LeetCode #2516 — MEDIUM

Take K of Each Character From Left and Right

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

The Problem

Problem Statement

You are given a string s consisting of the characters 'a', 'b', and 'c' and a non-negative integer k. Each minute, you may take either the leftmost character of s, or the rightmost character of s.

Return the minimum number of minutes needed for you to take at least k of each character, or return -1 if it is not possible to take k of each character.

Example 1:

Input: s = "aabaaaacaabc", k = 2
Output: 8
Explanation: 
Take three characters from the left of s. You now have two 'a' characters, and one 'b' character.
Take five characters from the right of s. You now have four 'a' characters, two 'b' characters, and two 'c' characters.
A total of 3 + 5 = 8 minutes is needed.
It can be proven that 8 is the minimum number of minutes needed.

Example 2:

Input: s = "a", k = 1
Output: -1
Explanation: It is not possible to take one 'b' or 'c' so return -1.

Constraints:

1 <= s.length <= 10⁵
s consists of only the letters 'a', 'b', and 'c'.
0 <= k <= s.length

Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting of the characters 'a', 'b', and 'c' and a non-negative integer k. Each minute, you may take either the leftmost character of s, or the rightmost character of s. Return the minimum number of minutes needed for you to take at least k of each character, or return -1 if it is not possible to take k of each character.

Baseline thinking

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

Pattern signal: Hash Map · Sliding Window

Example 1

"aabaaaacaabc"
2

Example 2

"a"
1

Core Insight

What unlocks the optimal approach

Start by counting the frequency of each character and checking if it is possible.
If you take x characters from the left side, what is the minimum number of characters you need to take from the right side? Find this for all values of x in the range 0 ≤ x ≤ s.length.
Use a two-pointers approach to avoid computing the same information multiple times.

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 #2516: Take K of Each Character From Left and Right
class Solution {
    public int takeCharacters(String s, int k) {
        int[] cnt = new int[3];
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            ++cnt[s.charAt(i) - 'a'];
        }
        for (int x : cnt) {
            if (x < k) {
                return -1;
            }
        }
        int mx = 0, j = 0;
        for (int i = 0; i < n; ++i) {
            int c = s.charAt(i) - 'a';
            --cnt[c];
            while (cnt[c] < k) {
                ++cnt[s.charAt(j++) - 'a'];
            }
            mx = Math.max(mx, i - j + 1);
        }
        return n - mx;
    }
}

// Accepted solution for LeetCode #2516: Take K of Each Character From Left and Right
func takeCharacters(s string, k int) int {
	cnt := [3]int{}
	for _, c := range s {
		cnt[c-'a']++
	}
	for _, x := range cnt {
		if x < k {
			return -1
		}
	}
	mx, j := 0, 0
	for i, c := range s {
		c -= 'a'
		for cnt[c]--; cnt[c] < k; j++ {
			cnt[s[j]-'a']++
		}
		mx = max(mx, i-j+1)
	}
	return len(s) - mx
}

# Accepted solution for LeetCode #2516: Take K of Each Character From Left and Right
class Solution:
    def takeCharacters(self, s: str, k: int) -> int:
        cnt = Counter(s)
        if any(cnt[c] < k for c in "abc"):
            return -1
        mx = j = 0
        for i, c in enumerate(s):
            cnt[c] -= 1
            while cnt[c] < k:
                cnt[s[j]] += 1
                j += 1
            mx = max(mx, i - j + 1)
        return len(s) - mx

// Accepted solution for LeetCode #2516: Take K of Each Character From Left and Right
use std::collections::HashMap;

impl Solution {
    pub fn take_characters(s: String, k: i32) -> i32 {
        let mut cnt: HashMap<char, i32> = HashMap::new();
        for c in s.chars() {
            *cnt.entry(c).or_insert(0) += 1;
        }

        if "abc".chars().any(|c| cnt.get(&c).unwrap_or(&0) < &k) {
            return -1;
        }

        let mut mx = 0;
        let mut j = 0;
        let mut cs = s.chars().collect::<Vec<char>>();
        for i in 0..cs.len() {
            let c = cs[i];
            *cnt.get_mut(&c).unwrap() -= 1;
            while cnt.get(&c).unwrap() < &k {
                *cnt.get_mut(&cs[j]).unwrap() += 1;
                j += 1;
            }
            mx = mx.max(i - j + 1);
        }
        (cs.len() as i32) - (mx as i32)
    }
}

// Accepted solution for LeetCode #2516: Take K of Each Character From Left and Right
function takeCharacters(s: string, k: number): number {
    const idx = (c: string) => c.charCodeAt(0) - 97;
    const cnt: number[] = Array(3).fill(0);
    for (const c of s) {
        ++cnt[idx(c)];
    }
    if (cnt.some(v => v < k)) {
        return -1;
    }
    const n = s.length;
    let [mx, j] = [0, 0];
    for (let i = 0; i < n; ++i) {
        const c = idx(s[i]);
        --cnt[c];
        while (cnt[c] < k) {
            ++cnt[idx(s[j++])];
        }
        mx = Math.max(mx, i - j + 1);
    }
    return n - mx;
}

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(k)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

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.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.