047-perfect-even-count

Perfect Even Count

Problem Description

Xiao C has defined a "perfect even number." A positive integer x is considered a perfect even number if it satisfies both of the following conditions:

1. x is even
2. x falls within the range [l, r]

Given an array a of length n, determine how many perfect even numbers are in the array.

Requirements

• Input: an integer n (array length), integers l and r (range endpoints), and array a
• Where 1 <= n <= 100000, 1 <= l <= r <= 1000000
• Each element in array a satisfies: 1 <= a[i] <= 1000000

Constraints

• A number must satisfy both conditions simultaneously to be counted.

Examples

Example 1:

Input:

n = 5, l = 3, r = 8, a = [1, 2, 6, 8, 7]

Output:

2

Explanation: The numbers 6 and 8 in the array are perfect even numbers because they are both even and within the range [3, 8].

Example 2:

Input:

n = 4, l = 10, r = 20, a = [12, 15, 18, 9]

Output:

2

Explanation: The numbers 12 and 18 in the array are perfect even numbers because they are both even and within the range [10, 20].

Example 3:

Input:

n = 3, l = 1, r = 10, a = [2, 4, 6]

Output:

3

Explanation: All numbers 2, 4, and 6 in the array are perfect even numbers because they are all even and within the range [1, 10].

Solution Approach

1. Problem Analysis

   • Count the numbers in the array that satisfy two conditions:
   • Condition 1: The number is even
   • Condition 2: The number falls within the given range [l, r]

2. Solution

   • Use a counter (count) to track numbers that meet the conditions
   • Iterate through each number in the array and check both conditions:
     • Use the modulo operation (num % 2 == 0) to check if it is even
     • Use range checking (l <= num <= r) to confirm it is within the range
   • If both conditions are satisfied simultaneously, increment the counter
   • Return the counter value at the end

Implementation

def solution(n: int, l: int, r: int, a: list) -> int:
    count = 0

    for num in a:
        if num % 2 == 0 and l <= num <= r:
            count += 1

    return count

Complexity Analysis

• Time Complexity: O(n), where n is the length of the array. Only one pass through the array is needed.
• Space Complexity: O(1), only a single counter variable is used.

Summary

This is a straightforward filtering problem. The solution iterates through the array once, checking each element against two conditions (even parity and range membership). The modulo operation determines evenness, and a double inequality checks the range. Python's chained comparison (l <= num <= r) provides a clean and readable implementation.