most efficient way to solve the problem.
    • Consider edge cases and how your solution handles them.
    • Practice similar problems to improve your problem-solving skills.

    Conclusion

    In this blog post, we discussed how to find the maximum value in an array and count its occurrences efficiently using a single-pass algorithm in C++. Understanding and solving such problems is crucial for improving your algorithmic thinking and coding skills. Keep practicing and exploring further to master these concepts.

    Additional Resources

    For further reading and practice, consider the following resources:

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