Given an array of integers nums, count and return the number of occurrences of a given value.
Example:
Input: nums = [1, 4, 2, 2, 5, 2]
, value = 2
Output: 3
Explanation: the value 2 appears 3 times in the array
The core challenge of this problem is to count the number of times a specific value appears in an array. This is a common problem in data processing and analysis, where you need to determine the frequency of certain elements.
Potential pitfalls include not correctly iterating through the entire array or miscounting the occurrences due to logic errors.
To solve this problem, we can use a simple linear scan of the array. This approach ensures that we check each element exactly once, making it efficient with a time complexity of O(n), where n is the number of elements in the array.
Here is a step-by-step approach:
Let's break down the algorithm step-by-step:
count
to 0.nums
.value
.value
, increment count
.count
after the loop ends.
#include <iostream>
#include <vector>
int countOccurrences(const std::vector<int>& nums, int value) {
// Initialize the counter to 0
int count = 0;
// Iterate through each element in the array
for (int num : nums) {
// If the current element matches the value, increment the counter
if (num == value) {
count++;
}
}
// Return the total count of occurrences
return count;
}
int main() {
// Example usage
std::vector<int> nums = {1, 4, 2, 2, 5, 2};
int value = 2;
// Call the function and print the result
std::cout << "The value " << value << " appears " << countOccurrences(nums, value) << " times in the array." << std::endl;
return 0;
}
The time complexity of this approach is O(n), where n is the number of elements in the array. This is because we are iterating through the array once. The space complexity is O(1) since we are using only a constant amount of extra space for the counter.
Consider the following edge cases:
Example edge cases:
Input: nums = [], value = 2 Output: 0 Input: nums = [1, 3, 4], value = 2 Output: 0 Input: nums = [2, 2, 2], value = 2 Output: 3
To test the solution comprehensively, consider the following test cases:
Example test cases:
Input: nums = [1, 4, 2, 2, 5, 2], value = 2 Output: 3 Input: nums = [1, 1, 1, 1, 1], value = 1 Output: 5 Input: nums = [1, 2, 3, 4, 5], value = 6 Output: 0
When approaching such problems, consider the following tips:
To improve problem-solving skills, practice regularly on coding challenge platforms and study different algorithms and data structures.
In this blog post, we discussed how to count the number of occurrences of a given value in an array using a simple and efficient approach. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing strong problem-solving skills in programming.
For further reading and practice, consider the following resources: