LeetCode Weekly Contest 483 Discussion and Solution

Problem-by-Problem Breakdown

Problem 1 - Largest Even Number

Approach

Code

class Solution {
    public String largestEven(String s) {
        int n = s.length();

        int i = n - 1;
        
        while(i >= 0 && s.charAt(i) == '1'){
            i--;
        }
        
        StringBuilder ans = new StringBuilder();
        
        for(int j = 0; j <= i; j++){
            ans.append(s.charAt(j));    
        }
        
        return ans.toString();
    }
}

Problem 2 - Word Squares II

• top forms the top row.
• bottom forms the bottom row.
• left forms the left column (top to bottom).
• right forms the right column (top to bottom).

• top[0] == left[0], top[3] == right[0]
• bottom[0] == left[3], bottom[3] == right[3]

Approach

Code

class Solution {
    public boolean isValid(List<String> list){
        String top = list.get(0);
        String left = list.get(1);
        String right = list.get(2);
        String bottom = list.get(3);

        if(top.charAt(0) == left.charAt(0) && top.charAt(3) == right.charAt(0) && bottom.charAt(0) == left.charAt(3) && bottom.charAt(3) == right.charAt(3)){
            return true;
        }

        return false;
    }
    
    public void permute(String[] strs, int idx, List<List<String>> comb){
        if(idx == strs.length){
            comb.add(new ArrayList<>(Arrays.asList(strs)));
            return;
        }

        for(int i = idx; i < strs.length; i++){
            swap(strs, idx, i);
            permute(strs, idx + 1, comb);
            swap(strs, idx, i);
        }
    }

    public void swap(String[] strs, int i, int j){
        String temp = strs[i];
        strs[i] = strs[j];
        strs[j] = temp;
    }
    
    public List<List<String>> wordSquares(String[] words) {
        int n = words.length;

        List<List<String>> ans = new ArrayList<>();

        for(int i = 0; i < n; i++){
            for(int j = i + 1; j < n; j++){
                for(int k = j + 1; k < n; k++){
                    for(int l = k + 1; l < n; l++){
                        String[] strs = {words[i], words[j], words[k], words[l]};
                        List<List<String>> comb = new ArrayList<>();

                        permute(strs, 0, comb);

                        for(int x = 0; x < comb.size(); x++){
                            if(isValid(comb.get(x))){
                                ans.add(comb.get(x));
                            }
                        }
                    }
                }
            }
        }

        Collections.sort(ans, (a, b) -> a.get(0).compareTo(b.get(0)) != 0 ? a.get(0).compareTo(b.get(0)) :
                        a.get(1).compareTo(b.get(1)) != 0 ? a.get(1).compareTo(b.get(1)) :
                        a.get(2).compareTo(b.get(2)) != 0 ? a.get(2).compareTo(b.get(2)) :
                        a.get(3).compareTo(b.get(3)));


        return ans;
    }
}

Problem 3 - Minimum Cost to Make Two Binary Strings Equal

• Choose any index i and flip s[i] or t[i] (change '0' to '1' or '1' to '0'). The cost of this operation is flipCost.
• Choose two distinct indices i and j, and swap either s[i] and s[j] or t[i] and t[j]. The cost of this operation is swapCost.
• Choose an index i and swap s[i] with t[i]. The cost of this operation is crossCost.

Approach

Problem 4 - Minimum Cost to Merge Sorted Lists

Approach

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