FindNoWithEvenNoOfDigits.java

/*Problem Statement
// 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] <= 105
*/




// Approach 1
publicintfindNumbers(int[] nums) {
intcount = 0, evenNo = 0;
for(inti = 0; i < nums.length; i++){
intj = nums[i];
while(j > 0){
count++;
j/= 10;
 }
if(count %2 == 0)
evenNo++;
count = 0;
 }
returnevenNo;
 }

// Approach 2
publicintfindNumbers(int[] nums){
inteven = 0;
for(inti = 0; i < nums.length; i++){
Stringa = String.valueOf(nums[i]);
if(a.length()%2 == 0)
even++;
 }
returneven;
 }


// Approach 3

publicintfindNumbers(int[] nums){
inteven = 0;
for(inti = 0; i < nums.length; i++){
if(nums[i]>9 && nums[i]<100 || nums[i]>999 && nums[i]<10000 || nums[i] == 100000)
even++;
 } 
returneven;
 }