LeetCode #2531 — MEDIUM

Make Number of Distinct Characters Equal

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

The Problem

Problem Statement

You are given two 0-indexed strings word1 and word2.

A move consists of choosing two indices i and j such that 0 <= i < word1.length and 0 <= j < word2.length and swapping word1[i] with word2[j].

Return true if it is possible to get the number of distinct characters in word1 and word2 to be equal with exactly one move. Return false otherwise.

Example 1:

Input: word1 = "ac", word2 = "b"
Output: false
Explanation: Any pair of swaps would yield two distinct characters in the first string, and one in the second string.

Example 2:

Input: word1 = "abcc", word2 = "aab"
Output: true
Explanation: We swap index 2 of the first string with index 0 of the second string. The resulting strings are word1 = "abac" and word2 = "cab", which both have 3 distinct characters.

Example 3:

Input: word1 = "abcde", word2 = "fghij"
Output: true
Explanation: Both resulting strings will have 5 distinct characters, regardless of which indices we swap.

Constraints:

1 <= word1.length, word2.length <= 10⁵
word1 and word2 consist of only lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given two 0-indexed strings word1 and word2. A move consists of choosing two indices i and j such that 0 <= i < word1.length and 0 <= j < word2.length and swapping word1[i] with word2[j]. Return true if it is possible to get the number of distinct characters in word1 and word2 to be equal with exactly one move. Return false otherwise.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"ac"
"b"

Example 2

"abcc"
"aab"

Example 3

"abcde"
"fghij"

Core Insight

What unlocks the optimal approach

Create a frequency array of the letters of each string.
There are 26*26 possible pairs of letters to swap. Can we try them all?
Iterate over all possible pairs of letters and check if swapping them will yield two strings that have the same number of distinct characters. Use the frequency array for the check.

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 #2531: Make Number of Distinct Characters Equal
class Solution {
    public boolean isItPossible(String word1, String word2) {
        int[] cnt1 = new int[26];
        int[] cnt2 = new int[26];
        int x = 0, y = 0;
        for (int i = 0; i < word1.length(); ++i) {
            if (++cnt1[word1.charAt(i) - 'a'] == 1) {
                ++x;
            }
        }
        for (int i = 0; i < word2.length(); ++i) {
            if (++cnt2[word2.charAt(i) - 'a'] == 1) {
                ++y;
            }
        }
        for (int i = 0; i < 26; ++i) {
            for (int j = 0; j < 26; ++j) {
                if (cnt1[i] > 0 && cnt2[j] > 0) {
                    if (i == j) {
                        if (x == y) {
                            return true;
                        }
                    } else {
                        int a = x - (cnt1[i] == 1 ? 1 : 0) + (cnt1[j] == 0 ? 1 : 0);
                        int b = y - (cnt2[j] == 1 ? 1 : 0) + (cnt2[i] == 0 ? 1 : 0);
                        if (a == b) {
                            return true;
                        }
                    }
                }
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #2531: Make Number of Distinct Characters Equal
func isItPossible(word1 string, word2 string) bool {
	cnt1 := [26]int{}
	cnt2 := [26]int{}
	x, y := 0, 0
	for _, c := range word1 {
		cnt1[c-'a']++
		if cnt1[c-'a'] == 1 {
			x++
		}
	}
	for _, c := range word2 {
		cnt2[c-'a']++
		if cnt2[c-'a'] == 1 {
			y++
		}
	}
	for i := range cnt1 {
		for j := range cnt2 {
			if cnt1[i] > 0 && cnt2[j] > 0 {
				if i == j {
					if x == y {
						return true
					}
				} else {
					a := x
					if cnt1[i] == 1 {
						a--
					}
					if cnt1[j] == 0 {
						a++
					}

					b := y
					if cnt2[j] == 1 {
						b--
					}
					if cnt2[i] == 0 {
						b++
					}

					if a == b {
						return true
					}
				}
			}
		}
	}
	return false
}

# Accepted solution for LeetCode #2531: Make Number of Distinct Characters Equal
class Solution:
    def isItPossible(self, word1: str, word2: str) -> bool:
        cnt1 = Counter(word1)
        cnt2 = Counter(word2)
        x, y = len(cnt1), len(cnt2)
        for c1, v1 in cnt1.items():
            for c2, v2 in cnt2.items():
                if c1 == c2:
                    if x == y:
                        return True
                else:
                    a = x - (v1 == 1) + (cnt1[c2] == 0)
                    b = y - (v2 == 1) + (cnt2[c1] == 0)
                    if a == b:
                        return True
        return False

// Accepted solution for LeetCode #2531: Make Number of Distinct Characters Equal
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2531: Make Number of Distinct Characters Equal
// class Solution {
//     public boolean isItPossible(String word1, String word2) {
//         int[] cnt1 = new int[26];
//         int[] cnt2 = new int[26];
//         int x = 0, y = 0;
//         for (int i = 0; i < word1.length(); ++i) {
//             if (++cnt1[word1.charAt(i) - 'a'] == 1) {
//                 ++x;
//             }
//         }
//         for (int i = 0; i < word2.length(); ++i) {
//             if (++cnt2[word2.charAt(i) - 'a'] == 1) {
//                 ++y;
//             }
//         }
//         for (int i = 0; i < 26; ++i) {
//             for (int j = 0; j < 26; ++j) {
//                 if (cnt1[i] > 0 && cnt2[j] > 0) {
//                     if (i == j) {
//                         if (x == y) {
//                             return true;
//                         }
//                     } else {
//                         int a = x - (cnt1[i] == 1 ? 1 : 0) + (cnt1[j] == 0 ? 1 : 0);
//                         int b = y - (cnt2[j] == 1 ? 1 : 0) + (cnt2[i] == 0 ? 1 : 0);
//                         if (a == b) {
//                             return true;
//                         }
//                     }
//                 }
//             }
//         }
//         return false;
//     }
// }

// Accepted solution for LeetCode #2531: Make Number of Distinct Characters Equal
function isItPossible(word1: string, word2: string): boolean {
    const cnt1: number[] = Array(26).fill(0);
    const cnt2: number[] = Array(26).fill(0);
    let [x, y] = [0, 0];

    for (const c of word1) {
        if (++cnt1[c.charCodeAt(0) - 'a'.charCodeAt(0)] === 1) {
            ++x;
        }
    }

    for (const c of word2) {
        if (++cnt2[c.charCodeAt(0) - 'a'.charCodeAt(0)] === 1) {
            ++y;
        }
    }

    for (let i = 0; i < 26; ++i) {
        for (let j = 0; j < 26; ++j) {
            if (cnt1[i] > 0 && cnt2[j] > 0) {
                if (i === j) {
                    if (x === y) {
                        return true;
                    }
                } else {
                    const a = x - (cnt1[i] === 1 ? 1 : 0) + (cnt1[j] === 0 ? 1 : 0);
                    const b = y - (cnt2[j] === 1 ? 1 : 0) + (cnt2[i] === 0 ? 1 : 0);
                    if (a === b) {
                        return true;
                    }
                }
            }
        }
    }

    return false;
}

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)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.