LeetCode #76 — HARD

Minimum Window Substring

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

The Problem

Problem Statement

Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "".

The testcases will be generated such that the answer is unique.

Example 1:

Input: s = "ADOBECODEBANC", t = "ABC"
Output: "BANC"
Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.

Example 2:

Input: s = "a", t = "a"
Output: "a"
Explanation: The entire string s is the minimum window.

Example 3:

Input: s = "a", t = "aa"
Output: ""
Explanation: Both 'a's from t must be included in the window.
Since the largest window of s only has one 'a', return empty string.

Constraints:

m == s.length
n == t.length
1 <= m, n <= 10⁵
s and t consist of uppercase and lowercase English letters.

Follow up: Could you find an algorithm that runs in O(m + n) time?

Step 01

Brute Force Baseline

Problem summary: Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "". The testcases will be generated such that the answer is unique.

Baseline thinking

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

Pattern signal: Hash Map · Sliding Window

Example 1

"ADOBECODEBANC"
"ABC"

Example 2

"a"
"a"

Example 3

"a"
"aa"

Core Insight

What unlocks the optimal approach

Use two pointers to create a window of letters in s, which would have all the characters from t.
Expand the right pointer until all the characters of t are covered.
Once all the characters are covered, move the left pointer and ensure that all the characters are still covered to minimize the subarray size.
Continue expanding the right and left pointers until you reach the end of s.

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

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

class Solution {
    public String minWindow(String s, String t) {
        if (t.length() > s.length()) return "";

        int[] need = new int[128];
        for (char ch : t.toCharArray()) need[ch]++;

        int required = t.length();
        int left = 0;
        int bestLen = Integer.MAX_VALUE, bestStart = 0;

        for (int right = 0; right < s.length(); right++) {
            char rc = s.charAt(right);
            if (need[rc] > 0) required--;
            need[rc]--;

            while (required == 0) {
                if (right - left + 1 < bestLen) {
                    bestLen = right - left + 1;
                    bestStart = left;
                }

                char lc = s.charAt(left++);
                need[lc]++;
                if (need[lc] > 0) required++;
            }
        }

        return bestLen == Integer.MAX_VALUE ? "" : s.substring(bestStart, bestStart + bestLen);
    }
}

func minWindow(s string, t string) string {
    if len(t) > len(s) {
        return ""
    }

    need := [128]int{}
    for i := 0; i < len(t); i++ {
        need[t[i]]++
    }

    required := len(t)
    left := 0
    bestLen := len(s) + 1
    bestStart := 0

    for right := 0; right < len(s); right++ {
        rc := s[right]
        if need[rc] > 0 {
            required--
        }
        need[rc]--

        for required == 0 {
            if right-left+1 < bestLen {
                bestLen = right - left + 1
                bestStart = left
            }

            lc := s[left]
            left++
            need[lc]++
            if need[lc] > 0 {
                required++
            }
        }
    }

    if bestLen == len(s)+1 {
        return ""
    }
    return s[bestStart : bestStart+bestLen]
}

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        if len(t) > len(s):
            return ""

        need = {}
        for ch in t:
            need[ch] = need.get(ch, 0) + 1

        required = len(t)
        left = 0
        best_len = float('inf')
        best_start = 0

        for right, ch in enumerate(s):
            if need.get(ch, 0) > 0:
                required -= 1
            need[ch] = need.get(ch, 0) - 1

            while required == 0:
                if right - left + 1 < best_len:
                    best_len = right - left + 1
                    best_start = left

                left_ch = s[left]
                left += 1
                need[left_ch] = need.get(left_ch, 0) + 1
                if need[left_ch] > 0:
                    required += 1

        return "" if best_len == float('inf') else s[best_start:best_start + best_len]

impl Solution {
    pub fn min_window(s: String, t: String) -> String {
        if t.len() > s.len() {
            return String::new();
        }

        let mut need = [0i32; 128];
        for &b in t.as_bytes() {
            need[b as usize] += 1;
        }

        let bytes = s.as_bytes();
        let mut required = t.len() as i32;
        let mut left = 0usize;
        let mut best_len = usize::MAX;
        let mut best_start = 0usize;

        for right in 0..bytes.len() {
            let rc = bytes[right] as usize;
            if need[rc] > 0 {
                required -= 1;
            }
            need[rc] -= 1;

            while required == 0 {
                if right - left + 1 < best_len {
                    best_len = right - left + 1;
                    best_start = left;
                }

                let lc = bytes[left] as usize;
                left += 1;
                need[lc] += 1;
                if need[lc] > 0 {
                    required += 1;
                }
            }
        }

        if best_len == usize::MAX {
            String::new()
        } else {
            s[best_start..best_start + best_len].to_string()
        }
    }
}

function minWindow(s: string, t: string): string {
  if (t.length > s.length) return '';

  const need: number[] = Array(128).fill(0);
  for (const ch of t) need[ch.charCodeAt(0)]++;

  let required = t.length;
  let left = 0;
  let bestLen = Infinity;
  let bestStart = 0;

  for (let right = 0; right < s.length; right++) {
    const rc = s.charCodeAt(right);
    if (need[rc] > 0) required--;
    need[rc]--;

    while (required === 0) {
      if (right - left + 1 < bestLen) {
        bestLen = right - left + 1;
        bestStart = left;
      }

      const lc = s.charCodeAt(left++);
      need[lc]++;
      if (need[lc] > 0) required++;
    }
  }

  return bestLen === Infinity ? '' : s.slice(bestStart, bestStart + bestLen);
}

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(m + n)

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.