LeetCode #2953 — HARD

Count Complete Substrings

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given a string word and an integer k.

A substring s of word is complete if:

  • Each character in s occurs exactly k times.
  • The difference between two adjacent characters is at most 2. That is, for any two adjacent characters c1 and c2 in s, the absolute difference in their positions in the alphabet is at most 2.

Return the number of complete substrings of word.

A substring is a non-empty contiguous sequence of characters in a string.

Example 1:

Input: word = "igigee", k = 2
Output: 3
Explanation: The complete substrings where each character appears exactly twice and the difference between adjacent characters is at most 2 are: igigee, igigee, igigee.

Example 2:

Input: word = "aaabbbccc", k = 3
Output: 6
Explanation: The complete substrings where each character appears exactly three times and the difference between adjacent characters is at most 2 are: aaabbbccc, aaabbbccc, aaabbbccc, aaabbbccc, aaabbbccc, aaabbbccc.

Constraints:

  • 1 <= word.length <= 105
  • word consists only of lowercase English letters.
  • 1 <= k <= word.length
Patterns Used
🪟Sliding Window→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given a string word and an integer k. A substring s of word is complete if: Each character in s occurs exactly k times. The difference between two adjacent characters is at most 2. That is, for any two adjacent characters c1 and c2 in s, the absolute difference in their positions in the alphabet is at most 2. Return the number of complete substrings of word. A substring is a non-empty contiguous sequence of characters in a string.

Baseline thinking

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

Pattern signal: Hash Map · Sliding Window

Example 1

"igigee"
2

Example 2

"aaabbbccc"
3

Related Problems

  • Number of Substrings Containing All Three Characters (number-of-substrings-containing-all-three-characters)
  • Count Substrings Without Repeating Character (count-substrings-without-repeating-character)
Step 02

Core Insight

What unlocks the optimal approach

  • There are at most 26 different lengths of the complete substrings: <code>k *1, k * 2, … k * 26</code>.****
  • For each length, we can use sliding window to count the frequency of each letter in the window.
  • We still need to check for all characters in the window that <code>abs(word[i] - word[i - 1]) <= 2</code>. We do this by maintaining the values of <code>abs(word[i] - word[i - 1])</code> in the sliding window dynamically in an ordered multiset or priority queue, so that we know the maximum value at each iteration.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #2953: Count Complete Substrings
class Solution {
    public int countCompleteSubstrings(String word, int k) {
        int n = word.length();
        int ans = 0;
        for (int i = 0; i < n;) {
            int j = i + 1;
            while (j < n && Math.abs(word.charAt(j) - word.charAt(j - 1)) <= 2) {
                ++j;
            }
            ans += f(word.substring(i, j), k);
            i = j;
        }
        return ans;
    }

    private int f(String s, int k) {
        int m = s.length();
        int ans = 0;
        for (int i = 1; i <= 26 && i * k <= m; ++i) {
            int l = i * k;
            int[] cnt = new int[26];
            for (int j = 0; j < l; ++j) {
                ++cnt[s.charAt(j) - 'a'];
            }
            Map<Integer, Integer> freq = new HashMap<>();
            for (int x : cnt) {
                if (x > 0) {
                    freq.merge(x, 1, Integer::sum);
                }
            }
            if (freq.getOrDefault(k, 0) == i) {
                ++ans;
            }
            for (int j = l; j < m; ++j) {
                int a = s.charAt(j) - 'a';
                int b = s.charAt(j - l) - 'a';
                freq.merge(cnt[a], -1, Integer::sum);
                ++cnt[a];
                freq.merge(cnt[a], 1, Integer::sum);

                freq.merge(cnt[b], -1, Integer::sum);
                --cnt[b];
                freq.merge(cnt[b], 1, Integer::sum);
                if (freq.getOrDefault(k, 0) == i) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × |\Sigma|)
Space
O(|\Sigma|)

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.

Related Problems
#30Substring with Concat of All Wordshard→#76Minimum Window Substringhard→#187Repeated DNA Sequencesmedium→#219Contains Duplicate IIeasy→#395Longest Substring with At Least K Repeating Charactersmedium→