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flip-string-to-monotone-increasing | Memo -> Tabulation -> Space Optimization -> Code Refactor | memo-tabulation-space-optimization-code-wbwu7 | \n# Memo -> Tabulation -> Space Optimization -> Code Refactor\n# Best Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space | mr_optimizer | NORMAL | 2023-01-17T13:33:00.746356+00:00 | 2023-01-17T13:33:00.746391+00:00 | 417 | false | \n# Memo -> Tabulation -> Space Optimization -> Code Refactor\n# Best Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Memo Code\n```\n int memo[100001][2];\n int dp(string &s, int i, int j){\n if(i == s.size()){\n return 0;\n }\n if(memo[i][j] != -1)return memo[i][j];\n if(j == 0){\n if(s[i] == \'1\'){\n return memo[i][j] = min(1 + dp(s, i + 1, 0), dp(s, i + 1, 1));\n }else{\n return memo[i][j] = min(1 + dp(s, i + 1, 1), dp(s, i + 1, 0));\n }\n }else{\n if(s[i] == \'1\'){\n return memo[i][j] = dp(s, i + 1, 1);\n }else{\n return memo[i][j] = 1 + dp(s, i + 1, 1);\n }\n }\n }\n int minFlipsMonoIncr(string s) {\n memset(memo, -1, sizeof(memo));\n return dp(s, 0, 0);\n }\n```\n# Tabulation Code\n```\n int minFlipsMonoIncr(string s) {\n int n = s.size();\n int dp[n + 1][2];\n dp[n][0] = dp[n][1] = 0;\n for(int i = n - 1; i >= 0; i--){\n for(int j = 0; j < 2; j++){\n if(j == 0){\n if(s[i] == \'1\'){\n dp[i][j] = min(1 + dp[i + 1][0], dp[i + 1][1]);\n }else{\n dp[i][j] = min(1 + dp[i + 1][1], dp[i + 1][0]);\n }\n }else{\n if(s[i] == \'1\'){\n dp[i][j] = dp[i + 1][1];\n }else{\n dp[i][j] = 1 + dp[i + 1][1];\n }\n }\n }\n }\n return dp[0][0];\n }\n```\n# Space Optimization\n```\nint minFlipsMonoIncr(string s) {\n int prev1 = 0, prev0 = 0;\n for(int i = s.size() - 1; i >= 0; i--){\n for(int j = 0; j < 2; j++){\n if(j == 0){\n if(s[i] == \'1\'){\n prev0 = min(1 + prev0, prev1);\n }else{\n prev0 = min(1 + prev1, prev0);\n }\n }else{\n if(s[i] != \'1\')prev1++;\n }\n }\n }\n return prev0;\n}\n```\n# Code Refactor\n```\nint minFlipsMonoIncr(string s) {\n int prev1 = 0, prev0 = 0;\n for(int i = s.size() - 1; i >= 0; i--){\n prev0 = s[i] == \'1\' ? min(1 + prev0, prev1) : min(1 + prev1, prev0);\n if(s[i] != \'1\')prev1++;\n }\n return prev0;\n}\n```\n\nAsk if you have any doubt | 6 | 0 | ['Dynamic Programming', 'Memoization', 'C++'] | 2 |
flip-string-to-monotone-increasing | Python - Picture/Video Solution | python-picturevideo-solution-by-cheatcod-vwb5 | I have explained the entire intuition here.\n\nIf this was helpful, please Upvote, like the video and subscribe for more such content.\n\n# Memoization\n\n\n\n\ | cheatcode-ninja | NORMAL | 2023-01-17T13:20:49.077850+00:00 | 2023-01-17T13:26:40.968331+00:00 | 222 | false | I have explained the entire intuition [here](https://youtu.be/t5-RdgOkYe4).\n\nIf this was helpful, please Upvote, like the video and subscribe for more such content.\n\n# Memoization\n\n\n\n\nThese are the valid patterns\n\n\n\nWhenever we encounter a new digit, this can fall into these 4 categories.\n\nThe problem is, when we have a streak of **0s**, we encounter a **1**, we\'ll have two choices. \nHence a Dynamic Programming approach.\n\n**Time:** `O(n)`\n\n**Space** `O(n)`\n\n```\nclass Solution:\n def minFlipsMonoIncr(self, s: str) -> int:\n \n @cache\n def dfs(i, streak=\'0\'):\n if i==len(s):\n return 0\n if s[i]==streak:\n return dfs(i+1, streak)\n elif s[i]==\'1\':\n return min(1+dfs(i+1, \'0\'), dfs(i+1, \'1\'))\n else:\n return 1 + dfs(i+1, \'1\')\n return dfs(0)\n```\n\n# Counting Prefix\n\nWhenever we encounter a **1**, if we assume our previous string to be valid.\n**1** can never cause a violation because, \n1. if we had a previous streak of **0**, this can be start of streak of **1**\n2. if we had a previous streak of **1**, just continue.\n\nSo we only have to handle **0**.\n\n\n\nWhenever we encounter a **0**, we have to either flip:\n1. the previous **1** -> **0** and keep our current char **0**\n2. the current **0** & previous **0** after first occurence of **1** --> **1**.\n\nWhichever is minimum, is the answer.\n\n**Time:** `O(n)`\n\n**Space** `O(n)`\n\n```\nclass Solution:\n def minFlipsMonoIncr(self, s: str) -> int:\n \n count_1 = 0\n res = 0\n \n for i in range(len(s)):\n count_1 += s[i]==\'1\'\n if s[i]==\'0\':\n res = min(count_1, res + 1)\n return res\n```\n\n\n\n | 6 | 0 | ['Dynamic Programming', 'Python3'] | 0 |
flip-string-to-monotone-increasing | 100% javascript fast very easy to understand with video explanation! | 100-javascript-fast-very-easy-to-underst-2flp | Here is video for explain if it is helpful please subscribe! :\n\nhttps://youtu.be/-m47DQxavnY\n\n\n\n# Code\n\n/**\n * @param {string} s\n * @return {number}\n | rlawnsqja850 | NORMAL | 2023-01-17T01:29:27.131620+00:00 | 2023-01-17T01:29:27.131660+00:00 | 772 | false | Here is video for explain if it is helpful please subscribe! :\n\nhttps://youtu.be/-m47DQxavnY\n\n\n\n# Code\n```\n/**\n * @param {string} s\n * @return {number}\n */\nvar minFlipsMonoIncr = function(s) {\n let result = 0; \n let count = 0;\n\n for (const str of s) {\n if(str== \'1\') {\n count++\n }else if(str ==\'0\' && count> 0){\n result++\n count--\n }\n }\n return result;\n};\n``` | 6 | 0 | ['JavaScript'] | 3 |
flip-string-to-monotone-increasing | Easy C++ solution in O(N) | easy-c-solution-in-on-by-tejas702-10cs | \nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int res = 0;\n int c1 = 0;\n for(char c : s){\n if(c == \'0\ | tejas702 | NORMAL | 2021-08-10T07:47:23.787903+00:00 | 2021-08-10T07:47:39.547778+00:00 | 501 | false | ```\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int res = 0;\n int c1 = 0;\n for(char c : s){\n if(c == \'0\'){\n if(c1>0)res+=1;\n }else{\n c1+=1; \n }\n res=min(res,c1);\n }\n return res;\n }\n};\n```\n**Time Complexity: O(N)** where N is length of the string.\n\n**Feel free to share improvements/suggestions. Questions/Discussions are welcome.** | 6 | 2 | ['String', 'C'] | 0 |
flip-string-to-monotone-increasing | Java Solution - O(N) time and O(1) space | java-solution-on-time-and-o1-space-by-at-ors0 | \nclass Solution {\n public int minFlipsMonoIncr(String S) {\n int zeroOnRight = 0;\n for(int i = 0; i < S.length(); i++){\n if(S.ch | itsmastersam | NORMAL | 2020-05-11T17:57:08.752730+00:00 | 2020-05-11T17:57:08.752782+00:00 | 521 | false | ```\nclass Solution {\n public int minFlipsMonoIncr(String S) {\n int zeroOnRight = 0;\n for(int i = 0; i < S.length(); i++){\n if(S.charAt(i) == \'0\'){\n zeroOnRight++;\n }\n }\n // partition at beginning\n int ans = zeroOnRight;\n int oneOnLeft = 0;\n for(int i = 0; i < S.length(); i++){\n if(S.charAt(i) == \'0\'){\n zeroOnRight--;\n }\n else{\n oneOnLeft++;\n }\n ans = Math.min(ans, oneOnLeft + zeroOnRight);\n }\n // partition at ending\n ans = Math.min(ans, oneOnLeft);\n return ans;\n }\n}\n``` | 6 | 0 | ['Java'] | 2 |
flip-string-to-monotone-increasing | dp c++ O(n) | dp-c-on-by-savecancel-5t6c | Runtime: 4 ms, faster than 99.25% of C++ online submissions for Flip String to Monotone Increasing.\nMemory Usage: 9.5 MB, less than 60.00% of C++ online submis | savecancel | NORMAL | 2019-12-27T04:41:04.823810+00:00 | 2019-12-27T04:41:04.823850+00:00 | 449 | false | Runtime: 4 ms, faster than 99.25% of C++ online submissions for Flip String to Monotone Increasing.\nMemory Usage: 9.5 MB, less than 60.00% of C++ online submissions for Flip String to Monotone Increasing.\n\n\nwe can optimize further by only cosidering previous locations \n\n\n```\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) \n {\n int n=s.size();\n int dp[n][2];\n memset(dp,0,sizeof(dp));\n \n \n int i=0;\n \n if(s[i]==\'0\')\n {\n dp[i][0]=0;\n dp[i][1]=1;\n }\n else\n {\n dp[i][1]=0;\n dp[i][0]=1;\n }\n \n for(i=1;i<n;i++)\n {\n \n if(s[i]==\'0\')\n {\n // not flip\n dp[i][0]=dp[i-1][0];\n // if flip\n dp[i][1]=min(dp[i-1][0],dp[i-1][1])+1;\n }\n else\n {\n // if current char is \'1\'\n \n //not flip\n dp[i][1]=min(dp[i-1][0],dp[i-1][1]);\n \n // flip\n dp[i][0]=dp[i-1][0]+1;\n }\n }\n \n return min(dp[n-1][0],dp[n-1][1]);\n \n \n }\n};\n```\n | 6 | 1 | ['Dynamic Programming'] | 1 |
flip-string-to-monotone-increasing | Java Super Easy to Understand O(n) Solution | java-super-easy-to-understand-on-solutio-dwqw | Just do two pass can do the Job.\nFirst pass: from left to right. if we want to convert all the numbers to 0, how many flips we need, save it to zero[i].\nSeco | x372p | NORMAL | 2018-10-21T04:41:30.827437+00:00 | 2018-10-21T04:41:30.827482+00:00 | 243 | false | Just do two pass can do the Job.\nFirst pass: from left to right. if we want to convert all the numbers to 0, how many flips we need, save it to zero[i].\nSecond pass: from right to left. if we want to convert all the numbers to 1, how many flips we need, save it to one[i].\n\nThen the answer is just min(zero[i] + one[i]), for each i in the array.\n(The third pass in my code can be combined with the second one. But I didn\'t do it to make it easier to read.)\n\n```\nclass Solution {\n public int minFlipsMonoIncr(String S) {\n if (S.length() == 0)\n return 0;\n int[] zero = new int[S.length()+1];//to flip before i (not included) to zero.\n int[] one = new int[S.length()+1];//to flip after i (included) to one\n zero[0] = 0;\n one[S.length()] = 0;\n for (int i = 1; i <= S.length(); i++){\n if (S.charAt(i-1) == \'1\')\n zero[i] = zero[i-1]+1;\n else\n zero[i] = zero[i-1];\n }\n for (int i = S.length()-1; i >= 0; i--){\n if (S.charAt(i) == \'0\')\n one[i] = one[i+1] + 1;\n else\n one[i] = one[i+1];\n }\n //for each i, the solution is zero[i] + one[i]\n int min = Integer.MAX_VALUE;\n for (int i = 0; i < zero.length; i++){\n min = Math.min(min, zero[i] + one[i]);\n }\n return min;\n }\n}\n``` | 6 | 1 | [] | 1 |
flip-string-to-monotone-increasing | C++ || No DP || Prefix & suffix sum | c-no-dp-prefix-suffix-sum-by-a__ky25-fdtf | Intuition\n Describe your first thoughts on how to solve this problem. \n In every point in string i.e. every index we have to check if you\n can miximize | A__ky25 | NORMAL | 2023-01-20T06:24:35.729063+00:00 | 2023-01-20T06:24:35.729108+00:00 | 75 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n In every point in string i.e. every index we have to check if you\n can miximize the flips.\n for example if string is 101010\n In i=k\n we will check if its possible of flip all 1\'s into 0\'s from\n 0 to k-1. \n and flip all k+1 to n-1 0\'s to 1\'s\n\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n O(n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n o(2n)\n# Code\n```\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int n=s.size();\n vector<int>pre(n),suf(n);\n int ans = n;\n for(int i=0;i<n;i++){\n if(s[i]==\'1\')\n pre[i]=i>0?pre[i-1]+1:1;\n else\n pre[i]=i>0?pre[i-1]:0;\n }\n for(int i=n-1;i>=0;i--){\n if(s[i]==\'0\')\n suf[i]=i<n-1?suf[i+1]+1:1;\n else\n suf[i]=i<n-1?suf[i+1]:0;\n }\n for(int i=0;i<n;i++){\n int x = 0;\n // from i to n make it to all 1 ie convert all 0 to 1;\n if(i<n-1)\n x= suf[i+1];\n //from 0 to index making all 0 ie convert all 1 to 0;\n if(i>0)\n x+=pre[i-1];\n \n ans = min(ans,x);\n }\n return ans;\n }\n};\n``` | 5 | 0 | ['Suffix Array', 'Prefix Sum', 'C++'] | 0 |
flip-string-to-monotone-increasing | Two Most Intuitive Approaches✅ || Full Explanation✅||C++ | two-most-intuitive-approaches-full-expla-h4nx | Approach 1 :\n# Intuition (DP + Memoization)\nTo find minimum number of flips we would try to explore all possiblities i.e. whether we can make string monotone | Pushkar2111 | NORMAL | 2023-01-17T18:06:08.488454+00:00 | 2023-01-17T18:06:08.488497+00:00 | 313 | false | ### Approach 1 :\n# Intuition (DP + Memoization)\nTo find minimum number of flips we would try to explore all possiblities i.e. whether we can make string monotone by flipping the current bit or not. \n\n# Approach\nThere can be two possible cases :-\n - If the previous character is \'0\':\n Now this has also two possible subcases :-\n - If curr character is \'0\' then \n don\'t need to do flip and can move to the next index.\n **OR** \n flip this bit to \'1\' and increse count by 1.\n\n - If curr character is \'1\' then \n don\'t need to do flip and can move to the next index as 0 \n followed by 1 is still monotone.\n **OR** \n flip this bit to \'0 and increse count by 1.\n\n- If the previous character is \'1\':\n Possible subcases :-\n - If curr character is \'0\' then \n we have flip this bit anyhow else it can not be made monotone.\n Thereafter count is increased. \n\n - If curr character is \'1\' then \n nothing need to done as monotonicity is being maintained.\n\n \n### Approach 2 (Better in terms of Time Complexity):- \n- We iterate through the string from left till we don\'t encounter a \'1\'.\n- Now we maintain zeroCount and oneCount in the rest of the substring.\n If at any point zeroCount is greater than oneCount then we store\n zeroCount in flips as we want to find minimum flips (rather \n than flipping zero we would flip ones till this point).\n\n Else we if after reaching end, oneCount is more zeroCount we return \n zeroCount as minimum flips.\n\n# Please Upvote if it helps !!! \n\n \n# Code (Approach 1) :\n```\nclass Solution {\npublic:\n \n int helper(string &s, int idx, char prev, vector<vector<int>> &dp){\n if(idx==s.length()) \n return 0;\n if(dp[idx][prev-\'0\'] != -1) \n return dp[idx][prev-\'0\'];\n int ans1 = 1e9, ans2 = 1e9;\n if(prev==\'0\'){\n if(s[idx]==\'0\'){\n ans1 = helper(s,idx+1,\'0\',dp);\n ans2 = 1 + helper(s,idx+1,\'1\',dp);\n }\n else{\n ans2 = helper(s,idx+1,\'1\',dp);\n ans1 = 1 + helper(s,idx+1,\'0\',dp);\n }\n }\n else{\n if(s[idx]==\'0\')\n ans1 = 1 + helper(s,idx+1,\'1\',dp);\n else \n ans2 = helper(s,idx+1,\'1\',dp);\n }\n return dp[idx][prev-\'0\'] = min(ans1, ans2);\n }\n\n int minFlipsMonoIncr(string s) {\n vector<vector<int>> dp(s.size(),vector<int>(2,-1));\n return helper(s, 0, \'0\', dp);\n }\n};\n```\n\n# Code (Approach 2):\n```\nclass Solution {\npublic:\n\n int minFlipsMonoIncr(string s) {\n int oneCount=0,zeroCount=0;\n int i=0;\n \n while(i<s.length() && s[i]==\'0\')i++;\n for(;i<s.length();i++){\n char c=s[i];\n if(c==\'0\') zeroCount++;\n else oneCount++;\n if(zeroCount > oneCount) zeroCount=oneCount;\n }\n return zeroCount;\n }\n};\n```\n\n | 5 | 0 | ['Dynamic Programming', 'C++'] | 1 |
flip-string-to-monotone-increasing | 🧑🏫 Java | DP | Fully Explained [Article] | java-dp-fully-explained-article-by-nadar-hg7x | Introduction\nIn contrast to solutions that post the code with a few words, I walk through my thinking process. The goal is to share my thoughts and hopefully t | nadaralp | NORMAL | 2023-01-17T09:29:09.519545+00:00 | 2023-01-17T09:55:23.870603+00:00 | 314 | false | # Introduction\nIn contrast to solutions that post the code with a few words, I walk through my thinking process. The goal is to share my thoughts and hopefully teach you something new today.\n\nNadar\n\n# Intuition\nInitially, I thought of approaching the problem using a greedy method. We have two choices to make at any given index `i`. We can either flip/keep it to \'1\' or flip/keep it to \'0\'.\n\nIf we flip/keep a number into \'1\' we know we must flip all following 0\'s to keep the monotonic property intact, so we could use a hash map to count the local cost of that move. The problem is that we don\'t know the local cost of flipping/keeping \'0\'. So there is no easy way to compare local moves to achieve a global best, hence greedy algorithm with this approach won\'t work.\n\nWhen facing an optimization problem (finding minimum), we can opt for greedy or dynamic programming. As we contradicted the greedy approach, we are left with DP.\n\nWhen using DP, we need to declare the states and transitions of the DP. There is no right or wrong declaration. There are sometimes a few ways to declare a DP state to achieve the result. The important thing is to maintain valid state transitions. If the state transition violates the problem constraints, we will fail to achieve the correct result.\n\nState declaration:\n`dp[i][0]` - the cost of flipping/keeping a 0 in the `ith` index.\n`dp[i][1]` - the cost of flipping/keeping a 1 in the `ith` index.\n\nTransitions:\nIf `s[i] == \'1\'`, then we can turn into 0 by taking the previous cost of turning into 0 and adding 1 for the flip (we can\'t take the previous cost of dp[i-1][1], because if we had a 1 previously will break the monotonic property). Or we can keep 1 and take the minimum cost of `Math.min(dp[i-1][0] + 1, dp[i-1][1])` because we can have a 1 after 0 OR 1 previously. note the +1 in dp[i-1][0] is the cost of flipping.\n\nWe do the inverse for `s[i] == 0`.\n\nIn the end, we return the cheapest cost of having 1 or 0 in the end, as `Math.min(dp[n][0], dp[n][1])`\n\nImplementation note: I initialize dp with dp[i+1][2] to have a one index buffer, making dp[i-1] valid in all the cases and not going out of bounds.\n\n# Post Edit Notes\nThere is a way to solve the problem greedily, similar to the way that my initial thought went. Think of every possible prefix/suffix configuration having only 0s in the prefix, and 1s in the suffix. Then, count how many flips you need to make the prefix full of 0s and the suffix full of 1s. I will not add implementation details here.\n\nAlso, the DP implementation uses `dp` array. It is possible to optimize to `O(1)` space because we depend only on the previous two variables. Hence, we can store the previously calculated results in two variables instead of an array.\n\n# Code\n```\nclass Solution {\n public int minFlipsMonoIncr(String s) {\n int n = s.length();\n int[][] dp = new int[n+1][2];\n for(int i = 1; i <= n; i++) {\n if(s.charAt(i-1) == \'1\') {\n dp[i][1] = Math.min(dp[i-1][1], dp[i-1][0] + 1); // to keep 1\n dp[i][0] = dp[i-1][0] + 1; // to turn into 0\n } else { // s.charAt(i-1) == \'0\'\n dp[i][0] = dp[i-1][0]; // to keep 0\n dp[i][1] = Math.min(dp[i-1][1] + 1, dp[i-1][0]); // to turn 1\n }\n }\n\n return Math.min(dp[n][0], dp[n][1]);\n }\n}\n``` | 5 | 0 | ['Java'] | 1 |
flip-string-to-monotone-increasing | [C++/Java] Full Explanation🔥||Easy||TC O(n) || SC O(1) | cjava-full-explanationeasytc-on-sc-o1-by-j0ib | Problem Statement:\n> We will be given a string \u2018s\u2019 containing \u20180\u2019s and \u20181\u2019s arranged in some order,so we need to find out the NoO | abdul_muqeet715 | NORMAL | 2023-01-17T04:46:06.430400+00:00 | 2023-01-17T04:46:06.430448+00:00 | 102 | false | **Problem Statement:**\n> We will be given a string \u2018s\u2019 containing \u20180\u2019s and \u20181\u2019s arranged in some order,so we need to find out the NoOfFlips required for the given to string to make it monotonically increasing sequence(eg: **\u20180010\u2019** then the monotonically increasing sequence may be **\u20180000\u2019**, **\u20190011\u2019**, **\u20190001\u2019** ...etc but the minimum NoOfFlips is done in **\u20180000\u2019** & **\u20180011\u2019** because only 1 flip is required either at index 2 or at index at index 3.\n\n**Approach**:\n1. Firstly we will have to maintain two variables namely countFlips and countOnes.\n2. Assign `countFlips=0` & `countOnes=0.`\n3. countFlips will count the number of flips required and countOnes will count the number of \u20181\u2019s and update the corresponding in each iteration.\n4. Iterate over the string \u2018s\u2019 and check whether the following conditions:\n5. If the character at that particular index is \u20181\u2019 and yes if it is \u20181\u2019 then update `countOnes=countOnes+1`\n6. If the character at that particular index is 0 then now here is the important step to take care about as we need the minimum no of flips,so let us understand with the help of example:\n\t\t**Example**:\nLet us take the string as \u20180010\u2019 so as we discussed earlier that there are many possible ways to make it a monotonically increasing subsequence but here we need to make it under minimum no of flips.\nThe possible ways are for **\u20180010\u2019** are:\n**0000**,**0011**,**0001**,**0111**,**1111**\n\n\n\n \nAccording to the above table we can see that the minimum no of flips required is 1\n\n> Now the basic approach will be that whenever we encounter a \u20181\u2019 in a string we will increase a count of \u20181\u2019s and similarly whenever we encounter a \u20180\u2019 then we will check for the minimum as \u201C*is it minimum to take the \u20181\u2019s count and replace it with \u20180\u2019*" or "*is it minimum to make the current \u20180\u2019 as \u20181\u2019*" so as to make it monotonically increasing.\n> So condition comes up to be \n```\ncountFlips=min(countOnes,countFlips+1)\n```\nIf it is minimum to make \u20181\u2019s as \u20180\u2019s then it choose it or else if it is minimum to make the existing \u20180\u2019s as \u20181\u2019s including the current one then that will be updated.\n\nThe below table demonstrate the procedure during each iteration for the string \'**0010**\'\n\n\n\n\n\n\n> So finally after all the iteration we will get our result in countFlips and simply return countFlips\n\nPlease do upvote it takes lot of efforts to make it easier.\n\n# Complexity\n- Time complexity:O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```C++ []\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int countFlips=0,countOnes=0;\n for(char c:s){\n if(c==\'0\') countFlips=min(countOnes,countFlips+1);\n else countOnes++;\n }\n return countFlips;\n }\n};\n```\n```Java []\nclass Solution {\n public int minFlipsMonoIncr(String s) {\n int countFlips=0,countOnes=0;\n for(char c:s.toCharArray()){\n if(c==\'0\') countFlips=Math.min(countOnes,countFlips+1);\n else countOnes++;\n }\n return countFlips;\n }\n}\n```\n\n | 5 | 0 | ['Java'] | 3 |
flip-string-to-monotone-increasing | SIMPLE PYTHON SOLUTION | simple-python-solution-by-beneath_ocean-yj1d | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | beneath_ocean | NORMAL | 2023-01-17T04:30:00.537747+00:00 | 2023-01-17T04:42:58.085557+00:00 | 1,654 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n \n def minFlipsMonoIncr(self, s: str) -> int:\n n=len(s)\n ones=[0]*(n)\n ct=0\n for i in range(n):\n if s[i]==\'1\':\n ct+=1\n ones[i]=ct\n prev=0\n for i in range(1,n+1):\n if s[i-1]==\'0\':\n curr=min(prev+1,ones[i-1])\n else:\n curr=min(prev,ones[i-1])\n prev=curr\n return prev\n``` | 5 | 0 | ['Python3'] | 2 |
flip-string-to-monotone-increasing | python3 || Keep track of 1s and 0s | python3-keep-track-of-1s-and-0s-by-iter_-kgnl | O(n), O(1)\n\n```\nclass Solution:\n def minFlipsMonoIncr(self, s: str) -> int:\n S, ans = len(s), len(s)\n zero_count = s.count("0")\n | iter_next | NORMAL | 2023-01-17T01:40:21.180741+00:00 | 2023-01-17T01:40:21.180789+00:00 | 667 | false | O(n), O(1)\n\n```\nclass Solution:\n def minFlipsMonoIncr(self, s: str) -> int:\n S, ans = len(s), len(s)\n zero_count = s.count("0")\n ones_count = 0\n\n for j in range(S):\n ans = min(ans, zero_count+ones_count)\n \n if s[j] == "1":\n ones_count += 1\n else:\n zero_count -= 1\n \n return min(ans, ones_count) | 5 | 0 | ['Iterator', 'Python', 'Python3'] | 1 |
flip-string-to-monotone-increasing | Java O(N) | One Pass | O(1) Space | Prefix-Suffix | java-on-one-pass-o1-space-prefix-suffix-u1wid | Intuition: We are returning the minimum of min( no. of 1s flipped to 0s in prefix , no. of 0s in suffix )\nAlgo:\n1. Right after we encounter our first 1, we st | shivam_taneja | NORMAL | 2022-07-14T19:23:18.875370+00:00 | 2022-07-21T17:49:15.958941+00:00 | 265 | false | **Intuition:** We are returning the minimum of **min(** no. of 1s flipped to 0s in prefix **,** no. of 0s in suffix **)**\n**Algo:**\n1. Right after we encounter our first 1, we start counting the **number of flips**\n```Java \nif(charAt == 0) countFlip++\nelse countOnes++;\n```\n2. Now at every iteration we make the `countFlips = min(countOnes, countFlips)`. So at every iteration we are checking if the number of 0s till now are more than the number of 1s till now.\n\n```Java\nclass Solution \n{\n public int minFlipsMonoIncr(String s) \n {\n int first = 0;\n int i = 0;\n for(i = 0; i < s.length(); i++)\n {\n if(s.charAt(i)==\'1\')\n break;\n }\n int countflips = 0;\n int countOnes = 0;\n for(; i < s.length(); i++)\n {\n if(s.charAt(i)==\'0\')\n countflips++;\n else\n countOnes++;\n countflips = Math.min(countflips, countOnes);\n } \n return countflips;\n }\n}\n```\n\n**Dry Runs:\neg. "0101100011"**\n```\ni = 1 char is 1, countFlips = 0, countOnes = 1\ni = 2 char is 0, countFlips = 1, countOnes = 1\ni = 3 char is 1, countFlips = 1, countOnes = 2\ni = 4 char is 1, countFlips = 1, countOnes = 3\ni = 5 char is 0, countFlips = 2, countOnes = 3\ni = 6 char is 0, countFlips = 3, countOnes = 3\ni = 7 char is 0, countFlips = 3, countOnes = 3\ni = 8 char is 1, countFlips = 3, countOnes = 4\ni = 9 char is 1, countFlips = 3, countOnes = 5\n```\n**Ans: 0000000011**\n\n**eg. 010110**\n```\ni = 1 char is 1, countFlips = 0, countOnes = 1\ni = 2 char is 0, countFlips = 1, countOnes = 1\ni = 3 char is 1, countFlips = 1, countOnes = 2\ni = 4 char is 1, countFlips = 1, countOnes = 3\ni = 5 char is 0, countFlips = 2, countOnes = 3\n```\n**Ans: 000111**(In this case the entire part after the first 1 was the suffix, we never reached the prefix, instead we create it). | 5 | 0 | ['Suffix Array', 'Java'] | 0 |
flip-string-to-monotone-increasing | Python O(n) time | python-on-time-by-yorkshire-2bt1 | Iterate over S. For each index, find the number of zeros after that index (that need to be flipped to ones) and the number of ones before and including that ind | yorkshire | NORMAL | 2018-10-21T03:02:12.200507+00:00 | 2018-10-23T05:50:33.833585+00:00 | 420 | false | Iterate over S.
For each index, find the number of zeros after that index (that need to be flipped to ones) and the number of ones before and including that index (that need to be flipped to zeros).
Return the lowest number of flips required for any index.
```
def minFlipsMonoIncr(self, S):
n = len(S)
zeros = sum(c == "0" for c in S) # count of zeros after current index, initially all zeros in S
if zeros == 0 or zeros == n:
return 0
result = zeros # flip all zeros to ones
ones = 0
for i, c in enumerate(S):
if c == "0":
zeros -= 1 # decrement the count of zeros to be flipped
else:
ones += 1 # increment the count of ones to be flipped
result = min(result, zeros + ones)
return result | 5 | 1 | [] | 0 |
flip-string-to-monotone-increasing | 6 line python solution with explanation | 6-line-python-solution-with-explanation-a5ytq | Intuition\nMy first thgouht was to use dynamic programming to calculate a sliding window of zeroflips from left to right as well as oneflips from right to left | smanian | NORMAL | 2023-01-17T09:03:51.599321+00:00 | 2023-01-18T01:35:52.669661+00:00 | 183 | false | # Intuition\nMy first thgouht was to use dynamic programming to calculate a sliding window of zeroflips from left to right as well as oneflips from right to left and then to find the location in the middle that minimizes the number of flips. \n\nAfter working through an elaborate solution I learned that this problem can be solved in a much simpler way by observing that a flip from a 1 to a zero is equivalent to flipping the next 0 to a 1. Therefore for the solution it doesn\'t matter which zeroes and ones actually get flipped. It only matters how many there are in total.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUsing the observation that a flip from a 1 to a 0 is the same as flipping the next 0 to a 1. We can solve this problem using two variables. One that keeps track of the number of flips required for either flipping to 1, or flipping to zero, and another that keeps track of the minimum number of flips required up to that point. The end result is the number of flips required. \n\nIf you find this logic confusing, here\'s a video explanation breaking the problem down. \n[Video Explanation](https://youtu.be/Z_yrYmrkIvw)\nhttps://youtu.be/Z_yrYmrkIvw\n\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def minFlipsMonoIncr(self, s: str) -> int:\n zeroflips = oneflips = 0\n for c in s:\n if c == \'1\':\n oneflips += 1\n else:\n zeroflips += 1\n zeroflips = min(oneflips, zeroflips)\n return zeroflips\n``` | 4 | 0 | ['Math', 'Python', 'Python3'] | 1 |
flip-string-to-monotone-increasing | Easiest to understand find the pt where max difference of 0 and 1 count exist | easiest-to-understand-find-the-pt-where-6tec2 | Intuition\n Describe your first thoughts on how to solve this problem. \nFinding the point where last 0 should be\n# Approach\n Describe your approach to solvin | Subhrajeet_Biswal | NORMAL | 2023-01-17T07:14:06.204702+00:00 | 2023-01-17T07:14:06.204744+00:00 | 49 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n**Finding the point where last 0 should be**\n# Approach\n<!-- Describe your approach to solving the problem. -->\nwe can do it by finding the place where there is max difference between c0&c1 ie **max(c0-c1)**\n# Complexity\n- Time complexity:O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int n=s.size();\n vector<int>v;\n int c1=0,c0=0;\n for(int i=0;i<n;i++)\n {\n if(s[i]==\'0\')c0++;\n else\n c1++;\n v.push_back(c0-c1);\n }\n int index=-1,mx=0;\n for(int i=0;i<n;i++)\n {\n if(v[i]>mx)\n {index=i;mx=v[i];}\n }\n int ans=0;\n for(int i=0;i<=index;i++)\n {\n if(s[i]==\'1\')ans++;\n }\n for(int i=index+1;i<n;i++)\n {\n if(s[i]==\'0\')ans++;\n }\n cout<<index;\n return ans;\n }\n};\n``` | 4 | 0 | ['C++'] | 0 |
flip-string-to-monotone-increasing | C++ ||Begineer Friendly|| Easy-Understanding|| No DP || Video solution | c-begineer-friendly-easy-understanding-n-afyz | Intuition\n Describe your first thoughts on how to solve this problem. \nC++ Clear Explaination ,Please support if you find it usefull. Can give me feedback in | CodeWithSky | NORMAL | 2023-01-17T05:48:17.583180+00:00 | 2023-01-17T05:50:44.480913+00:00 | 350 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n**C++ Clear Explaination ,Please support if you find it usefull. Can give me feedback in comment for improvement.,will be very thankfull.**\n<iframe width="560" height="315" src="https://www.youtube.com/embed/MSXfEZC-pcw" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" allowfullscreen></iframe>\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int countflip = 0; // 0->1 or 1->0\n int countOne = 0;\n\n for(auto x:s){\n if(x == \'1\'){countOne++;}\n else{\n // at any x == 0 calcualte whether we need it to flip it or\n // can flip all the 1 upto that point to 0. One with min val is our required flip count.\n countflip = min(++countflip,countOne);\n }\n }\n return countflip;\n }\n};\n``` | 4 | 0 | ['String', 'Dynamic Programming', 'Bit Manipulation', 'C++', 'Java'] | 0 |
flip-string-to-monotone-increasing | ✅C++ | easy solution | prefix sum✅ | c-easy-solution-prefix-sum-by-suraj_jha9-fxc0 | \n# Approach\nwe have three choices ==> all 0\'s,all 1\'s or (0\'s and 1\'s)\nout of these choices take min of them.\n\nfor third case : 0\'s followed by 1\'s\n | suraj_jha989 | NORMAL | 2023-01-17T05:00:51.960525+00:00 | 2023-01-17T05:00:51.960568+00:00 | 69 | false | \n# Approach\nwe have three choices ==> all 0\'s,all 1\'s or (0\'s and 1\'s)\nout of these choices take min of them.\n\nfor third case : 0\'s followed by 1\'s\nwe will try out all possible strings that we can make\n\nfor each idx i, we will form a string that has all zeros till "i\'th" idx(including i idx) \nand all 1\'s after i\'th idx.\n\nflips req. for each such oprn = no. of 1\'s till i\'th idx[including i\'th idx] + no. of 0\'s after i\'th idx.\n\ntry all possible ways and take min. of all cases.\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1), since,we are only using map of size 2, we could also use two variables for that.\n\n# Code\n```\nclass Solution {\npublic:\n\n int minFlipsMonoIncr(string s) {\n unordered_map<char,int> cnt;\n for(char &ch: s) cnt[ch]++;\n\n int ans=min(cnt[\'0\'],cnt[\'1\']);\n \n int ones=0;//rep cnt of ones till "ith" idx\n for(int i=0; i<s.length(); i++){\n int flips=0;\n \n if(s[i] == \'1\') ones++;\n\n //update map : we basically care about no. of 0\'s after ith idx\n //cnt[\'0\'] tells us no. of 0\'s after i\'th idx\n if(s[i] == \'0\') cnt[s[i]]--;\n\n //make cur char 0 if it is one, and all zeros after ith idx to ones\n flips += (ones + cnt[\'0\']);\n\n ans = min(ans,flips);\n }\n \n return ans;\n }\n};\n```\n# Do Upvote, if it\'s been any help to you! \uD83D\uDE80 \uD83D\uDE80\n | 4 | 0 | ['C++'] | 0 |
flip-string-to-monotone-increasing | ✅ easy and fast code | C++ | easy-and-fast-code-c-by-coding_menance-kmlx | Here is the solution for the question:\n\nC++ []\nclass Solution\n{\npublic:\n int minFlipsMonoIncr(string S)\n {\n int count_flip = 0, count_one = | coding_menance | NORMAL | 2023-01-17T03:32:26.342692+00:00 | 2023-01-17T03:32:26.342795+00:00 | 322 | false | Here is the solution for the question:\n\n``` C++ []\nclass Solution\n{\npublic:\n int minFlipsMonoIncr(string S)\n {\n int count_flip = 0, count_one = 0;\n for (auto i : S)\n { \n //keep track of all one (we will use count_one in else condition if we need) \n//if we want flip one into 0\n if (i == \'1\')\n count_one++;\n else{\n count_flip++;\n count_flip = min(count_flip, count_one);\n }\n }\n return count_flip;\n }\n};\n```\n\n\n*Upvote if helped* | 4 | 0 | ['C++'] | 3 |
flip-string-to-monotone-increasing | C++ || Easy to understand || PostFix Sum | c-easy-to-understand-postfix-sum-by-moha-ajoq | Intuition\n\n1) Traverse the string from left to right\n2) Whenever you encounter a \'1\' there would be 2 cases possible \n\t you can keep this as \'1\' , the | mohakharjani | NORMAL | 2023-01-17T02:07:07.919475+00:00 | 2023-01-17T03:17:57.042383+00:00 | 1,520 | false | Intuition\n\n1) Traverse the string from left to right\n2) Whenever you encounter a \'1\' there would be 2 cases possible \n\t* you **can keep this as \'1\'** , then all the chars at right should be \'1\' to make it increasing,\n\t so flip all the \'0\'s (towards right) to \'1\'s\n\t * **or flip the curr \'1\' to \'0\'** and move on\n```\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) \n {\n int n = s.size();\n vector<int>post(n);//keeps the count of number of \'0\'s to the right\n post[n - 1] = (s[n - 1] == \'0\')? 1 : 0;\n for (int i = n - 2; i >= 0; i--) post[i] = post[i + 1] + (s[i] == \'0\'? 1 : 0);\n //============================================================\n int currFlips = 0, ans = n;\n for (int i = 0; i < n; i++)\n {\n if (s[i] == \'1\')\n {\n int toFlip = (i + 1 == n)? 0 : post[i + 1];\n int totalFlips = currFlips + toFlip; //make all the 0\'s to 1\'s towards right\n ans = min(ans, totalFlips); //record the answer\n \n //or\n currFlips++; //just flip the current \'1\' to \'0\' and move\n }\n }\n //=====================================================================\n ans = min(ans, currFlips);\n return ans;\n \n }\n};\n``` | 4 | 0 | ['C', 'C++'] | 2 |
flip-string-to-monotone-increasing | 3 Solution || Brute to Best || Complexity | 3-solution-brute-to-best-complexity-by-s-02wu | Intuition\n Describe your first thoughts on how to solve this problem. \nspecial mention : prathz for this approach\nPrefix sum \n\n# Approach\n Describe your | sunnyjha1512002 | NORMAL | 2023-01-17T02:02:02.621321+00:00 | 2023-01-17T02:02:02.621357+00:00 | 472 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nspecial mention : [prathz](https://leetcode.com/problems/flip-string-to-monotone-increasing/solutions/3061177/c-prefix-sum/?orderBy=hot) for this approach\nPrefix sum \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nSkip the loop until we find the first \'1\'\nafter finding first occurrence of \'1\' we have to keep a count of number of \'1\'s in countOnes\n\nif we find any \'0\' after occurrence of \'1\' it means turning that \'0\' to one may be one possible solution so we have to keep number of flip needed in countFlips variable\n\nif value of countFlips is greater than countOnes(number of ones) then converting all \'1\' to \'0\' would be better option so assign value of countOnes to the countFlips\n\nin the end we will return countFlips\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(n)$$\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n$$O(1)$$\n\n# Code\n```\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int flip=0,one=0,n = s.size();\n for(int i=0;i<n;i++){\n if(s[i]==\'1\'){\n one++;\n }else{\n if(one==0)continue;\n else flip++;\n }\n if(one<flip)flip = one;\n }\n return flip;\n }\n};\n\n```\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nCounting the no. of 1\'s to flip on left side + no. of 0\'s to flip on right side of each index i and then select minimum flips from it.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nMaintain a prefix and suffix array to count 1 on left side and 0 on right side.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(n)$$\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n$$O(n)$$\n\n# Code\n```\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int n = s.size(),ans = INT_MAX;\n vector<int> onecount(n,0),zerocount(n,0);\n onecount[0] = 0,zerocount[n-1]=0;\n for(int i=1;i<n;i++){\n if(s[i-1]==\'1\')onecount[i] = onecount[i-1]+1;\n else onecount[i] = onecount[i-1];\n if(s[n-i]==\'0\')zerocount[n-i-1] = zerocount[n-i]+1;\n else zerocount[n-i-1] = zerocount[n-i];\n }\n for(int i=0;i<n;i++){\n ans = min(ans,zerocount[i]+onecount[i]);\n }\n return ans;\n }\n};\n```\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nCounting the no. of 1\'s to flip on left side + no. of 0\'s to flip on right side of each index i and then select minimum flips from it.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nBrute \n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(n^2)$$\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n$$O(n)$$\n\n### TLE on 73rd test case \n# Code\n```\nclass Solution {\nprivate:\nint onecount(string s){\n int n = s.size();\n int cnt=0;\n for(int i=0;i<n;i++){\n if(s[i]==\'1\')cnt++;\n }\n return cnt;\n}\nint zerocount(string s){\n int n = s.size();\n int cnt=0;\n for(int i=0;i<n;i++){\n if(s[i]==\'0\')cnt++;\n }\n return cnt;\n}\npublic:\n int minFlipsMonoIncr(string s) {\n int n = s.size();\n int mini = INT_MAX;\n vector<int> ans(n,0);\n for(int i=0;i<n;i++){\n ans[i]+= onecount(s.substr(0,i));\n ans[i]+= zerocount(s.substr(i+1));\n mini = min(mini,ans[i]);\n }\n return mini;\n }\n};\n```\n | 4 | 0 | ['Array', 'Suffix Array', 'Prefix Sum', 'C++'] | 3 |
flip-string-to-monotone-increasing | DAILY LEETCODE SOLUTION || EASY C++ SOLUTION | daily-leetcode-solution-easy-c-solution-kkcn1 | \n//Memoisation\nclass Solution {\npublic:\n int solve(int idx,int prev,string &s,vector<vector<int>>& dp){\n if(idx==s.size()){\n return 0 | pankaj_777 | NORMAL | 2023-01-17T00:33:52.916770+00:00 | 2023-01-17T01:05:18.706180+00:00 | 639 | false | ```\n//Memoisation\nclass Solution {\npublic:\n int solve(int idx,int prev,string &s,vector<vector<int>>& dp){\n if(idx==s.size()){\n return 0;\n }\n if(dp[idx][prev]!=-1) return dp[idx][prev];\n if(prev){\n return dp[idx][prev]=(s[idx]==\'0\')+solve(idx+1,prev,s,dp);\n }\n else{\n return dp[idx][prev]=min((s[idx]==\'1\')+solve(idx+1,0,s,dp),(s[idx]==\'0\')+solve(idx+1,1,s,dp));\n }\n }\n int minFlipsMonoIncr(string s) {\n vector<vector<int>> dp(s.size(),vector<int>(2,-1));\n return solve(0,0,s,dp);\n }\n};\n\n```\n```\n//Tabulation\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int n=s.size();\n vector<vector<int>> dp(n+1,vector<int>(2,0));\n for(int idx=n-1;idx>=0;idx--){\n for(int prev=1;prev>=0;prev--){\n if(prev){\n dp[idx][prev]=(s[idx]==\'0\')+dp[idx+1][prev];\n }\n else{\n dp[idx][prev]=min((s[idx]==\'1\')+dp[idx+1][0],(s[idx]==\'0\')+dp[idx+1][1]);\n }\n }\n }\n return dp[0][0];\n }\n};\n```\n```\n//Space Optimization\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int n=s.size();\n vector<int> dp(2,0);\n for(int idx=n-1;idx>=0;idx--){\n vector<int> curr(2,0);\n for(int prev=1;prev>=0;prev--){\n if(prev){\n curr[prev]=(s[idx]==\'0\')+dp[prev];\n }\n else{\n curr[prev]=min((s[idx]==\'1\')+dp[0],(s[idx]==\'0\')+dp[1]);\n }\n }\n dp=curr;\n }\n return dp[0];\n }\n};\n``` | 4 | 0 | ['Dynamic Programming', 'Memoization', 'C++'] | 2 |
flip-string-to-monotone-increasing | C++ | Memoization | Easy Code | c-memoization-easy-code-by-letthecodebe-euky | Code\n\nclass Solution {\npublic:\n int helper(int i, int prev, string &s, int n, vector<vector<int>> &dp){\n if(i>=n) return 0;\n\n if(dp[i][p | letthecodebe | NORMAL | 2023-01-12T12:46:29.272726+00:00 | 2023-01-12T12:46:29.272765+00:00 | 578 | false | # Code\n```\nclass Solution {\npublic:\n int helper(int i, int prev, string &s, int n, vector<vector<int>> &dp){\n if(i>=n) return 0;\n\n if(dp[i][prev] != -1) return dp[i][prev];\n\n int ans = INT_MAX;\n\n if(s[i]==\'0\'){\n if(prev == 0){\n ans = min(ans,min(1 + helper(i+1,1,s,n,dp), helper(i+1,0,s,n,dp)));\n }else{\n ans = 1 + helper(i+1,1,s,n,dp);\n }\n }else {\n if(prev == 0){\n ans = min(ans,min(1 + helper(i+1,0,s,n,dp), helper(i+1,1,s,n,dp)));\n }else{\n ans = helper(i+1,1,s,n,dp);\n }\n }\n\n return dp[i][prev] = ans;\n }\n\n int minFlipsMonoIncr(string s) {\n int n = s.size();\n vector<vector<int>> dp(n+1, vector<int> (2,-1));\n\n int res = helper(0,0,s,n,dp);\n return res;\n }\n};\n``` | 4 | 0 | ['String', 'Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 1 |
flip-string-to-monotone-increasing | [JAVA] BEATS 100.00% MEMORY/SPEED 0ms // APRIL 2022 | java-beats-10000-memoryspeed-0ms-april-2-rysz | \n\tclass Solution {\n\n public int minFlipsMonoIncr(String s) {\n int ones = 0, flips = 0;\n for(char c: s.toCharArray()) {\n if (c | darian-catalin-cucer | NORMAL | 2022-04-22T18:52:13.093258+00:00 | 2022-04-22T18:52:13.093306+00:00 | 316 | false | \n\tclass Solution {\n\n public int minFlipsMonoIncr(String s) {\n int ones = 0, flips = 0;\n for(char c: s.toCharArray()) {\n if (c == \'1\') {\n ones++;\n } else {\n flips = Math.min(ones, flips + 1);\n }\n }\n return flips;\n }\n\t} | 4 | 0 | ['Java'] | 0 |
flip-string-to-monotone-increasing | Python simple | python-simple-by-ploypaphat-ukn3 | Keep track of zero and one. Since this is an increasing, the left side would likely be \'zero\', hence while iterating through the string, all the leading zero( | ploypaphat | NORMAL | 2022-03-27T23:27:35.827217+00:00 | 2022-04-03T02:10:48.285234+00:00 | 403 | false | Keep track of zero and one. Since this is an increasing, the left side would likely be \'zero\', hence while iterating through the string, all the leading zero(s) will be ignored (set count_zero to count_one, which is zero because we have not hit any \'one\' yet). Once we starts hitting \'one\', then the logic will compute which one has the majority of the population. The minority should be flipped accordingly. \n\n```\n def minFlipsMonoIncr(self, s: str) -> int:\n \n count_one = 0\n count_zero = 0\n\n for i in range(len(s)):\n if s[i] == \'0\':\n count_zero += 1\n else:\n count_one += 1\n # ignore leading \'zero\' while counting. Once hits \'one\', count population and return the minority to get the minimum flipping needed. \n if count_zero > count_one:\n count_zero = count_one\n \n return count_zero\n``` | 4 | 0 | ['Python', 'Python3'] | 0 |
flip-string-to-monotone-increasing | C++||Recursion+Memoization|| | crecursionmemoization-by-rohitsharma8-85eq | \n vector<vector<int>>dp;\n int util(string&s,int i,bool flag){\n if(i>=s.size())\n return 0;\n if(dp[i][int(flag)]!=-1)\n | rohitsharma8 | NORMAL | 2022-03-24T09:04:33.337801+00:00 | 2022-03-24T09:13:29.448618+00:00 | 433 | false | ```\n vector<vector<int>>dp;\n int util(string&s,int i,bool flag){\n if(i>=s.size())\n return 0;\n if(dp[i][int(flag)]!=-1)\n return dp[i][int(flag)];\n if(s[i]==\'0\' and flag){//if we already come across 1 and now "i" is at 0 then we have to flip it\n return dp[i][int(flag)]=1+util(s,i+1,true);\n }\n if(s[i]==\'0\')//we have 2 options either flip it or continue with "0" only\n return dp[i][int(flag)]=min(util(s,i+1,false),1+util(s,i+1,true));\n if(s[i]==\'1\' and !flag){//consider test case like this "10000" for this we need following condition\n return dp[i][int(flag)]=min(util(s,i+1,true),1+util(s,i+1,false));\n }\n return dp[i][int(flag)]=util(s,i+1,true);\n }\n int minFlipsMonoIncr(string s) {\n dp.resize(s.size()+1,vector<int>(2,-1));\n return util(s,0,false);\n }\n``` | 4 | 0 | ['Recursion', 'Memoization', 'C'] | 1 |
flip-string-to-monotone-increasing | C++ | Detailed Explanation | Time O(N) | Space O(1) | c-detailed-explanation-time-on-space-o1-nm34i | Let\'s understand by an example:\nSuppose,\ns = "101"\nn = s.length() = 3\n\nPossible monotonic sequences for a string of length 3 are:\n\n\t"000"\n\t"001"\n\t" | joshiatharva | NORMAL | 2021-11-08T06:08:01.255548+00:00 | 2021-11-08T06:19:10.634393+00:00 | 311 | false | Let\'s understand by an example:\nSuppose,\n`s = "101"`\n`n = s.length() = 3`\n\nPossible monotonic sequences for a string of length 3 are:\n```\n\t"000"\n\t"001"\n\t"011"\n\t"111"\n```\n\t\nWe will achieve minimum flips by converting given string "s" to one of these possible monotonic sequences. But, at the moment we do not know which monotonic sequence we should consider.\n\nSo, let\'s convert our string to these sequences\n\n"101" to "000" => flips = 2\n"101" to "001" => flips = 1\n"101" to "011" => flips = 2\n"101" to "111" => flips = 1\n\n Hence, minmum flips = min({2, 1, 2, 1}) = 1\n \n**Algorithm**\n```\nflips = 0; \nminflips = 0;\n```\n\n**"101" to "000"** \n` flips = all ones in given string need to be flipped = 2`\n\n**"101" to "001"**\n```\nif(last character in given string is \'1\')\n\tflips = flips - 1; // because we considered it as flip in previous sequence but in this sequence we should not flip it \nelse\n\tflips = flips + 1; // we considered it as not a flip in previous sequence but in this sequence we should consider it as a flip\n```\n\tHence, here flips = flips - 1 = 1\n\t\n**"101" to "011"**\n```\nif(second last character given string \'1\')\n\tflips = flips - 1; // we considered it as a flip in last sequence but in this sequence it should not be a flip\nelse\n\tflips = flips + 1; // Not a flip in last sequence but should be flipped in this\n```\n\tHence here, flips = flips + 1 = 2\n\t\n**"101" to "111"**\n```\nif(third last character in given string is \'1\')\n\tflips = flips - 1; // flipped in last sequence but not in this\nelse\n\tflips = flips + 1; // not flipped in last but needs to flip for this\n```\n\tHence, here flips = flips - 1 = 1\n\t\t\t\t\t\t\t\n`Minimum flips = min({2, 1, 2, 1}) = 1`\n\nSo, for a string of length n we will have (n+1) monotonic sequences.\nAt first we will have to iterate the whole string to initialize flips\nFor \'r\'th monotonic sequences we will just have to check (n-r)th character to determine flips\n\n**Code**\n\n```\nint minFlipsMonoIncr(string s) {\n int flips = 0, i, n = s.length(), minflips = 0;\n for(i = 0; i < n; i += 1){\n flips += (s[i]-\'0\'); // count \'1\'s in given string\n }\n minflips = flips;\n for(i = n-1; i > -1; i -= 1){\n if(s[i] == \'1\') // considered flip in last sequence\n flips -= 1;\n else // not considered flip in last sequence\n flips += 1;\n\t\t\t\t\n minflips = min(minflips, flips); // check minimum\n }\n return minflips;\n }\n```\n\n**Time Complexity**\n\t`O(N + 1* N) = O(2*N) = O(N)`\n\t\n**Space Complexity**\n`\tO(1)` | 4 | 0 | ['C'] | 0 |
flip-string-to-monotone-increasing | [Python 3] Single pass without extra memory | python-3-single-pass-without-extra-memor-4sd1 | \nclass Solution(object):\n def minFlipsMonoIncr(self, s):\n ones = 0\n flips = 0\n \n for c in s:\n if c == \'0\'and | abstractart | NORMAL | 2021-09-16T06:06:18.110811+00:00 | 2021-11-28T09:10:09.197904+00:00 | 352 | false | ```\nclass Solution(object):\n def minFlipsMonoIncr(self, s):\n ones = 0\n flips = 0\n \n for c in s:\n if c == \'0\'and ones > 0:\n flips += 1\n ones -= 1\n elif c == \'1\':\n ones += 1\n \n return flips\n``` | 4 | 0 | ['Python', 'Python3'] | 2 |
flip-string-to-monotone-increasing | C++ RECURSIVE,DP O(N)|O(1) Auxiliary space sol | c-recursivedp-ono1-auxiliary-space-sol-b-kozu | 1.Recursive solution\nThe recursive idea is taking the last character of the string.We can have two cases:-\n##### 1)If s[last]==\'1\' \nIn this case we can sim | jayadeepbose18 | NORMAL | 2021-08-11T05:19:59.267149+00:00 | 2022-01-18T14:43:56.434448+00:00 | 125 | false | **1.Recursive solution**\nThe recursive idea is taking the last character of the string.We can have two cases:-\n##### 1)*If s[last]==\'1\'* \nIn this case we can simply ignore the character and move on to the previous character.Bcoz this doesnt effect the monotonicty.\n**f(s,l)=f(s,l-1);**\n##### 2)*If s[last]==\'0\'*\nIn this case we can either make our current character to \'1\' (or) flip all the previous characters which are \'1\' to \'0\' to maintain the monotonicty. We have to take the minimum of two choices.\n**f(s,l)=min(1+f(s,l-1) , nofones(s,l-1));**\n#### Base case:\nIf the nof characters is one then our answer will be 0.Bcoz we dont need any flips to make it montonically increasing.\n\n### code:-\n\n```\nint nofones(string s,int idx){\n \n int c=0;\n for(int i=0;i<=idx;i++){\n if(s[i]==\'1\'){\n c++;\n }\n }\n return c;\n }\n int rec(string s,int h){\n //base case\n if(h==0){\n return 0;\n } \n if(s[h]==\'1\'){\n return rec(s,h-1);\n }\n else {\n return min((1+rec(s,h-1)),nofones(s,h-1));\n }\n \n }\n int minFlipsMonoIncr(string s) {\n int sz=s.size();\n //cout<<size;\n return rec(s,sz-1);\n }\n``` \n\n \n \n#### **2)DP APPROACH**\nAs we are recomputing the nof ones everytime we can optimize it by storing the values in an array.Also we can use another array for the nof flips required for a string upto the given index.\n\n#### *Code*:\n\n```\nint minFlipsMonoIncr(string s) {\n int sz=s.size();\n int dp[sz+1];\n int dpones[sz+1];\n\t\t//Initially we assign the values which are not possible\n for(int i=0;i<=sz;i++){\n dp[i]=INT_MAX;\n }\n for(int i=0;i<=sz;i++){\n dpones[i]=INT_MAX;\n }\n dp[0]=0;\n dpones[0]=0;\n dp[1]=0;\n if(s[0]==\'1\'){\n dpones[1]=1;\n }\n else{\n dpones[1]=0;\n }\n //Filling up our dp arrays\n for(int i=2;i<=sz;i++){\n \n if(s[i-1]==\'1\'){\n dpones[i]=dpones[i-1]+1;\n dp[i]=dp[i-1];\n }\n else{\n //if(1+dp[i-1]<dpones[i-1])\n dpones[i]=dpones[i-1];\n dp[i]=min(1+dp[i-1],dpones[i-1]);\n }\n }\n\t\t//Our last value in the dp array is the min nof flips required\n return dp[sz];\n```\n##### T.C:-O(SIZE) , A.S:-O(SIZE);\n\n\n\n#### **2)Space optimized solution**\nWe can observe that in our dp sol we only use the previous values in both cases of updating the nof ones , min.flips required for an index.\nso we can just use varaibles to store them.\n\n##### *CODE*:\n```\nint minFlipsMonoIncr(string s) {\n int sz=s.size();\n int x=0,y=0;\n if(s[0]==\'1\'){\n x=1;\n }\n else{\n x=0;\n } \n int x1,y1;\n for(int i=2;i<=sz;i++){ \n if(s[i-1]==\'1\'){ \n x1=1+x;\n y1=y;\n }\n else{\t\t\t \n x1=x;\n y1=min(1+y,x); \n }\n x=x1;\n y=y1;\n }\n return y;\n```\n##### T.C:-O(SIZE) , A.S:-O(1);\n\t\t | 4 | 0 | [] | 0 |
flip-string-to-monotone-increasing | Easy Solution , detailed explanation (with eg) | easy-solution-detailed-explanation-with-e66zz | class Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n /\n\t\t\n ans can be either flip all 1 to 0 000000000 | nish2447 | NORMAL | 2021-08-10T10:11:59.086840+00:00 | 2021-08-10T10:18:22.188896+00:00 | 194 | false | class Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n /*\n\t\t\n ans can be either flip all 1 to 0 000000000000\n\t or flip all 0 to one 111111111111\n or traverse and find the terms to be flipped 0010101010011\n or no change 000000011111\n \n After a "1" all zeroes should be flipped to 1 only\n 010101001111\n ->\n 011111111111\n \n OR\n \n 010101001111\n ->\n 000001111111\n \n so at each i we check if all the preceding 1\'s should be flipped or the remaining 0\'s \n */ \n int ze=0,on=0;\n for(int i=0;i<s.length();i++)\n {\n if(s[i]==\'1\')\n on++;\n else\n ze++;\n \n ze=min(ze,on);\n }\n return ze;\n \n }\n}; | 4 | 0 | [] | 1 |
flip-string-to-monotone-increasing | C++ one-pass DP in 4 lines | c-one-pass-dp-in-4-lines-by-mrsuyi-cho6 | d0: minimum moves to flip S[0, i) to array of all \'0\'.\nd1: minimum moves to flip S[0, i) to array with some \'0\' in front and some \'1\' (possibly none) in | mrsuyi | NORMAL | 2020-04-16T17:49:05.435508+00:00 | 2020-04-16T17:49:05.435559+00:00 | 446 | false | d0: minimum moves to flip S[0, i) to array of all \'0\'.\nd1: minimum moves to flip S[0, i) to array with some \'0\' in front and some \'1\' (possibly none) in back.\n```\nclass Solution {\npublic:\n int minFlipsMonoIncr(string S) {\n int d0 = 0, d1 = 0;\n for (int i = 0; i < S.size(); ++i)\n d0 += S[i] == \'1\', d1 = min(d0, d1 + (S[i] == \'0\'));\n return min(d0, d1);\n }\n};\n``` | 4 | 0 | ['Dynamic Programming', 'C'] | 1 |
flip-string-to-monotone-increasing | DP | Explain | Recursion | Simple | CPP | OnePass | dp-explain-recursion-simple-cpp-onepass-w3hrf | Calculating the number of fllips required till current position \'i\':\n\nKeep maintaining count of number of one\'s till current index, in some variable say "o | nitishal | NORMAL | 2020-04-15T13:11:32.137138+00:00 | 2020-04-15T13:17:28.838696+00:00 | 300 | false | **Calculating the number of fllips required till current position \'i\':**\n\nKeep maintaining count of number of one\'s till current index, in some variable say "one", \nNow the problem at hand can be split into following two cases :\n* **If current index value is one \'1\' :**\n\t1. Increment the count of ones "one"\n\t2. number of flips will be same as for the previous index (considering the monotonicity is maintained till last index)\n* **if current index value is zero \'0\' :**\n\t* \tnumber of flips will be minimum of **"sum of 1 (considering flipping the current bit, + 1) and considering the number of flips till previous index"**, and **"flipping all ones encountered till now to zero, using maintained count of ones till now \'one\'"**.\n\t\t\n#### Recursion:\n \t\tnflips(i) = {\n \t\t\t\tnflips(i - 1); if S[i] == \'1\'\n \t\t\t\tmin(1 + nflips(i - 1), one); if S[i] == \'0\'\n \t\t}\n```\none : count(1\'s) till index i\nnflips(i) : being number of flips required to maintaint monotonicity till index i\n```\n\nclass Solution {\npublic:\n int minFlipsMonoIncr(string S) {\n \n int n = S.length();\n int one, dp[n];\n dp[0] = 0;\n one = (S[0] == \'1\');\n for(int i = 1; i < n; i++) {\n if(S[i] == \'0\')\n dp[i] = min(dp[i - 1] + 1, one);\n else {\n one += 1;\n dp[i] = dp[i - 1];\n }\n }\n return dp[n - 1];\n }\n}; | 4 | 0 | [] | 1 |
flip-string-to-monotone-increasing | javascript beat 100% | javascript-beat-100-by-huihancarl188-atcj | \nvar minFlipsMonoIncr = function(S) {\n let cur = 0;\n for(let i = 0 ; i < S.length; i++) {\n if(S.charAt(i) === \'0\') {\n // first fl | huihancarl188 | NORMAL | 2019-05-17T04:52:54.314566+00:00 | 2019-05-17T04:52:54.314596+00:00 | 248 | false | ```\nvar minFlipsMonoIncr = function(S) {\n let cur = 0;\n for(let i = 0 ; i < S.length; i++) {\n if(S.charAt(i) === \'0\') {\n // first flip all 0 into 1;\n cur++;\n }\n }\n let res = cur;\n for(let i = 0 ; i < S.length; i++) {\n if(S.charAt(i) === \'0\') {\n cur--;\n } else {\n cur++;\n }\n // change into 000111 case, \n res = Math.min(res, cur);\n }\n \n return res;\n};\n``` | 4 | 0 | [] | 0 |
flip-string-to-monotone-increasing | Clean DP Python - O(n)/O(1) | clean-dp-python-ono1-by-wxy9018-6951 | for each position, we will rember the cost for maintaining an array that ends with \'0\' and ends with \'1\'\n\n\n\'\'\'\nclass Solution:\n def minFlipsMonoI | wxy9018 | NORMAL | 2019-04-25T03:25:13.005419+00:00 | 2019-04-25T03:25:13.005490+00:00 | 274 | false | for each position, we will rember the cost for maintaining an array that ends with \'0\' and ends with \'1\'\n\n```\n\'\'\'\nclass Solution:\n def minFlipsMonoIncr(self, S: str) -> int:\n dp0, dp1 = 0, 0 # dp0: maintain a all-0 array; dp1: maintain an array that end with 1\n for c in S:\n if c == \'0\':\n dp0, dp1 = dp0, min(dp0, dp1) + 1\n else:\n dp0, dp1 = dp0 + 1, min(dp0, dp1)\n return min(dp0, dp1)\n\'\'\'\n``` | 4 | 0 | [] | 0 |
flip-string-to-monotone-increasing | O(n) Java with O(1) space with Intuitive Explanation | on-java-with-o1-space-with-intuitive-exp-czkq | The basic idea is to travel from left to right. At every step i:\n(1) flip every 1 to the left of i(including) to 0\n(2) flip every 0 to the right of i(excludin | yuanb10 | NORMAL | 2018-10-21T03:12:31.990857+00:00 | 2018-10-21T16:26:01.418348+00:00 | 918 | false | The basic idea is to travel from left to right. At every step i:\n(1) flip every 1 to the left of i(including) to 0\n(2) flip every 0 to the right of i(excluding) to 1\nFind the minimum along the way.\n\n```\nclass Solution {\n public int minFlipsMonoIncr(String S) {\n int r0 = 0, l1 = 0;\n \n for (int i = 0; i < S.length(); i++) {\n if (S.charAt(i) == \'0\') {\n r0++;\n }\n }\n \n int res = r0;\n \n for (int j = 0; j < S.length(); j++) {\n if (S.charAt(j) == \'0\') {\n r0--;\n l0++;\n } else {\n r1--;\n l1++;\n }\n res = Math.min(l1 + r0, res);\n }\n \n return res;\n }\n}\n``` | 4 | 0 | [] | 3 |
flip-string-to-monotone-increasing | PREFIX 1's & SUFFIX 0's || C++ | prefix-1s-suffix-0s-c-by-ganeshkumawat87-9wyj | Code\n\nclass Solution {\npublic:\n int minFlipsMonoIncr(string str) {\n int n = str.length(),i;\n vector<int> p(n,0),s(n,0);\n p[0] = ( | ganeshkumawat8740 | NORMAL | 2023-06-22T03:48:19.848801+00:00 | 2023-06-22T03:48:19.848827+00:00 | 328 | false | # Code\n```\nclass Solution {\npublic:\n int minFlipsMonoIncr(string str) {\n int n = str.length(),i;\n vector<int> p(n,0),s(n,0);\n p[0] = (str[0]==\'1\');\n for(i = 1; i < n; i++){\n if(str[i]==\'1\'){\n p[i] = p[i-1]+1;\n }else{\n p[i] = p[i-1];\n }\n }\n s[n-1] = (str[n-1]==\'0\');\n for(i = n-2; i >= 0; i--){\n if(str[i]==\'0\'){\n s[i] = s[i+1]+1;\n }else{\n s[i] = s[i+1];\n }\n }\n int ans = n;\n ans = min(s[0],p[n-1]);\n for(i=1;i<n;i++){\n ans = min(ans,p[i-1]+s[i]);\n }\n return ans;\n }\n};\n``` | 3 | 0 | ['String', 'Dynamic Programming', 'C++'] | 0 |
flip-string-to-monotone-increasing | Explanation in laymens terms [Easy to understand] | explanation-in-laymens-terms-easy-to-und-n8a5 | Intuition\nBasically the idea is that once we reach a \'1\' we will start flipping \'0\'s to 1s until the count for zeros is less than ones but as an when the c | darkmatter404 | NORMAL | 2023-01-17T21:20:57.580255+00:00 | 2023-01-17T21:40:36.802960+00:00 | 320 | false | # Intuition\nBasically the idea is that once we reach a \'1\' we will start flipping \'0\'s to 1s until the count for zeros is less than ones but as an when the count for zero increases more than that of ones as we are logically supposed to flip the 1s instead of zeroes. \n\nThe way we will be able to acheive this is by considering count of 1 and number of flips(initially with the thought of flipping at 0s) and then we take a minimum of these two values to transition between these two flips min(counter_right, flips). \n\n```\n E.g. - \n s ="00110100" \n\n # string = "0" counter = 0, flips = 1 -> flips = min(0, 1) = 0\n - as bit is not \'0\' counter will be 0 and flips will be reset to zero as we don\'t need to flip anything yet\n\n # string = "00" counter = 0, flips = 1 -> flips = min(0, 1) = 0\n - as 1 is not reached counter will be 0 and flips will be reset to zero as we don\'t need to flip anything yet\n\n # string = "001" counter = 1, flips = 0 -> flips = min(1, 0) = 0\n - as we have a 1 now counter will be changed to 1 and the flip will still be zero as there is not to flip yet\n\n # string = "0011" counter = 2, flips = 0 -> flips = min(2, 0) = 0\n - as we have a 1 now counter will be changed to 2 and the flip will still be zero as there is not to flip yet\n\n # string = "00110" counter = 2, flips = 1 -> flips = min(2, 1) = 1\n - as bit is \'0\' counter will stay at 2 and the flip will changed to 1 with the idea that we need to have all 1s after the first 1 and as the minimum of counter and flips is 1 it will be saved\n\n # string = "001100" counter = 2, flips = 2 -> flips = min(2, 2) = 2\n - as bit is \'0\' counter will stay at 2 and the flip will changed to 2 and as the minimum of counter and flips is 2 it will be saved\n\n # string = "0011001" counter = 3, flips = 2 -> flips = min(3, 2) = 2\n - as now we encountered one more 1 our idea still holds true that we have to flip 0s to 1s so the value for flip will stay as 2 because of min and will be saved\n\n # string = "00110010" counter = 3, flips = 3 -> flips = min(3, 3) = 3\n - now we encountered \'0\' making its count same as the count for \'1\' (which means we will not care about if we are fliping 0 or 1) so we will increase the flips to 3. As min of counter and flips would be 3 which will be saved\n\n # string = "001100100" counter = 3, flips = 4 -> flips = min(3, 4) = 3\n - another \'0\' which makes its count more than that of \'1\' which will reset our idea that we have to flip 1s now. As we have the counter for 1s our min condition will support our idea and set the flips to min(3, 4) = 3\n\n\n This lands us to the conclusion that we had to flip 3 1s to make it binary monotonic and adds up to our solutions \n```\n\n# Approach\n- Take counter for 1s and flips\n- Increase counter for 1s at every \'1\' encountered\n- Increase the flips for every \'0\' encountered\n- reset flips to minimum of flips and counter for 1s(this will help us transition between fliping 0s ar 1s)\n- return flips\n\n> Note: Try to write your own code after this if you get the idea. :-)\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution:\n def minFlipsMonoIncr(self, s: str) -> int:\n counter = 0\n flips = 0\n for ch in s:\n if ch == \'1\':\n counter += 1\n else:\n flips += 1\n flips = min(counter, flips)\n return flips\n``` | 3 | 0 | ['Python', 'C++', 'Java', 'Python3', 'C#'] | 2 |
flip-string-to-monotone-increasing | Simple solution without DP | simple-solution-without-dp-by-aadarshpnd-2ah3 | Intuition\nNo need to think much, stand at any index, count number of 1s to the left of it, count numbr of 0s as flips. At the end, u either have to replace 1s | aadarshpndey | NORMAL | 2023-01-17T17:50:08.912748+00:00 | 2023-01-17T17:50:08.912804+00:00 | 22 | false | # Intuition\nNo need to think much, stand at any index, count number of 1s to the left of it, count numbr of 0s as flips. At the end, u either have to replace 1s with 0 or otherwise. Store the min of them as ur ans(flips). \n\nThis que is tagged as DP only bcoz its updating ans at a particular state using previous state.\n\n# Approach\nSimply iterate through the string. Rest is briefly explained in the code.\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\n# Code\n```\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int oneCnt = 0;//to maintain number of 1s before a particular index\n int flipCnt = 0;//this will keep record of flips\n\n for(auto it : s){\n if(it == \'1\') oneCnt++;\n else{\n flipCnt++;\n flipCnt = min(flipCnt, oneCnt); //we\'ll be having minimum of both the cases as our ans\n }\n }\n return flipCnt;\n }\n};\n``` | 3 | 0 | ['C++'] | 0 |
flip-string-to-monotone-increasing | intuitive way to solve| TC-O(n) and SC-O(1). | intuitive-way-to-solve-tc-on-and-sc-o1-b-w5gv | \n# Code\n\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s)\n {\n int cnt_flip = 0, cnt_one = 0;\n for (auto i : s)\n { \n | divyam_04 | NORMAL | 2023-01-17T11:01:55.460389+00:00 | 2023-01-17T11:01:55.460436+00:00 | 26 | false | \n# Code\n```\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s)\n {\n int cnt_flip = 0, cnt_one = 0;\n for (auto i : s)\n { \n\n if (i == \'0\')\n cnt_flip++;\n else{\n cnt_one++; \n }\n cnt_flip = min(cnt_flip, cnt_one);\n }\n return cnt_flip;\n }\n};\n``` | 3 | 0 | ['C++'] | 0 |
flip-string-to-monotone-increasing | C++ O(N) Solution|| Simple Explanation with example | c-on-solution-simple-explanation-with-ex-89vg | Approach\n Describe your approach to solving the problem. \nFor monotone increasing string all the zeros are on the left side (possibly 0) and all the ones are | umangrathi110 | NORMAL | 2023-01-17T10:21:08.755595+00:00 | 2023-01-17T10:21:08.755640+00:00 | 24 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nFor monotone increasing string all the zeros are on the left side (possibly 0) and all the ones are on the right side(possibly 0). \n(000111 , 00000 , 11111 all are monotone increasing)\n1. So, all the zeroes occur before the first one are useless for us, and we skip those zeros. (**0 0** 1 0 1 0 --> first two zeros are useless)\n2. After frst one, we find the number of ones occur and number of zeros occur that need to be converted into ones. (0 1 1 **0** 1 **0** 1 --> after first one there are only two zeros that need to be converted into one to get the monotone increasing string.)\n3. When the number of zero count is greater than the one count we update the zero count to one count because if the number of one are less then we convert it into the zero.\n(**1** 0 0 1 1 1 --> in this case we convert the first one to zero instead of converting the all the zeros to ones.)\n4. At the end return the number of zero count. \n\n\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int i = 0, n = s.size();\n int zto = 0, oc = 0;\n while(i < n && s[i] == \'0\')\n i++;\n for (int k = i ;k < n ; k++){\n if (s[k] == \'0\')\n zto++;\n else\n oc++;\n if (zto > oc)\n zto = oc;\n }\n return zto;\n }\n};\n``` | 3 | 0 | ['C++'] | 0 |
flip-string-to-monotone-increasing | Easy Understanding Solution | JAVA | Beats 92% ✅✅ | easy-understanding-solution-java-beats-9-su06 | \n# Code\n\nclass Solution {\n public int minFlipsMonoIncr(String s) {\n\n //Initialization\n int flips=0,ones=0;\n\n //flipping zeros a | s147 | NORMAL | 2023-01-17T06:57:59.983782+00:00 | 2023-01-17T06:57:59.983833+00:00 | 78 | false | \n# Code\n```\nclass Solution {\n public int minFlipsMonoIncr(String s) {\n\n //Initialization\n int flips=0,ones=0;\n\n //flipping zeros and oes to get monotone\n for(int i=0;i<s.length();i++)\n {\n //checking character is \'0\'\n if(s.charAt(i)==\'0\')\n {\n if(ones==0)\n continue;\n flips++; //incrementing flips\n }\n\n //if character is not \'0\' increment ones\n else\n ones++;\n\n //if ones are lesser than flips, equalize them\n if(ones<flips)\n flips=ones;\n }\n\n //return the final count of flips\n return flips;\n }\n}\n```\n**Please UPVOTE if you like the Solution**\n\n | 3 | 0 | ['Java'] | 0 |
flip-string-to-monotone-increasing | ✅Accepted| | ✅Easy solution || ✅Short & Simple || ✅Best Method | accepted-easy-solution-short-simple-best-hqr8 | \n# Code\n\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int n = s.size();\n int one_counts = 0;\n int dp = 0;\n | sanjaydwk8 | NORMAL | 2023-01-17T06:17:06.912817+00:00 | 2023-01-17T06:17:06.912922+00:00 | 27 | false | \n# Code\n```\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int n = s.size();\n int one_counts = 0;\n int dp = 0;\n for(int i = 1; i <= n; i++) {\n if(s[i - 1] == \'1\')\n one_counts++;\n else\n dp = min(dp + 1, one_counts);\n }\n return dp;\n }\n};\n```\nPlease **UPVOTE** if it helps \u2764\uFE0F\uD83D\uDE0A\nThank You and Happy To Help You!! | 3 | 0 | ['C++'] | 0 |
flip-string-to-monotone-increasing | Easiest Approach || Dynamic Programming | easiest-approach-dynamic-programming-by-kgzty | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ankiit_k_ | NORMAL | 2023-01-17T05:15:05.670421+00:00 | 2023-01-17T05:15:05.670452+00:00 | 25 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int dp[100001][2];\n int minFlipsMonoIncr(string s) {\n int n=s.length();\n if(s[0]==\'0\')\n {\n dp[0][0]=0;\n dp[0][1]=1;\n }\n else\n {\n dp[0][0]=1;\n dp[0][1]=0;\n }\n for(int i=1;i<n;i++)\n {\n if(s[i]==\'0\')\n {\n dp[i][0]=dp[i-1][0];\n dp[i][1]=1+min(dp[i-1][0],dp[i-1][1]);\n }\n else\n {\n dp[i][1]=min(dp[i-1][1],dp[i-1][0]);\n dp[i][0]=1+dp[i-1][0];\n }\n }\n int ans=min(dp[n-1][0],dp[n-1][1]);\n return ans;\n }\n};\n``` | 3 | 0 | ['Dynamic Programming', 'C++'] | 1 |
flip-string-to-monotone-increasing | ✅ [Java/C++] 100% Solution using Dynamic Programming (Flip String to Monotone Increasing) | javac-100-solution-using-dynamic-program-9etq | Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space complexity here, e.g. O(n) \n\n# Co | arnavsharma2711 | NORMAL | 2023-01-17T05:11:01.508505+00:00 | 2023-01-17T05:11:01.508538+00:00 | 175 | false | # Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```Java []\nclass Solution {\n public int minFlipsMonoIncr(String s) {\n int countOne = 0, countFlip = 0; \n for(int i=0;i<s.length();i++)\n {\n if(s.charAt(i)==\'1\')\n ++countOne;\n else if(countOne>0)\n ++countFlip;\n countFlip = Math.min(countFlip,countOne);\n } \n return countFlip;\n }\n}\n```\n```C++ []\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int countOne = 0, countFlip = 0; \n for(int i=0;i<s.length();i++)\n {\n if(s[i]==\'1\')\n ++countOne;\n else if(countOne>0)\n ++countFlip;\n countFlip = min(countFlip,countOne);\n } \n \n return countFlip;\n }\n};\n```\nUpvote and Code By: [Arnav Sharma](https://www.linkedin.com/in/arnavsharma2711/) | 3 | 0 | ['String', 'Dynamic Programming', 'C++', 'Java'] | 0 |
flip-string-to-monotone-increasing | python, dynamic programming, 5 lines of code | python-dynamic-programming-5-lines-of-co-2y2o | ERROR: type should be string, got "https://leetcode.com/submissions/detail/879635531/ \\nRuntime: 187 ms, faster than 84.62% of Python3 online submissions for Flip String to Monotone Increasing. " | nov05 | NORMAL | 2023-01-17T03:55:35.229155+00:00 | 2023-01-24T17:02:06.582761+00:00 | 59 | false | ERROR: type should be string, got "https://leetcode.com/submissions/detail/879635531/ \\nRuntime: **187 ms**, faster than 84.62% of Python3 online submissions for Flip String to Monotone Increasing. \\nMemory Usage: 14.9 MB, less than 72.76% of Python3 online submissions for Flip String to Monotone Increasing. \\n```\\nclass Solution:\\n def minFlipsMonoIncr(self, s: str) -> int:\\n minFlips = n = s.count(\\'0\\') ## flip to all \\'1\\'s\\n for c in s: ## flip to one \"0\", two \"0\"s, three \"0\"s...\\n n += 1 if c==\\'1\\' else -1\\n minFlips = min(minFlips, n)\\n return minFlips\\n```\\n**A large test case** \\nhttps://leetcode.com/submissions/detail/879579954/testcase/" | 3 | 0 | ['Dynamic Programming', 'Python'] | 0 |
flip-string-to-monotone-increasing | Day 17 || O(N) Time and O(1) space || Easiest DP Logic Based Solution | day-17-on-time-and-o1-space-easiest-dp-l-6p94 | \nIf my solution is helpful to you then please upvote me.\n# Intuition\n1. If s[i - 1] = \'1\', then we have dp[i] = dp[i - 1], since we can always append a cha | singhabhinash | NORMAL | 2023-01-17T02:50:18.625324+00:00 | 2023-01-17T04:24:32.374992+00:00 | 222 | false | \n**If my solution is helpful to you then please upvote me.**\n# Intuition\n1. If s[i - 1] = \'1\', then we have dp[i] = dp[i - 1], since we can always append a character \'1\' to the end of a monotone increasing string and it\'s still monotone increasing.\n2. If s[i - 1] = \'0\', let\'s consider whether we flip it or not.\n - If we don\'t flip it, we have to flip all the \'1\'s in s before it.\n - If we flip it, then we can treat it as the above case where s[i - 1] = \'1\' with one more flip.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Initialize two variables: countOnes, which is the number of 1\'s in the string until the current index, and countFlip, which is the number of flips required to make the string monotonically increasing until the current index.\n2. Iterate through the characters in the input string.\n3. For each character, check if it is equal to \'1\'.\n4. If it is, increment the countOnes variable.\n5. If it is not, increment the countFlip variable by 1 and update it to the minimum of itself and the current value of countOnes.\n6. Return the final value of countFlip.\n<!-- Describe your approach to solving the problem. -->\n\n# Code\n```C++ []\nclass Solution {\npublic:\n int minFlipsMonoIncr(string s) {\n int countOnes = 0; // where countOnes is the number of 1\'s till length i\n int countFlip = 0; // where countFlip is the number of flips require to make string monotonic incresing till lenght i;\n for(char ch : s){\n if(ch == \'1\')\n countOnes++;\n else{\n countFlip = min(countFlip + 1, countOnes);\n }\n }\n return countFlip;\n }\n};\n```\n```Java []\nclass Solution {\n public int minFlipsMonoIncr(String s) {\n int countOnes = 0;\n int countFlip = 0;\n for(char ch : s.toCharArray()){\n if(ch == \'1\')\n countOnes++;\n else{\n countFlip = Math.min(countFlip + 1, countOnes);\n }\n }\n return countFlip;\n }\n}\n\n```\n```Python []\nclass Solution:\n def minFlipsMonoIncr(self, s: str) -> int:\n countOnes = 0\n countFlip = 0\n for ch in s:\n if ch == \'1\':\n countOnes += 1\n else:\n countFlip = min(countFlip + 1, countOnes)\n return countFlip\n\n```\n# Complexity\n- Time complexity: **O(n)** // where n is the length of the string\n\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: **O(1)** // bcz no extra space used\n<!-- Add your space complexity here, e.g. $$O(n)$$ --> | 3 | 0 | ['Dynamic Programming', 'C++'] | 1 |
flip-string-to-monotone-increasing | 𝗝𝗮𝘃𝗮𝗦𝗰𝗿𝗶𝗽𝘁 | 𝗦𝗶𝗺𝗽𝗹𝗲 𝗩𝗶𝗱𝗲𝗼 𝗦𝗼𝗹𝘂𝘁𝗶𝗼𝗻 | DP | javascript-simple-video-solution-dp-by-m-mh04 | Intuition & Approach\n Describe your approach to solving the problem. \n\uD835\uDDD7\uD835\uDDF2\uD835\uDE01\uD835\uDDEE\uD835\uDDF6\uD835\uDDF9\uD835\uDDF2\uD | monuchaudhary1 | NORMAL | 2023-01-17T02:44:30.087665+00:00 | 2023-01-17T02:44:30.087711+00:00 | 418 | false | # Intuition & Approach\n<!-- Describe your approach to solving the problem. -->\n\uD835\uDDD7\uD835\uDDF2\uD835\uDE01\uD835\uDDEE\uD835\uDDF6\uD835\uDDF9\uD835\uDDF2\uD835\uDDF1 \uD835\uDDD4\uD835\uDDFD\uD835\uDDFD\uD835\uDDFF\uD835\uDDFC\uD835\uDDEE\uD835\uDDF0\uD835\uDDF5 \uD835\uDDD8\uD835\uDE05\uD835\uDDFD\uD835\uDDF9\uD835\uDDEE\uD835\uDDFB\uD835\uDDEE\uD835\uDE01\uD835\uDDF6\uD835\uDDFC\uD835\uDDFB\n\uD835\uDDD6\uD835\uDDFC\uD835\uDDFA\uD835\uDDFA\uD835\uDDF2\uD835\uDDFB\uD835\uDE01\uD835\uDE00 \uD835\uDDEE\uD835\uDDFB\uD835\uDDF1 \uD835\uDDF3\uD835\uDDF2\uD835\uDDF2\uD835\uDDF1\uD835\uDDEF\uD835\uDDEE\uD835\uDDF0\uD835\uDDF8 \uD835\uDDEE\uD835\uDDFF\uD835\uDDF2 \uD835\uDDEE\uD835\uDDFD\uD835\uDDFD\uD835\uDDFF\uD835\uDDF2\uD835\uDDF0\uD835\uDDF6\uD835\uDDEE\uD835\uDE01\uD835\uDDF2\uD835\uDDF1\nhttps://youtu.be/6MY8XYH1mrY\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nvar minFlipsMonoIncr = function(s) {\n let res = 0\n let n = 0\n for (let c of s) {\n if (c == \'0\') {\n res = Math.min(n, res + 1)\n }\n else n++\n }\n return res\n};\n``` | 3 | 0 | ['Dynamic Programming', 'JavaScript'] | 2 |
flip-string-to-monotone-increasing | Easy to understand - Intuitive DP using recursion | easy-to-understand-intuitive-dp-using-re-s8pn | https://github.com/Freeze777/SDE-Interviewer-Notes/blob/main/LeetCodeKotlin/src/main/kotlin/leetcode/medium/dp/FlipStringMonotoneIncrease.kt | freeze_francis | NORMAL | 2023-01-17T01:29:34.291240+00:00 | 2023-01-17T01:29:34.291281+00:00 | 611 | false | https://github.com/Freeze777/SDE-Interviewer-Notes/blob/main/LeetCodeKotlin/src/main/kotlin/leetcode/medium/dp/FlipStringMonotoneIncrease.kt | 3 | 0 | ['Kotlin'] | 0 |
flip-string-to-monotone-increasing | Python3 || 79 ms, faster than 100.00% of Python3 || Clean and Easy to Understand | python3-79-ms-faster-than-10000-of-pytho-e1zy | \ndef minFlipsMonoIncr(self, s: str) -> int:\n result = 0\n ones = 0\n for num in s:\n if num ==\'1\':\n ones += | harshithdshetty | NORMAL | 2023-01-17T01:22:14.168981+00:00 | 2023-01-17T01:22:14.169028+00:00 | 195 | false | ```\ndef minFlipsMonoIncr(self, s: str) -> int:\n result = 0\n ones = 0\n for num in s:\n if num ==\'1\':\n ones += 1\n else:\n result += 1\n if result > ones:\n result = ones \n return result\n``` | 3 | 1 | ['Dynamic Programming', 'Prefix Sum', 'Python', 'Python3'] | 0 |
roman-to-integer | ✅Best Method || C++ || JAVA || PYTHON || Beginner Friendly🔥🔥🔥 | best-method-c-java-python-beginner-frien-pmoe | Certainly! Let\'s break down the code and provide a clear intuition and explanation, using the examples "IX" and "XI" to demonstrate its functionality.\n\n# Int | rahulvarma5297 | NORMAL | 2023-06-18T08:25:56.677662+00:00 | 2023-06-26T09:36:55.834340+00:00 | 566,677 | false | **Certainly! Let\'s break down the code and provide a clear intuition and explanation, using the examples "IX" and "XI" to demonstrate its functionality.**\n\n# Intuition:\nThe key intuition lies in the fact that in Roman numerals, when a smaller value appears before a larger value, it represents subtraction, while when a smaller value appears after or equal to a larger value, it represents addition.\n\n# Explanation:\n\n1. The unordered map `m` is created and initialized with mappings between Roman numeral characters and their corresponding integer values. For example, \'I\' is mapped to 1, \'V\' to 5, \'X\' to 10, and so on.\n2. The variable `ans` is initialized to 0. This variable will accumulate the final integer value of the Roman numeral string.\n3. The for loop iterates over each character in the input string `s`.\n **For the example "IX":**\n \n - When `i` is 0, the current character `s[i]` is \'I\'. Since there is a next character (\'X\'), and the value of \'I\' (1) is less than the value of \'X\' (10), the condition `m[s[i]] < m[s[i+1]]` is true. In this case, we subtract the value of the current character from `ans`.\n \n `ans -= m[s[i]];`\n `ans -= m[\'I\'];`\n `ans -= 1;`\n `ans` becomes -1.\n \n - When `i` is 1, the current character `s[i]` is \'X\'. This is the last character in the string, so there is no next character to compare. Since there is no next character, we don\'t need to evaluate the condition. In this case, we add the value of the current character to `ans`.\n \n `ans += m[s[i]];`\n `ans += m[\'X\'];`\n `ans += 10;`\n `ans` becomes 9.\n \n **For the example "XI":**\n \n - When `i` is 0, the current character `s[i]` is \'X\'. Since there is a next character (\'I\'), and the value of \'X\' (10) is greater than the value of \'I\' (1), the condition `m[s[i]] < m[s[i+1]]` is false. In this case, we add the value of the current character to `ans`.\n \n `ans += m[s[i]];`\n `ans += m[\'X\'];`\n `ans += 10;`\n `ans` becomes 10.\n \n - When `i` is 1, the current character `s[i]` is \'I\'. This is the last character in the string, so there is no next character to compare. Since there is no next character, we don\'t need to evaluate the condition. In this case, we add the value of the current character to `ans`.\n \n `ans += m[s[i]];`\n `ans += m[\'I\'];`\n `ans += 1;`\n `ans` becomes 11.\n\n4. After the for loop, the accumulated value in `ans` represents the integer conversion of the Roman numeral string, and it is returned as the result.\n\n# Code\n```C++ []\nclass Solution {\npublic:\n int romanToInt(string s) {\n unordered_map<char, int> m;\n \n m[\'I\'] = 1;\n m[\'V\'] = 5;\n m[\'X\'] = 10;\n m[\'L\'] = 50;\n m[\'C\'] = 100;\n m[\'D\'] = 500;\n m[\'M\'] = 1000;\n \n int ans = 0;\n \n for(int i = 0; i < s.length(); i++){\n if(m[s[i]] < m[s[i+1]]){\n ans -= m[s[i]];\n }\n else{\n ans += m[s[i]];\n }\n }\n return ans;\n }\n};\n```\n```Java []\nclass Solution {\n public int romanToInt(String s) {\n Map<Character, Integer> m = new HashMap<>();\n \n m.put(\'I\', 1);\n m.put(\'V\', 5);\n m.put(\'X\', 10);\n m.put(\'L\', 50);\n m.put(\'C\', 100);\n m.put(\'D\', 500);\n m.put(\'M\', 1000);\n \n int ans = 0;\n \n for (int i = 0; i < s.length(); i++) {\n if (i < s.length() - 1 && m.get(s.charAt(i)) < m.get(s.charAt(i + 1))) {\n ans -= m.get(s.charAt(i));\n } else {\n ans += m.get(s.charAt(i));\n }\n }\n \n return ans;\n }\n}\n\n```\n```Python3 []\nclass Solution:\n def romanToInt(self, s: str) -> int:\n m = {\n \'I\': 1,\n \'V\': 5,\n \'X\': 10,\n \'L\': 50,\n \'C\': 100,\n \'D\': 500,\n \'M\': 1000\n }\n \n ans = 0\n \n for i in range(len(s)):\n if i < len(s) - 1 and m[s[i]] < m[s[i+1]]:\n ans -= m[s[i]]\n else:\n ans += m[s[i]]\n \n return ans\n```\n\n\n**If you are a beginner solve these problems which makes concepts clear for future coding:**\n1. [Two Sum](https://leetcode.com/problems/two-sum/solutions/3619262/3-method-s-c-java-python-beginner-friendly/)\n2. [Roman to Integer](https://leetcode.com/problems/roman-to-integer/solutions/3651672/best-method-c-java-python-beginner-friendly/)\n3. [Palindrome Number](https://leetcode.com/problems/palindrome-number/solutions/3651712/2-method-s-c-java-python-beginner-friendly/)\n4. [Maximum Subarray](https://leetcode.com/problems/maximum-subarray/solutions/3666304/beats-100-c-java-python-beginner-friendly/)\n5. [Remove Element](https://leetcode.com/problems/remove-element/solutions/3670940/best-100-c-java-python-beginner-friendly/)\n6. [Contains Duplicate](https://leetcode.com/problems/contains-duplicate/solutions/3672475/4-method-s-c-java-python-beginner-friendly/)\n7. [Add Two Numbers](https://leetcode.com/problems/add-two-numbers/solutions/3675747/beats-100-c-java-python-beginner-friendly/)\n8. [Majority Element](https://leetcode.com/problems/majority-element/solutions/3676530/3-methods-beats-100-c-java-python-beginner-friendly/)\n9. [Remove Duplicates from Sorted Array](https://leetcode.com/problems/remove-duplicates-from-sorted-array/solutions/3676877/best-method-100-c-java-python-beginner-friendly/)\n10. **Practice them in a row for better understanding and please Upvote for more questions.**\n\n\n**If you found my solution helpful, I would greatly appreciate your upvote, as it would motivate me to continue sharing more solutions.**\n\n | 3,188 | 9 | ['Hash Table', 'Math', 'C++', 'Java', 'Python3'] | 122 |
roman-to-integer | Clean Python, beats 99.78%. | clean-python-beats-9978-by-hgrsd-axkt | The Romans would most likely be angered by how it butchers their numeric system. Sorry guys.\n\nPython\nclass Solution:\n def romanToInt(self, s: str) -> int | hgrsd | NORMAL | 2019-03-29T20:58:38.986386+00:00 | 2019-03-29T20:58:38.986426+00:00 | 242,494 | false | The Romans would most likely be angered by how it butchers their numeric system. Sorry guys.\n\n```Python\nclass Solution:\n def romanToInt(self, s: str) -> int:\n translations = {\n "I": 1,\n "V": 5,\n "X": 10,\n "L": 50,\n "C": 100,\n "D": 500,\n "M": 1000\n }\n number = 0\n s = s.replace("IV", "IIII").replace("IX", "VIIII")\n s = s.replace("XL", "XXXX").replace("XC", "LXXXX")\n s = s.replace("CD", "CCCC").replace("CM", "DCCCC")\n for char in s:\n number += translations[char]\n return number\n``` | 2,330 | 10 | ['Python', 'Python3'] | 291 |
roman-to-integer | java 90% faster solution | java-90-faster-solution-by-sd98754-re1o | \n public int romanToInt(String s) {\n int ans = 0, num = 0;\n for (int i = s.length()-1; i >= 0; i--) {\n switch(s.charAt(i)) {\n | sd98754 | NORMAL | 2022-09-27T22:03:31.255244+00:00 | 2022-09-27T22:03:31.255283+00:00 | 270,451 | false | ```\n public int romanToInt(String s) {\n int ans = 0, num = 0;\n for (int i = s.length()-1; i >= 0; i--) {\n switch(s.charAt(i)) {\n case \'I\': num = 1; break;\n case \'V\': num = 5; break;\n case \'X\': num = 10; break;\n case \'L\': num = 50; break;\n case \'C\': num = 100; break;\n case \'D\': num = 500; break;\n case \'M\': num = 1000; break;\n }\n if (4 * num < ans) ans -= num;\n else ans += num;\n }\n return ans;\n }\n```\n\nplease upvote if you liked the solution.\n | 1,181 | 1 | ['Java'] | 76 |
roman-to-integer | My Straightforward Python Solution | my-straightforward-python-solution-by-we-qxzz | \n class Solution:\n # @param {string} s\n # @return {integer}\n def romanToInt(self, s):\n roman = {'M': 1000,'D': 500 ,'C': 100,'L': 50,'X' | wenfengqiu | NORMAL | 2015-06-25T20:20:14+00:00 | 2018-10-26T00:54:16.503090+00:00 | 109,705 | false | \n class Solution:\n # @param {string} s\n # @return {integer}\n def romanToInt(self, s):\n roman = {'M': 1000,'D': 500 ,'C': 100,'L': 50,'X': 10,'V': 5,'I': 1}\n z = 0\n for i in range(0, len(s) - 1):\n if roman[s[i]] < roman[s[i+1]]:\n z -= roman[s[i]]\n else:\n z += roman[s[i]]\n return z + roman[s[-1]]\n\n\n*Note: The trick is that the last letter is always added. Except the last one, if one letter is less than its latter one, this letter is subtracted. | 723 | 5 | [] | 49 |
roman-to-integer | 【Video】Looping two characters at a time. | video-looping-two-characters-at-a-time-b-squ4 | Intuition\nLooping two characters at a time.\n\n# Solution Video\n\nhttps://youtu.be/yUYp0LlVXH4\n\n### \u2B50\uFE0F\u2B50\uFE0F Don\'t forget to subscribe to m | niits | NORMAL | 2024-09-29T15:35:29.663498+00:00 | 2024-11-29T13:11:19.369139+00:00 | 78,729 | false | # Intuition\nLooping two characters at a time.\n\n# Solution Video\n\nhttps://youtu.be/yUYp0LlVXH4\n\n### \u2B50\uFE0F\u2B50\uFE0F Don\'t forget to subscribe to my channel! \u2B50\uFE0F\u2B50\uFE0F\n\n**\u25A0 Subscribe URL**\nhttp://www.youtube.com/channel/UC9RMNwYTL3SXCP6ShLWVFww?sub_confirmation=1\n\nSubscribers: 9,135\nThank you for your support!\n\n---\n\n# Approach\n\nThe description includes the example of `27`. **If the values are written from left to right in descending order, simply adding each number one by one will give the result.**\n\n```\nXXVII\n\u2193\n10(X) + 10(X) + 5(V) + 1(I) + 1(I) = 27\n```\n```\nLVIII\n\u2193\n50(L) + 5(V) + 1(I) + 1(I) + 1(I) = 58\n```\nBut numbers like `4` and `9` are represented as a single number **using two Roman numerals**.\n\n```\nIV = 4\nIX = 9\n```\n---\n\n\u2B50\uFE0F Points\n\nBasically, you just need to convert either one or two characters into a number to get the answer.\n\n---\n\n##### How do we determine if it\u2019s one or two characters?\n\nSince a single number can use up to two Roman characters, **we loop through the string two characters at a time, shifting by one each time**.\n\nLet\u2019s take a look at this example.\n```\nInput: s = "XIV"\n```\nIn the first loop, we use `X` and `I`. In this case, the second character `I` is less than `X`. `I` is `1` and `X` is `10`. That means we can simply add `X` to result variable, **because `10 \u2192 1` is a descending order**. It\'s the same as `27` or `58` above.\n\n```\nres = 10\n```\nNext, we use `I` and `V`. In this case, the second character `V` is greater than `I`, because `V` is `5` and `I` is `1`.\n\nThe description says "there are six instances where subtraction is used."\n\n```\nIV = 4\nIX = 9\nXL = 40\nXC = 90\nCD = 400\nCM = 900\n```\nLook at all the first characters and the second characters. The first characters are less than the second characters.\n\n```\nI(1) + V(5) = 4\nI(1) + X(10) = 9\nX(10) + L(50) = 40\nX(10) + C(100) = 90\nC(100) + D(500) = 400\nC(100) + M(1000) = 900\n```\nThe current first character is `I` and the second character is `V`, so we are using two Roman characters to form a single number.\n\nIn this case, we subtract `I`(= `1`) from result variable, because in the next loop, we will add `V`(= `5`), so the total of `I` and `V` will be `4` in the end.\n\n```\nres = 9\n```\nThe next two number is `V` and nothing(= out of bounds), so we stop iteration.\n\nThe last point is that as I told you, we should add `V` before return the answer, because one of values was out of bounds, that\'s why we couldn\'t add the last Roman chartacter which is `V`.\n\n```\nreturn res + 5\n```\nBefore we start looping, we have `HashMap` to combine Roman and Integer, so that we can convert Roman to Integer easily.\n\nIn the solution code,\n```\nreturn res + roman[s[-1]]\n\ns is input string\nroman is HashMap\n```\n\n---\n\nhttps://youtu.be/bU_dXCOWHls\n\n# Complexity\n- Time complexity: $$O(n)$$\n`n` is the length of the input string.\n\n- Space complexity: $$O(1)$$\nThe dictionary `roman` always has a fixed size of seven key-value pairs, regardless of the input size, so it uses constant space.\n\n# Code\n```python []\nclass Solution:\n def romanToInt(self, s: str) -> int:\n res = 0\n roman = {\n \'I\': 1,\n \'V\': 5,\n \'X\': 10,\n \'L\': 50,\n \'C\': 100,\n \'D\': 500,\n \'M\': 1000\n }\n\n for a, b in zip(s, s[1:]):\n if roman[a] < roman[b]:\n res -= roman[a]\n else:\n res += roman[a]\n\n return res + roman[s[-1]] \n```\n```javascript []\nvar romanToInt = function(s) {\n let res = 0;\n const roman = {\n \'I\': 1,\n \'V\': 5,\n \'X\': 10,\n \'L\': 50,\n \'C\': 100,\n \'D\': 500,\n \'M\': 1000\n };\n\n for (let i = 0; i < s.length - 1; i++) {\n if (roman[s[i]] < roman[s[i + 1]]) {\n res -= roman[s[i]];\n } else {\n res += roman[s[i]];\n }\n }\n\n return res + roman[s[s.length - 1]]; \n};\n```\n```java []\nclass Solution {\n public int romanToInt(String s) {\n int res = 0;\n Map<Character, Integer> roman = new HashMap<>();\n roman.put(\'I\', 1);\n roman.put(\'V\', 5);\n roman.put(\'X\', 10);\n roman.put(\'L\', 50);\n roman.put(\'C\', 100);\n roman.put(\'D\', 500);\n roman.put(\'M\', 1000);\n\n for (int i = 0; i < s.length() - 1; i++) {\n if (roman.get(s.charAt(i)) < roman.get(s.charAt(i + 1))) {\n res -= roman.get(s.charAt(i));\n } else {\n res += roman.get(s.charAt(i));\n }\n }\n\n return res + roman.get(s.charAt(s.length() - 1)); \n }\n}\n```\n```C++ []\nclass Solution {\npublic:\n int romanToInt(string s) {\n int res = 0;\n unordered_map<char, int> roman = {\n {\'I\', 1},\n {\'V\', 5},\n {\'X\', 10}, \n {\'L\', 50},\n {\'C\', 100},\n {\'D\', 500},\n {\'M\', 1000}\n };\n\n for (int i = 0; i < s.size() - 1; i++) {\n if (roman[s[i]] < roman[s[i + 1]]) {\n res -= roman[s[i]];\n } else {\n res += roman[s[i]];\n }\n }\n\n return res + roman[s[s.size() - 1]]; \n }\n};\n```\n\n---\n\nThank you for reading my post. Please upvote it and don\'t forget to subscribe to my channel!\n\n### \u2B50\uFE0F Subscribe URL\nhttp://www.youtube.com/channel/UC9RMNwYTL3SXCP6ShLWVFww?sub_confirmation=1\n\n### \u2B50\uFE0F Twitter\nhttps://twitter.com/CodingNinjaAZ\n\n### \u2B50\uFE0F Related question \n#12 - Integer To Roman\n\nhttps://youtu.be/SFDETXM6-6Y\n | 702 | 0 | ['Hash Table', 'String', 'C++', 'Java', 'Python3', 'JavaScript'] | 14 |
roman-to-integer | Clean O(n) c++ solution | clean-on-c-solution-by-wsugrad77-53hk | Problem is simpler to solve by working the string from back to front and using a map. Runtime speed is 88 ms.\n\n\n\n int romanToInt(string s) \n {\n | wsugrad77 | NORMAL | 2015-01-23T03:32:10+00:00 | 2018-10-22T13:36:56.935703+00:00 | 136,002 | false | Problem is simpler to solve by working the string from back to front and using a map. Runtime speed is 88 ms.\n\n\n\n int romanToInt(string s) \n {\n unordered_map<char, int> T = { { 'I' , 1 },\n { 'V' , 5 },\n { 'X' , 10 },\n { 'L' , 50 },\n { 'C' , 100 },\n { 'D' , 500 },\n { 'M' , 1000 } };\n \n int sum = T[s.back()];\n for (int i = s.length() - 2; i >= 0; --i) \n {\n if (T[s[i]] < T[s[i + 1]])\n {\n sum -= T[s[i]];\n }\n else\n {\n sum += T[s[i]];\n }\n }\n \n return sum;\n } | 677 | 7 | [] | 68 |
roman-to-integer | Easy C++ solution with short code | easy-c-solution-with-short-code-by-jeeta-nro5 | \n# Approach\nIf there are Numbers such as XL, IV, etc. \nSimply subtract the smaller number and add the larger number in next step. \nFor example, if there is | Jeetaksh | NORMAL | 2023-01-14T17:26:29.362602+00:00 | 2023-01-15T06:50:37.303305+00:00 | 119,430 | false | \n# Approach\nIf there are Numbers such as XL, IV, etc. \nSimply subtract the smaller number and add the larger number in next step. \nFor example, if there is XL, ans =-10 in first step, ans=-10+50=40 in next step.\n\nOtherwise, just add the numbers. \n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n\n\n# Code\n```\nclass Solution {\npublic:\n int romanToInt(string s) {\n int ans=0;\n unordered_map <char,int> mp{\n {\'I\',1},{\'V\',5},{\'X\',10},{\'L\',50},{\'C\',100},{\'D\',500},{\'M\',1000}};\n\n for(int i=0;i<s.size();i++){\n if(mp[s[i]]<mp[s[i+1]]){\n //for cases such as IV,CM, XL, etc...\n ans=ans-mp[s[i]];\n }\n else{\n ans=ans+mp[s[i]];\n }\n }\n return ans;\n }\n};\n```\n**Please upvote if it helped. Happy Coding!** | 447 | 1 | ['C++'] | 18 |
roman-to-integer | 7ms solution in Java. easy to understand | 7ms-solution-in-java-easy-to-understand-zul35 | public int romanToInt(String s) {\n int nums[]=new int[s.length()];\n for(int i=0;i<s.length();i++){\n switch (s.charAt(i)){\n | yangneu2015 | NORMAL | 2015-10-31T00:34:12+00:00 | 2018-10-21T22:33:37.060446+00:00 | 106,144 | false | public int romanToInt(String s) {\n int nums[]=new int[s.length()];\n for(int i=0;i<s.length();i++){\n switch (s.charAt(i)){\n case 'M':\n nums[i]=1000;\n break;\n case 'D':\n nums[i]=500;\n break;\n case 'C':\n nums[i]=100;\n break;\n case 'L':\n nums[i]=50;\n break;\n case 'X' :\n nums[i]=10;\n break;\n case 'V':\n nums[i]=5;\n break;\n case 'I':\n nums[i]=1;\n break;\n }\n }\n int sum=0;\n for(int i=0;i<nums.length-1;i++){\n if(nums[i]<nums[i+1])\n sum-=nums[i];\n else\n sum+=nums[i];\n }\n return sum+nums[nums.length-1];\n } | 425 | 2 | [] | 56 |
roman-to-integer | 🔥 Python || Easily Understood ✅ || Faster than 98% || Less than 76% || O(n) | python-easily-understood-faster-than-98-gcdnn | Code:\n\nclass Solution:\n def romanToInt(self, s: str) -> int:\n roman_to_integer = {\n \'I\': 1,\n \'V\': 5,\n \'X\ | wingskh | NORMAL | 2022-08-15T10:51:54.609226+00:00 | 2022-08-15T10:51:54.609270+00:00 | 130,440 | false | Code:\n```\nclass Solution:\n def romanToInt(self, s: str) -> int:\n roman_to_integer = {\n \'I\': 1,\n \'V\': 5,\n \'X\': 10,\n \'L\': 50,\n \'C\': 100,\n \'D\': 500,\n \'M\': 1000,\n }\n s = s.replace("IV", "IIII").replace("IX", "VIIII").replace("XL", "XXXX").replace("XC", "LXXXX").replace("CD", "CCCC").replace("CM", "DCCCC")\n return sum(map(lambda x: roman_to_integer[x], s))\n```\n**Time Complexity**: `O(n)`\n**Space Complexity**: `O(1)`\n<br/>\n | 412 | 1 | ['Python', 'Python3'] | 54 |
roman-to-integer | My solution for this question but I don't know is there any easier way? | my-solution-for-this-question-but-i-dont-s7ic | count every Symbol and add its value to the sum, and minus the extra part of special cases. \n\n public int romanToInt(String s) {\n int sum=0;\n | hongbin2 | NORMAL | 2014-01-26T05:53:45+00:00 | 2018-10-24T18:26:52.487861+00:00 | 173,255 | false | count every Symbol and add its value to the sum, and minus the extra part of special cases. \n\n public int romanToInt(String s) {\n int sum=0;\n if(s.indexOf("IV")!=-1){sum-=2;}\n if(s.indexOf("IX")!=-1){sum-=2;}\n if(s.indexOf("XL")!=-1){sum-=20;}\n if(s.indexOf("XC")!=-1){sum-=20;}\n if(s.indexOf("CD")!=-1){sum-=200;}\n if(s.indexOf("CM")!=-1){sum-=200;}\n \n char c[]=s.toCharArray();\n int count=0;\n \n for(;count<=s.length()-1;count++){\n if(c[count]=='M') sum+=1000;\n if(c[count]=='D') sum+=500;\n if(c[count]=='C') sum+=100;\n if(c[count]=='L') sum+=50;\n if(c[count]=='X') sum+=10;\n if(c[count]=='V') sum+=5;\n if(c[count]=='I') sum+=1;\n \n }\n \n return sum;\n \n } | 410 | 20 | [] | 111 |
roman-to-integer | Hash Table Concept Python3 | hash-table-concept-python3-by-ganjinavee-uvmm | \n# Hash Table in python\n\nclass Solution:\n def romanToInt(self, s: str) -> int:\n roman={"I":1,"V":5,"X":10,"L":50,"C":100,"D":500,"M":1000}\n | GANJINAVEEN | NORMAL | 2023-04-16T03:53:56.762583+00:00 | 2023-04-16T03:53:56.762632+00:00 | 79,928 | false | \n# Hash Table in python\n```\nclass Solution:\n def romanToInt(self, s: str) -> int:\n roman={"I":1,"V":5,"X":10,"L":50,"C":100,"D":500,"M":1000}\n number=0\n for i in range(len(s)-1):\n if roman[s[i]] < roman[s[(i+1)]]:\n number-=roman[s[i]]\n else:\n number+=roman[s[i]]\n return number+roman[s[-1]]\n\n\n```\n# please upvote me it would encourage me alot\n | 403 | 0 | ['Python3'] | 26 |
roman-to-integer | [C++] ✅ || O(n) || Short Code 🔥 || Clean code || fast and easy | c-on-short-code-clean-code-fast-and-easy-ouyj | Intuitive\nRoman numerals are usually written largest to smallest from left to right, for example: XII (7), XXVII (27), III (3)...\nIf a small value is placed b | hoshang2900 | NORMAL | 2022-08-15T00:04:28.367892+00:00 | 2023-03-12T09:37:32.486373+00:00 | 65,394 | false | **Intuitive**\nRoman numerals are usually written largest to smallest from left to right, for example: XII (7), XXVII (27), III (3)...\nIf a small value is placed before a bigger value then it\'s a combine number, for exampe: IV (4), IX (9), XIV (14)...\nIV = -1 + 5\nVI = 5 + 1\nXL = -10 + 50\nLX = 50 + 10\nSo, if a smaller number appears before a larger number, it indicates that the number is smaller by a quantity used as a suffix to it, which made going backwards seem easy.\n\n\t\n\t\n\t\n\tclass Solution {\n\tpublic:\n int romanToInt(string s) {\n unordered_map<char,int> mp{\n {\'I\',1},\n {\'V\',5},\n {\'X\',10},\n {\'L\',50},\n {\'C\',100},\n {\'D\',500},\n {\'M\',1000},\n };\n int ans =0;\n for(int i=0;i<s.size();i++){\n if(mp[s[i]]<mp[s[i+1]])\n ans-=mp[s[i]];\n else\n ans+=mp[s[i]];\n }\n return ans;\n \n }\n\t};\n\t\n**Please upvote to motivate me in my quest of documenting all leetcode solutions(to help the community). \nHAPPY CODING:)**\n\n*Any suggestions and improvements are always welcome*\n\n\t | 370 | 5 | ['C'] | 26 |
roman-to-integer | JS | Hash Table | With exlanation | js-hash-table-with-exlanation-by-karina_-16lt | To solve this problem, we need to create a hash table, the characters in which will correspond to a certain number. Passing along the line, we will check the cu | Karina_Olenina | NORMAL | 2022-10-15T17:05:42.231625+00:00 | 2022-10-20T12:38:19.879082+00:00 | 54,381 | false | To solve this problem, we need to create a hash table, the characters in which will correspond to a certain number. Passing along the line, we will check the current and the next character at once, if the current one is greater than the next one, then everything is fine, we add it to the result (it is initially equal to 0), otherwise, we subtract the current value from the next value (for example, the current 10, and the next 100 , we do 100 - 10, and we get 90), we also add to the result and additionally increase the index **by 1**. As a result, at the end of the loop, we will get the result we need.\n\nI hope the picture below will help you understand in more detail!)\n\n\n\n```\nvar romanToInt = function(s) {\n const sym = {\n \'I\': 1,\n \'V\': 5,\n \'X\': 10,\n \'L\': 50,\n \'C\': 100,\n \'D\': 500,\n \'M\': 1000\n }\n\n let result = 0;\n\n for (let i = 0; i < s.length; i++) {\n const cur = sym[s[i]];\n const next = sym[s[i + 1]];\n\n if (cur < next) {\n result += next - cur;\n i++;\n } else {\n result += cur;\n }\n }\n\n return result;\n};\n```\n\nI hope I was able to explain clearly.\n**Happy coding!** \uD83D\uDE43\n | 347 | 0 | ['Hash Table', 'Ordered Set', 'JavaScript'] | 60 |
roman-to-integer | Simple JavaScript Solution Easy Understanding | simple-javascript-solution-easy-understa-uxop | \nsymbols = {\n "I": 1,\n "V": 5,\n "X": 10,\n "L": 50,\n "C": 100,\n "D": 500,\n "M": 1000\n};\n\nvar romanToInt = function(s) {\n valu | garyguan0713 | NORMAL | 2019-07-03T17:49:00.961763+00:00 | 2019-07-03T17:49:00.961804+00:00 | 39,622 | false | ```\nsymbols = {\n "I": 1,\n "V": 5,\n "X": 10,\n "L": 50,\n "C": 100,\n "D": 500,\n "M": 1000\n};\n\nvar romanToInt = function(s) {\n value = 0;\n for(let i = 0; i < s.length; i+=1){\n symbols[s[i]] < symbols[s[i+1]] ? value -= symbols[s[i]]: value += symbols[s[i]]\n }\n return value\n};\n``` | 338 | 1 | ['JavaScript'] | 30 |
roman-to-integer | 4 lines in Python | 4-lines-in-python-by-agave-dywz | d = {'M':1000, 'D':500, 'C':100, 'L':50, 'X':10, 'V':5, 'I':1}\n \n def romanToInt(self, s):\n res, p = 0, 'I'\n for c in s[::-1]:\n | agave | NORMAL | 2016-06-04T03:44:43+00:00 | 2018-10-06T23:14:27.296354+00:00 | 33,490 | false | d = {'M':1000, 'D':500, 'C':100, 'L':50, 'X':10, 'V':5, 'I':1}\n \n def romanToInt(self, s):\n res, p = 0, 'I'\n for c in s[::-1]:\n res, p = res - d[c] if d[c] < d[p] else res + d[c], c\n return res | 195 | 2 | [] | 31 |
roman-to-integer | JAVA----------------Easy Version To Understand!!!! | java-easy-version-to-understand-by-hello-ak0n | \tpublic static int romanToInt(String s) {\n\t\tif (s == null || s.length() == 0)\n\t\t\treturn -1;\n\t\tHashMap<Character, Integer> map = new HashMap<Character | helloworld123456 | NORMAL | 2016-01-01T02:40:23+00:00 | 2018-10-14T20:29:16.666415+00:00 | 34,238 | false | \tpublic static int romanToInt(String s) {\n\t\tif (s == null || s.length() == 0)\n\t\t\treturn -1;\n\t\tHashMap<Character, Integer> map = new HashMap<Character, Integer>();\n\t\tmap.put('I', 1);\n\t\tmap.put('V', 5);\n\t\tmap.put('X', 10);\n\t\tmap.put('L', 50);\n\t\tmap.put('C', 100);\n\t\tmap.put('D', 500);\n\t\tmap.put('M', 1000);\n\t\tint len = s.length(), result = map.get(s.charAt(len - 1));\n\t\tfor (int i = len - 2; i >= 0; i--) {\n\t\t\tif (map.get(s.charAt(i)) >= map.get(s.charAt(i + 1)))\n\t\t\t\tresult += map.get(s.charAt(i));\n\t\t\telse\n\t\t\t\tresult -= map.get(s.charAt(i));\n\t\t}\n\t\treturn result;\n\t} | 184 | 1 | [] | 24 |
roman-to-integer | Python Beginner [98% fast 100% Memo] | python-beginner-98-fast-100-memo-by-dxbk-slag | \ndef romanToInt(self, s: str) -> int:\n\tres, prev = 0, 0\n\tdict = {\'I\':1, \'V\':5, \'X\':10, \'L\':50, \'C\':100, \'D\':500, \'M\':1000}\n\tfor i in s[::-1 | dxbking | NORMAL | 2019-12-13T06:03:44.512362+00:00 | 2019-12-13T06:03:44.512414+00:00 | 19,354 | false | ```\ndef romanToInt(self, s: str) -> int:\n\tres, prev = 0, 0\n\tdict = {\'I\':1, \'V\':5, \'X\':10, \'L\':50, \'C\':100, \'D\':500, \'M\':1000}\n\tfor i in s[::-1]: # rev the s\n\t\tif dict[i] >= prev:\n\t\t\tres += dict[i] # sum the value iff previous value same or more\n\t\telse:\n\t\t\tres -= dict[i] # substract when value is like "IV" --> 5-1, "IX" --> 10 -1 etc \n\t\tprev = dict[i]\n\treturn res\n```\n | 156 | 1 | ['Python', 'Python3'] | 13 |
roman-to-integer | 💻✅BEATS 100%⌛⚡[C++/Java/Py3/JS]🍨| EASY n CLEAN EXPLANATION⭕💌 | beats-100cjavapy3js-easy-n-clean-explana-l3c8 | IntuitionRoman numerals use specific rules:
If a smaller numeral comes before a larger one, it is subtracted (e.g., IV = 4).
Otherwise, the values are added (e. | Fawz-Haaroon | NORMAL | 2025-01-26T22:46:14.216333+00:00 | 2025-01-26T22:46:14.216333+00:00 | 31,985 | false | # Intuition
Roman numerals use specific rules:
- If a smaller numeral comes before a larger one, it is subtracted (e.g., `IV = 4`).
- Otherwise, the values are added (e.g., `VI = 6`).
We can traverse the string, compare each numeral with the next, and decide whether to add or subtract.
# Approach
1. Create a helper function to map Roman numerals (`I`, `V`, `X`, etc.) to their integer values.
2. Initialize a variable `result` to store the final integer value.
3. Loop through the string:
- If the current numeral is smaller than the next numeral, subtract its value from `result`.
- Otherwise, add its value to `result`.
4. Return the final `result`.
# Complexity
- **Time Complexity**: `O(n)`
The string is traversed once, where `n` is its length.
- **Space Complexity**: `O(1)`
Only a few variables are used; no additional data structures.
# Code
```cpp []
class Solution {
public:
int char2num(char a) {
switch (a) {
case 'I': return 1;
case 'V': return 5;
case 'X': return 10;
case 'L': return 50;
case 'C': return 100;
case 'D': return 500;
case 'M': return 1000;
default: return 0;
}
}
int romanToInt(string s) {
int result = 0;
for (int i = 0; i < s.length(); i++) {
if (i + 1 < s.length() && char2num(s[i]) < char2num(s[i + 1])) {
result -= char2num(s[i]);
} else {
result += char2num(s[i]);
}
}
return result;
}
};
```
```java []
class Solution {
public int romanToInt(String s) {
int ans = 0, num = 0;
for (int i = s.length() - 1; i >= 0; i--) {
switch (s.charAt(i)) {
case 'I':
num = 1;
break;
case 'V':
num = 5;
break;
case 'X':
num = 10;
break;
case 'L':
num = 50;
break;
case 'C':
num = 100;
break;
case 'D':
num = 500;
break;
case 'M':
num = 1000;
break;
}
if (4 * num < ans)
ans -= num;
else
ans += num;
}
return ans;
}
}
```
```python []
class Solution:
def romanToInt(self, s: str) -> int:
roman_to_int = {
'I': 1, 'V': 5, 'X': 10, 'L': 50,
'C': 100, 'D': 500, 'M': 1000
}
result = 0
for i in range(len(s)):
if i + 1 < len(s) and roman_to_int[s[i]] < roman_to_int[s[i + 1]]:
result -= roman_to_int[s[i]]
else:
result += roman_to_int[s[i]]
return result
```
```javascript []
/**
* @param {string} s
* @return {number}
*/
var romanToInt = function(s) {
const romanToInt = {
'I': 1, 'V': 5, 'X': 10, 'L': 50,
'C': 100, 'D': 500, 'M': 1000
};
let result = 0;
for (let i = 0; i < s.length; i++) {
if (i + 1 < s.length && romanToInt[s[i]] < romanToInt[s[i + 1]]) {
result -= romanToInt[s[i]];
} else {
result += romanToInt[s[i]];
}
}
return result;
};
```
```
✨ AN UPVOTE WILL BE APPRECIATED ^_~ ✨
``` | 137 | 2 | ['C', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 9 |
roman-to-integer | ✅ [C] || 100% Hashtable Solution || O(n) | c-100-hashtable-solution-on-by-bezlant-bieo | \nint romanToInt(char * s)\n{\n int t[\'X\' + 1] = {\n [\'I\'] = 1,\n [\'V\'] = 5,\n [\'X\'] = 10,\n [\'L\'] = 50,\n [\'C\ | bezlant | NORMAL | 2022-04-18T13:53:15.808297+00:00 | 2022-04-18T13:54:28.982948+00:00 | 10,810 | false | ```\nint romanToInt(char * s)\n{\n int t[\'X\' + 1] = {\n [\'I\'] = 1,\n [\'V\'] = 5,\n [\'X\'] = 10,\n [\'L\'] = 50,\n [\'C\'] = 100,\n [\'D\'] = 500,\n [\'M\'] = 1000,\n };\n int res = 0;\n for (int i = 0; s[i]; i++) {\n if (t[s[i]] < t[s[i+1]])\n res -= t[s[i]];\n else\n res += t[s[i]];\n }\n return res;\n}\n```\n**If this was helpful, don\'t hesitate to upvote! :)**\nHave a nice day!\n\n | 116 | 0 | ['C'] | 13 |
roman-to-integer | Nice and clean TS/JS | nice-and-clean-tsjs-by-pirastrino-lt5z | You don\'t need to map CM, XC, "IX", "IV" separately. Eg.\nwhen you parse "MCMXCIV" to an array of integers you get:\n[1000, 100, 1000, 10, 100, 1, 5], if you a | pirastrino | NORMAL | 2022-06-04T11:16:21.716788+00:00 | 2022-06-04T13:19:17.200355+00:00 | 6,779 | false | You don\'t need to map `CM`, `XC`, `"IX"`, `"IV"` separately. Eg.\nwhen you parse `"MCMXCIV"` to an array of integers you get:\n`[1000, 100, 1000, 10, 100, 1, 5]`, if you add them together (from left to right),\nyou just need to check if next number is bigger, if so then you simply subtract\nthat number: `1000 - 100 + 1000 - 10 + 100 - 1 + 5 = 1994`.\n\n```typescript\nconst roman = {\n I: 1,\n V: 5,\n X: 10,\n L: 50,\n C: 100,\n D: 500,\n M: 1000,\n};\n\nfunction romanToInt(s: string): number {\n const integers = s.split(\'\').map(c => roman[c]);\n \n return integers.reduce((acc, x, i) => x < integers[i+1] ? acc - x : acc + x, 0);\n};\n\n``` | 96 | 0 | ['TypeScript', 'JavaScript'] | 17 |
roman-to-integer | [C++] Concise Solution | c-concise-solution-by-kasracode-k4z1 | \nint romanToInt(string s) {\n\tunordered_map<char, int> mp = {{\'M\', 1000}, {\'D\', 500}, {\'C\', 100}, {\'L\', 50}, {\'X\', 10}, {\'V\', 5}, {\'I\', 1}};\n\t | KasraCode | NORMAL | 2019-06-23T00:31:16.737437+00:00 | 2019-06-23T00:31:50.076597+00:00 | 13,861 | false | ```\nint romanToInt(string s) {\n\tunordered_map<char, int> mp = {{\'M\', 1000}, {\'D\', 500}, {\'C\', 100}, {\'L\', 50}, {\'X\', 10}, {\'V\', 5}, {\'I\', 1}};\n\tint res = mp[s.back()];\n\tfor(int i = 0; i < s.size() - 1; i++) {\n\t\tif(mp[s[i]] < mp[s[i + 1]]) res -= mp[s[i]];\n\t\telse res += mp[s[i]];\n\t}\n\treturn res;\n}\n``` | 96 | 1 | ['C', 'C++'] | 10 |
roman-to-integer | ✅☑️ Best C++ Solution || Hash Table || Math || String || One Stop Solution. | best-c-solution-hash-table-math-string-o-1tbv | Intuition\n Describe your first thoughts on how to solve this problem. \nWe can solve this problem using String + Hash Table + Math.\n\n# Approach\n Describe yo | its_vishal_7575 | NORMAL | 2023-02-18T18:21:49.038787+00:00 | 2023-02-18T19:27:38.304904+00:00 | 13,607 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can solve this problem using String + Hash Table + Math.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe can easily understand the approach by seeing the code which is easy to understand with comments.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nTime Complexity : O(1), The maximum length of the string(s) can be 15 (as per the Constgraint), therefore, the worst case time complexity can be O(15) or O(1).\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nSpace Complexity : O(1), We are using unordered_map(map) to store the Roman symbols and their corresponding integer values but there are only 7 symbols hence the worst case space complexity can be O(7) which is equivalent to O(1).\n\n# Code\n```\n/*\n\n Time Complexity : O(1), The maximum length of the string(s) can be 15 (as per the Constgraint), therefore, the\n worst case time complexity can be O(15) or O(1).\n\n Space Complexity : O(1), We are using unordered_map(map) to store the Roman symbols and their corresponding\n integer values but there are only 7 symbols hence the worst case space complexity can be O(7) which is\n equivalent to O(1).\n\n Solved using String + Hash Table + Math.\n\n*/\n\nclass Solution {\npublic:\n int romanToInt(string s) {\n unordered_map<char, int> map;\n\n map[\'I\'] = 1;\n map[\'V\'] = 5;\n map[\'X\'] = 10;\n map[\'L\'] = 50;\n map[\'C\'] = 100;\n map[\'D\'] = 500;\n map[\'M\'] = 1000;\n \n int ans = 0;\n \n for(int i=0; i<s.length(); i++){\n if(map[s[i]] < map[s[i+1]]){\n ans -= map[s[i]];\n }\n else{\n ans += map[s[i]];\n }\n }\n return ans;\n }\n};\n\n```\n\n***IF YOU LIKE THE SOLUTION THEN PLEASE UPVOTE MY SOLUTION BECAUSE IT GIVES ME MOTIVATION TO REGULARLY POST THE SOLUTION.***\n\n | 88 | 0 | ['Hash Table', 'Math', 'String', 'C++'] | 4 |
roman-to-integer | Java efficient, easy to understand, with explaination | java-efficient-easy-to-understand-with-e-zq8u | First we use a hashmap to map the conversions of roman digits to integer.\nNow, if a numeral with smaller value precedes one with a larger value, we subtract th | shivamshah | NORMAL | 2020-04-14T17:31:20.885668+00:00 | 2020-04-14T17:31:20.885720+00:00 | 9,604 | false | First we use a hashmap to map the conversions of roman digits to integer.\nNow, if a numeral with smaller value precedes one with a larger value, we subtract the value from the total, otherwise we add the value to the total.\nAt the end, we still have to add the last character value.\n```\nclass Solution {\n public int romanToInt(String s) {\n \n Map<Character, Integer> map = new HashMap();\n map.put(\'I\', 1);\n map.put(\'V\', 5);\n map.put(\'X\', 10);\n map.put(\'L\', 50);\n map.put(\'C\', 100);\n map.put(\'D\', 500);\n map.put(\'M\', 1000);\n \n int res = 0;\n for(int i = 0; i < s.length()-1; i++){\n if(map.get(s.charAt(i)) < map.get(s.charAt(i+1))) res -= map.get(s.charAt(i));\n else res += map.get(s.charAt(i));\n }\n return res + map.get(s.charAt(s.length()-1));\n }\n}\n``` | 87 | 0 | ['Java'] | 4 |
roman-to-integer | ✅ [Accepted] Solution for Swift | accepted-solution-for-swift-by-asahiocea-ufj1 | \nDisclaimer: By using any content from this post or thread, you release the author(s) from all liability and warranty of any kind. You are free to use the cont | AsahiOcean | NORMAL | 2021-03-31T20:25:43.447934+00:00 | 2023-12-27T22:10:16.746363+00:00 | 8,991 | false | <blockquote>\n<b>Disclaimer:</b> By using any content from this post or thread, you release the author(s) from all liability and warranty of any kind. You are free to use the content freely and as you see fit. Any suggestions for improvement are welcome and greatly appreciated! Happy coding!\n</blockquote>\n\n```swift\nclass Solution {\n func romanToInt(_ s: String) -> Int {\n let dict: [Character:Int] = ["I":1,"V":5,"X":10,"L":50,"C":100,"D":500,"M":1000]\n var out = 0, prev = 0\n for i in s {\n let val = dict[i] ?? 0\n out += (val <= prev) ? prev : -prev\n prev = val\n }\n out += prev\n return out\n }\n}\n```\n\n---\n\n```swift\nimport XCTest\n\nclass Tests: XCTestCase {\n \n private let solution = Solution()\n \n // III = 3\n func test0() {\n let value = solution.romanToInt("III")\n XCTAssertEqual(value, 3)\n }\n \n // L = 50, V = 5, III = 3\n func test1() {\n let value = solution.romanToInt("LVIII")\n XCTAssertEqual(value, 58)\n }\n \n // M = 1000, CM = 900, XC = 90 and IV = 4\n func test2() {\n let value = solution.romanToInt("MCMXCIV")\n XCTAssertEqual(value, 1994)\n }\n}\n\nTests.defaultTestSuite.run()\n``` | 84 | 2 | ['Swift'] | 5 |
roman-to-integer | ✅🔥Beats 99.99% || 📈Easy solution with O(n) Approach ✔| C++ | Python | Java | Javascript | beats-9999-easy-solution-with-on-approac-rzlv | \u2705Do upvote if you like the solution and explanation please \uD83D\uDE80\n\n# \u2705Intuition\n Describe your first thoughts on how to solve this problem. | anshuP_cs24 | NORMAL | 2023-10-04T06:48:21.254011+00:00 | 2023-10-20T07:11:34.281842+00:00 | 15,726 | false | # \u2705Do upvote if you like the solution and explanation please \uD83D\uDE80\n\n# \u2705Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this code is to iterate through the Roman numeral string from right to left, converting each symbol to its corresponding integer value. If the current symbol has a smaller value than the symbol to its right, we subtract its value from the result; otherwise, we add its value to the result. By processing the string from right to left, we can handle cases where subtraction is required (e.g., IV for 4) effectively.\n\n---\n\n# \u2705Approach for code \n<!-- Describe your approach to solving the problem. -->\n1. Create an unordered map (romanValues) to store the integer values corresponding to each Roman numeral symbol (\'I\', \'V\', \'X\', \'L\', \'C\', \'D\', \'M\').\n\n2. Initialize a variable result to 0 to store the accumulated integer value.\n\n3. Iterate through the input string s from right to left (starting from the last character).\n\n4. For each character at index i, get its integer value (currValue) from the romanValues map.\n\n5. Check if the current symbol has a smaller value than the symbol to its right (i.e., currValue < romanValues[s[i + 1]]) using the condition i < s.length() - 1. If true, subtract currValue from the result; otherwise, add it to the result.\n\n6. Update the result accordingly for each symbol as you iterate through the string.\n\n7. Finally, return the accumulated result as the integer equivalent of the Roman numeral.\n\n---\n\n# \u2705Complexity\n## 1. Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of this code is O(n), where \'n\' is the length of the input string s. This is because we iterate through the entire string once from right to left.\n## 2. Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity is O(1) because the size of the romanValues map is fixed and does not depend on the input size. We use a constant amount of additional space to store the result and loop variables, so the space complexity is constant.\n\n---\n\n# \uD83D\uDCA1"If you have made it this far, I would like to kindly request that you upvote this solution to ensure its reach extends to others as well."\u2763\uFE0F\uD83D\uDCA1\n\n---\n\n# \u2705Code\n```C++ []\nclass Solution {\npublic:\n int romanToInt(string s) {\n unordered_map<char, int> romanValues = {\n {\'I\', 1},\n {\'V\', 5},\n {\'X\', 10},\n {\'L\', 50},\n {\'C\', 100},\n {\'D\', 500},\n {\'M\', 1000}\n };\n\n int result = 0;\n\n for (int i = s.length() - 1; i >= 0; i--) {\n int currValue = romanValues[s[i]];\n\n if (i < s.length() - 1 && currValue < romanValues[s[i + 1]]) {\n result -= currValue;\n } else {\n result += currValue;\n }\n }\n\n return result;\n }\n};\n\n```\n```Python []\nclass Solution:\n def romanToInt(self, s: str) -> int:\n romanValues = {\n \'I\': 1,\n \'V\': 5,\n \'X\': 10,\n \'L\': 50,\n \'C\': 100,\n \'D\': 500,\n \'M\': 1000\n }\n\n result = 0\n\n for i in range(len(s) - 1, -1, -1):\n currValue = romanValues[s[i]]\n\n if i < len(s) - 1 and currValue < romanValues[s[i + 1]]:\n result -= currValue\n else:\n result += currValue\n\n return result\n\n```\n```Java []\nimport java.util.HashMap;\nimport java.util.Map;\n\nclass Solution {\n public int romanToInt(String s) {\n Map<Character, Integer> romanValues = new HashMap<>();\n romanValues.put(\'I\', 1);\n romanValues.put(\'V\', 5);\n romanValues.put(\'X\', 10);\n romanValues.put(\'L\', 50);\n romanValues.put(\'C\', 100);\n romanValues.put(\'D\', 500);\n romanValues.put(\'M\', 1000);\n\n int result = 0;\n\n for (int i = s.length() - 1; i >= 0; i--) {\n int currValue = romanValues.get(s.charAt(i));\n\n if (i < s.length() - 1 && currValue < romanValues.get(s.charAt(i + 1))) {\n result -= currValue;\n } else {\n result += currValue;\n }\n }\n\n return result;\n }\n}\n\n```\n```Javascript []\nvar romanToInt = function(s) {\n const romanValues = {\n \'I\': 1,\n \'V\': 5,\n \'X\': 10,\n \'L\': 50,\n \'C\': 100,\n \'D\': 500,\n \'M\': 1000\n };\n\n let result = 0;\n\n for (let i = s.length - 1; i >= 0; i--) {\n const currValue = romanValues[s[i]];\n\n if (i < s.length - 1 && currValue < romanValues[s[i + 1]]) {\n result -= currValue;\n } else {\n result += currValue;\n }\n }\n\n return result;\n};\n\n```\n | 79 | 0 | ['Hash Table', 'Math', 'Two Pointers', 'String', 'C', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 9 |
roman-to-integer | Python simple solution | python-simple-solution-by-tovam-n98l | Python :\n\n\ndef romanToInt(self, s: str) -> int:\n\troman_dict = {\n\t\t\'I\' :1,\n\t\t\'V\' :5,\n\t\t\'X\' :10,\n\t\t\'L\' :50,\n\t\t\'C\' :100,\n\t\t\'D\' : | TovAm | NORMAL | 2021-10-30T22:20:35.389680+00:00 | 2021-10-30T22:20:35.389714+00:00 | 7,145 | false | **Python :**\n\n```\ndef romanToInt(self, s: str) -> int:\n\troman_dict = {\n\t\t\'I\' :1,\n\t\t\'V\' :5,\n\t\t\'X\' :10,\n\t\t\'L\' :50,\n\t\t\'C\' :100,\n\t\t\'D\' :500,\n\t\t\'M\' :1000\n\t}\n\n\ts = s.replace("IV", "IIII").replace("IX", "IIIIIIIII")\n\ts = s.replace("XL", "XXXX").replace("XC", "XXXXXXXXX")\n\ts = s.replace("CD", "CCCC").replace("CM", "CCCCCCCCC")\n\n\tres = 0\n\tfor char in s:\n\t\tres += roman_dict[char]\n\n\treturn res\n```\n\n**Like it ? please upvote !** | 79 | 1 | ['Python', 'Python3'] | 12 |
roman-to-integer | C solution | c-solution-by-thedrafttin-29jj | Runtime: 12 ms, faster than 100.00% of C online submissions for Roman to Integer.\nMemory Usage: 7 MB, less than 100.00% of C online submissions for Roman to In | thedrafttin | NORMAL | 2019-04-09T15:12:34.161604+00:00 | 2019-04-09T15:12:34.161691+00:00 | 6,326 | false | Runtime: 12 ms, faster than 100.00% of C online submissions for Roman to Integer.\nMemory Usage: 7 MB, less than 100.00% of C online submissions for Roman to Integer\n```\nint romanToInt(char* s) {\n int value[100];\n value[\'I\'] = 1;\n value[\'V\'] = 5;\n value[\'X\'] = 10;\n value[\'L\'] = 50;\n value[\'C\'] = 100;\n value[\'D\'] = 500;\n value[\'M\'] = 1000;\n value[\'\\0\'] = 0;\n int sum = 0;\n for(int i = 0; s[i] !=\'\\0\'; i++){\n if(value[s[i]] < value[s[i + 1]])\n sum = sum - value[s[i]];\n else\n sum += value[s[i]];\n }\n return sum;\n} | 69 | 2 | [] | 10 |
roman-to-integer | 0ms • 1LINER • 100% • Fastest Solution Explained • O(n) Time Complexity • O(n) Space Complexity | 0ms-1liner-100-fastest-solution-explaine-m9ia | 0ms \u2022 Beats 100% Time Complexity \u2022 Beats 100% Space Complexity\n\n### Kotlin\n\nclass Solution {\n fun romanToInt(s: String): Int {\n // 1. | darian-catalin-cucer | NORMAL | 2022-05-20T14:30:27.487958+00:00 | 2024-06-02T09:10:19.057092+00:00 | 15,750 | false | # 0ms \u2022 Beats 100% Time Complexity \u2022 Beats 100% Space Complexity\n\n### Kotlin\n```\nclass Solution {\n fun romanToInt(s: String): Int {\n // 1. Create a Map for Roman Numeral Values\n val translations = mapOf(\n "I" to 1,\n "V" to 5,\n "X" to 10,\n "L" to 50,\n "C" to 100,\n "D" to 500,\n "M" to 1000\n )\n\n // 2. Preprocess the Input String (Subtractive Notation Handling)\n var modifiedString = s \n .replace("IV", "IIII")\n .replace("IX", "VIIII")\n .replace("XL", "XXXX")\n .replace("XC", "LXXXX")\n .replace("CD", "CCCC")\n .replace("CM", "DCCCC")\n\n // 3. Initialize the Result Variable\n var number = 0\n\n // 4. Iterate Through the String and Sum Values\n for (char in modifiedString) {\n number += translations.getValue(char.toString()) // Get value from map and add to total\n }\n\n // 5. Return the Final Result\n return number\n }\n}\n```\n\n### Java\n```\nclass Solution {\n public int romanToInt(String s) {\n // 1. Create a Map for Roman Numeral Values\n Map<String, Integer> translations = new HashMap<>();\n translations.put("I", 1);\n translations.put("V", 5);\n translations.put("X", 10);\n translations.put("L", 50);\n translations.put("C", 100);\n translations.put("D", 500);\n translations.put("M", 1000);\n\n // 2. Preprocess the Input String (Subtractive Notation Handling)\n s = s.replace("IV", "IIII").replace("IX", "VIIII");\n s = s.replace("XL", "XXXX").replace("XC", "LXXXX");\n s = s.replace("CD", "CCCC").replace("CM", "DCCCC");\n\n // 3. Initialize the Result Variable\n int number = 0;\n\n // 4. Iterate Through the String and Sum Values\n for (int i = 0; i < s.length(); i++) {\n String currentSymbol = String.valueOf(s.charAt(i)); \n number += translations.get(currentSymbol); \n }\n\n // 5. Return the Final Result\n return number;\n }\n}\n```\n\n### Python / Python3\n```\nclass Solution:\n def romanToInt(self, s: str) -> int:\n # 1. Create a Dictionary for Roman Numeral Values\n # This dictionary maps each Roman numeral symbol to its corresponding integer value.\n translations = {\n "I": 1,\n "V": 5,\n "X": 10,\n "L": 50,\n "C": 100,\n "D": 500,\n "M": 1000\n }\n\n # 2. Preprocess the Input String (Subtractive Notation Handling)\n # Replace the special cases of subtractive notation ("IV", "IX", "XL", etc.) \n # with their equivalent additive forms to simplify the conversion later.\n s = s.replace("IV", "IIII").replace("IX", "VIIII")\n s = s.replace("XL", "XXXX").replace("XC", "LXXXX")\n s = s.replace("CD", "CCCC").replace("CM", "DCCCC")\n\n # 3. Initialize the Result Variable\n # Start with a total of zero, which will be accumulated as we process the numerals.\n number = 0\n\n # 4. Iterate Through the String and Sum Values\n # Loop over each character (Roman numeral) in the preprocessed string.\n for char in s:\n # Look up the integer value of the current character in the \'translations\' dictionary.\n # Add this value to the \'number\' variable to build the total.\n number += translations[char]\n\n # 5. Return the Final Result\n # After processing all characters, return the accumulated integer value.\n return number\n``` | 67 | 3 | ['Array', 'Hash Table', 'Math', 'String', 'Combinatorics', 'Python', 'C++', 'Java', 'Python3', 'Kotlin'] | 11 |
roman-to-integer | ⚠️Easiest 😎 FAANG Method Ever !!! 💥 | easiest-faang-method-ever-by-adityabhate-jtsg | \n# \uD83D\uDDEF\uFE0FComplexity :-\n- Time complexity:O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(1)\n Add your space complexity h | AdityaBhate | NORMAL | 2022-12-02T11:26:39.397719+00:00 | 2023-01-05T12:10:11.153469+00:00 | 18,385 | false | \n# \uD83D\uDDEF\uFE0FComplexity :-\n- Time complexity:O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# \uD83D\uDDEF\uFE0FCode :-\n```\nclass Solution {\npublic:\n int romanToInt(string s) {\n int res=0;\n s+=\' \';\n for(int i=0;i<s.size();){\n if(s[i]==\'I\' && s[i+1]==\'V\') { res+=4; i+=2;}\n else if(s[i]==\'I\' && s[i+1]==\'X\') { res+=9; i+=2;}\n else if(s[i]==\'I\') { res++; i++;}\n else if(s[i]==\'V\') { res+=5; i++;}\n else if(s[i]==\'X\' && s[i+1]==\'L\') { res+=40; i+=2;}\n else if(s[i]==\'X\' && s[i+1]==\'C\') { res+=90; i+=2;}\n else if(s[i]==\'X\') { res+=10; i++;} \n else if(s[i]==\'L\') { res+=50; i++;}\n else if(s[i]==\'C\' && s[i+1]==\'M\') { res+=900; i+=2;}\n else if(s[i]==\'C\' && s[i+1]==\'D\') { res+=400; i+=2;}\n else if(s[i]==\'C\') { res+=100; i++;}\n else if(s[i]==\'D\') { res+=500; i++;}\n else if(s[i]==\'M\') { res+=1000; i++;}\n else if(s[i]==\' \') break;\n }\n return res;\n }\n};\n```\n# ***Please Upvote if it helps :) \u2764\uFE0F***\n\n | 65 | 3 | ['Hash Table', 'Math', 'String', 'C++', 'Java'] | 21 |
roman-to-integer | JavaScript Clean Solution | javascript-clean-solution-by-control_the-w9hg | javascript\nvar romanToInt = function(s) {\n const map = { \'I\': 1, \'V\': 5, \'X\': 10, \'L\': 50, \'C\': 100, \'D\': 500, \'M\': 1000};\n let num = 0;\ | control_the_narrative | NORMAL | 2020-08-20T02:08:16.786358+00:00 | 2020-08-20T02:08:16.786407+00:00 | 6,365 | false | ```javascript\nvar romanToInt = function(s) {\n const map = { \'I\': 1, \'V\': 5, \'X\': 10, \'L\': 50, \'C\': 100, \'D\': 500, \'M\': 1000};\n let num = 0;\n \n for(let i = 0; i < s.length; i++) {\n const curr = map[s[i]], next = map[s[i+1]];\n if(curr < next) num -= curr;\n else num += curr;\n }\n return num; \n};\n``` | 63 | 1 | ['JavaScript'] | 5 |
roman-to-integer | My easy-to-understand C++ solutions | my-easy-to-understand-c-solutions-by-xia-rr64 | class Solution {\n public:\n int romanToInt(string s) {\n int num = 0;\n int size = s.size();\n | xiaohui7 | NORMAL | 2014-12-17T16:30:21+00:00 | 2018-09-25T05:29:26.405360+00:00 | 11,598 | false | class Solution {\n public:\n int romanToInt(string s) {\n int num = 0;\n int size = s.size();\n \n for (int i = 0; i < size; i++) {\n \tif (i < (size - 1) && romanCharToInt(s[i]) < romanCharToInt(s[i + 1])) {\n \t\tnum -= romanCharToInt(s[i]);\n \t} else {\n \t\t\t\tnum += romanCharToInt(s[i]);\n \t\t\t}\n }\n return num;\n }\n \n int romanCharToInt(char c) {\n \tswitch (c) {\n \t\tcase 'I': \treturn 1;\n \t\tcase 'V':\treturn 5;\n \t\tcase 'X':\treturn 10;\n \t\tcase 'L':\treturn 50;\n \t\tcase 'C':\treturn 100;\n \t\tcase 'D':\treturn 500;\n \t\tcase 'M':\treturn 1000;\n \t\tdefault:\treturn 0;\n \t}\n }\n };\n\n[The code is faster][1] if the body of the for loop is replaced with:\n\n \tif (i < (size - 1) && (\n \t\t'I' == s[i] && ('V' == s[i + 1] || 'X' == s[i + 1]) ||\n \t\t'X' == s[i] && ('L' == s[i + 1] || 'C' == s[i + 1]) ||\n \t\t'C' == s[i] && ('D' == s[i + 1] || 'M' == s[i + 1]) )) {\n \t\tnum -= romanCharToInt(s[i]);\n \t} else {\n\t\t\tnum += romanCharToInt(s[i]);\n\t\t}\n\n \n\n\n [1]: http://xiaohuiliucuriosity.blogspot.com/2014/12/problem-given-roman-numeral-convert-it.html | 57 | 1 | [] | 3 |
roman-to-integer | Very Simple Python Solution | very-simple-python-solution-by-user9872y-bsck | Intuition\n Describe your first thoughts on how to solve this problem. \nWe essentially start scanning adding all of the corresponding values for each character | user9872yq | NORMAL | 2023-01-07T22:15:43.895614+00:00 | 2023-01-07T22:15:43.895665+00:00 | 26,438 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe essentially start scanning adding all of the corresponding values for each character regardless of order. (e.g. "IX" is 11 not 9) Then, we check the order of the elements, and if we find that the order is reversed (i.e. "IX"), we make the necessary adjustment (e.g. for "IX," we subtract 2)\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1) Introduce the total sum variable, and the dictionary containing the corresponding number values for each Roman letter\n2) We add all of the corresponding number values together regardless of order of elements\n3) We check if certain ordered pairs are in the Roman number and make the adjustments if necessary\n\n\n# Code\n```\nclass Solution:\n def romanToInt(self, s: str) -> int:\n total = 0\n theDict = {"I": 1, "V": 5, "X": 10, "L": 50, "C": 100, "D": 500, "M": 1000}\n\n for i in s:\n total += theDict[i]\n\n if "IV" in s:\n total -= 2\n if "IX" in s:\n total -= 2\n if "XL" in s:\n total -= 20\n if "XC" in s:\n total -= 20\n if "CD" in s:\n total -= 200\n if "CM" in s:\n total -= 200\n\n \n return total\n``` | 53 | 0 | ['Python3'] | 10 |
roman-to-integer | Easiest Beginner Friendly Sol || HashMap || C ++, Java, Python | easiest-beginner-friendly-sol-hashmap-c-koqgg | Intuition of this Problem:\nThe given code is implementing the conversion of a Roman numeral string into an integer. It uses an unordered map to store the mappi | singhabhinash | NORMAL | 2023-02-22T02:31:03.383970+00:00 | 2023-02-22T02:31:03.384013+00:00 | 13,934 | false | # Intuition of this Problem:\nThe given code is implementing the conversion of a Roman numeral string into an integer. It uses an unordered map to store the mapping between Roman numerals and their corresponding integer values. The algorithm takes advantage of the fact that in a valid Roman numeral string, the larger numeral always appears before the smaller numeral if the smaller numeral is subtracted from the larger one. Thus, it starts with the last character in the string and adds the corresponding value to the integer variable. Then, it iterates through the remaining characters from right to left and checks whether the current numeral is greater than or equal to the previous numeral. If it is greater, then it adds the corresponding value to the integer variable, otherwise, it subtracts the corresponding value from the integer variable.\n<!-- Describe your first thoughts on how to solve this problem. -->\n**NOTE - PLEASE READ APPROACH FIRST THEN SEE THE CODE. YOU WILL DEFINITELY UNDERSTAND THE CODE LINE BY LINE AFTER SEEING THE APPROACH.**\n\n# Approach for this Problem:\n1. Create an unordered map and store the mapping between Roman numerals and their corresponding integer values.\n2. Reverse the input Roman numeral string.\n3. Initialize an integer variable to 0.\n4. Add the integer value corresponding to the last character in the string to the integer variable.\n5. Iterate through the remaining characters from right to left.\n6. Check whether the integer value corresponding to the current character is greater than or equal to the integer value corresponding to the previous character.\n7. If it is greater, add the integer value to the integer variable.\n8. If it is smaller, subtract the integer value from the integer variable.\n9. Return the final integer variable.\n<!-- Describe your approach to solving the problem. -->\n\n# Humble Request:\n- If my solution is helpful to you then please **UPVOTE** my solution, your **UPVOTE** motivates me to post such kind of solution.\n- Please let me know in comments if there is need to do any improvement in my approach, code....anything.\n- **Let\'s connect on** https://www.linkedin.com/in/abhinash-singh-1b851b188\n\n\n\n# Code:\n```C++ []\nclass Solution {\npublic:\n int romanToInt(string s) {\n unordered_map<char, int> storeKeyValue;\n storeKeyValue[\'I\'] = 1;\n storeKeyValue[\'V\'] = 5;\n storeKeyValue[\'X\'] = 10;\n storeKeyValue[\'L\'] = 50;\n storeKeyValue[\'C\'] = 100;\n storeKeyValue[\'D\'] = 500;\n storeKeyValue[\'M\'] = 1000;\n reverse(s.begin(), s.end());\n int integer = 0;\n integer += storeKeyValue[s[0]];\n for (int i = 1; i < s.length(); i++) {\n if(storeKeyValue[s[i]] >= storeKeyValue[s[i-1]])\n integer += storeKeyValue[s[i]];\n else\n integer -= storeKeyValue[s[i]];\n }\n return integer;\n }\n};\n```\n```Java []\nclass Solution {\n public int romanToInt(String s) {\n HashMap<Character, Integer> storeKeyValue = new HashMap<>();\n storeKeyValue.put(\'I\', 1);\n storeKeyValue.put(\'V\', 5);\n storeKeyValue.put(\'X\', 10);\n storeKeyValue.put(\'L\', 50);\n storeKeyValue.put(\'C\', 100);\n storeKeyValue.put(\'D\', 500);\n storeKeyValue.put(\'M\', 1000);\n int integer = 0;\n integer += storeKeyValue.get(s.charAt(0));\n for (int i = 1; i < s.length(); i++) {\n if(storeKeyValue.get(s.charAt(i)) >= storeKeyValue.get(s.charAt(i-1)))\n integer += storeKeyValue.get(s.charAt(i));\n else\n integer -= storeKeyValue.get(s.charAt(i));\n }\n return integer;\n }\n}\n\n```\n```Python []\nclass Solution:\n def romanToInt(self, s: str) -> int:\n storeKeyValue = {}\n storeKeyValue[\'I\'] = 1\n storeKeyValue[\'V\'] = 5\n storeKeyValue[\'X\'] = 10\n storeKeyValue[\'L\'] = 50\n storeKeyValue[\'C\'] = 100\n storeKeyValue[\'D\'] = 500\n storeKeyValue[\'M\'] = 1000\n s = s[::-1]\n integer = 0\n integer += storeKeyValue[s[0]]\n for i in range(1, len(s)):\n if storeKeyValue[s[i]] >= storeKeyValue[s[i-1]]:\n integer += storeKeyValue[s[i]]\n else:\n integer -= storeKeyValue[s[i]]\n return integer\n\n```\n\n# Time Complexity and Space Complexity:\n- Time complexity: **O(s.length()) = O(15) = O(1)**\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: **O(7) = O(1)**\n<!-- Add your space complexity here, e.g. $$O(n)$$ --> | 50 | 0 | ['Hash Table', 'Math', 'C++', 'Java', 'Python3'] | 4 |
roman-to-integer | Python | python-by-lividsu-5oi8 | Python\n\n rd = {\n "I" : 1,\n "V" : 5,\n "X" : 10,\n "L" : 50,\n "C" : 100,\n "D" : 50 | lividsu | NORMAL | 2020-12-09T09:09:06.022453+00:00 | 2020-12-09T09:09:06.022483+00:00 | 6,103 | false | Python\n```\n rd = {\n "I" : 1,\n "V" : 5,\n "X" : 10,\n "L" : 50,\n "C" : 100,\n "D" : 500,\n "M" : 1000\n }\n \n n = len(s)\n rt = 0\n for i in range(n):\n if i==n-1 or rd[s[i]] >= rd[s[i+1]] :\n rt += rd[s[i]]\n else :\n rt -= rd[s[i]]\n \n return rt\n``` | 50 | 1 | [] | 8 |
roman-to-integer | Python 3 -> Simple and detailed explanation | python-3-simple-and-detailed-explanation-3ux8 | Suggestions to make it better are always welcomed.\n\nFirst things first: Because we need to look up the value of each roman charater multiple times, let\'s sto | mybuddy29 | NORMAL | 2022-02-22T15:09:48.606565+00:00 | 2022-02-24T20:55:25.794610+00:00 | 2,350 | false | **Suggestions to make it better are always welcomed.**\n\nFirst things first: Because we need to look up the value of each roman charater multiple times, let\'s store them in a dictionary called sym to have O(1) lookup.\n\nExample:\nLooking at these characters is confusing. To come up with an algorithm, FIRST replace these characters with their numeral values and then forget about the roman letters.\nLVIII: 50, 5, 1, 1, 1\nXIV: 10, 1, 5\nMCMXCIV: 1000, 100, 1000, 10, 100, 1, 5\n\nJust keep referencing the numbers and you\'ll see that you have 2 choices to add the numbers:\n1. left-to-right. I\'ll leave left-to-right for you to solve and debate.\n2. right-to-left.\n\nWhen we go right-to-left, we see the pattern:\n1. we need to add current number if it is bigger than previous number or if both are same.\n2. we subtract current number if current number is smaller than the previous number.\n\nExample: XIV (10, 1, 5)\n1. We first add 5\n2. In the next iteration, current is 1 and previous is 5. So, we subtract 1 from result\n3. In next iteration, current is 10 and previous is 1. So, we add 10 to the result\n\nNow try this with other example: MCMXCIV.\n\n```\nclass Solution:\n def romanToInt(self, s: str) -> int:\n sym = {\n "I" : 1,\n "V" : 5,\n "X" : 10,\n "L" : 50,\n "C" : 100,\n "D" : 500,\n "M" : 1000\n }\n \n result = 0\n prev = 0\n \n for c in reversed(s):\n if sym[c] >= prev:\n result += sym[c]\n else:\n result -= sym[c]\n prev = sym[c]\n \n return result\n```\n\n**I hope that you\'ve found this useful.\nIn that case, please upvote. It only motivates me to write more such posts\uD83D\uDE03** | 49 | 1 | ['Python3'] | 3 |
roman-to-integer | JavaScript Solution | javascript-solution-by-jeantimex-qtsc | javascript\nconst romanToInt = s => {\n if (!s || s.length === 0) {\n return 0;\n }\n\n const map = new Map([[\'I\', 1], [\'V\', 5], [\'X\', 10], [\'L\', | jeantimex | NORMAL | 2018-09-22T21:57:23.175267+00:00 | 2018-09-22T21:57:23.175306+00:00 | 6,440 | false | ```javascript\nconst romanToInt = s => {\n if (!s || s.length === 0) {\n return 0;\n }\n\n const map = new Map([[\'I\', 1], [\'V\', 5], [\'X\', 10], [\'L\', 50], [\'C\', 100], [\'D\', 500], [\'M\', 1000]]);\n\n let i = s.length - 1;\n let result = map.get(s[i]);\n\n while (i > 0) {\n const curr = map.get(s[i]);\n const prev = map.get(s[i - 1]);\n\n if (prev >= curr) {\n result += prev;\n } else {\n result -= prev;\n }\n\n i--;\n }\n\n return result;\n};\n``` | 49 | 0 | [] | 5 |
roman-to-integer | Java Solution - Clean and Simple :) ( 7 ms ) | java-solution-clean-and-simple-7-ms-by-e-u7l2 | public int romanToInt(String str) {\n int[] a = new int[26];\n a['I' - 'A'] = 1;\n a['V' - 'A'] = 5;\n a['X' - 'A'] = 10;\n a | earlme | NORMAL | 2015-10-13T05:09:41+00:00 | 2018-08-28T01:30:03.210701+00:00 | 19,128 | false | public int romanToInt(String str) {\n int[] a = new int[26];\n a['I' - 'A'] = 1;\n a['V' - 'A'] = 5;\n a['X' - 'A'] = 10;\n a['L' - 'A'] = 50;\n a['C' - 'A'] = 100;\n a['D' - 'A'] = 500;\n a['M' - 'A'] = 1000;\n char prev = 'A';\n int sum = 0;\n for(char s : str.toCharArray()) {\n if(a[s - 'A'] > a[prev - 'A']) {\n sum = sum - 2 * a[prev - 'A'];\n }\n sum = sum + a[s - 'A'];\n prev = s;\n }\n return sum;\n } | 49 | 6 | ['Java'] | 4 |
roman-to-integer | C# Runtime 94% and Memory 98% O(n) | c-runtime-94-and-memory-98-on-by-koliter-lqlr | This solution uses a switch case to reduce memory usage as well as a method scope currentValue Variable which enables the code to reuse the memory and thus dras | Koliter | NORMAL | 2021-07-25T19:47:11.529470+00:00 | 2022-11-04T13:22:28.336419+00:00 | 9,163 | false | This solution uses a switch case to reduce memory usage as well as a method scope `currentValue` Variable which enables the code to reuse the memory and thus drastically reduce memory usage. Simple for loop means solution is O(n), n being the length of the string.\n\nThe Results : \nRuntime: 80 ms, faster than 93.83% of C# online submissions for Roman to Integer.\nMemory Usage: 25.9 MB, less than 97.87% of C# online submissions for Roman to Integer.\n\n```\npublic class Solution {\n public int RomanToInt(string s) {\n\n var chars = s.ToCharArray();\n var result = 0;\n var currentValue = 0;\n for(var i = 0; i < chars.Length - 1; i++){\n currentValue = RomanNumerals(chars[i]);\n result += (RomanNumerals(chars[i+1]) > currentValue ? -1 : 1) * currentValue;\n }\n \n return result + RomanNumerals(chars[chars.Length - 1]);\n }\n \n public int RomanNumerals(char c)\n {\n switch(c){\n case \'I\' : return 1;\n case \'V\' : return 5;\n case \'X\' : return 10;\n case \'L\' : return 50;\n case \'C\' : return 100;\n case \'D\' : return 500;\n case \'M\' : return 1000;\n }; \n return 0;\n }\n}\n``` | 47 | 0 | ['C#'] | 8 |
roman-to-integer | PYTHON FASTEST SOLUTION | python-fastest-solution-by-coder_hash-6z69 | IV and VI both were being treated as 6. \nSimilarly IX and XI both were treated as 11.\n\nSo I put a condition whenever it encounters IV or IX then subtract 2\n | coder_hash | NORMAL | 2022-01-15T09:12:21.492140+00:00 | 2022-01-15T09:12:21.492180+00:00 | 2,146 | false | IV and VI both were being treated as 6. \nSimilarly IX and XI both were treated as 11.\n\nSo I put a condition whenever it encounters IV or IX then subtract 2\n\nSimilarily if XL or XC then subtract 20, \'CD\' or \'CM\' then subtract 200\n\nLike my logic? An upvote won\'t cost you anything ;)\n```\nclass Solution(object):\n def romanToInt(self, s):\n """\n :type s: str\n :rtype: int\n """\n roman_dict={\'I\':1,\'V\':5,\'X\':10,\'L\':50,\'C\':100,\'D\':500,\'M\':1000}\n x=0\n \n for i in s:\n x+=roman_dict[i]\n \n \n if \'IV\' in s or \'IX\' in s:\n \n x-=2\n if \'XL\' in s or \'XC\' in s:\n \n x-=20\n if \'CD\' in s or \'CM\' in s:\n \n x-=200\n \n return x\n```\nLike my logic? An upvote won\'t cost you anything | 43 | 2 | ['Python'] | 3 |
roman-to-integer | Javascript | javascript-by-rbwn-p8ky | Runtime: 172 ms, faster than 48.97% of JavaScript online submissions for Roman to Integer.\nMemory Usage: 43.7 MB, less than 97.40% of JavaScript online submiss | rbwn | NORMAL | 2020-12-11T18:08:47.579002+00:00 | 2020-12-11T18:08:47.579036+00:00 | 8,436 | false | Runtime: 172 ms, faster than 48.97% of JavaScript online submissions for Roman to Integer.\nMemory Usage: 43.7 MB, less than 97.40% of JavaScript online submissions for Roman to Integer.\n\n```function romanToInt(s) {\n const legend = {\n I:1,\n V:5,\n X:10,\n L:50,\n C:100,\n D:500,\n M:1000\n };\n let total = 0;\n\n for (let i = 0; i < s.length; i++) {\n if (legend[s[i]] < legend[s[i+1]]) {\n total += legend[s[i+1]] - legend[s[i]];\n i++;\n } else {\n total += legend[s[i]];\n }\n }\n\n return total;\n}; | 43 | 0 | [] | 5 |
roman-to-integer | Easy C# Solution | easy-c-solution-by-erenyeagertatakae-kjk3 | We can parse the string from right to left and then subtract from grand total if the last value parsed is bigger than the current value.\n\n public int Roman | erenyeagertatakae | NORMAL | 2022-04-05T14:24:48.539893+00:00 | 2022-04-05T14:24:48.539937+00:00 | 8,007 | false | We can parse the string from right to left and then subtract from grand total if the last value parsed is bigger than the current value.\n```\n public int RomanToInt(string s) {\n var map = new Dictionary<char, int>();\n map.Add(\'I\', 1);\n map.Add(\'V\', 5);\n map.Add(\'X\', 10);\n map.Add(\'L\', 50);\n map.Add(\'C\', 100);\n map.Add(\'D\', 500);\n map.Add(\'M\', 1000);\n int sum = 0;\n int last = 0;\n for (int i = s.Length - 1; i >= 0; i--)\n {\n int current = map[s[i]];\n if ( current < last)\n {\n sum -= current;\n }\n else\n {\n sum += current;\n }\n\n last = current;\n }\n return sum;\n }\n``` | 42 | 0 | ['C#'] | 3 |
roman-to-integer | Golang solution (0ms) | golang-solution-0ms-by-klakovskiy-1d7r | To avoid many check I use var lv (last value) for define sign of current operations.\n\nRuntime: 0 ms, faster than 100.00% \nMemory Usage: 3 MB, less than 40.11 | klakovskiy | NORMAL | 2021-12-23T08:39:28.484107+00:00 | 2021-12-23T08:59:41.474161+00:00 | 6,448 | false | To avoid many check I use ```var lv``` (last value) for define sign of current operations.\n\nRuntime: 0 ms, faster than 100.00% \nMemory Usage: 3 MB, less than 40.11% (removing usages ```h,lv,cv``` variables doesn\'t help to improve memory usage)\n\n```\nfunc romanToInt(s string) int {\n\tvar v, lv, cv int\n\th := map[uint8]int{\n\t\t\'I\': 1,\n\t\t\'V\': 5,\n\t\t\'X\': 10,\n\t\t\'L\': 50,\n\t\t\'C\': 100,\n\t\t\'D\': 500,\n\t\t\'M\': 1000,\n\t}\n\n for i := len(s) - 1; i >= 0; i-- {\n\t\tcv = h[s[i]]\n\t\tif cv < lv {\n\t\t\tv -= cv\n\t\t} else {\n\t\t\tv += cv\n\t\t}\n\t\tlv = cv\n\t}\n\n\treturn v\n}\n``` | 39 | 0 | ['Go'] | 9 |
roman-to-integer | Simple Python solution | simple-python-solution-by-lokeshsk1-4e78 | \nclass Solution:\n def romanToInt(self, s: str) -> int:\n r={\'I\':1,\'V\':5,\'X\':10,\'L\':50,\'C\':100,\'D\':500,\'M\':1000}\n tot=0\n | lokeshsk1 | NORMAL | 2020-07-20T10:04:25.847343+00:00 | 2021-01-04T11:52:24.939068+00:00 | 3,522 | false | ```\nclass Solution:\n def romanToInt(self, s: str) -> int:\n r={\'I\':1,\'V\':5,\'X\':10,\'L\':50,\'C\':100,\'D\':500,\'M\':1000}\n tot=0\n for i in range(len(s)-1):\n if r[s[i]] < r[s[i+1]]:\n tot-=r[s[i]]\n else:\n tot+=r[s[i]]\n tot+=r[s[-1]]\n return tot\n``` | 39 | 1 | ['Python', 'Python3'] | 2 |
roman-to-integer | [C++] Best Methods || Clean Code || Straightforward Solutions | c-best-methods-clean-code-straightforwar-m3s3 | Using ASCII equivalent of Roman numeralsThe implementation uses "Key" and "value" to map the Roman numeral characters to their corresponding integer values.The | nitishhsinghhh | NORMAL | 2023-03-22T09:56:37.031414+00:00 | 2025-01-29T09:03:38.718588+00:00 | 7,374 | false | # Using ASCII equivalent of Roman numerals
The implementation uses "Key" and "value" to map the Roman numeral characters to their corresponding integer values.
The code converts Roman numerals to integers. It iterates over the input string from right to left and uses the arrays to calculate the integer value of each character. If a character has a lower index than the previous character, its value is subtracted from the result, otherwise it is added.
**Time complexity:** O(n)
**Space complexity:** O(1)
```C++ []
#include <algorithm>
#include <iostream>
#include <vector>
#include <cassert>
using std::cout;
using std::endl;
using std::find_if;
using std::pair;
using std::vector;
using std::string;
/**
* @brief Solution class to convert Roman numerals to integers.
*/
class Solution {
public:
/**
* @brief Converts a Roman numeral string to an integer.
* @param s The Roman numeral string.
* @return The integer representation of the Roman numeral.
*/
int romanToInt(const string& s) {
vector<pair<int, int>> keyValues = {
{'I' - '0', 1}, {'V' - '0', 5}, {'X' - '0', 10},
{'L' - '0', 50}, {'C' - '0', 100}, {'D' - '0', 500},
{'M' - '0', 1000}};
int res = 0, tKey = 0;
for (int i = s.size() - 1; i >= 0; --i) {
auto cKeyValue = find_if(keyValues.begin(), keyValues.end(),
& {
return p.first == (s[i] - '0');
});
if (cKeyValue != keyValues.end()) {
int index = std::distance(keyValues.begin(), cKeyValue);
if (tKey > index)
res -= keyValues[index].second;
else
res += keyValues[index].second;
tKey = index;
}
}
return res;
}
};
/**
* @brief Function to test the romanToInt method.
*/
void testRomanToInt() {
Solution sol;
// Unit test cases to validate the romanToInt method
assert(sol.romanToInt("III") == 3 && "Test case 1 failed"); // Test case should pass
assert(sol.romanToInt("LVIII") == 58 && "Test case 2 failed"); // Test case should pass
assert(sol.romanToInt("MCMXCIV") == 1994 && "Test case 3 failed"); // Test case should pass
assert(sol.romanToInt("MCXCIV") == 1194 && "Test case 4 failed"); // Test case should pass
cout << "All test cases passed!" << endl;
}
/**
* @brief Main function to run the test cases.
* @return Exit status.
*/
int main() {
// Run the test cases
testRomanToInt();
return 0;
}
```
```cpp []
/**
* @brief Solution class to convert Roman numerals to integers.
*/
class Solution {
public:
/**
* @brief Converts a Roman numeral string to an integer.
* @param s The Roman numeral string.
* @return The integer representation of the Roman numeral.
*/
int romanToInt(string s) {
vector<pair<char, int>> KeyValues = {
{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50},
{'C', 100}, {'D', 500}, {'M', 1000}};
int res = 0, tKey = 0;
for (int i = s.size() - 1; i >= 0; i--) {
auto cKeyValue = find_if(
KeyValues.begin(), KeyValues.end(),
& { return p.first == s[i]; });
if (cKeyValue != KeyValues.end()) {
int index = distance(KeyValues.begin(), cKeyValue);
if (tKey > index)
res -= KeyValues[index].second;
else
res += KeyValues[index].second;
tKey = index;
}
}
return res;
}
};
```
# Using std::unordered_map
This C++ function uses an unordered map to store the values of each Roman numeral character. The function iterates through the input string from right to left, comparing the values of adjacent characters to determine whether to add or subtract from the total sum. The final sum is returned as the integer equivalent of the Roman numeral string provided.
**Time complexity:** O(n)
**Space complexity:** O(1).
```C++ []
/**
* @brief Solution class to convert Roman numerals to integers.
*/
class Solution {
public:
/**
* @brief Converts a Roman numeral string to an integer.
* @param s The Roman numeral string.
* @return The integer representation of the Roman numeral.
*/
int romanToInt(string s) {
unordered_map<char, int> map = {{'I', 1}, {'V', 5}, {'X', 10},
{'L', 50}, {'C', 100}, {'D', 500},
{'M', 1000}};
int sum = map[s.back()]; ///< Initialize sum with the value of the last character.
for (int i = s.length() - 2; i >= 0; --i) {
if (map[s[i]] < map[s[i + 1]])
sum -= map[s[i]]; ///< Subtract value if the current character is less than the next character.
else
sum += map[s[i]]; ///< Add value otherwise.
}
return sum;
}
};
```
# Using std::regex_replace
This C++ function uses an unordered map to store the values of each Roman numeral character. The function first replaces specific Roman numeral substrings (like "CM" with "DCCCC") to standardize the representation. Then, it iterates through the modified string, adding up the corresponding integer values from the map to calculate the total sum. Finally, it returns the integer equivalent of the simplified Roman numeral string provided.
**Time complexity:** O(n)
**Space complexity:** O(1).
```C++ []
/**
* @brief Solution class to convert Roman numerals to integers.
*/
class Solution {
public:
/**
* @brief Converts a Roman numeral string to an integer.
* @param s The Roman numeral string.
* @return The integer representation of the Roman numeral.
*/
int romanToInt(std::string s) {
std::unordered_map<char, int> map = {{'I', 1}, {'V', 5}, {'X', 10},
{'L', 50}, {'C', 100}, {'D', 500},
{'M', 1000}};
int sum = 0;
// Replace subtractive combinations with additive equivalents
s = regex_replace(
regex_replace(
regex_replace(
regex_replace(
regex_replace(
regex_replace(s, regex("CM"), "DCCCC"),
regex("CD"), "CCCC"),
regex("XC"), "LXXXX"),
regex("XL"), "XXXX"),
regex("IX"), "VIIII"),
regex("IV"), "IIII");
// Sum the values of the characters
for (auto c : s)
sum += map[c];
return sum;
}
};
```
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
## Frequently encountered in technical interviews
The provided code contains a list of companies and their interview frequencies, which has been a frequently asked question in technical interviews.
```
std::vector<std::pair<std::string, int>> interview_frequency = {
{"Amazon", 27}, {"Adobe", 16}, {"Apple", 13},
{"Google", 10}, {"Bloomberg ", 10}, {"Facebook", 8},
{"Microsoft", 6}, {"tiktok ", 5}, {"Yahoo ", 4}};
```
# Big O Notation 101: The Secret to Writing Efficient Algorithms
From simple array operations to complex sorting algorithms, understanding the Big O Notation is critical for building high-performance software solutions.
1 - **O(1):** *This is the constant time notation*. The runtime remains steady regardless of input size. For example, accessing an element in an array by index and inserting/deleting an element in a hash table.
2 - **O(n):** *Linear time notation*. The runtime grows in direct proportion to the input size. For example, finding the max or min element in an unsorted array.
3 - **O(log n):** *Logarithmic time notation*. The runtime increases slowly as the input grows. For example, a binary search on a sorted array and operations on balanced binary search trees.
4 - **O(n^2):** *Quadratic time notation*. The runtime grows exponentially with input size. For example, simple sorting algorithms like bubble sort, insertion sort, and selection sort.
5 - **O(n^3):** *Cubic time notation*. The runtime escalates rapidly as the input size increases. For example, multiplying two dense matrices using the naive algorithm.
PC:Algomaster
6 - **O(n logn):** *Linearithmic time notation*. This is a blend of linear and logarithmic growth. For example, efficient sorting algorithms like merge sort, quick sort, and heap sort
7 - **O(2^n):** *Exponential time notation*. The runtime doubles with each new input element. For example, recursive algorithms solve problems by dividing them into multiple subproblems.
8 - **O(n!):** *Factorial time notation*. Runtime skyrockets with input size. For example, permutation-generation problems.
9 - **O(sqrt(n)):** *Square root time notation*. Runtime increases relative to the input’s square root. For example, searching within a range such as the Sieve of Eratosthenes for finding all primes up to n.
PC:ByteByteGo
| 36 | 0 | ['Hash Table', 'Math', 'String', 'C++'] | 0 |
roman-to-integer | Java clean and fast solution | java-clean-and-fast-solution-by-qiyidk-b4ud | public int romanToInt(String s) {\n int num = 0;\n int l = s.length();\n int last = 1000;\n for (int i = 0; i < | qiyidk | NORMAL | 2016-01-11T05:23:46+00:00 | 2018-10-04T13:43:59.356961+00:00 | 10,645 | false | public int romanToInt(String s) {\n int num = 0;\n int l = s.length();\n int last = 1000;\n for (int i = 0; i < l; i++){\n int v = getValue(s.charAt(i));\n if (v > last) num = num - last * 2;\n num = num + v;\n last = v;\n }\n return num;\n }\n private int getValue(char c){\n switch(c){\n case 'I' : return 1;\n case 'V' : return 5;\n case 'X' : return 10;\n case 'L' : return 50;\n case 'C' : return 100;\n case 'D' : return 500;\n case 'M' : return 1000;\n default : return 0;\n }\n } | 35 | 1 | ['Java'] | 3 |
roman-to-integer | Python 3 solution less than 98% Memory Usage with explanation. | python-3-solution-less-than-98-memory-us-c5sf | Take one example:\nPre-calculate for "IV" which represent -2.\nNote: If you traversal from string, you will count "I", "V", "IV" => 5 + 1 + (-2) = 4.\nSo I put | wabesasa | NORMAL | 2022-04-13T04:38:33.444858+00:00 | 2022-04-13T04:39:16.604495+00:00 | 2,049 | false | Take one example:\nPre-calculate for "IV" which represent -2.\nNote: If you traversal from string, you will count "I", "V", "IV" => 5 + 1 + (-2) = 4.\nSo I put "IV" to -2.\n\n```\nclass Solution:\n def romanToInt(self, s: str) -> int:\n mapping = {\n "I": 1,\n "V": 5,\n "X": 10,\n "L": 50,\n "C": 100,\n "D": 500,\n "M": 1000,\n "IV": -2,\n "IX": -2,\n "XL": -20,\n "XC": -20,\n "CD": -200,\n "CM": -200,\n }\n sum = 0\n for symbol, val in mapping.items():\n sum += s.count(symbol) * val\n return sum\n```\n\nIf you like it please upvote!\n | 34 | 0 | ['Python3'] | 2 |
roman-to-integer | [Rust] pattern matching without extra allocation | rust-pattern-matching-without-extra-allo-kjej | With only 2.3MiB memory usage and 100% ranking\n- Only use pattern guard expression (MatchArmGuard)\n\nrust\nimpl Solution {\n pub fn roman_to_int(s: String) | leetcodebot168 | NORMAL | 2019-09-04T14:42:08.516228+00:00 | 2019-09-04T14:51:23.972409+00:00 | 2,033 | false | - With only 2.3MiB memory usage and 100% ranking\n- Only use pattern guard expression (MatchArmGuard)\n\n```rust\nimpl Solution {\n pub fn roman_to_int(s: String) -> i32 {\n s.chars().rfold(0, |acc, c| {\n acc + match c {\n \'I\' if acc >= 5 => -1,\n \'I\' => 1,\n \'V\' => 5,\n \'X\' if acc >= 50 => -10,\n \'X\' => 10,\n \'L\' => 50,\n \'C\' if acc >= 500 => -100,\n \'C\' => 100,\n \'D\' => 500,\n \'M\' => 1000,\n _ => 0,\n }\n })\n }\n}\n``` | 34 | 0 | ['Rust'] | 2 |
roman-to-integer | Simple java solution, 100% time | simple-java-solution-100-time-by-andreaz-twhh | \npublic int romanToInt(String s) {\n int n = 0;\n char prev = \' \';\n for (byte i = 0; i < s.length(); i++) {\n char c = s.cha | andreazube | NORMAL | 2019-04-24T23:49:32.379855+00:00 | 2019-04-24T23:49:32.379896+00:00 | 5,112 | false | ```\npublic int romanToInt(String s) {\n int n = 0;\n char prev = \' \';\n for (byte i = 0; i < s.length(); i++) {\n char c = s.charAt(i);\n n += getValue(c, prev);\n prev = c;\n }\n \n return n;\n }\n \n private int getValue(char c, char prev) {\n switch (c) {\n case \'I\':\n return 1;\n case \'V\':\n return prev == \'I\' ? 3 : 5;\n case \'X\':\n return prev == \'I\' ? 8 : 10;\n case \'L\':\n return prev == \'X\' ? 30 : 50;\n case \'C\':\n return prev == \'X\' ? 80 : 100;\n case \'D\':\n return prev == \'C\' ? 300 : 500;\n case \'M\':\n return prev == \'C\' ? 800 : 1000;\n }\n \n throw new IllegalArgumentException();\n }\n``` | 33 | 0 | ['Java'] | 5 |
roman-to-integer | 13. Roman to Integer ✅ | Golang | 4ms 2MB | Simple readable solution | 13-roman-to-integer-golang-4ms-2mb-simpl-96jj | \n\n# Code\n\nfunc romanToInt(s string) int {\nsum := 0\n\n\trim := map[string]int{\n\t\t"I": 1,\n\t\t"V": 5,\n\t\t"X": 10,\n\t\t"L": 50,\n\t\t"C": 100,\n\t\t"D | realtemirov | NORMAL | 2023-11-30T15:34:27.340543+00:00 | 2023-11-30T15:34:27.340566+00:00 | 2,483 | false | \n\n# Code\n```\nfunc romanToInt(s string) int {\nsum := 0\n\n\trim := map[string]int{\n\t\t"I": 1,\n\t\t"V": 5,\n\t\t"X": 10,\n\t\t"L": 50,\n\t\t"C": 100,\n\t\t"D": 500,\n\t\t"M": 1000,\n\t}\n\n\tfor i, v := range s {\n\t\tsum += rim[string(v)]\n\t\tif i != 0 {\n\t\t\tif rim[string(s[i-1])] < rim[string(v)] {\n\t\t\t\tsum -= 2 * rim[string(s[i-1])]\n\t\t\t}\n\t\t}\n\t}\n\n\treturn sum\n}\n``` | 31 | 0 | ['Go'] | 6 |
roman-to-integer | Golang simplest and efficient solution-- 100% faster | golang-simplest-and-efficient-solution-1-5nqd | \nfunc romanToInt(s string) int {\n var romanMap = map[byte]int{\'I\':1, \'V\':5, \'X\':10, \'L\':50, \'C\':100, \'D\':500, \'M\':1000}\n var result = rom | punitpandey | NORMAL | 2020-04-15T05:46:21.855996+00:00 | 2020-04-15T05:47:02.617369+00:00 | 3,505 | false | ```\nfunc romanToInt(s string) int {\n var romanMap = map[byte]int{\'I\':1, \'V\':5, \'X\':10, \'L\':50, \'C\':100, \'D\':500, \'M\':1000}\n var result = romanMap[s[len(s)-1]]\n \n for i := len(s)-2; i >= 0; i-- {\n if romanMap[s[i]] < romanMap[s[i+1]] {\n result -= romanMap[s[i]]\n } else {\n result += romanMap[s[i]]\n }\n\t}\n\treturn result\n}\n``` | 31 | 0 | ['Go'] | 1 |
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