LeetCode #1400 — MEDIUM

Construct K Palindrome Strings

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

Given a string s and an integer k, return true if you can use all the characters in s to construct non-empty 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

Step 01

Brute Force Baseline

Problem summary: Given a string s and an integer k, return true if you can use all the characters in s to construct non-empty k palindrome strings or false otherwise.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Greedy

Example 1

"annabelle"
2

Example 2

"leetcode"
3

Example 3

"true"
4

Step 02

Core Insight

What unlocks the optimal approach

If the s.length < k we cannot construct k strings from s and answer is false.
If the number of characters that have odd counts is > k then the minimum number of palindrome strings we can construct is > k and answer is false.
Otherwise you can construct exactly k palindrome strings and answer is true (why ?).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1400: Construct K Palindrome Strings
class Solution {
    public boolean canConstruct(String s, int k) {
        int n = s.length();
        if (n < k) {
            return false;
        }
        int[] cnt = new int[26];
        for (int i = 0; i < n; ++i) {
            ++cnt[s.charAt(i) - 'a'];
        }
        int x = 0;
        for (int v : cnt) {
            x += v & 1;
        }
        return x <= k;
    }
}

// Accepted solution for LeetCode #1400: Construct K Palindrome Strings
func canConstruct(s string, k int) bool {
	if len(s) < k {
		return false
	}
	cnt := [26]int{}
	for _, c := range s {
		cnt[c-'a']++
	}
	x := 0
	for _, v := range cnt {
		x += v & 1
	}
	return x <= k
}

# Accepted solution for LeetCode #1400: Construct K Palindrome Strings
class Solution:
    def canConstruct(self, s: str, k: int) -> bool:
        if len(s) < k:
            return False
        cnt = Counter(s)
        return sum(v & 1 for v in cnt.values()) <= k

// Accepted solution for LeetCode #1400: Construct K Palindrome Strings
struct Solution;

impl Solution {
    fn can_construct(s: String, k: i32) -> bool {
        let k = k as usize;
        let n = s.len();
        if n < k {
            return false;
        }
        let mut count = vec![0; 26];
        for c in s.bytes() {
            count[(c - b'a') as usize] += 1;
        }
        let mut odd = 0;
        for v in count {
            if v % 2 != 0 {
                odd += 1;
            }
        }
        odd <= k
    }
}

#[test]
fn test() {
    let s = "annabelle".to_string();
    let k = 2;
    let res = true;
    assert_eq!(Solution::can_construct(s, k), res);
    let s = "leetcode".to_string();
    let k = 3;
    let res = false;
    assert_eq!(Solution::can_construct(s, k), res);
    let s = "true".to_string();
    let k = 4;
    let res = true;
    assert_eq!(Solution::can_construct(s, k), res);
    let s = "yzyzyzyzyzyzyzy".to_string();
    let k = 2;
    let res = true;
    assert_eq!(Solution::can_construct(s, k), res);
    let s = "cr".to_string();
    let k = 7;
    let res = false;
    assert_eq!(Solution::can_construct(s, k), res);
    let s = "aaa".to_string();
    let k = 2;
    let res = true;
    assert_eq!(Solution::can_construct(s, k), res);
}

// Accepted solution for LeetCode #1400: Construct K Palindrome Strings
function canConstruct(s: string, k: number): boolean {
    if (s.length < k) {
        return false;
    }
    const cnt: number[] = new Array(26).fill(0);
    for (const c of s) {
        ++cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)];
    }
    let x = 0;
    for (const v of cnt) {
        x += v & 1;
    }
    return x <= k;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(C)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.