O(N) JAVA Sliding Window solution with explanation

  • 7

    Sliding Window Solution:

    1) Spread the right pointer until it satisfies the requirement  
    2) Shrink the left pointer to get the minimum range 
    3) Keep the above steps.

    Time complexity = O(2n) = O(n)

    There're 2 loops: for loop of i, while loop of j. As j only steps forward, never steps backward. So time complexity = O(n) + O(n) = O(2n) = O(n)

    JAVA Code:

    A little trick is using two arrays to count the characters in s and t, instead of HashMap, to avoid TLE.

    boolean sContainsT(int mapS[], int mapT[]) {// Runtime = O(256) = O(1)
        for (int i = 0; i < mapT.length; i++) {// s should cover all characters in t
            if (mapT[i] > mapS[i])
                return false; 
        return true;
    public String minWindow(String s, String t) {   
        int mapS[] = new int[256];// Count characters in s
        int mapT[] = new int[256];// Count characters in t      
        for (char ch : t.toCharArray())
        String res = "";
        int right = 0, min = Integer.MAX_VALUE;         
        for (int i = 0; i < s.length(); i++) {// Two pointers of the sliding window: i(left), right
            while (right < s.length() && !sContainsT(mapS, mapT)) {// Extend the right pointer of the sliding window
            if (sContainsT(mapS, mapT) && min > right - i + 1) {
                res = s.substring(i, right);
                min = right - i + 1;
            mapS[s.charAt(i)]--;// Shrink the left pointer from i to i + 1
        return res;

  • 0

    obviously this is not O(N), since for each i you call isCovered(), which takes O(len(T)) time.

  • -5

    I think you cannot assume that the characters are ascii.

  • 0

    Looks like O(N^2) to me. In the worst case, for every i, you could go to the end of string s.

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