JAVA || SIMPLE AND EASY SOLUTION ||  oddCounts logic || TRY WITH forEach LOOP 
IntuitionUSING forEach loop it will reduce the runtime.ApproachComplexity
Time complexity:

Space complexity:
Code
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
USING forEach loop it will reduce the runtime.
# Approach
<!-- Describe your approach to solving the problem. -->

# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```java []
class Solution {
    public boolean canConstruct(String s, int k) {
        if(s.length() < k) return false;
        int oddCount = 0;
        int frq [] = new int [26];
        for(char i : s.toCharArray())
        {
            frq[i -'a']++;
        }
        for(int i : frq)
        {
            if(i > 0)
            {
                 if(i % 2 == 1)
                {
                     oddCount++;
                }
            }
        }
       if(oddCount > k)
       {
        return false;
       }
       return true;
    }
}
```
['Java']