C++ 100% EASY TO UNDERSTAND TLDE;
IntuitionCHECK CODE COMMENTSApproachComplexity
Time complexity: O(∣s∣)

Space complexity: O(1)
Code
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
CHECK CODE COMMENTS
# Approach
<!-- Describe your approach to solving the problem. -->

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

- Space complexity: $$O(1)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```cpp []
class Solution {
public:
    bool canConstruct(string s, int k) {
        /* 
            TC - O(|s|)
            SC - O(1)

            1. IF NO OF CHARACTERS ARE LESS THAN K THEN RETURN FALSE
            2. IF NO OF CHARACTERS ARE EQUAL TO K RETURN TRUE SINCE A CHARACTER IS A WORD OF LENGHT 1, WHICH IS A PALINDROME
            3. IF NO OF CHARACTERS ARE MORE
                A. COUNT NO OF ODD FREQUENT CHRACTERS IN THE STRING
                B. IF IT IS MORE THAN K RETURN FALSE
                C. ELSE IT IS A VALID CASE FOR EVERY OTHER SCENARIO
                    EX. AAAAAA  3
                        ALL ARE SAME SO oddFreq = 0
                            AAA | AA | A -> VALID ANS
                    EX. ABCD 3
                        oddFreq = 4
                        A | B | C | D ATLEAST 4 DIFFERENT PALINDROMES WILL BE MADE SO RETURN FALSE;
                    EX. AAB 3
                        oddFreq = 2
                        A | A | B -> VALID ANS
                        SINCE LENGTH OF STRING IS ALWAYS > K (PRE CHECKED)
                        IT FORMS A VALID ANSWER TO ALL CASES
        */
        if(s.size() < k) return false;
        if(s.size() == k) return true;
        vector<int> charFreq(26, 0);
        for(auto i: s) charFreq[i - 'a'] ^= 1;
        int oddFreq = 0;
        for(auto i : charFreq) oddFreq += i;
        if(oddFreq > k) return false;
        return true;
    }
};
```
['Array', 'C++']