**Intuition:**\n*Simple brute force for every pair of characters in both strings.*\n\n**Steps with explanation:** \n1. Store frequency of characters of both strings.\n2. Count no of unique characters in each string.\n3. If no of unique characters in both strings are equal and both the string lengths are equal, then it is always possible to get equal no of unique characters after one swap.\n\n\tLets understand this step by step with few examples:\n\n\t* Example 1:\n\t\tword1 = `"abc"`\n\t\tword2 = `"def"`\n\n\t\tLets say, we take `"a"` and `"d"`. After swapping, word1 = `"dbc"` and word2 = `"aef"`\n\t\tHence, no of unique characters are same after swapping.\n\n\t* Example 2:\n\t\tword1 = `"aaa"`\n\t\tword2 = `"bbb"`\n\n\t\tLets say, we take any `"a"` and any `"b"`. After swapping, word1 = `"aab"` and word2 = `"bba"`\n\t\tHence, no of unique characters are same after swapping.\n\n\t* Example 3:\n\t\tword1 = `"aab"`\n\t\tword2 = `"bba"`\n\n\t\tLets say, we take `"a"` from word1 and `"b"` from word2. After swapping, word1 = `"bab"` and word2 = `"aba"`\n\t\tHence, no of unique characters are same after swapping.\n\n\t* Example 4:\n\t\tword1 = `"aab"`\n\t\tword2 = `"bbc"`\n\n\t\tLets say, we take `"b"` from word1 and `"b"` from word3. After swapping, word1 = `"aab"` and word2 = `"bbc"`.\n\t\tHence, no of unique characters are same after swapping.\n\n4. Now if the two string lengths are not equal, we need to check for each pair of different characters from both strings.\n\n\tHere, we do not need to actually swap the characters. We can simply simulate this swapping just by changing the frequency of characters in our frequency map. Lets understand this with an example.\n\n\tExample:\n\tword1 = `"aaabb"`\n\tword2 = `"ccdd"`\n\n\tFrequency map of word1, map1 = `[\'a\' = 3, \'b\' = 2]` and count of unique characters = `2`\n\tFrequency map of word2, map2 = `[\'c\' = 2, \'d\' = 2]` and count of unique characters = `2`\n\n\tLets say, we need to swap \'a\' with \'c\'. So after swapping, map1 = `[\'a\' = 2, \'b\' = 2, \'c\' = 1]`, and map2 = `[\'a\' = 1, \'c\' = 1, \'d\' = 2]`.\n\n\tSo for any char `i` from word1 and char `j` from word2, we are increasing freq of `j` in map1 and decreasing freq of `i` in map1. In the same way, we are also increasing freq of `i` in map2 and decreasing freq of `j` in map2. Now this change of freq may change the no of unique characters as we saw in the example.\n\n\tWhile doing this process, we can also keep track of change of unique chars by changing count1 and count2.\n\n\tLets say, characters are `[\'a\', \'b\', \'c\']` and we want to decrease freq of `\'a\'`. The count of unique chars will only change (decrease in this case) if count of `\'a\'` is `1` initially.\n\n\tLets say, characters are `[\'a\', \'b\']` and we want to increase freq of `\'c\'`. The count of unique chars will only change (increase in this case) if `\'c\'` is not present in characters initially.\n\n5. Check if the count of unique chars becomes equal in any situation. If so, return true.\n\n6. We need to revert all changes done to frequency maps while simulating the swappping process.\n\n\n**Java Code:**\n```java\nclass Solution {\n public boolean isItPossible(String word1, String word2) {\n int[] map1 = new int[26];\n int[] map2 = new int[26];\n \n\t\t// store frequency of characters\n for (int i = 0; i < word1.length(); i++) map1[word1.charAt(i)-\'a\']++;\n for (int i = 0; i < word2.length(); i++) map2[word2.charAt(i)-\'a\']++;\n\t\t\n\t\tint count1 = 0, count2 = 0;\n\t\t// count no of distinct characters in each string\n for (int i = 0; i < 26; i++) {\n if (map1[i] > 0) count1++;\n if (map2[i] > 0) count2++;\n }\n \n\t\t// as explained in step 3\n if (count1 == count2 && word1.length() == word2.length()) return true;\n \n for (int i = 0; i < 26; i++) {\n \tfor (int j = 0; j < 26; j++) {\n if (map1[i] == 0 || map2[j] == 0) continue;\n \n int uniqueCharInMap1 = count1, uniqueCharInMap2 = count2;\n \n\t\t\t\t// changing char count for map1\n if (map1[j] == 0) uniqueCharInMap1++;\n map1[j]++;\n \n if (map1[i] == 1) uniqueCharInMap1--;\n map1[i]--;\n \n\t\t\t\t// changing char count for map2\n if (map2[i] == 0) uniqueCharInMap2++;\n map2[i]++;\n \n if (map2[j] == 1) uniqueCharInMap2--;\n map2[j]--;\n \n\t\t\t\t// if count of unique chars are same, return true\n if (uniqueCharInMap1 == uniqueCharInMap2) return true;\n \n map1[j]--;\n map1[i]++;\n map2[i]--;\n map2[j]++;\n }\n }\n \n return false;\n }\n}\n```\n\n**C++ Code:**\n```cpp\nclass Solution {\npublic:\n bool isItPossible(string word1, string word2) {\n vector map1(26, 0);\n vector map2(26, 0);\n \n // store frequency of characters\n for (int i = 0; i < word1.length(); i++) map1[word1[i]-\'a\']++;\n for (int i = 0; i < word2.length(); i++) map2[word2[i]-\'a\']++;\n\n int count1 = 0, count2 = 0;\n\t\t// count no of distinct characters in each string\n for (int i = 0; i < 26; i++) {\n if (map1[i] > 0) count1++;\n if (map2[i] > 0) count2++;\n }\n \n // as explained in step 3\n if (count1 == count2 && word1.length() == word2.length()) return true;\n \n for (int i = 0; i < 26; i++) {\n \tfor (int j = 0; j < 26; j++) {\n if (map1[i] == 0 || map2[j] == 0) continue;\n \n int uniqueCharInMap1 = count1, uniqueCharInMap2 = count2;\n \n\t\t\t\t// changing char count for map1\n if (map1[j] == 0) uniqueCharInMap1++;\n map1[j]++;\n \n if (map1[i] == 1) uniqueCharInMap1--;\n map1[i]--;\n \n\t\t\t\t// changing char count for map2\n if (map2[i] == 0) uniqueCharInMap2++;\n map2[i]++;\n \n if (map2[j] == 1) uniqueCharInMap2--;\n map2[j]--;\n \n\t\t\t\t// if count of unique chars are same, return true\n if (uniqueCharInMap1 == uniqueCharInMap2) return true;\n \n map1[j]--;\n map1[i]++;\n map2[i]--;\n map2[j]++;\n }\n }\n \n return false;\n }\n};\n```\n\n**Complexity Analysis:**\n*Time Complexity:* **O(26 * 26 + (m+n)) --> O(m+n) --> O(max(m,n))**\n*Space Complexity:* **O(26+26) --> O(1)**

Make Number of Distinct Characters Equal

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Problem Description

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