g1301_1400.s1400_construct_k_palindrome_strings.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1301_1400.s1400_construct_k_palindrome_strings;

// #Medium #String #Hash_Table #Greedy #Counting
// #2022_03_12_Time_5_ms_(94.62%)_Space_53.8_MB_(27.48%)

/**
 * 1400 - Construct K Palindrome Strings\.
 *
 * Medium
 *
 * Given a string `s` and an integer `k`, return `true` _if you can use all the characters in_ `s` _to construct_ `k` _palindrome strings or_ `false` _otherwise_.
 *
 * **Example 1:**
 *
 * **Input:** s = "annabelle", k = 2
 *
 * **Output:** true
 *
 * **Explanation:** You can construct two palindromes using all characters in s. Some possible constructions "anna" + "elble", "anbna" + "elle", "anellena" + "b"
 *
 * **Example 2:**
 *
 * **Input:** s = "leetcode", k = 3
 *
 * **Output:** false
 *
 * **Explanation:** It is impossible to construct 3 palindromes using all the characters of s.
 *
 * **Example 3:**
 *
 * **Input:** s = "true", k = 4
 *
 * **Output:** true
 *
 * **Explanation:** The only possible solution is to put each character in a separate string.
 *
 * **Constraints:**
 *
 * *   1 <= s.length <= 10⁵
 * *   `s` consists of lowercase English letters.
 * *   1 <= k <= 10⁵
**/
public class Solution {
    public boolean canConstruct(String s, int k) {
        if (s.length() == k) {
            // if size is same as k we can separate out all letters
            return true;
        }
        if (s.length() < k) {
            // if size is less than it is not possible
            return false;
        }
        // count occurrence of each letter
        int[] count = new int[26];
        for (char curr : s.toCharArray()) {
            count[curr - 'a']++;
        }
        // reduce k whenever count is odd
        for (int i = 0; i < 26; i++) {
            if (count[i] % 2 != 0) {
                k--;
            }
        }
        // we can have max k odd characters
        return k >= 0;
    }
}