Minimum Window Substring

Leet Code link - https://leetcode.com/problems/minimum-window-substring/description/

Complexity - Hard

Description

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?

Solution

The question asks us to return the minimum window from the string S which has all the characters of the string T. Let us call a window desirable if it has all the characters from T.

We can use a simple sliding window approach to solve this problem.

In any sliding window based problem we have two pointers. One right pointer whose job is to expand the current window and then we have the left pointer whose job is to contract a given window. At any point in time only one of these pointers move and the other one remains fixed.

The solution is pretty intuitive. We keep expanding the window by moving the right pointer. When the window has all the desired characters, we contract (if possible) and save the smallest window till now.

The answer is the smallest desirable window.

For eg. S = "ABAACBAB" T = "ABC". Then our answer window is "ACB" and shown below is one of the possible desirable windows

We start with two pointers, left and right initially pointing to the first element of the string S.

We use the right pointer to expand the window until we get a desirable window i.e. a window that contains all of the characters of T.

Once we have a window with all the characters, we can move the left pointer ahead one by one. If the window is still a desirable one we keep on updating the minimum window size.

If the window is not desirable any more, we repeat step2 onwards.

The above steps are repeated until we have looked at all the windows. The smallest window is returned.

class Solution {
    public String minWindow(String s, String t) {
        if (s.length() == 0 || t.length() == 0) {
            return "";
        }

        // Dictionary which keeps a count of all the unique characters in t.
        Map<Character, Integer> dictT = new HashMap<Character, Integer>();
        for (int i = 0; i < t.length(); i++) {
            int count = dictT.getOrDefault(t.charAt(i), 0);
            dictT.put(t.charAt(i), count + 1);
        }

        // Number of unique characters in t, which need to be present in the desired window.
        int required = dictT.size();

        // Left and Right pointer
        int l = 0, r = 0;

        // formed is used to keep track of how many unique characters in t
        // are present in the current window in its desired frequency.
        // e.g. if t is "AABC" then the window must have two A's, one B and one C.
        // Thus formed would be = 3 when all these conditions are met.
        int formed = 0;

        // Dictionary which keeps a count of all the unique characters in the current window.
        Map<Character, Integer> windowCounts = new HashMap<
            Character,
            Integer
        >();

        // ans list of the form (window length, left, right)
        int[] ans = { -1, 0, 0 };

        while (r < s.length()) {
            // Add one character from the right to the window
            char c = s.charAt(r);
            int count = windowCounts.getOrDefault(c, 0);
            windowCounts.put(c, count + 1);

            // If the frequency of the current character added equals to the
            // desired count in t then increment the formed count by 1.
            if (
                dictT.containsKey(c) &&
                windowCounts.get(c).intValue() == dictT.get(c).intValue()
            ) {
                formed++;
            }

            // Try and contract the window till the point where it ceases to be 'desirable'.
            while (l <= r && formed == required) {
                c = s.charAt(l);
                // Save the smallest window until now.
                if (ans[0] == -1 || r - l + 1 < ans[0]) {
                    ans[0] = r - l + 1;
                    ans[1] = l;
                    ans[2] = r;
                }

                // The character at the position pointed by the
                // `Left` pointer is no longer a part of the window.
                windowCounts.put(c, windowCounts.get(c) - 1);
                if (
                    dictT.containsKey(c) &&
                    windowCounts.get(c).intValue() < dictT.get(c).intValue()
                ) {
                    formed--;
                }

                // Move the left pointer ahead, this would help to look for a new window.
                l++;
            }

            // Keep expanding the window once we are done contracting.
            r++;
        }

        return ans[0] == -1 ? "" : s.substring(ans[1], ans[2] + 1);
    }
}

Time complexity: O ( n ) O(n) Space complexity: O ( m ) O(m) Where n n is the length of the string s s and m m is the total number of unique characters in the strings t t and s s.