LeetCode #1170 — MEDIUM

Compare Strings by Frequency of the Smallest Character

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Let the function f(s) be the frequency of the lexicographically smallest character in a non-empty string s. For example, if s = "dcce" then f(s) = 2 because the lexicographically smallest character is 'c', which has a frequency of 2.

You are given an array of strings words and another array of query strings queries. For each query queries[i], count the number of words in words such that f(queries[i]) < f(W) for each W in words.

Return an integer array answer, where each answer[i] is the answer to the i^th query.

Example 1:

Input: queries = ["cbd"], words = ["zaaaz"]
Output: [1]
Explanation: On the first query we have f("cbd") = 1, f("zaaaz") = 3 so f("cbd") < f("zaaaz").

Example 2:

Input: queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
Output: [1,2]
Explanation: On the first query only f("bbb") < f("aaaa"). On the second query both f("aaa") and f("aaaa") are both > f("cc").

Constraints:

1 <= queries.length <= 2000
1 <= words.length <= 2000
1 <= queries[i].length, words[i].length <= 10
queries[i][j], words[i][j] consist of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Let the function f(s) be the frequency of the lexicographically smallest character in a non-empty string s. For example, if s = "dcce" then f(s) = 2 because the lexicographically smallest character is 'c', which has a frequency of 2. You are given an array of strings words and another array of query strings queries. For each query queries[i], count the number of words in words such that f(queries[i]) < f(W) for each W in words. Return an integer array answer, where each answer[i] is the answer to the ith query.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Binary Search

Example 1

["cbd"]
["zaaaz"]

Example 2

["bbb","cc"]
["a","aa","aaa","aaaa"]

Step 02

Core Insight

What unlocks the optimal approach

For each string from words calculate the leading count and store it in an array, then sort the integer array.
For each string from queries calculate the leading count "p" and in base of the sorted array calculated on the step 1 do a binary search to count the number of items greater than "p".

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1170: Compare Strings by Frequency of the Smallest Character
class Solution {
    public int[] numSmallerByFrequency(String[] queries, String[] words) {
        int n = words.length;
        int[] nums = new int[n];
        for (int i = 0; i < n; ++i) {
            nums[i] = f(words[i]);
        }
        Arrays.sort(nums);
        int m = queries.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            int x = f(queries[i]);
            int l = 0, r = n;
            while (l < r) {
                int mid = (l + r) >> 1;
                if (nums[mid] > x) {
                    r = mid;
                } else {
                    l = mid + 1;
                }
            }
            ans[i] = n - l;
        }
        return ans;
    }

    private int f(String s) {
        int[] cnt = new int[26];
        for (int i = 0; i < s.length(); ++i) {
            ++cnt[s.charAt(i) - 'a'];
        }
        for (int x : cnt) {
            if (x > 0) {
                return x;
            }
        }
        return 0;
    }
}

// Accepted solution for LeetCode #1170: Compare Strings by Frequency of the Smallest Character
func numSmallerByFrequency(queries []string, words []string) (ans []int) {
	f := func(s string) int {
		cnt := [26]int{}
		for _, c := range s {
			cnt[c-'a']++
		}
		for _, x := range cnt {
			if x > 0 {
				return x
			}
		}
		return 0
	}
	n := len(words)
	nums := make([]int, n)
	for i, w := range words {
		nums[i] = f(w)
	}
	sort.Ints(nums)
	for _, q := range queries {
		x := f(q)
		ans = append(ans, n-sort.SearchInts(nums, x+1))
	}
	return
}

# Accepted solution for LeetCode #1170: Compare Strings by Frequency of the Smallest Character
class Solution:
    def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
        def f(s: str) -> int:
            cnt = Counter(s)
            return next(cnt[c] for c in ascii_lowercase if cnt[c])

        n = len(words)
        nums = sorted(f(w) for w in words)
        return [n - bisect_right(nums, f(q)) for q in queries]

// Accepted solution for LeetCode #1170: Compare Strings by Frequency of the Smallest Character
struct Solution;

impl Solution {
    fn f(s: &str) -> usize {
        let mut count = vec![0; 26];
        let mut min = b'z';
        for b in s.bytes() {
            min = min.min(b);
            count[(b - b'a') as usize] += 1;
        }
        count[(min - b'a') as usize]
    }
    fn num_smaller_by_frequency(queries: Vec<String>, words: Vec<String>) -> Vec<i32> {
        let f_words: Vec<usize> = words.iter().map(|s| Self::f(s)).collect();
        let mut counts = vec![0; 12];
        for f in f_words {
            counts[f] += 1;
        }
        for i in (1..10).rev() {
            counts[i] += counts[i + 1];
        }
        queries
            .iter()
            .map(|s| Self::f(s))
            .map(|f| counts[f + 1])
            .collect()
    }
}

#[test]
fn test() {
    let queries = vec_string!["cbd"];
    let words = vec_string!["zaaaz"];
    let res = vec![1];
    assert_eq!(Solution::num_smaller_by_frequency(queries, words), res);
    let queries = vec_string!["bbb", "cc"];
    let words = vec_string!["a", "aa", "aaa", "aaaa"];
    let res = vec![1, 2];
    assert_eq!(Solution::num_smaller_by_frequency(queries, words), res);
}

// Accepted solution for LeetCode #1170: Compare Strings by Frequency of the Smallest Character
function numSmallerByFrequency(queries: string[], words: string[]): number[] {
    const f = (s: string): number => {
        const cnt = new Array(26).fill(0);
        for (const c of s) {
            cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)]++;
        }
        return cnt.find(x => x > 0);
    };
    const nums = words.map(f).sort((a, b) => a - b);
    const ans: number[] = [];
    for (const q of queries) {
        const x = f(q);
        let l = 0,
            r = nums.length;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (nums[mid] > x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        ans.push(nums.length - l);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O((n + q)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.