[Leetcode] 1170. Compare Strings by Frequency of the Smallest Character

1. Pre-calculated on words.

2. Counting sort and built-in quick-sort (need to know the average/worst time complexity of quick sort.)

3. Binary search of time complexity O(log(n)). Stick on single implementation.

• left-closed and right-open interval. Same as for loop.
• < in while loop.
• mid = lower + (upper - lower) / 2 to avoid overflow.
• return lower.

You can develop other various implementations based on this one.

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Source code:

class Solution {
    public int[] numSmallerByFrequency(String[] queries, String[] words) {
        int[] sorted = new int[words.length];
        for (int i = 0; i < words.length; i++) {
            sorted[i] = calculate(words[i]);
        }
        Arrays.sort(sorted);

        int[] nums = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int target = calculate(queries[i]);
            int lower = 0;
            int upper = words.length;
            while (lower < upper) {
                int mid = lower + (upper - lower) / 2;
                if (sorted[mid] <= target) {
                    lower = mid + 1;
                } else {
                    upper = mid;
                }
            }
            nums[i] = words.length - lower;
        }
        return nums;
    }
    
    private int calculate(String word) {
        int[] counting = new int[26];
        for (char c : word.toCharArray()) {
            counting[c - 'a']++;
        }
        for (int i = 0; i < 26; i++) {
            if (counting[i] > 0) {
                return counting[i];
            }
        }
        return 0;
    }
}