76. Minimum Window Substring LeetCode Solution

In this guide, you will get 76. Minimum Window Substring LeetCode Solution with the best time and space complexity. The solution to Minimum Window Substring problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Minimum Window Substring solution in C++
Minimum Window Substring solution in Java
Minimum Window Substring solution in Python
Additional Resources

76. Minimum Window Substring LeetCode Solution image

Problem Statement of Minimum Window Substring

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?

Complexity Analysis

Time Complexity: O(m + n)
Space Complexity: O(128) = O(1)

76. Minimum Window Substring LeetCode Solution in C++

class Solution {
 public:
  string minWindow(string s, string t) {
    vector<int> count(128);
    int required = t.length();
    int bestLeft = -1;
    int minLength = s.length() + 1;

    for (const char c : t)
      ++count[c];

    for (int l = 0, r = 0; r < s.length(); ++r) {
      if (--count[s[r]] >= 0)
        --required;
      while (required == 0) {
        if (r - l + 1 < minLength) {
          bestLeft = l;
          minLength = r - l + 1;
        }
        if (++count[s[l++]] > 0)
          ++required;
      }
    }

    return bestLeft == -1 ? "" : s.substr(bestLeft, minLength);
  }
};
/* code provided by PROGIEZ */

76. Minimum Window Substring LeetCode Solution in Java

class Solution {
  public String minWindow(String s, String t) {
    int[] count = new int[128];
    int required = t.length();
    int bestLeft = -1;
    int minLength = s.length() + 1;

    for (final char c : t.toCharArray())
      ++count[c];

    for (int l = 0, r = 0; r < s.length(); ++r) {
      if (--count[s.charAt(r)] >= 0)
        --required;
      while (required == 0) {
        if (r - l + 1 < minLength) {
          bestLeft = l;
          minLength = r - l + 1;
        }
        if (++count[s.charAt(l++)] > 0)
          ++required;
      }
    }

    return bestLeft == -1 ? "" : s.substring(bestLeft, bestLeft + minLength);
  }
}
// code provided by PROGIEZ

76. Minimum Window Substring LeetCode Solution in Python

class Solution:
  def minWindow(self, s: str, t: str) -> str:
    count = collections.Counter(t)
    required = len(t)
    bestLeft = -1
    minLength = len(s) + 1

    l = 0
    for r, c in enumerate(s):
      count[c] -= 1
      if count[c] >= 0:
        required -= 1
      while required == 0:
        if r - l + 1 < minLength:
          bestLeft = l
          minLength = r - l + 1
        count[s[l]] += 1
        if count[s[l]] > 0:
          required += 1
        l += 1

    return '' if bestLeft == -1 else s[bestLeft: bestLeft + minLength]
 # code by PROGIEZ