LeetCode #2506 — EASY

Count Pairs Of Similar Strings

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 0-indexed string array words.

Two strings are similar if they consist of the same characters.

For example, "abca" and "cba" are similar since both consist of characters 'a', 'b', and 'c'.
However, "abacba" and "bcfd" are not similar since they do not consist of the same characters.

Return the number of pairs (i, j) such that 0 <= i < j <= word.length - 1 and the two strings words[i] and words[j] are similar.

Example 1:

Input: words = ["aba","aabb","abcd","bac","aabc"]
Output: 2
Explanation: There are 2 pairs that satisfy the conditions:
- i = 0 and j = 1 : both words[0] and words[1] only consist of characters 'a' and 'b'. 
- i = 3 and j = 4 : both words[3] and words[4] only consist of characters 'a', 'b', and 'c'.

Example 2:

Input: words = ["aabb","ab","ba"]
Output: 3
Explanation: There are 3 pairs that satisfy the conditions:
- i = 0 and j = 1 : both words[0] and words[1] only consist of characters 'a' and 'b'. 
- i = 0 and j = 2 : both words[0] and words[2] only consist of characters 'a' and 'b'.
- i = 1 and j = 2 : both words[1] and words[2] only consist of characters 'a' and 'b'.

Example 3:

Input: words = ["nba","cba","dba"]
Output: 0
Explanation: Since there does not exist any pair that satisfies the conditions, we return 0.

Constraints:

1 <= words.length <= 100
1 <= words[i].length <= 100
words[i] consist of only lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed string array words. Two strings are similar if they consist of the same characters. For example, "abca" and "cba" are similar since both consist of characters 'a', 'b', and 'c'. However, "abacba" and "bcfd" are not similar since they do not consist of the same characters. Return the number of pairs (i, j) such that 0 <= i < j <= word.length - 1 and the two strings words[i] and words[j] are similar.

Baseline thinking

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

Pattern signal: Array · Hash Map · Bit Manipulation

Example 1

["aba","aabb","abcd","bac","aabc"]

Example 2

["aabb","ab","ba"]

Example 3

["nba","cba","dba"]

Core Insight

What unlocks the optimal approach

How can you check if two strings are similar?
Use a hashSet to store the character of each string.

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 #2506: Count Pairs Of Similar Strings
class Solution {
    public int similarPairs(String[] words) {
        int ans = 0;
        Map<Integer, Integer> cnt = new HashMap<>();
        for (var s : words) {
            int x = 0;
            for (char c : s.toCharArray()) {
                x |= 1 << (c - 'a');
            }
            ans += cnt.merge(x, 1, Integer::sum) - 1;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2506: Count Pairs Of Similar Strings
func similarPairs(words []string) (ans int) {
	cnt := map[int]int{}
	for _, s := range words {
		x := 0
		for _, c := range s {
			x |= 1 << (c - 'a')
		}
		ans += cnt[x]
		cnt[x]++
	}
	return
}

# Accepted solution for LeetCode #2506: Count Pairs Of Similar Strings
class Solution:
    def similarPairs(self, words: List[str]) -> int:
        ans = 0
        cnt = Counter()
        for s in words:
            x = 0
            for c in map(ord, s):
                x |= 1 << (c - ord("a"))
            ans += cnt[x]
            cnt[x] += 1
        return ans

// Accepted solution for LeetCode #2506: Count Pairs Of Similar Strings
use std::collections::HashMap;

impl Solution {
    pub fn similar_pairs(words: Vec<String>) -> i32 {
        let mut ans = 0;
        let mut cnt: HashMap<i32, i32> = HashMap::new();
        for s in words {
            let mut x = 0;
            for c in s.chars() {
                x |= 1 << ((c as u8) - b'a');
            }
            ans += cnt.get(&x).unwrap_or(&0);
            *cnt.entry(x).or_insert(0) += 1;
        }
        ans
    }
}

// Accepted solution for LeetCode #2506: Count Pairs Of Similar Strings
function similarPairs(words: string[]): number {
    let ans = 0;
    const cnt = new Map<number, number>();
    for (const s of words) {
        let x = 0;
        for (const c of s) {
            x |= 1 << (c.charCodeAt(0) - 97);
        }
        ans += cnt.get(x) || 0;
        cnt.set(x, (cnt.get(x) || 0) + 1);
    }
    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(L)

Space

O(n)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

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.