Make Number of Distinct Characters Equal - Practice Coding Problems

Make Number of Distinct Characters Equal - Problem

You are given two strings word1 and word2. Your goal is to equalize the number of distinct characters in both strings using exactly one character swap between them.

A move consists of choosing any character at position i in word1 and any character at position j in word2, then swapping them. You must determine if it's possible to make both strings have the same number of unique characters with exactly one such swap.

Example: If word1 = "ac" has 2 distinct characters and word2 = "b" has 1 distinct character, can we swap one character to make them equal?

Return true if such a swap exists, false otherwise.

Input & Output

example_1.py — Basic case

$ Input: word1 = "ac", word2 = "b"

› Output: false

💡 Note: No matter which characters we swap, we can't make both strings have the same number of distinct characters. Initially: word1 has 2 distinct, word2 has 1 distinct.

example_2.py — Successful swap

$ Input: word1 = "abcc", word2 = "aab"

› Output: true

💡 Note: We can swap the 'c' from word1 with 'a' from word2. After swap: word1 becomes "abca" (3 distinct: a,b,c), word2 becomes "cab" (3 distinct: a,b,c).

example_3.py — Already equal

$ Input: word1 = "abc", word2 = "def"

› Output: true

💡 Note: Both strings already have 3 distinct characters. Any swap maintains this equality (though the specific characters change).

Visualization

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Understanding the Visualization

Count Your Cards

First, count how many of each type you have in both collections

Analyze Trade Impact

For each possible trade, predict how it affects unique card counts

Find Valid Trade

Look for any trade that equalizes the unique card counts

Make Decision

Return true if such a trade exists, false otherwise

Key Takeaway

🎯 Key Insight: Instead of trying every possible trade, we can mathematically predict the outcome by analyzing character frequencies and the four possible swap scenarios.

Time & Space Complexity

Time Complexity

⏱️

O(n + m)

Single pass to count frequencies, then O(1) analysis for each unique character pair

✓ Linear Growth

Space Complexity

O(1)

Fixed size arrays for character frequencies (26 letters max)

✓ Linear Space

Constraints

1 ≤ word1.length, word2.length ≤ 10⁵
word1 and word2 consist only of lowercase English letters
Exactly one swap must be performed

Asked in

G Google 35 a Amazon 28 f Meta 22 ⊞ Microsoft 18

The optimal approach uses character frequency analysis to determine valid swaps without actually performing them. By counting frequencies and analyzing the four possible swap scenarios, we can predict the outcome in O(n+m) time, making it much more efficient than the brute force O(n²m) approach.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Swaps)	O(n²m)	O(1)	Try every possible character swap and count distinct characters
Character Frequency Analysis (Optimal)	O(n + m)	O(1)	Analyze character frequencies to determine valid swaps mathematically

Brute Force (Try All Swaps) — Algorithm Steps

Iterate through every position i in word1
For each i, iterate through every position j in word2
Simulate swapping word1[i] with word2[j]
Count distinct characters in both modified strings
If counts are equal, return true
If no valid swap found, return false

Visualization

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Step-by-Step Walkthrough

Initial State

word1='ac', word2='b', distinct counts: 2 vs 1

Try Swap (0,0)

Swap 'a' and 'b': word1='bc', word2='a', counts: 2 vs 1

Try Swap (1,0)

Swap 'c' and 'b': word1='ab', word2='c', counts: 2 vs 1

Result

No valid swap found, return false

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

int countDistinct(char* s, int len) {
    bool seen[256] = {false};
    int count = 0;
    for (int i = 0; i < len; i++) {
        if (!seen[(unsigned char)s[i]]) {
            seen[(unsigned char)s[i]] = true;
            count++;
        }
    }
    return count;
}

bool closeStrings(char* word1, char* word2) {
    int n1 = strlen(word1), n2 = strlen(word2);
    
    for (int i = 0; i < n1; i++) {
        for (int j = 0; j < n2; j++) {
            // Swap characters
            char temp = word1[i];
            word1[i] = word2[j];
            word2[j] = temp;
            
            // Check if distinct counts are equal
            if (countDistinct(word1, n1) == countDistinct(word2, n2)) {
                return true;
            }
            
            // Swap back
            temp = word1[i];
            word1[i] = word2[j];
            word2[j] = temp;
        }
    }
    
    return false;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²m)

n*m possible swaps, each requiring O(n+m) time to count distinct characters

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables, not counting input strings

✓ Linear Space

Constraints

1 ≤ word1.length, word2.length ≤ 10⁵
word1 and word2 consist only of lowercase English letters
Exactly one swap must be performed

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Try All Swaps) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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