Sliding Window

Table of Contents

• 209. Minimum Size Subarray Sum (Medium)
• 3. Longest Substring Without Repeating Characters (Medium)
• 30. Substring with Concatenation of All Words (Hard)
• 76. Minimum Window Substring (Hard)

209. Minimum Size Subarray Sum

• LeetCode | LeetCode CH (Medium)

• Tags: array, binary search, sliding window, prefix sum

209. Minimum Size Subarray Sum - Python Solution

import bisect
from typing import List


# Prefix Sum
def minSubArrayLenPS(target: int, nums: List[int]) -> int:
    n = len(nums)
    prefix_sums = [0] * (n + 1)

    for i in range(1, n + 1):
        prefix_sums[i] = prefix_sums[i - 1] + nums[i - 1]

    minLen = float("inf")

    for i in range(n + 1):
        new_target = target + prefix_sums[i]
        bound = bisect.bisect_left(prefix_sums, new_target)
        if bound != len(prefix_sums):
            minLen = min(minLen, bound - i)

    return 0 if minLen == float("inf") else minLen


# Sliding Window Variable Size
def minSubArrayLenSW(target: int, nums: List[int]) -> int:
    res, cur = float("inf"), 0
    left = 0

    for right in range(len(nums)):
        cur += nums[right]

        while cur >= target:
            res = min(res, right - left + 1)
            cur -= nums[left]
            left += 1

    return res if res != float("inf") else 0


target = 7
nums = [2, 3, 1, 2, 4, 3]
print(minSubArrayLenPS(target, nums))  # 2
print(minSubArrayLenSW(target, nums))  # 2

3. Longest Substring Without Repeating Characters

• LeetCode | LeetCode CH (Medium)

• Tags: hash table, string, sliding window

• Classic variable sliding window problem. Use a set to keep track of the characters in the current window.
• Return the length of the longest substring without repeating characters.

3. Longest Substring Without Repeating Characters - Python Solution

# Sliding Window Variable Max
def lengthOfLongestSubstring(s: str) -> int:
    n = len(s)
    if n <= 1:
        return n

    left = 0
    res = 0
    window = set()

    for right in range(n):
        while left < right and s[right] in window:
            window.remove(s[left])
            left += 1
        window.add(s[right])
        res = max(res, right - left + 1)

    return res


if __name__ == "__main__":
    s = "abcabcbb"
    assert lengthOfLongestSubstring(s) == 3

3. Longest Substring Without Repeating Characters - C++ Solution

#include <iostream>
#include <string>
#include <unordered_set>
using namespace std;

int lengthOfLongestSubstring(string s) {
    int n = s.length();
    int res = 0;
    int left = 0;
    unordered_set<char> window;

    for (int right = 0; right < n; right++) {
        char ch = s[right];

        while (window.find(ch) != window.end()) {
            window.erase(s[left]);
            left++;
        }

        window.insert(ch);
        res = max(res, right - left + 1);
    }
    return (int)res;
}

int main() {
    string s = "abcabcbb";
    cout << lengthOfLongestSubstring(s) << endl;  // 3
    s = "bbbbb";
    cout << lengthOfLongestSubstring(s) << endl;  // 1
    s = "pwwkew";
    cout << lengthOfLongestSubstring(s) << endl;  // 3
    return 0;
}

30. Substring with Concatenation of All Words

• LeetCode | LeetCode CH (Hard)

• Tags: hash table, string, sliding window

76. Minimum Window Substring

• LeetCode | LeetCode CH (Hard)

• Tags: hash table, string, sliding window

76. Minimum Window Substring - Python Solution

from collections import Counter


def minWindow(s: str, t: str) -> str:
    if not t or not s:
        return ""

    counts = Counter(t)
    required = len(counts)

    left, right = 0, 0
    formed = 0
    window_counts = dict()

    result = float("inf"), None, None

    while right < len(s):
        char = s[right]
        window_counts[char] = window_counts.get(char, 0) + 1
        if char in counts and window_counts[char] == counts[char]:
            formed += 1

        while left <= right and formed == required:
            char = s[left]
            if right - left + 1 < result[0]:
                result = (right - left + 1, left, right)
            window_counts[char] -= 1
            if char in counts and window_counts[char] < counts[char]:
                formed -= 1
            left += 1

        right += 1

    return "" if result[0] == float("inf") else s[result[1] : result[2] + 1]


s = "ADOBECODEBANC"
t = "ABC"
print(minWindow(s, t))  # BANC

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