LeetCode Weekly Contest 480 Discussion and Solution

Problem-by-Problem Breakdown

Problem 1 - Absolute Difference Between Maximum and Minimum K Elements

• the sum of the k largest elements in the array; and
• the sum of the k smallest elements in the array.

Approach

• Sort the array
• Take the sum of the first k elements
• Take the sum of the last k elements
• Return the absolute difference

Code

class Solution {
    public int absDifference(int[] nums, int k) {
        Arrays.sort(nums);

        int n = nums.length;

        int max = 0;
        int min = 0;

        for(int i = 0; i < k; i++){
            min += nums[i];
        }

        for(int i = n - 1; i >= n - k; i--){
            max += nums[i];
        }

        return Math.abs(max - min);
    }
}

Problem 2 - Reverse Words With Same Vowel Count

Approach

• First, count the number of vowels in the first word
• Then extract words one by one
• For each word, If its vowel count matches the first word’s vowel count, reverse it, Otherwise, append it as it is

Code

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

        StringBuilder ans = new StringBuilder();

        int count = 0;

        int i = 0;

        while(i < n && s.charAt(i) != ' '){
            char ch = s.charAt(i);

            if(ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u'){
                count++;
            }

            ans.append(ch);

            i++;
        }

        if(i < n && s.charAt(i) == ' ') ans.append(' ');
        i++;

        while(i < n){
            StringBuilder sb = new StringBuilder();
            int vowels = 0;

            int j = i;

            while(j < n && s.charAt(j) != ' '){
                char ch = s.charAt(j);

                if(ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u'){
                    vowels++;
                }

                sb.append(ch);

                j++;
            }

            if(vowels == count){
                for(int k = sb.length() - 1; k >= 0; k--){
                    char ch = sb.charAt(k);
                    ans.append(ch);
                }
            }
            else{
                for(int k = 0; k < sb.length(); k++){
                    char ch = sb.charAt(k);
                    ans.append(ch);
                }
            }

            if(j < n && s.charAt(j) == ' ') ans.append(' ');
            j++;
            i = j;
        }

        return ans.toString();
    }
}

Problem 3 - Minimum Moves to Balance Circular Array

Approach

Problem 4 - Minimum Deletions to Make Alternating Substring

• [1, j]: Flip the character at index j of s i.e. 'A' changes to 'B' (and vice versa). This operation mutates s and affects subsequent queries.
• [2, l, r]: Compute the minimum number of character deletions required to make the substring s[l..r] alternating. This operation does not modify s; the length of s remains n.

Approach

• [1, j]: Flip the character at index j
• [2, l, r]: Find the minimum number of deletions needed to make the substring alternating

My Contest Experience

• I solved 2 problems in 38 min 25 sec
• Had 10 minutes penalty
• Total time became 48 min 25 sec
• Rank was 5891

• I again solved only 2 problems
• But the total time was just 15 min 51 sec
• No wrong submissions, no penalty