Find Numbers with Even Number of Digits using the Reduce Function in Javascript, C++ and Python

Given an array nums of integers, return how many of them contain an even number of digits.

Example 1:
Input: nums = [12,345,2,6,7896]
Output: 2
Explanation:
12 contains 2 digits (even number of digits).
345 contains 3 digits (odd number of digits).
2 contains 1 digit (odd number of digits).
6 contains 1 digit (odd number of digits).
7896 contains 4 digits (even number of digits).
Therefore only 12 and 7896 contain an even number of digits.

Example 2:
Input: nums = [555,901,482,1771]
Output: 1
Explanation:
Only 1771 contains an even number of digits.

Constraints:
1 <= nums.length <= 500
1 <= nums[i] <= 10^5

Hints:
How to compute the number of digits of a number ?
Divide the number by 10 again and again to get the number of digits.

How to Compute the Number of Digits given a Integer?

You could, however, convert the number into string, then obtain its length. If you want to do it mathematically, you could do the following division by ten, until it is smaller than 10.

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int getNumberOfDigits(n) {
   int len = 1;
   while (n >= 10) {
      n /= 10;
      len ++;
   }
   return len;
}

int getNumberOfDigits(n) {
   int len = 1;
   while (n >= 10) {
      n /= 10;
      len ++;
   }
   return len;
}

Using Reduce In Python to Find Numbers with Even Number of Digits

The Reduce() helps to accumulate a value based on a lambda function, known as the reducer function.

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from functools import reduce
 
class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        return reduce(lambda s, x: s + (1 - len(str(x)) % 2), nums, 0)

from functools import reduce

class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        return reduce(lambda s, x: s + (1 - len(str(x)) % 2), nums, 0)

Using Reduce In Javascript to Find Numbers with Even Number of Digits

The reduce() can be invoked on the array in Javascript.

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/**
 * @param {number[]} nums
 * @return {number}
 */
var findNumbers = function(nums) {
    return nums.reduce((s, x) => s + (1 - ('' + x).length % 2), 0);
};

/**
 * @param {number[]} nums
 * @return {number}
 */
var findNumbers = function(nums) {
    return nums.reduce((s, x) => s + (1 - ('' + x).length % 2), 0);
};

Using std::accumulate() to Find Numbers with Even Number of Digits

In C++, the std::accumulate() is the reduce method.

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class Solution {
public:
    int findNumbers(vector<int>& nums) {
        return std::accumulate(begin(nums), end(nums), 0, [](auto &s, auto &n) {
            return s + 1 - std::to_string(n).size() % 2;
        });
    }
};

class Solution {
public:
    int findNumbers(vector<int>& nums) {
        return std::accumulate(begin(nums), end(nums), 0, [](auto &s, auto &n) {
            return s + 1 - std::to_string(n).size() % 2;
        });
    }
};

Using std::count_if() to Find Numbers with Even Number of Digits

Alternatively, you can count_if based on an (anonymous) function:

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class Solution {
public:
    int findNumbers(vector<int>& nums) {
        return std::count_if(begin(nums), end(nums), [](auto &num) {
            return std::to_string(num).size() % 2 == 0;
        });
    }
};

class Solution {
public:
    int findNumbers(vector<int>& nums) {
        return std::count_if(begin(nums), end(nums), [](auto &num) {
            return std::to_string(num).size() % 2 == 0;
        });
    }
};

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