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‎DetermineIfTwoStringsAreClose/determine_if_two_strings_are_close.dart

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/*
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-* 1657. Determine if Two Strings Are Close *-
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Two strings are considered close if you can attain one from the other using the following operations:
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Operation 1: Swap any two existing characters.
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For example, abcde -> aecdb
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Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character.
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For example, aacabb -> bbcbaa (all a's turn into b's, and all b's turn into a's)
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You can use the operations on either string as many times as necessary.
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Given two strings, word1 and word2, return true if word1 and word2 are close, and false otherwise.
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Example 1:
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Input: word1 = "abc", word2 = "bca"
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Output: true
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Explanation: You can attain word2 from word1 in 2 operations.
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Apply Operation 1: "abc" -> "acb"
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Apply Operation 1: "acb" -> "bca"
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Example 2:
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Input: word1 = "a", word2 = "aa"
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Output: false
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Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.
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Example 3:
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Input: word1 = "cabbba", word2 = "abbccc"
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Output: true
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Explanation: You can attain word2 from word1 in 3 operations.
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Apply Operation 1: "cabbba" -> "caabbb"
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Apply Operation 2: "caabbb" -> "baaccc"
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Apply Operation 2: "baaccc" -> "abbccc"
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Constraints:
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1 <= word1.length, word2.length <= 105
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word1 and word2 contain only lowercase English letters.
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*/
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import 'dart:collection';
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import 'dart:math';
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class A {
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bool closeStrings(String word1, String word2) {
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List<int> freq1 = List.filled(26, 0);
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List<int> freq2 = List.filled(26, 0);
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for (int i = 0; i < word1.length; i++) {
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freq1[word1.codeUnitAt(i) - 'a'.codeUnitAt(0)]++;
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}
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for (int i = 0; i < word2.length; i++) {
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freq2[word2.codeUnitAt(i) - 'a'.codeUnitAt(0)]++;
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}
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for (int i = 0; i < 26; i++) {
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if ((freq1[i] == 0 && freq2[i] != 0) || (freq1[i] != 0 && freq2[i] == 0))
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return false;
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}
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freq1.sort();
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freq2.sort();
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for (int i = 0; i < 26; i++) {
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if (freq1[i] != freq2[i]) return false;
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}
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return true;
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}
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}
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class B {
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bool closeStrings(String word1, String word2) {
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HashSet<String> set1 = HashSet();
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for (String c in word1.split("")) {
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set1.add(c);
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}
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for (String c in word2.split("")) {
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if (set1.contains(c)) {
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set1.remove(c);
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}
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}
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if (set1.length != 0) {
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return false;
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}
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List<int> l1 = [];
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List<int> l2 = [];
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Map<String, int> map1 = HashMap();
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Map<String, int> map2 = HashMap();
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for (String c in word1.split("")) {
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if (!map1.containsKey(c)) {
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map1[c] = 1;
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} else {
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map1[c] = (map1[c] ?? 0) + 1;
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}
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}
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for (String c in word2.split("")) {
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if (!map2.containsKey(c)) {
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map2[c] = 1;
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} else {
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map2[c] = (map2[c] ?? 0) + 1;
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}
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}
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for (int p in map1.values) {
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l1.add(p);
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}
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for (int p in map2.values) {
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l2.add(p);
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}
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// Collections.sort(l1);
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l1.sort();
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// Collections.sort(l2);
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l2.sort();
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StringBuffer sb1 = StringBuffer();
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StringBuffer sb2 = StringBuffer();
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for (int p in l1) {
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sb1.write(String.fromCharCode(p));
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}
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for (int p in l2) {
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sb2.write(String.fromCharCode(p));
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}
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return sb1.toString() == sb2.toString();
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}
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}
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class C {
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bool closeStrings(String word1, String word2) {
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if (word1.length > word2.length || word1.length < word2.length) {
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return false;
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}
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List<int> w1 = List.filled(26, 0);
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List<int> w2 = List.filled(26, 0);
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int maxi = 0;
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for (int i = 0; i < word1.length; i++) {
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w1[word1.codeUnitAt(i) - 'a'.codeUnitAt(0)] += 1;
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maxi = max(maxi, w1[word1.codeUnitAt(i) - 'a'.codeUnitAt(0)]);
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}
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for (int i = 0; i < word2.length; i++) {
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w2[word2.codeUnitAt(i) - 'a'.codeUnitAt(0)] += 1;
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maxi = max(maxi, w2[word2.codeUnitAt(i) - 'a'.codeUnitAt(0)]);
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}
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for (int i = 0; i < 26; i++) {
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if ((w1[i] != 0 && w2[i] == 0) || (w1[i] == 0 && w2[i] != 0)) {
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return false;
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}
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}
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List<int> f = List.filled(maxi + 1, 0);
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for (int i = 0; i < 26; i++) {
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f[w1[i]] += 1;
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}
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for (int i = 0; i < 26; i++) {
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f[w2[i]] -= 1;
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}
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for (int i = 1; i <= maxi; i++) {
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if (f[i] > 0) {
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return false;
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}
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}
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return true;
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}
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}

‎DetermineIfTwoStringsAreClose/determine_if_two_strings_are_close.go

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package main
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func closeStrings(word1 string, word2 string) bool {
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var ctr1, ctr2 [26]int
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if len(word1) != len(word2) {
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return false
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}
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for i := 0; i < len(word1); i++ {
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ctr1[word1[i]-byte('a')]++
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ctr2[word2[i]-byte('a')]++
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}
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for i := 0; i < len(ctr1); i++ {
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if (ctr1[i] != 0) != (ctr2[i] != 0) {
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return false
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}
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}
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cctr := make(map[int]int)
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for i := 0; i < len(ctr1); i++ {
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cctr[ctr1[i]]++
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cctr[ctr2[i]]--
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}
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for _, c := range cctr {
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if c != 0 {
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return false
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}
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}
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return true
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}

‎DetermineIfTwoStringsAreClose/determine_if_two_strings_are_close.md

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# 🔥 Determine if Two Strings Are Close 🔥 || 3 Approaches|| Simple Fast and Easy || with Explanation
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## Solution - 1
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```dart
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class Solution {
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bool closeStrings(String word1, String word2) {
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List<int> freq1 = List.filled(26, 0);
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List<int> freq2 = List.filled(26, 0);
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for (int i = 0; i < word1.length; i++) {
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freq1[word1.codeUnitAt(i) - 'a'.codeUnitAt(0)]++;
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}
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for (int i = 0; i < word2.length; i++) {
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freq2[word2.codeUnitAt(i) - 'a'.codeUnitAt(0)]++;
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}
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for (int i = 0; i < 26; i++) {
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if ((freq1[i] == 0 && freq2[i] != 0) || (freq1[i] != 0 && freq2[i] == 0))
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return false;
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}
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freq1.sort();
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freq2.sort();
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for (int i = 0; i < 26; i++) {
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if (freq1[i] != freq2[i]) return false;
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}
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return true;
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}
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}
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```
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## Solution - 2
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```dart
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class Solution {
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bool closeStrings(String word1, String word2) {
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HashSet<String> set1 = HashSet();
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for (String c in word1.split("")) {
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set1.add(c);
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}
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for (String c in word2.split("")) {
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if (set1.contains(c)) {
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set1.remove(c);
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}
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}
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if (set1.length != 0) {
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return false;
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}
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List<int> l1 = [];
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List<int> l2 = [];
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Map<String, int> map1 = HashMap();
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Map<String, int> map2 = HashMap();
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for (String c in word1.split("")) {
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if (!map1.containsKey(c)) {
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map1[c] = 1;
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} else {
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map1[c] = (map1[c] ?? 0) + 1;
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}
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}
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for (String c in word2.split("")) {
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if (!map2.containsKey(c)) {
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map2[c] = 1;
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} else {
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map2[c] = (map2[c] ?? 0) + 1;
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}
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}
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for (int p in map1.values) {
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l1.add(p);
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}
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for (int p in map2.values) {
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l2.add(p);
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}
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l1.sort();
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l2.sort();
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StringBuffer sb1 = StringBuffer();
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StringBuffer sb2 = StringBuffer();
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for (int p in l1) {
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sb1.write(String.fromCharCode(p));
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}
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for (int p in l2) {
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sb2.write(String.fromCharCode(p));
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}
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return sb1.toString() == sb2.toString();
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}
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}
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```
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## Solution - 3
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```dart
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class Solution {
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bool closeStrings(String word1, String word2) {
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if (word1.length > word2.length || word1.length < word2.length) {
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return false;
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}
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List<int> w1 = List.filled(26, 0);
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List<int> w2 = List.filled(26, 0);
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int maxi = 0;
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for (int i = 0; i < word1.length; i++) {
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w1[word1.codeUnitAt(i) - 'a'.codeUnitAt(0)] += 1;
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maxi = max(maxi, w1[word1.codeUnitAt(i) - 'a'.codeUnitAt(0)]);
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}
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for (int i = 0; i < word2.length; i++) {
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w2[word2.codeUnitAt(i) - 'a'.codeUnitAt(0)] += 1;
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maxi = max(maxi, w2[word2.codeUnitAt(i) - 'a'.codeUnitAt(0)]);
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}
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for (int i = 0; i < 26; i++) {
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if ((w1[i] != 0 && w2[i] == 0) || (w1[i] == 0 && w2[i] != 0)) {
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return false;
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}
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}
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List<int> f = List.filled(maxi + 1, 0);
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for (int i = 0; i < 26; i++) {
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f[w1[i]] += 1;
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}
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for (int i = 0; i < 26; i++) {
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f[w2[i]] -= 1;
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}
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for (int i = 1; i <= maxi; i++) {
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if (f[i] > 0) {
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return false;
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}
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}
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return true;
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}
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}
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```

‎README.md

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- [**1207.** Unique Number of Occurrences](UniqueNumberOfOccurrences/unique_number_of_occurrences.dart)
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- [**392.** Is Subsequence](IsSubsequence/is_subsequence.dart)
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- [**1704.** Determine if String Halves Are Alike](DetermineIfStringHalvesAreAlike/determine_if_string_halves_are_alike.dart)
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- [**1657.** Determine if Two Strings Are Close](DetermineIfTwoStringsAreClose/determine_if_two_strings_are_close.dart)
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## Reach me via
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