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2531. Make Number of Distinct Characters Equal

Description

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.

Solutions

  • class Solution {
    public boolean isItPossible(String word1, String word2) {
        int[] cnt1 = new int[26];
        int[] cnt2 = new int[26];
        for (int i = 0; i < word1.length(); ++i) {
            ++cnt1[word1.charAt(i) - 'a'];
        }
        for (int i = 0; i < word2.length(); ++i) {
            ++cnt2[word2.charAt(i) - 'a'];
        }
        for (int i = 0; i < 26; ++i) {
            for (int j = 0; j < 26; ++j) {
                if (cnt1[i] > 0 && cnt2[j] > 0) {
                    --cnt1[i];
                    --cnt2[j];
                    ++cnt1[j];
                    ++cnt2[i];
                    int d = 0;
                    for (int k = 0; k < 26; ++k) {
                        if (cnt1[k] > 0) {
                            ++d;
                        }
                        if (cnt2[k] > 0) {
                            --d;
                        }
                    }
                    if (d == 0) {
                        return true;
                    }
                    ++cnt1[i];
                    ++cnt2[j];
                    --cnt1[j];
                    --cnt2[i];
                }
            }
        }
        return false;
    }
}

  • class Solution {
public:
    bool isItPossible(string word1, string word2) {
        int cnt1[26]{};
        int cnt2[26]{};
        for (char& c : word1) {
            ++cnt1[c - 'a'];
        }
        for (char& c : word2) {
            ++cnt2[c - 'a'];
        }
        for (int i = 0; i < 26; ++i) {
            for (int j = 0; j < 26; ++j) {
                if (cnt1[i] > 0 && cnt2[j] > 0) {
                    --cnt1[i];
                    --cnt2[j];
                    ++cnt1[j];
                    ++cnt2[i];
                    int d = 0;
                    for (int k = 0; k < 26; ++k) {
                        if (cnt1[k] > 0) {
                            ++d;
                        }
                        if (cnt2[k] > 0) {
                            --d;
                        }
                    }
                    if (d == 0) {
                        return true;
                    }
                    ++cnt1[i];
                    ++cnt2[j];
                    --cnt1[j];
                    --cnt2[i];
                }
            }
        }
        return false;
    }
};

  • class Solution:
    def isItPossible(self, word1: str, word2: str) -> bool:
        cnt1 = [0] * 26
        cnt2 = [0] * 26
        for c in word1:
            cnt1[ord(c) - ord('a')] += 1
        for c in word2:
            cnt2[ord(c) - ord('a')] += 1
        for i, a in enumerate(cnt1):
            for j, b in enumerate(cnt2):
                if a and b:
                    cnt1[i], cnt2[j] = cnt1[i] - 1, cnt2[j] - 1
                    cnt1[j], cnt2[i] = cnt1[j] + 1, cnt2[i] + 1
                    if sum(v > 0 for v in cnt1) == sum(v > 0 for v in cnt2):
                        return True
                    cnt1[i], cnt2[j] = cnt1[i] + 1, cnt2[j] + 1
                    cnt1[j], cnt2[i] = cnt1[j] - 1, cnt2[i] - 1
        return False


  • func isItPossible(word1 string, word2 string) bool {
	cnt1 := [26]int{}
	cnt2 := [26]int{}
	for _, c := range word1 {
		cnt1[c-'a']++
	}
	for _, c := range word2 {
		cnt2[c-'a']++
	}
	for i := range cnt1 {
		for j := range cnt2 {
			if cnt1[i] > 0 && cnt2[j] > 0 {
				cnt1[i], cnt2[j] = cnt1[i]-1, cnt2[j]-1
				cnt1[j], cnt2[i] = cnt1[j]+1, cnt2[i]+1
				d := 0
				for k, a := range cnt1 {
					if a > 0 {
						d++
					}
					if cnt2[k] > 0 {
						d--
					}
				}
				if d == 0 {
					return true
				}
				cnt1[i], cnt2[j] = cnt1[i]+1, cnt2[j]+1
				cnt1[j], cnt2[i] = cnt1[j]-1, cnt2[i]-1
			}
		}
	}
	return false
}

  • 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;
}

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