Shortest, Easiest & Well Explained || To the Point & Beginners Friendly Approach (❤️ ω ❤️)
Welcome to My Coding Family\u30FE(\u2267 \u25BD \u2266)\u309D\nThis is my shortest, easiest & to the point approach. This solution mainly aims for purely beginn
# Welcome to My Coding Family\u30FE(\u2267 \u25BD \u2266)\u309D\n*This is my shortest, easiest & to the point approach. This solution mainly aims for purely beginners so if you\'re new here then too you\'ll be very easily be able to understand my this code. If you like my code then make sure to UpVOTE me. Let\'s start^_~*\n\n# Approach:-\n\n1. Firstly, create integer \'x\' to store answer.\n2. Now by using for loop, iterate from the 0th till the last index of given string \'s\' and check that the difference of ith index character\'s ASCII Value & (i+1)th index character\'s ASCII Value inside string \'s\' is smaller than or equal to 1.\n3. If yes then increment your answer \'x\' by 1 & also as we\'re checking two characters at the same time so we\'ve to also increment the following index \'i\' by 1.\n4. Lastly we return our answer \'x\'.\n\n**The code is given below for further understanding(p\u2267w\u2266q)**\n\n# Code\n```\nclass Solution {\npublic:\n    int removeAlmostEqualCharacters(string s) {\n        int x=0;\n        for(int i=0;i<s.length();i++){\n            if(abs((int)s[i]-(int)s[i+1])<=1){x++;i++;}\n        }\n        return x;\n\n// if my code helped you then please UpVOTE me, an UpVOTE encourages me to bring some more exciting codes for my coding family. Thank You o((>\u03C9< ))o\n    }\n};\n```\n\n![f5f.gif](https://assets.leetcode.com/users/images/0baec9b8-8c77-4920-96c5-aa051ddea6e4_1702180058.2605765.gif)\n![IUeYEqv.gif](https://assets.leetcode.com/users/images/9ce0a777-25fd-4677-8ccc-a885eb2b08f0_1702180061.2367256.gif)
['C++']