Lexicographically Smallest Generated String

class Solution {
    public String generateString(String str1, String str2) {
        int n = str1.length(), m = str2.length();
        char[] word = new char[n+m-1];

        // Step 1: Fill all the 'T' related values
        for(int i=0; i<n; i++) {
            if(str1.charAt(i)=='T') {
                for(int j=0; j<m; j++) {
                    // word[i+j] 
                    if(word[i+j]!='\u0000' && word[i+j]!=str2.charAt(j)) {
                        return "";
                    } else word[i+j] = str2.charAt(j);
                }
            }
        }

        // Step 2: filling remaining values
        for(int i=0; i<word.length; i++) {
            if(word[i] == '\u0000')
                word[i] = 'a';
        }

        // Step 3: Verify 'F' value index in the word
        for(int i=0; i<n; i++) {
            // O(m*n)
            if(str1.charAt(i)=='T')
                continue;
            boolean match = true;
            for(int j=0; j<m; j++) {
                if(word[i+j]!=str2.charAt(j)) {
                    match = false;
                    break;
                }                    
            }
            if(match==true) {
                boolean fixed = false;
                for(int j=m-1; j>=0; j--) {
                    if(setByT(i, j, str1, m)==false) {
                        char change = str2.charAt(j)=='a' ? 'b' : 'a';
                        word[i+j] = change;
                        fixed = true;
                        break;
                    }
                }
                if(fixed==false)
                    return "";
            }
        }

        return new String(word);
    }

    boolean setByT(int i, int j, String str1, int m) {
        // i+j
        int first = Math.max(0,i+j-m+1);
        int last = Math.min(i+j, str1.length()-1);

        for(int ind=first; ind<=last; ind++) {
            if(str1.charAt(ind)=='T')
                return true;
        }

        return false;
    }
}