diff --git a/solution/2500-2599/2531.Make Number of Distinct Characters Equal/README.md b/solution/2500-2599/2531.Make Number of Distinct Characters Equal/README.md index 562650aeba64f..a38fdaea70ed9 100644 --- a/solution/2500-2599/2531.Make Number of Distinct Characters Equal/README.md +++ b/solution/2500-2599/2531.Make Number of Distinct Characters Equal/README.md @@ -70,13 +70,15 @@ tags: ### 方法一：计数 + 枚举 -我们先用两个长度为 $26$ 的数组分别统计字符串 $word1$ 和 $word2$ 中每个字母的出现次数，记为 $cnt1$ 和 $cnt2$。 +我们先用两个长度为 $26$ 的数组 $\textit{cnt1}$ 和 $\textit{cnt2}$ 分别记录字符串 $\textit{word1}$ 和 $\textit{word2}$ 中每个字符的出现次数。 -然后我们枚举 $cnt1$ 中的每个字母，接着枚举 $cnt2$ 中的每个字母，如果 $cnt1[i]$ 和 $cnt2[j]$ 都不为 $0$，那么我们就可以交换 $word1$ 中的第 $i$ 个字母和 $word2$ 中的第 $j$ 个字母。如果交换后 $word1$ 和 $word2$ 中不同字母的数目相等，那么就返回 `true`，否则，我们撤销此次交换，继续枚举。 +然后我们分别统计 $\textit{word1}$ 和 $\textit{word2}$ 中不同字符的个数，分别记为 $x$ 和 $y$。 -如果枚举完所有的字母对，仍然没有找到一种交换方式，那么就返回 `false`。 +接下来我们枚举 $\textit{word1}$ 中的每个字符 $c1$ 和 $\textit{word2}$ 中的每个字符 $c2$，如果 $c1 = c2$，那么我们只需要判断 $x$ 和 $y$ 是否相等；否则，我们需要判断 $x - (\textit{cnt1}[c1] = 1) + (\textit{cnt1}[c2] = 0)$ 和 $y - (\textit{cnt2}[c2] = 1) + (\textit{cnt2}[c1] = 0)$ 是否相等。如果相等，那么我们就找到了一种方案，返回 $\text{true}$。 -时间复杂度 $O(n + C^3)$，空间复杂度 $O(C)$，其中 $n$ 是字符串的长度，而 $C$ 是字符集的大小。本题中 $C = 26$。 +如果我们枚举完所有的字符都没有找到合适的方案，那么我们就返回 $\text{false}$。 + +时间复杂度 $O(m + n + |\Sigma|^2)$，其中 $m$ 和 $n$ 分别是字符串 $\textit{word1}$ 和 $\textit{word2}$ 的长度，而 $\Sigma$ 是字符集，本题中字符集为小写字母，所以 $|\Sigma| = 26$。 @@ -85,21 +87,19 @@ tags: ```python 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): + 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 - cnt1[i], cnt2[j] = cnt1[i] + 1, cnt2[j] + 1 - cnt1[j], cnt2[i] = cnt1[j] - 1, cnt2[i] - 1 return False ``` @@ -110,35 +110,31 @@ 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) { - ++cnt1[word1.charAt(i) - 'a']; + if (++cnt1[word1.charAt(i) - 'a'] == 1) { + ++x; + } } for (int i = 0; i < word2.length(); ++i) { - ++cnt2[word2.charAt(i) - 'a']; + 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) { - --cnt1[i]; - --cnt2[j]; - ++cnt1[j]; - ++cnt2[i]; - int d = 0; - for (int k = 0; k < 26; ++k) { - if (cnt1[k] > 0) { - ++d; + if (i == j) { + if (x == y) { + return true; } - if (cnt2[k] > 0) { - --d; + } 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; } } - if (d == 0) { - return true; - } - ++cnt1[i]; - ++cnt2[j]; - --cnt1[j]; - --cnt2[i]; } } } @@ -155,35 +151,31 @@ public: bool isItPossible(string word1, string word2) { int cnt1[26]{}; int cnt2[26]{}; + int x = 0, y = 0; for (char& c : word1) { - ++cnt1[c - 'a']; + if (++cnt1[c - 'a'] == 1) { + ++x; + } } for (char& c : word2) { - ++cnt2[c - 'a']; + if (++cnt2[c - 'a'] == 1) { + ++y; + } } 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 (i == j) { + if (x == y) { + return true; } - if (cnt2[k] > 0) { - --d; + } 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; } } - if (d == 0) { - return true; - } - ++cnt1[i]; - ++cnt2[j]; - --cnt1[j]; - --cnt2[i]; } } } @@ -198,31 +190,47 @@ public: 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 { - 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 i == j { + if x == y { + return true } - if cnt2[k] > 0 { - d-- + } 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 } } - 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 } } } @@ -230,6 +238,48 @@ func isItPossible(word1 string, word2 string) bool { } ``` +#### TypeScript + +```ts +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; +} +``` + diff --git a/solution/2500-2599/2531.Make Number of Distinct Characters Equal/README_EN.md b/solution/2500-2599/2531.Make Number of Distinct Characters Equal/README_EN.md index 9a1bb1fc3f55c..2ab7bf03fb512 100644 --- a/solution/2500-2599/2531.Make Number of Distinct Characters Equal/README_EN.md +++ b/solution/2500-2599/2531.Make Number of Distinct Characters Equal/README_EN.md @@ -65,7 +65,17 @@ tags: -### Solution 1 +### Solution 1: Counting + Enumeration + +We first use two arrays $\textit{cnt1}$ and $\textit{cnt2}$ of length $26$ to record the frequency of each character in the strings $\textit{word1}$ and $\textit{word2}$, respectively. + +Then, we count the number of distinct characters in $\textit{word1}$ and $\textit{word2}$, denoted as $x$ and $y$ respectively. + +Next, we enumerate each character $c1$ in $\textit{word1}$ and each character $c2$ in $\textit{word2}$. If $c1 = c2$, we only need to check if $x$ and $y$ are equal; otherwise, we need to check if $x - (\textit{cnt1}[c1] = 1) + (\textit{cnt1}[c2] = 0)$ and $y - (\textit{cnt2}[c2] = 1) + (\textit{cnt2}[c1] = 0)$ are equal. If they are equal, then we have found a solution and return $\text{true}$. + +If we have enumerated all characters and have not found a suitable solution, we return $\text{false}$. + +The time complexity is $O(m + n + |\Sigma|^2)$, where $m$ and $n$ are the lengths of the strings $\textit{word1}$ and $\textit{word2}$, and $\Sigma$ is the character set. In this problem, the character set consists of lowercase letters, so $|\Sigma| = 26$. @@ -74,21 +84,19 @@ tags: ```python 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): + 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 - cnt1[i], cnt2[j] = cnt1[i] + 1, cnt2[j] + 1 - cnt1[j], cnt2[i] = cnt1[j] - 1, cnt2[i] - 1 return False ``` @@ -99,35 +107,31 @@ 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) { - ++cnt1[word1.charAt(i) - 'a']; + if (++cnt1[word1.charAt(i) - 'a'] == 1) { + ++x; + } } for (int i = 0; i < word2.length(); ++i) { - ++cnt2[word2.charAt(i) - 'a']; + 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) { - --cnt1[i]; - --cnt2[j]; - ++cnt1[j]; - ++cnt2[i]; - int d = 0; - for (int k = 0; k < 26; ++k) { - if (cnt1[k] > 0) { - ++d; + if (i == j) { + if (x == y) { + return true; } - if (cnt2[k] > 0) { - --d; + } 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; } } - if (d == 0) { - return true; - } - ++cnt1[i]; - ++cnt2[j]; - --cnt1[j]; - --cnt2[i]; } } } @@ -144,35 +148,31 @@ public: bool isItPossible(string word1, string word2) { int cnt1[26]{}; int cnt2[26]{}; + int x = 0, y = 0; for (char& c : word1) { - ++cnt1[c - 'a']; + if (++cnt1[c - 'a'] == 1) { + ++x; + } } for (char& c : word2) { - ++cnt2[c - 'a']; + if (++cnt2[c - 'a'] == 1) { + ++y; + } } 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 (i == j) { + if (x == y) { + return true; } - if (cnt2[k] > 0) { - --d; + } 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; } } - if (d == 0) { - return true; - } - ++cnt1[i]; - ++cnt2[j]; - --cnt1[j]; - --cnt2[i]; } } } @@ -187,31 +187,47 @@ public: 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 { - 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 i == j { + if x == y { + return true } - if cnt2[k] > 0 { - d-- + } 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 } } - 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 } } } @@ -219,6 +235,48 @@ func isItPossible(word1 string, word2 string) bool { } ``` +#### TypeScript + +```ts +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; +} +``` + diff --git a/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.cpp b/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.cpp index b309ff49b9c9f..dc00fc6f16854 100644 --- a/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.cpp +++ b/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.cpp @@ -3,35 +3,31 @@ class Solution { bool isItPossible(string word1, string word2) { int cnt1[26]{}; int cnt2[26]{}; + int x = 0, y = 0; for (char& c : word1) { - ++cnt1[c - 'a']; + if (++cnt1[c - 'a'] == 1) { + ++x; + } } for (char& c : word2) { - ++cnt2[c - 'a']; + if (++cnt2[c - 'a'] == 1) { + ++y; + } } 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 (i == j) { + if (x == y) { + return true; } - if (cnt2[k] > 0) { - --d; + } 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; } } - if (d == 0) { - return true; - } - ++cnt1[i]; - ++cnt2[j]; - --cnt1[j]; - --cnt2[i]; } } } diff --git a/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.go b/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.go index 06c7beb3201b6..30ce21c70eaa3 100644 --- a/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.go +++ b/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.go @@ -1,31 +1,47 @@ 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 { - 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 i == j { + if x == y { + return true } - if cnt2[k] > 0 { - d-- + } 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 } } - 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 } } } diff --git a/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.java b/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.java index 1d94b88e3e93b..1f994d523444b 100644 --- a/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.java +++ b/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.java @@ -2,35 +2,31 @@ 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) { - ++cnt1[word1.charAt(i) - 'a']; + if (++cnt1[word1.charAt(i) - 'a'] == 1) { + ++x; + } } for (int i = 0; i < word2.length(); ++i) { - ++cnt2[word2.charAt(i) - 'a']; + 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) { - --cnt1[i]; - --cnt2[j]; - ++cnt1[j]; - ++cnt2[i]; - int d = 0; - for (int k = 0; k < 26; ++k) { - if (cnt1[k] > 0) { - ++d; + if (i == j) { + if (x == y) { + return true; } - if (cnt2[k] > 0) { - --d; + } 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; } } - if (d == 0) { - return true; - } - ++cnt1[i]; - ++cnt2[j]; - --cnt1[j]; - --cnt2[i]; } } } diff --git a/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.py b/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.py index bc82518581c3b..e4df0cbf51759 100644 --- a/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.py +++ b/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.py @@ -1,18 +1,16 @@ 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): + 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 - cnt1[i], cnt2[j] = cnt1[i] + 1, cnt2[j] + 1 - cnt1[j], cnt2[i] = cnt1[j] - 1, cnt2[i] - 1 return False diff --git a/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.ts b/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.ts new file mode 100644 index 0000000000000..f88b74a61e25f --- /dev/null +++ b/solution/2500-2599/2531.Make Number of Distinct Characters Equal/Solution.ts @@ -0,0 +1,37 @@ +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; +}