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values | created_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | updated_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | hit_count int64 0 10.6M | has_video bool 2
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construct-string-with-repeat-limit | Pretty Straightforward || Using max heap | pretty-straightforward-using-max-heap-by-1zzp | Intuition\n Describe your first thoughts on how to solve this problem. \nStore the frequency of each character in the map. Push them into the heap pair wise {c | jeet_sankala | NORMAL | 2024-01-08T00:12:08.601197+00:00 | 2024-01-08T00:12:08.601245+00:00 | 236 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nStore the frequency of each character in the map. Push them into the heap pair wise {char,freq}.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nPop the elements and take care of their freq . if(freq>limit) then you... | 2 | 0 | ['Greedy', 'Heap (Priority Queue)', 'C++'] | 1 |
construct-string-with-repeat-limit | [Python3] Heap approach, beats 100% in time | python3-heap-approach-beats-100-in-time-puk1u | \n# Code\n\nclass Solution:\n def repeatLimitedString(self, s: str, repeatLimit: int) -> str:\n # Create max heap based on the lexigraphic order and n | pandede | NORMAL | 2024-01-05T16:53:56.487773+00:00 | 2024-01-05T16:53:56.487802+00:00 | 109 | false | \n# Code\n```\nclass Solution:\n def repeatLimitedString(self, s: str, repeatLimit: int) -> str:\n # Create max heap based on the lexigraphic order and number of occurrences \n heap = [\n ... | 2 | 0 | ['Heap (Priority Queue)', 'Python3'] | 1 |
construct-string-with-repeat-limit | Python 3 Linked list - O(n) time and space - if you want to avoid the O(logN) heap push/pop | python-3-linked-list-on-time-and-space-i-ckv9 | Approach\nThis approach is similar to that of a priority queue / heap.\n\nHowever, using a linked list avoid the O(logN) time operations of heappush() and heapp | leokln | NORMAL | 2023-10-25T15:08:49.505412+00:00 | 2023-10-25T15:08:49.505432+00:00 | 120 | false | # Approach\nThis approach is similar to that of a priority queue / heap.\n\nHowever, using a linked list avoid the O(logN) time operations of heappush() and heappop().\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\n# Code\n```\n"""\nImplementation of a Node in a linked list.\n"""\nclass Node:... | 2 | 0 | ['Python3'] | 2 |
construct-string-with-repeat-limit | Using priority queue (faster 100%) | using-priority-queue-faster-100-by-remed-13sb | Approach\nWe count every character and use\npriority queue to achieve lexicographical order\n\nE.g zzzzccca, repeatLimit=3\n0) we build a queue: [z:4, c:3, a:1] | remedydev | NORMAL | 2023-02-24T00:16:34.535535+00:00 | 2023-02-24T00:27:14.815199+00:00 | 193 | false | # Approach\nWe count every character and use\npriority queue to achieve lexicographical order\n```\nE.g zzzzccca, repeatLimit=3\n0) we build a queue: [z:4, c:3, a:1]\n1) Start iterating a queue, on every iteration we\n 2) dequeue {z:4} \n 3) generate "zzz" (now we left with {z:1})\n 4) pick next element in a queue {... | 2 | 0 | ['JavaScript'] | 0 |
construct-string-with-repeat-limit | c++ faster = and less memory than 90% with explanation | c-faster-and-less-memory-than-90-with-ex-6ib1 | Intuition\n Describe your first thoughts on how to solve this problem. \n 1. construct a map\n 2. then build a string using largest char first, wi | o2thief | NORMAL | 2023-02-19T19:32:31.634086+00:00 | 2023-02-19T19:32:31.634111+00:00 | 210 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n 1. construct a map\n 2. then build a string using largest char first, with in repeatlimit.\n 3. if reach limit, insert a second largest char and continue try using largest char\n# Approach\n<!-- Describe your approac... | 2 | 0 | ['C++'] | 1 |
construct-string-with-repeat-limit | Java | O(n) | No map or queue | Beats > 95% | java-on-no-map-or-queue-beats-95-by-judg-cyl5 | Intuition\n Describe your first thoughts on how to solve this problem. \nCreate a frequency count array for all letters in the input string. Iterate over the fr | judgementdey | NORMAL | 2023-01-12T22:22:00.847640+00:00 | 2023-01-12T22:22:45.570073+00:00 | 126 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nCreate a frequency count array for all letters in the input string. Iterate over the frequency count array from the count of `z` to `a`. Skip the letters for which the count is `0`. Append one letter at a time to the output string till th... | 2 | 0 | ['String', 'Greedy', 'Counting', 'Java'] | 0 |
construct-string-with-repeat-limit | python 3 || O(n)/O(1) || without priority queue | python-3-ono1-without-priority-queue-by-awudi | ```\nclass Solution:\n def repeatLimitedString(self, s: str, repeatLimit: int) -> str:\n count = collections.Counter(s)\n chrs = list(map(list, | derek-y | NORMAL | 2022-04-23T02:06:30.820359+00:00 | 2022-04-23T02:06:30.820406+00:00 | 228 | false | ```\nclass Solution:\n def repeatLimitedString(self, s: str, repeatLimit: int) -> str:\n count = collections.Counter(s)\n chrs = list(map(list, sorted(count.items(), reverse=True)))\n res = []\n first, second = 0, 1\n n = len(chrs)\n \n while second < n:\n ... | 2 | 0 | ['Greedy', 'Python', 'Python3'] | 1 |
construct-string-with-repeat-limit | C++ | Construct String with Repeat Limit. | O(n) approach using priority queue | c-construct-string-with-repeat-limit-on-nu6km | class Solution {\npublic:\n \n \n string repeatString(char ch,int num) {\n string res="";\n while(num--)\n res+=ch;\n r | vamsimudaliar | NORMAL | 2022-02-27T04:50:23.548755+00:00 | 2022-02-27T04:50:23.548785+00:00 | 161 | false | class Solution {\npublic:\n \n \n string repeatString(char ch,int num) {\n string res="";\n while(num--)\n res+=ch;\n return res;\n }\n \n string repeatLimitedString(string s, int repeatLimit) {\n \n priority_queue<pair<char,int>> pq;\n vector<int> ... | 2 | 0 | ['C', 'Heap (Priority Queue)'] | 0 |
construct-string-with-repeat-limit | Java | Time O(n) | Space O(1) | java-time-on-space-o1-by-williamhsucs-azbt | \nclass Solution {\n /**\n * Time O(n)\n * Space O(1)\n */\n public String repeatLimitedString(String s, int repeatLimit) {\n String dic = "abcdefghi | williamhsucs | NORMAL | 2022-02-24T16:13:29.946103+00:00 | 2022-02-24T16:13:29.946158+00:00 | 177 | false | ```\nclass Solution {\n /**\n * Time O(n)\n * Space O(1)\n */\n public String repeatLimitedString(String s, int repeatLimit) {\n String dic = "abcdefghijklmnopqrstuvwxyz";\n // Space O(26)\n int[] bucket = new int[26];\n // Time O(n)\n for (int i = 0; i < s.length(); i++) {\n bucket[s.charAt... | 2 | 0 | ['Java'] | 0 |
construct-string-with-repeat-limit | [Python] Simple solution faster than 100%- Using 2 pointers + sorting | python-simple-solution-faster-than-100-u-233l | Hi ,\n\nHere is a simple solution. Please Upvote if you like!\n\nNote: \n\t\tletters = sorted(list(c.keys())+[\'\']) is to take care of the edgecase when seco | pal1545 | NORMAL | 2022-02-21T10:11:49.462259+00:00 | 2022-02-21T10:11:49.462289+00:00 | 222 | false | Hi ,\n\nHere is a simple solution. Please **Upvote** if you like!\n\nNote: \n\t\tletters = sorted(list(c.keys())+[\'\']) is to take care of the edgecase when `second` is None\n\n```\n\n def repeatLimitedString(self, s: str, repeatLimit: int) -> str:\n\t\n c = collections.Counter(s)\n res = []\n ... | 2 | 0 | ['Python'] | 0 |
construct-string-with-repeat-limit | CONFUSION | HELP NEEDED | confusion-help-needed-by-aayushisingh170-y7an | Can someone explain why the answer returned for testcase #2 aababab is bbabaa and not bbaabaa even when the latter is lexicographically larger? | aayushisingh1703 | NORMAL | 2022-02-20T16:29:33.269234+00:00 | 2022-02-20T16:32:21.103160+00:00 | 57 | false | Can someone explain why the answer returned for testcase #2 ```aababab``` is ```bbabaa``` and not ```bbaabaa``` even when the latter is lexicographically larger? | 2 | 0 | [] | 1 |
construct-string-with-repeat-limit | [c++] fully commented with O(N) time complexity and O(1) space | c-fully-commented-with-on-time-complexit-zwj8 | class Solution {\npublic:\n \n string repeatLimitedString(string s, int l) {\n\t\tint m[26]; // variable for storing 26 characters\n\t\n for(char c | abhishekjha9909 | NORMAL | 2022-02-20T16:12:44.944169+00:00 | 2022-02-20T16:12:44.944205+00:00 | 81 | false | class Solution {\npublic:\n \n string repeatLimitedString(string s, int l) {\n\t\tint m[26]; // variable for storing 26 characters\n\t\n for(char c:s)m[c-\'a\']++; // frequncy count;\n \n int i=25,j; //i-> last character ie., \'z\' \n \n string res=""; //for storing ans \n ... | 2 | 1 | ['C', 'C++'] | 1 |
construct-string-with-repeat-limit | C++, Worst to best, Bruteforce (N^2), priorityQueue(O(NlogN)), Counting + greedy(O(N)) | c-worst-to-best-bruteforce-n2-priorityqu-kmfx | Bruteforce Solution\n#1 Bruteforce Approach O(N^2) not accepted (failed at 144th testcase were the limit is \'1\')\n1. first sort the string.\n2. now traverse t | Albatross912 | NORMAL | 2022-02-20T11:26:04.116278+00:00 | 2022-02-20T11:26:04.116312+00:00 | 209 | false | # Bruteforce Solution\n**#1 Bruteforce Approach O(N^2) not accepted (failed at 144th testcase were the limit is \'1\')**\n1. first sort the string.\n2. now traverse through the string and find the index were the *limit* is reached as soon as the limit reached find the next element which is not equal to the current elem... | 2 | 0 | ['Greedy', 'C', 'Heap (Priority Queue)', 'Counting Sort', 'C++'] | 0 |
construct-string-with-repeat-limit | priority queue C++ solution #ACCEPTED😍 | priority-queue-c-solution-accepted-by-ki-xra8 | class Solution {\npublic:\n\n string repeatLimitedString(string s, int k) {\n string ans = "";\n \n unordered_mapmp;\n \n | kinggaurav | NORMAL | 2022-02-20T11:04:02.197992+00:00 | 2022-02-20T11:04:02.198024+00:00 | 207 | false | class Solution {\npublic:\n\n string repeatLimitedString(string s, int k) {\n string ans = "";\n \n unordered_map<char , int>mp;\n \n for(int i = 0 ; i < s.size() ; i++){\n mp[s[i]]++;\n }\n \n priority_queue<pair<char , int>>pq;\n \n f... | 2 | 0 | ['C', 'Heap (Priority Queue)'] | 0 |
construct-string-with-repeat-limit | Simple javascript solution - 159 ms | simple-javascript-solution-159-ms-by-efo-znbq | \nvar repeatLimitedString = function(s, repeatLimit) {\n const aCode = "a".charCodeAt();\n const counts = new Array(26).fill(0);\n let result = "";\n | eforce | NORMAL | 2022-02-20T08:30:16.475626+00:00 | 2022-02-20T08:30:16.475664+00:00 | 63 | false | ```\nvar repeatLimitedString = function(s, repeatLimit) {\n const aCode = "a".charCodeAt();\n const counts = new Array(26).fill(0);\n let result = "";\n let cnt, takeOne, len, lastLetter;\n \n for (let i = 0; i < s.length; ++i) {\n ++counts[s.charCodeAt(i) - aCode];\n }\n \n while (res... | 2 | 0 | [] | 0 |
construct-string-with-repeat-limit | Easy to understand | c++ | easy-to-understand-c-by-smit3901-mmt2 | \n// Using priority queue\n string repeatLimitedString(string s, int k) {\n string res = "";\n \n map<char,int> mp;\n \n f | smit3901 | NORMAL | 2022-02-20T08:04:45.548390+00:00 | 2022-02-20T08:04:45.548417+00:00 | 148 | false | ```\n// Using priority queue\n string repeatLimitedString(string s, int k) {\n string res = "";\n \n map<char,int> mp;\n \n for(auto x:s)\n {\n mp[x]++;\n }\n \n priority_queue<pair<char,int>> pq;\n \n for(auto x:mp)\n {\n ... | 2 | 0 | ['C', 'Heap (Priority Queue)'] | 0 |
construct-string-with-repeat-limit | Priority Queue || Max Heap || Map || C++ | priority-queue-max-heap-map-c-by-krishna-y84c | \nclass Solution {\npublic:\n string repeatLimitedString(string s, int r) {\n unordered_map<char,int>mp;\n for(auto c:s)\n {\n | krishnakant01 | NORMAL | 2022-02-20T06:12:35.434358+00:00 | 2022-02-20T06:14:02.611216+00:00 | 40 | false | ```\nclass Solution {\npublic:\n string repeatLimitedString(string s, int r) {\n unordered_map<char,int>mp;\n for(auto c:s)\n {\n mp[c]++;\n }\n // Keep the characters and it\'s count in max heap, Charater as first element\n priority_queue<pair<char,int>>pq;\n ... | 2 | 0 | [] | 0 |
construct-string-with-repeat-limit | My Java solution with comments | my-java-solution-with-comments-by-zadelu-ozsf | The approach is to add as many copies of the "higher" characters as possible, up to repeatLimit. Then, add only 1 copy of the next highest character as a separa | zadeluca | NORMAL | 2022-02-20T04:36:10.372053+00:00 | 2022-02-20T05:32:45.979725+00:00 | 152 | false | The approach is to add as many copies of the "higher" characters as possible, up to repeatLimit. Then, add only 1 copy of the next highest character as a separator, and "go back" to the previous character.\n\nFor example, with ```s = "zzzzyy"``` and ```repeatLimit = 2```, we:\n1) Add 2 copies of \'z\'. Since we still h... | 2 | 0 | ['Java'] | 2 |
construct-string-with-repeat-limit | Java || Simple || Map || Greedy | java-simple-map-greedy-by-lagahuahubro-silm | ```\nclass Solution {\n public String repeatLimitedString(String s, int limit) {\n int[] map = new int[26];\n int n = s.length();\n for | lagaHuaHuBro | NORMAL | 2022-02-20T04:20:09.574602+00:00 | 2022-02-20T04:24:31.629157+00:00 | 72 | false | ```\nclass Solution {\n public String repeatLimitedString(String s, int limit) {\n int[] map = new int[26];\n int n = s.length();\n for (int i = 0; i < n; i++) {\n map[s.charAt(i) - \'a\']++;\n }\n return construct(map, limit);\n }\n \n public String construct(i... | 2 | 0 | ['Greedy'] | 0 |
construct-string-with-repeat-limit | javascript greedy 192ms | javascript-greedy-192ms-by-henrychen222-wcpq | Main idea: greedy.\nto make lexical greatest, so each time use out current largest char (search from end 25 -> 0)\n (1) freq <= limit, append (freq char) direc | henrychen222 | NORMAL | 2022-02-20T04:14:09.627336+00:00 | 2022-02-20T04:28:05.317182+00:00 | 248 | false | Main idea: greedy.\nto make lexical greatest, so each time use out current largest char (search from end 25 -> 0)\n (1) freq <= limit, append (freq char) directly, f[i] = 0\n (2) freq > limit, append (limit char), find a bridge (second largest char, to make lexical greatest), append it, f[i] -= limit,\n until f... | 2 | 0 | ['Greedy', 'JavaScript'] | 0 |
construct-string-with-repeat-limit | C++ | Easy to understand | Hash Map | c-easy-to-understand-hash-map-by-parth_c-mvhy | \nclass Solution {\npublic:\n string repeatLimitedString(string s, int repeatLimit) {\n vector<int> cnt(26, 0);\n for(auto &x : s){\n | parth_chovatiya | NORMAL | 2022-02-20T04:13:10.356204+00:00 | 2022-02-20T04:13:10.356242+00:00 | 74 | false | ```\nclass Solution {\npublic:\n string repeatLimitedString(string s, int repeatLimit) {\n vector<int> cnt(26, 0);\n for(auto &x : s){\n cnt[x-97]++;\n }\n \n string ans = "";\n int j = 25;\n while(j>=0 and cnt[j]==0){\n j--;\n }\n ... | 2 | 2 | [] | 0 |
construct-string-with-repeat-limit | Java heap | simple | java-heap-simple-by-xavi_an-8g6x | ```\nclass Solution {\n public String repeatLimitedString(String s, int r) {\n int [] f = new int[26];\n for(int i=0;i<s.length();i++){\n | xavi_an | NORMAL | 2022-02-20T04:04:13.618306+00:00 | 2022-02-20T04:51:24.230624+00:00 | 110 | false | ```\nclass Solution {\n public String repeatLimitedString(String s, int r) {\n int [] f = new int[26];\n for(int i=0;i<s.length();i++){\n f[s.charAt(i)-\'a\']++; // frequency map with char count\n }\n PriorityQueue<int []> queue = new PriorityQueue<>((a, b) -> b[0] - a[0]); // ... | 2 | 1 | [] | 0 |
construct-string-with-repeat-limit | Easy Java Solution || Priority Queue | easy-java-solution-priority-queue-by-ver-usu8 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | vermaanshul975 | NORMAL | 2025-03-28T09:46:33.462330+00:00 | 2025-03-28T09:46:33.462330+00:00 | 9 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 1 | 0 | ['Java'] | 0 |
construct-string-with-repeat-limit | 🔥🔥🔥🔥O(N) solution 🔥🔥🔥🔥 | on-solution-by-vadapally_varshitha-ktea | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | vadapally_varshitha | NORMAL | 2025-01-12T17:52:00.838040+00:00 | 2025-01-12T17:52:00.838040+00:00 | 12 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 1 | 0 | ['Counting', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Dynamic programming and space optimized. Beats 100%, easy to understand. | dynamic-programming-and-space-optimized-3by5a | Intuition\n Describe your first thoughts on how to solve this problem. \nKind of dynamic programming\n\n# Approach\n\nFor each number we have 2 choices, add or | null_pointer_exception | NORMAL | 2024-06-23T04:02:27.369280+00:00 | 2024-06-23T04:51:04.422506+00:00 | 4,600 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nKind of dynamic programming\n\n# Approach\n\nFor each number we have 2 choices, add or subtract to previous result.\nWe will maintain 2 result states, **addResult** and **subResult**.\n\nNow coming to current element, if we want to\n- **A... | 53 | 5 | ['Dynamic Programming', 'Java', 'Python3'] | 7 |
maximize-total-cost-of-alternating-subarrays | Recursion + Memo | recursion-memo-by-donpaul271-fc4i | Function used in the recursion will have the following parameters.\n\npos - postion in the array\nstat - whether a subarray is already under construction or no | donpaul271 | NORMAL | 2024-06-23T04:03:48.946430+00:00 | 2024-06-26T04:19:09.576400+00:00 | 3,285 | false | Function used in the recursion will have the following parameters.\n\npos - postion in the array\nstat - whether a subarray is already under construction or not.\nsign - The sign of the current element if its added to the subarray \n\n-----\n\nWe can use a memo of the form dp[100001][2][2]\n(make sure the dp is initia... | 27 | 5 | ['C++'] | 6 |
maximize-total-cost-of-alternating-subarrays | Take not take dp! | take-not-take-dp-by-minato_10-ghe8 | Intuition\n Describe your first thoughts on how to solve this problem. \n -> This problem is similar to take not take dp the only twist\nit that we have to take | Minato_10 | NORMAL | 2024-06-23T04:02:50.242722+00:00 | 2024-06-23T04:11:16.156866+00:00 | 2,468 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n -> This problem is similar to take not take dp the only twist\nit that we have to take all the elements but we can change their sign based on the previous element.\n -> Here we are using f variable to check if we can change the sign of t... | 23 | 8 | ['Dynamic Programming', 'C++'] | 3 |
maximize-total-cost-of-alternating-subarrays | Striver's appraoch take or nottake for beginners!!!! | strivers-appraoch-take-or-nottake-for-be-ioaa | Intuition\nThe intution is simple just here I have just taken the index which represents the maximum value of cost and boolean flag represents the state wether | aero_coder | NORMAL | 2024-06-23T04:10:56.535098+00:00 | 2024-07-04T20:31:07.912353+00:00 | 2,076 | false | # Intuition\nThe intution is simple just here I have just taken the index which represents the maximum value of cost and boolean flag represents the state wether we have to multipy the nums[ind] by 1 or -1.\n\n# Approach\nHere as you can see that I have taken a 2d dp array.\nIn the find function, if we reach the base c... | 21 | 3 | ['Dynamic Programming', 'Memoization', 'C++'] | 5 |
maximize-total-cost-of-alternating-subarrays | [Java/C++/Python] Easy and Concise, O(1) Space | javacpython-easy-and-concise-o1-space-by-a12s | Explanation\npos is last element is positive sign.\nneg is last element is negative sign.\n\nAfter the neg, we need to append a positive A[i]\nAfter the pos, we | lee215 | NORMAL | 2024-06-23T04:09:29.028863+00:00 | 2024-06-23T07:44:47.596975+00:00 | 1,794 | false | # **Explanation**\n`pos` is last element is positive sign.\n`neg` is last element is negative sign.\n\nAfter the `neg`, we need to append a positive `A[i]`\nAfter the `pos`, we can append a positive/negative `A[i]`\n<br>\n\n# **Complexity**\nTime `O(n)`\nSpace `O(n)`\n<br>\n\n**Java**\n```java\n public long maximumT... | 21 | 6 | ['C', 'Python', 'Java'] | 6 |
maximize-total-cost-of-alternating-subarrays | Masked House Robber | masked-house-robber-by-votrubac-cj9z | I did not recognize the hourse robber at first.\n\nWhen we see a positive number, we just take it (starting a new subarray).\n\nFor a streak of negative numbers | votrubac | NORMAL | 2024-06-23T04:08:12.046038+00:00 | 2024-06-23T04:32:32.637973+00:00 | 2,299 | false | I did not recognize the hourse robber at first.\n\nWhen we see a positive number, we just take it (starting a new subarray).\n\nFor a streak of negative numbers, we can flip some of them to positive. \n\nHowever, we cannot flip two numbers if they are next to each other.\n- The house robber algorithm can find the maxim... | 19 | 3 | ['C'] | 2 |
maximize-total-cost-of-alternating-subarrays | Easy Video Solution 🔥 || How to 🤔 in Interview || Dynamic programming || Memoization 🔥 | easy-video-solution-how-to-in-interview-zo50z | If you like the solution Please Upvote and subscribe to my youtube channel\n\n# Easy Video Explanation\n\nhttps://youtu.be/ZzOdbGx3KAs\n\n# Code\n\nclass Soluti | ayushnemmaniwar12 | NORMAL | 2024-06-23T05:50:18.922565+00:00 | 2024-06-23T07:08:01.932793+00:00 | 459 | false | ***If you like the solution Please Upvote and subscribe to my youtube channel***\n\n# ***Easy Video Explanation***\n\nhttps://youtu.be/ZzOdbGx3KAs\n\n# Code\n```\nclass Solution {\npublic:\n long long solve(int i,int f,int n,vector<int>&v,vector<vector<long long>>&dp) {\n if(i==n)\n return 0;\n ... | 10 | 3 | ['Dynamic Programming', 'Memoization', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | [Java/Python 3] DP O(n) codes, from space O(n) to O(1). | javapython-3-dp-on-codes-from-space-on-t-puos | Except nums[0], we can always choose an item in the input either as original value, or flip its sign plus keep previous item as original value, and there is no | rock | NORMAL | 2024-06-23T04:02:17.847474+00:00 | 2024-06-28T18:09:35.484054+00:00 | 459 | false | Except `nums[0]`, we can always choose an item in the input either as original value, or flip its sign plus keep previous item as original value, and there is no other options; Therefore, starting from `i = 1, nums[i]`, our state transition function is `dp[i + 1] = max(dp[i] + nums[i], dp[i - 1] + nums[i - 1] - nums[i]... | 9 | 0 | ['Dynamic Programming', 'Java', 'Python3'] | 1 |
maximize-total-cost-of-alternating-subarrays | Python 3 || 4 lines, iteration with dp || T/S: 99% / 98% | python-3-4-lines-iteration-with-dp-ts-99-alae | Here\'s the plan:\n1. We initialize dp1 and dp2 with the first element of nums. These integer variables keep track of the current maximum total that can be obta | Spaulding_ | NORMAL | 2024-06-23T18:16:33.468146+00:00 | 2024-07-02T15:03:49.360495+00:00 | 49 | false | Here\'s the plan:\n1. We initialize `dp1` and `dp2` with the first element of `nums`. These integer variables keep track of the current maximum total that can be obtained by either adding(`dp1`) or subtracting(`dp2`) the current element to the previous maximum total.\n\n1. We iterate though `nums`, updating `dp1` and ... | 8 | 0 | ['Python', 'Python3'] | 0 |
maximize-total-cost-of-alternating-subarrays | Solution By Dare2Solve | Detailed Explanation | 1D DP | solution-by-dare2solve-detailed-explanat-c23m | Explanation []\nauthorslog.vercel.app/blog/fwKyrRoXX5\n\n\n# Code\n\ncpp []\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n | Dare2Solve | NORMAL | 2024-06-23T04:06:01.352072+00:00 | 2024-06-23T04:07:24.518423+00:00 | 739 | false | ```Explanation []\nauthorslog.vercel.app/blog/fwKyrRoXX5\n```\n\n# Code\n\n```cpp []\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n int n = nums.size();\n vector<long long> dp(n + 1, -LLONG_MAX);\n dp[0] = 0;\n dp[1] = nums[0];\n for (int i = 1; i < ... | 7 | 0 | ['Dynamic Programming', 'Python', 'C++', 'Java', 'Go', 'Python3', 'JavaScript', 'C#'] | 0 |
maximize-total-cost-of-alternating-subarrays | DP: Recursion Memo-> tabular with space O(1)||46ms beats 100% | dp-recursion-memo-tabular-with-space-o14-w5mh | Intuition\n Describe your first thoughts on how to solve this problem. \nTry DP.\nUse Recursion with Memo-> Tabular with optimized space.\n# Approach\n Describe | anwendeng | NORMAL | 2024-06-23T07:50:49.951834+00:00 | 2024-06-23T07:50:49.951866+00:00 | 476 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTry DP.\nUse Recursion with Memo-> Tabular with optimized space.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Define recursive function `long long f(int i, bool minus, vector<int>& nums)` with cache `long long... | 6 | 0 | ['Dynamic Programming', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Simple DP Approach || Memorization || Java || C++ || Python | simple-dp-approach-memorization-java-c-p-pthc | Intuition\n Describe your first thoughts on how to solve this problem. \nWe need to consider only subarrays of size 1 and 2 to maximize the cost.\n\n\n\n\n\n\n\ | jeevanraj73 | NORMAL | 2024-06-23T05:55:11.410942+00:00 | 2024-06-23T05:55:11.410970+00:00 | 652 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe need to consider only subarrays of size 1 and 2 to maximize the cost.\n\n\n\n\n\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe proposed solution uses a dynamic programming approach with space optimization... | 6 | 0 | ['Dynamic Programming', 'Memoization', 'C++', 'Java', 'Python3'] | 1 |
maximize-total-cost-of-alternating-subarrays | DP || Pick/ Not Pick || O(n) | dp-pick-not-pick-on-by-thiennk-5ohm | 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 | Thiennk | NORMAL | 2024-06-23T04:02:11.494428+00:00 | 2024-06-23T07:09:13.469545+00:00 | 336 | 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(2*n)\n<!-- Add your space complexity here, e.g.... | 6 | 0 | ['Dynamic Programming', 'Kotlin'] | 0 |
maximize-total-cost-of-alternating-subarrays | C++ || Simple DP Solution | c-simple-dp-solution-by-_potter-t0lk | Code\n\nclass Solution {\npublic:\n // Consider index starts with 1\n long long dp[100001][2];\n long long topDown(int i, int oddIndex, vector<int> &nu | _Potter_ | NORMAL | 2024-06-23T04:31:44.459153+00:00 | 2024-06-23T04:38:17.067336+00:00 | 321 | false | # Code\n```\nclass Solution {\npublic:\n // Consider index starts with 1\n long long dp[100001][2];\n long long topDown(int i, int oddIndex, vector<int> &nums){\n if(i == nums.size()){\n return 0;\n }\n if(dp[i][oddIndex] != -1) {\n return dp[i][oddIndex];\n }\... | 5 | 0 | ['Dynamic Programming', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | [Beats 100%] 4 approaches HOW TO BUILD UP TO OPTIMAL SOLUTION + Full thought process | beats-100-4-approaches-how-to-build-up-t-z7jv | Approach\nThe goal is to maximize the sum of alternating subarrays.\n\nAn alternating subarray starts by adding the first element, then subtracts the 2nd, adds | AlecLC | NORMAL | 2024-06-23T04:17:21.766916+00:00 | 2024-06-23T04:17:21.766938+00:00 | 205 | false | # Approach\nThe goal is to maximize the sum of alternating subarrays.\n\nAn alternating subarray starts by adding the first element, then subtracts the 2nd, adds the third, etc.\nIn other words, add up all the even indices and subtract the odd ones.\n\nThe first thing to do is look at the constraints. Solutions TLE aro... | 5 | 0 | ['Python3'] | 0 |
maximize-total-cost-of-alternating-subarrays | Take not take dp | take-not-take-dp-by-techtinkerer-2nk7 | \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 | TechTinkerer | NORMAL | 2024-06-23T07:39:18.315330+00:00 | 2024-06-23T07:40:05.226438+00:00 | 256 | false | \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```\n#define ll long long\nclass Solution {\npublic:\n vector<int> nums;\n int n;\n ll dp[100001][2];\n \n ll solve... | 4 | 0 | ['Dynamic Programming', 'Greedy', 'C++'] | 1 |
maximize-total-cost-of-alternating-subarrays | C++/Python3/Java Solution | Beats 100% | 4 Line ✔️ | cpython3java-solution-beats-100-4-line-b-5036 | Intuition\nUpon first glance, the problem seems to involve a pattern of alternating +ve and -ve signs. It appears as if we start with a positive sign and altern | satyam2001 | NORMAL | 2024-06-24T13:38:04.100549+00:00 | 2024-09-21T09:39:37.680342+00:00 | 174 | false | # Intuition\nUpon first glance, the problem seems to involve a pattern of alternating `+ve and -ve signs`. It appears as if we start with a positive sign and alternate signs thereafter, and we can switch all the subsequent signs if the current sign is negative. To maximize the cost, we need to determine the optimal pos... | 3 | 0 | ['Dynamic Programming', 'C++', 'Java', 'Python3'] | 0 |
maximize-total-cost-of-alternating-subarrays | ✅✅ Detailed explanation || Recursive, Memoized Solution 😱😃 | detailed-explanation-recursive-memoized-7pzfn | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires splitting the array into subarrays to maximize the total cost, whe | Rishu_Raj5938 | NORMAL | 2024-06-23T13:00:16.151254+00:00 | 2024-06-23T13:02:56.564457+00:00 | 160 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires splitting the array into subarrays to maximize the total cost, where the cost of each subarray alternates between adding and subtracting elements. We use a recursive approach with memoization to solve this efficiently... | 3 | 0 | ['Memoization', 'C++', 'Java', 'Python3'] | 1 |
maximize-total-cost-of-alternating-subarrays | C++ O(N) solution . | c-on-solution-by-deleted_user-1eq7 | Intuition\n Describe your first thoughts on how to solve this problem. \nSince in a subarray, sum is el1 - el2 + el3 - el4 ....\nWe can use subarray of size ei | deleted_user | NORMAL | 2024-06-23T04:23:00.183449+00:00 | 2024-06-23T04:23:00.183467+00:00 | 43 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSince in a subarray, sum is el1 - el2 + el3 - el4 ....\nWe can use subarray of size either 1 or 2 .\nRest is DP.\n# Complexity\n- Time complexity:$$O(N)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(N... | 3 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | 2d DP memoized solution do upvote me 😎😎 | 2d-dp-memoized-solution-do-upvote-me-by-21hns | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | rahulX64 | NORMAL | 2024-06-26T08:47:49.659066+00:00 | 2024-06-26T08:47:49.659094+00:00 | 33 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Simple || DP | Easy | C++ | O(N) | Seven line code | optimal | simple-dp-easy-c-on-seven-line-code-opti-nsgh | Intuition\n Describe your first thoughts on how to solve this problem. \nTo maximize the total cost of subarrays, we need to strategically decide where to split | gavnish_kumar | NORMAL | 2024-06-24T08:46:53.279079+00:00 | 2024-06-24T08:46:53.279117+00:00 | 91 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo maximize the total cost of subarrays, we need to strategically decide where to split the array. The cost of a subarray nums[l..r] alternates between adding and subtracting the elements based on their positions (even-indexed elements ar... | 2 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | DP. One line solution. Time O(n). Space O(1) | dp-one-line-solution-time-on-space-o1-by-jfeh | \nclass Solution:\n def maximumTotalCost(self, n: List[int]) -> int:\n return reduce(lambda d, m:(d[1], max(d[0]+m[0]-m[1], d[1]+m[1])), pairwise(n), | xxxxkav | NORMAL | 2024-06-23T22:26:37.485132+00:00 | 2024-06-23T23:35:58.247570+00:00 | 38 | false | ```\nclass Solution:\n def maximumTotalCost(self, n: List[int]) -> int:\n return reduce(lambda d, m:(d[1], max(d[0]+m[0]-m[1], d[1]+m[1])), pairwise(n), (0,n[0]))[1]\n```\n> More readable\n```\nclass Solution:\n def maximumTotalCost(self, nums: List[int]) -> int:\n dp = (0, nums[0])\n for a, ... | 2 | 0 | ['Dynamic Programming', 'Python3'] | 1 |
maximize-total-cost-of-alternating-subarrays | Two Approaches - Recursive and Dynamic Programming Solutions | two-approaches-recursive-and-dynamic-pro-6q1z | Intuition\nThe problem is similar to jump by 1 step and jump by 2 step, whereas in this case, we are referring it to select the subarray of length 1 or of lengt | nanda_kumar_2005 | NORMAL | 2024-06-23T06:27:16.954123+00:00 | 2024-06-23T06:27:16.954157+00:00 | 79 | false | # Intuition\nThe problem is similar to jump by 1 step and jump by 2 step, whereas in this case, we are referring it to select the subarray of length 1 or of length 2.\n\n# Complexity\nBetter complexity when DP array is used over recursion, to pre solve the subproblems in recursion.\n\n# Tip\nBoth the solutions are give... | 2 | 0 | ['Dynamic Programming', 'Recursion', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | C++ Recursion, Memoization , Tabulation ✅ | c-recursion-memoization-tabulation-by-ya-r56y | 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 | yashwanthreddi | NORMAL | 2024-06-23T04:04:10.747164+00:00 | 2024-07-14T05:44:10.701700+00:00 | 448 | 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)$$ --... | 2 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 1 |
maximize-total-cost-of-alternating-subarrays | C++ | | Easy Solution | | Top-Down DP | | With Explanation | c-easy-solution-top-down-dp-with-explana-et0z | Approach\nAt each i, we have 2 options:\n Just take nums[i] in the subarray.\n Take nums[i] and nums[i+1] in the subarray\n\ndp[i]: Maximum sum of subarrays fro | yash8589 | NORMAL | 2024-06-23T04:03:48.878199+00:00 | 2024-06-23T04:03:48.878228+00:00 | 76 | false | # Approach\nAt each `i`, we have 2 options:\n* Just take `nums[i]` in the subarray.\n* Take `nums[i]` and `nums[i+1]` in the subarray\n\n`dp[i]`: Maximum sum of subarrays from `i` to `n-1`.\n\nBe sure to check for bounds in case of `i+1`\n\n# Complexity\n- Time complexity: $O(n)$\n\n# Code\n```\nclass Solution {\npubli... | 2 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | 🔥Clean DP Code✅ || ⚡Memoization | clean-dp-code-memoization-by-adish_21-t4xo | \n\n# Complexity\n\n- Time complexity:\nO(n * 2)\n\n- Space complexity:\nO(n * 2)\n\n\n# Code\n## Please Upvote if it helps\uD83E\uDD17\n\nclass Solution {\npub | aDish_21 | NORMAL | 2024-06-23T04:02:31.441455+00:00 | 2024-06-23T08:22:09.392160+00:00 | 460 | false | \n\n# Complexity\n```\n- Time complexity:\nO(n * 2)\n\n- Space complexity:\nO(n * 2)\n```\n\n# Code\n## Please Upvote if it helps\uD83E\uDD17\n```\nclass Solution {\npublic:\n long dp[100001][2];\n long helper(int ind, int n, vector<int>& nums, bool ch){\n if(ind == n)\n return 0;\n if(dp... | 2 | 0 | ['Dynamic Programming', 'Memoization', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | c++ dp | c-dp-by-pb_matrix-61b9 | \nclass Solution {\npublic:\n vector<int>v;\n int n;\n long long dp[100001][2][2];\n long long fn(int ind,int st,int o){\n if(ind>=n) return | pb_matrix | NORMAL | 2024-09-29T11:18:27.198475+00:00 | 2024-09-29T11:18:27.198506+00:00 | 23 | false | ```\nclass Solution {\npublic:\n vector<int>v;\n int n;\n long long dp[100001][2][2];\n long long fn(int ind,int st,int o){\n if(ind>=n) return 0;\n if(dp[ind][st][o]!=-1) return dp[ind][st][o];\n long long x;\n if(st) x=v[ind]+max(fn(ind+1,0,1),fn(ind+1,1,0));\n else{\n ... | 1 | 0 | ['Dynamic Programming', 'C'] | 1 |
maximize-total-cost-of-alternating-subarrays | beats 100% || considers all conditions seperately | beats-100-considers-all-conditions-seper-njwb | Intuition\nfirstly tried greedy,though of binary search,etc but a regular dp question\n Describe your first thoughts on how to solve this problem. \n\n\n\n# Com | mzip19 | NORMAL | 2024-06-23T20:04:14.533896+00:00 | 2024-06-23T20:04:14.533928+00:00 | 201 | false | # Intuition\nfirstly tried greedy,though of binary search,etc but a regular dp question\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\n\n# Complexity\n- Time complexity:\n- $$O(n)$$ dp array of 2*n size\n\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n- $$O(n)$... | 1 | 0 | ['Dynamic Programming', 'Java'] | 2 |
maximize-total-cost-of-alternating-subarrays | Easy to Understand (intuitive) , Tabulation | easy-to-understand-intuitive-tabulation-nvkpq | Intuition\n Describe your first thoughts on how to solve this problem. \n1. There is no constraints on number of partitions.\n2. we can create as many as we wan | quack_quack | NORMAL | 2024-06-23T18:29:07.178004+00:00 | 2024-06-24T13:16:03.602509+00:00 | 244 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1. There is no constraints on number of partitions.\n2. we can create as many as we want , think in group of 2 ,at any index i , think about i , and i - 1 only\n3. Think about these case:\n a. [-1 , -2] => 1\n b. [-1 , -2 , -5] ... | 1 | 0 | ['Dynamic Programming', 'C++'] | 2 |
maximize-total-cost-of-alternating-subarrays | Top down DP | | Space optimization(O(1)) | top-down-dp-space-optimizationo1-by-rake-mgwc | \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(1 | Rakesh_kumar_reddy | NORMAL | 2024-06-23T15:49:07.585082+00:00 | 2024-06-23T15:49:07.585117+00:00 | 5 | false | \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(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int... | 1 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Dynamic Programming || Easy Approach with intuition || Memoization || C++ | dynamic-programming-easy-approach-with-i-ip7i | Intuition\nFor each element in the array, we have two possibilities:\n\n1. Start a new subarray from the current element.\n2. Continue the current subarray by a | arshika3086 | NORMAL | 2024-06-23T10:38:32.510008+00:00 | 2024-06-23T10:38:32.510035+00:00 | 134 | false | # Intuition\nFor each element in the array, we have two possibilities:\n\n1. Start a new subarray from the current element.\n2. Continue the current subarray by adding the current element.\n\nWe use a dynamic programming approach to keep track of these possibilities and make optimal decisions at each step.\n\n# Approac... | 1 | 0 | ['Dynamic Programming', 'Memoization', 'C++'] | 1 |
maximize-total-cost-of-alternating-subarrays | 3Dp with state , index and minus 1 as parameters | 3dp-with-state-index-and-minus-1-as-para-4hr7 | State represents whether you are in continuation of the previous array or not (1 or 0 respectively).\n\nState is either 0 or 1 we will have 2 choices in both ca | Tilak27 | NORMAL | 2024-06-23T09:31:45.169123+00:00 | 2024-06-23T09:31:45.169147+00:00 | 21 | false | State represents whether you are in continuation of the previous array or not (1 or 0 respectively).\n\nState is either 0 or 1 we will have 2 choices in both cases i.e whether to continue the same state or not.\nend_array means we are ending the array at that index.\ncurr_array that the next call will be in continuatio... | 1 | 0 | ['C++'] | 1 |
maximize-total-cost-of-alternating-subarrays | Explained tabulation dynamic programming. | explained-tabulation-dynamic-programming-a2h8 | Intuition\n Describe your first thoughts on how to solve this problem. \n\nYou are given an integer array nums with length n.\n\nThe cost of a subarray nums[l.. | ddhira123 | NORMAL | 2024-06-23T08:26:27.542484+00:00 | 2024-06-23T08:26:27.542508+00:00 | 56 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nYou are given an integer array `nums` with length $$n$$.\n\nThe cost of a subarray `nums[l..r]`, where $$0 \\leq l \\leq r < n$$, is defined as:\n\n$$\\text{cost}(l, r) = nums[l] - nums[l + 1] + ... + nums[r] \\times (\u22121)^{r \u2212... | 1 | 0 | ['Dynamic Programming', 'Java'] | 1 |
maximize-total-cost-of-alternating-subarrays | Rust/Python 3 lines solution with explanation how to get there | rustpython-3-lines-solution-with-explana-ngd6 | Intuition\n\nFirst it does not name sense to have an division in which we have more than 3 elements. For example if one of the divisions is a1, a2, a3, you can | salvadordali | NORMAL | 2024-06-23T07:53:30.825045+00:00 | 2024-06-23T07:53:30.825074+00:00 | 61 | false | # Intuition\n\nFirst it does not name sense to have an division in which we have more than 3 elements. For example if one of the divisions is `a1, a2, a3`, you can do the same with divison `a1, a2` and another one `a3`.\n\nNow we can look at the problem recursively.\n\n```\nf(arr) = max(\n arr[i] + f(arr[i + 1:]),\n ... | 1 | 0 | ['Python3', 'Rust'] | 0 |
maximize-total-cost-of-alternating-subarrays | Python solution using Recursion, Dynamic Programming (DP) | python-solution-using-recursion-dynamic-9s8wx | Intuition\n Describe your first thoughts on how to solve this problem. \ngiven a nums array, calculate the maximum total of subarray, \n- when split, start the | nvhp46 | NORMAL | 2024-06-23T07:48:21.468491+00:00 | 2024-06-23T07:48:21.468522+00:00 | 82 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\ngiven a nums array, calculate the maximum total of subarray, \n- when split, start the sign as +\n- when not split, flip the sign after each element.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUsing recursive ... | 1 | 0 | ['Dynamic Programming', 'Recursion', 'Python3'] | 0 |
maximize-total-cost-of-alternating-subarrays | easy dp c++ | easy-dp-c-by-shivanshu0287-x8vw | Intuition\ncase 1-when you multiply a element by -1 then you have only one option on next element to multiply it by 1(continue in same subarray or switch the su | shivanshu0287 | NORMAL | 2024-06-23T06:21:43.201023+00:00 | 2024-06-23T06:21:43.201053+00:00 | 8 | false | # Intuition\ncase 1-when you multiply a element by -1 then you have only one option on next element to multiply it by 1(continue in same subarray or switch the subarray).\n\n\ncase 2-when you multiply a element by 1 then you have 2 options:-\na-multiply by 1 (means swith the subarray).\nb-multiply by -1 (means continue... | 1 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Recursive + meme | TLE | recursive-meme-tle-by-vaibhav1701-2q53 | Intuition\n- At any index we have two option to be part of previous subarray or start a new subarray.\n\n# Code\n\nclass Solution {\n public static long maxi | vaibhav1701 | NORMAL | 2024-06-23T05:35:56.525985+00:00 | 2024-06-23T05:35:56.526008+00:00 | 39 | false | # Intuition\n- At any index we have two option to be part of previous subarray or start a new subarray.\n\n# Code\n```\nclass Solution {\n public static long maximumTotalCost(int[] nums) {\n HashMap<String, Long> map = new HashMap<>();\n\n return helper(0, false, nums, 0, map);\n }\n\n public sta... | 1 | 0 | ['Hash Table', 'Recursion', 'Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | Simple DP Similar to Take/Not-take✅ | simple-dp-similar-to-takenot-take-by-har-bhds | 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 | Harshsharma08 | NORMAL | 2024-06-23T05:04:28.623230+00:00 | 2024-06-23T05:04:28.623259+00:00 | 46 | 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)$$ --... | 1 | 0 | ['Array', 'Dynamic Programming', 'Memoization', 'C++', 'Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | dp || c++ || 2D DP | dp-c-2d-dp-by-aditya_kunwar-x65m | Intuition\ndekhlo code accha se intutively likhe hai\n\n# Code\n\nclass Solution {\npublic:\nlong long dp[100001][2];\nlong long f(vector<int>&nums,int flag,int | aditya_kunwar | NORMAL | 2024-06-23T04:51:33.492103+00:00 | 2024-06-23T04:51:33.492136+00:00 | 10 | false | # Intuition\ndekhlo code accha se intutively likhe hai\n\n# Code\n```\nclass Solution {\npublic:\nlong long dp[100001][2];\nlong long f(vector<int>&nums,int flag,int ind){\n if(nums.size()<=ind)return 0;\n if(dp[ind][flag]!=-1)return dp[ind][flag];\n long long split=0,skip=0;\n if(flag==0)\n split=1LL*nu... | 1 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | ✔️ Beats 100% : 1D -DP : Detailed Step-by-Step Explanation - C++/Java/Python/C#/JS/Ruby/Go/Dart | beats-100-1d-dp-detailed-step-by-step-ex-njhu | Intuition\nWe need to find the maximum cost of a subarray where the cost is calculated by flipping the sign of every second element. This means alternating sign | Cregan_wolf | NORMAL | 2024-06-23T04:47:05.493829+00:00 | 2024-06-23T04:48:26.964617+00:00 | 87 | false | # Intuition\nWe need to find the maximum cost of a subarray where the cost is calculated by flipping the sign of every second element. This means alternating signs as we move through the array. Dynamic Programming (DP) is suitable to solve this problem efficiently.\n\n# Approach\n1. **Dynamic Programming Table:** Use a... | 1 | 0 | ['Dynamic Programming', 'C++', 'Java', 'Go', 'Python3', 'JavaScript', 'C#', 'Dart'] | 1 |
maximize-total-cost-of-alternating-subarrays | Recursion+Memo-Easy Solution-O(n)-Dynamic Programming | recursionmemo-easy-solution-on-dynamic-p-cxz3 | Intuition\nSign and index represents the states of dp and if the sign remains 0 means we take the next subarray.\n\n# Approach\nTake or not take approach\n\n# C | DivyenduVimal | NORMAL | 2024-06-23T04:37:53.412344+00:00 | 2024-06-23T04:37:53.412369+00:00 | 91 | false | # Intuition\nSign and index represents the states of dp and if the sign remains 0 means we take the next subarray.\n\n# Approach\nTake or not take approach\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\n# Code\n```\nusing ll=long long;\nclass Solution {\npublic:\n ll dp[100001][2];\n lo... | 1 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Recursion -> memoization || Python code | recursion-memoization-python-code-by-rin-k4s8 | \n# Code\n\nclass Solution:\n def asd(self,i,nums,parity,dp):\n if(i>=len(nums)):\n return 0\n if(dp[i][parity]!=-1):\n r | rini03 | NORMAL | 2024-06-23T04:35:39.342892+00:00 | 2024-06-23T04:35:39.342916+00:00 | 82 | false | \n# Code\n```\nclass Solution:\n def asd(self,i,nums,parity,dp):\n if(i>=len(nums)):\n return 0\n if(dp[i][parity]!=-1):\n return dp[i][parity]\n add=nums[i]\n if(parity==1):\n dp[i][parity]=self.asd(i+1,nums,0,dp)-nums[i]\n else:\n dp[i]... | 1 | 0 | ['Recursion', 'Memoization', 'Python3'] | 2 |
maximize-total-cost-of-alternating-subarrays | DP Approach | Memoization | Tabulation | dp-approach-memoization-tabulation-by-ta-5wm2 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nWe can consider this problem as a dp problem of Pattern Take/notTake, whe | tanishq0_0 | NORMAL | 2024-06-23T04:26:46.119836+00:00 | 2024-06-23T04:26:46.119859+00:00 | 89 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nWe can consider this problem as a dp problem of Pattern Take/notTake, where our Take is Positive value and not Take is our negative value. We can use a bool variable to check whether our curr idx can have negative or not.\n\... | 1 | 0 | ['Dynamic Programming', 'Memoization', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Kotlin | O(n) time, O(n) space | DP | explained (655ms, 161MB) | kotlin-on-time-on-space-dp-explained-655-zlig | Intuition\n\nThe problem is equal to "you can change any two consequent items from a, b to a, -b multiple times (or zero), but each original item could be chang | tonycode88 | NORMAL | 2024-06-23T04:23:15.344064+00:00 | 2024-07-06T16:30:44.790423+00:00 | 16 | false | # Intuition\n\nThe problem is equal to "you can change any two consequent items from `a, b` to `a, -b` multiple times (or zero), but each original item could be changed only once. Maximize the sum of items."\n\n# Approach\n\nDynamic programming:\n\n- $$maxCost[0] = nums[0]$$\n- $$maxCost[i] = max\\left( maxCost[i-1] + ... | 1 | 0 | ['Dynamic Programming', 'Kotlin'] | 0 |
maximize-total-cost-of-alternating-subarrays | Easy Python solution. O(n) time, O(1) space | easy-python-solution-on-time-o1-space-by-tj61 | Intuition\nThere is no reason to split it up into groups larger than 2 because it is alternating. We can either choose to keep the number, or flip the number (a | tyler__ | NORMAL | 2024-06-23T04:21:18.045428+00:00 | 2024-06-23T04:21:18.045491+00:00 | 10 | false | # Intuition\nThere is no reason to split it up into groups larger than 2 because it is alternating. We can either choose to keep the number, or flip the number (and not have flipped the one before it).\n\n# Approach\nIf we don\'t flip it, then we can just add it to the max of our 2 previous scenarios. If we do flip it,... | 1 | 0 | ['Python3'] | 1 |
maximize-total-cost-of-alternating-subarrays | Easy Iterative dynamic programming | easy-iterative-dynamic-programming-by-om-6vc3 | \n\n# Code\n\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n \n int n=nums.size();\n vector<long long>dp1(num | omrajhalwa | NORMAL | 2024-06-23T04:14:45.291131+00:00 | 2024-06-23T04:14:45.291148+00:00 | 15 | false | \n\n# Code\n```\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n \n int n=nums.size();\n vector<long long>dp1(nums.size(),0);\n vector<long long>dp2(nums.size(),0);\n dp1[0]=nums[0];dp2[0]=nums[0];\n for(int i=1;i<nums.size();i++){\n dp1[... | 1 | 0 | ['Dynamic Programming', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Beginner friendly approach...Take Not take approach using DP | beginner-friendly-approachtake-not-take-78cd2 | Intuition\n Describe your first thoughts on how to solve this problem. \nAssume u have taken the 1st element\nnow u can include the 2nd element as part of 1st e | Tanooj13 | NORMAL | 2024-06-23T04:11:32.420065+00:00 | 2024-06-23T04:11:32.420088+00:00 | 30 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAssume u have taken the 1st element\nnow u can include the 2nd element as part of 1st element subarray if u multiply it with negative sign\nor\nu can make the 2nd element a new subarray if u dont multiply it with negative sign\n\nFor exam... | 1 | 0 | ['Dynamic Programming', 'Memoization', 'Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | EZ DP 🔥🔥 | ez-dp-by-chitraksh24-ywev | Code\n\nclass Solution:\n def maximumTotalCost(self, nums: List[int]) -> int: \n @cache\n def solve(curr,sign):\n if curr | chitraksh24 | NORMAL | 2024-06-23T04:02:23.584221+00:00 | 2024-06-23T04:02:23.584258+00:00 | 161 | false | # Code\n```\nclass Solution:\n def maximumTotalCost(self, nums: List[int]) -> int: \n @cache\n def solve(curr,sign):\n if curr>=len(nums):\n return 0\n \n if sign:\n return max(-nums[curr]+solve(curr+1,not sign),nums[curr]+solve(... | 1 | 0 | ['Dynamic Programming', 'Memoization', 'Python3'] | 2 |
maximize-total-cost-of-alternating-subarrays | Beats 100%, Intuitive approach with explanation, in python | beats-100-intuitive-approach-with-explan-0i73 | Intuition\n\nEach element will either have a positive multiplier (+1) or a negative multiplier(-1). We just need to consider multiplier sign of an element, with | kardeepakkumar | NORMAL | 2024-06-23T04:02:09.381450+00:00 | 2024-06-23T04:02:09.381486+00:00 | 103 | false | # Intuition\n\nEach element will either have a positive multiplier (+1) or a negative multiplier(-1). We just need to consider multiplier sign of an element, with regard to the multiplier sign of the previous element:\n- If the previous element is negative multiplied, then current has to be positive multiplied\n- If th... | 1 | 0 | ['Dynamic Programming', 'Python3'] | 0 |
maximize-total-cost-of-alternating-subarrays | DP solution memoization C++ | dp-solution-memoization-c-by-yashgarala2-rdbc | \n# Code\n\nclass Solution {\npublic:\n // Function to calculate the maximum total cost using recursion and memoization\n long long calculate(vector<int> | yashgarala29 | NORMAL | 2024-06-23T04:02:00.744506+00:00 | 2024-06-23T04:02:00.744546+00:00 | 196 | false | \n# Code\n```\nclass Solution {\npublic:\n // Function to calculate the maximum total cost using recursion and memoization\n long long calculate(vector<int> &nums, int current, int sign, vector<vector<long long>> &dp) {\n int n = nums.size();\n // Base case: If we have processed all elements\n ... | 1 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | C++ || EASY DP | c-easy-dp-by-abhishek6487209-ii5l | 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 | Abhishek6487209 | NORMAL | 2024-06-23T04:01:45.118590+00:00 | 2024-06-23T04:01:45.118619+00:00 | 57 | 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. $$O(n)$$ --... | 1 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Easy to understand DP | O(n) beats 100% | easy-to-understand-dp-on-beats-100-by-pr-zfis | Intuition\nKey idea: All subarrays will either be of length $1$ or length $2$. This is because any subarray length greater than $2$ can be made with various sub | Pras28 | NORMAL | 2024-06-23T04:00:57.916373+00:00 | 2024-06-23T04:49:25.380509+00:00 | 104 | false | # Intuition\n**Key idea:** All subarrays will either be of length $1$ or length $2$. This is because any subarray length greater than $2$ can be made with various subarrays of lengths $1$ and $2$, and the sum of the scores of these smaller subarrays will be $\\geq$ the score of the larger subarray.\n\nNow, we simply ha... | 1 | 0 | ['Dynamic Programming', 'Python', 'Python3'] | 0 |
maximize-total-cost-of-alternating-subarrays | Simple recursion + memo | simple-recursion-memo-by-joe54-bjp8 | 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 | joe54 | NORMAL | 2024-06-23T04:00:55.399096+00:00 | 2024-06-23T04:00:55.399139+00:00 | 145 | 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. $$O(n)$$ --... | 1 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Shortest 2 line solution | shortest-2-line-solution-by-karan_udaypi-32yp | IntuitionRefer to this video in order to learn to write tabular solutions from memo solutions :- ApproachComplexity
Time complexity:
Space complexity:
Code | karan_udaypie | NORMAL | 2025-03-24T07:38:23.400744+00:00 | 2025-03-24T07:38:23.400744+00:00 | 2 | false | # Intuition
Refer to this video in order to learn to write tabular solutions from memo solutions :- https://www.youtube.com/watch?v=5p4M8RmeQCU
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Kadane's Algorithm (sort of...) | kadanes-algorithm-sort-of-by-vijayshanka-ksxz | Code | vijayshankarkumar | NORMAL | 2025-03-08T21:50:28.393448+00:00 | 2025-03-08T21:50:28.393448+00:00 | 1 | false | # Code
```cpp []
class Solution {
public:
long long maximumTotalCost(vector<int>& nums) {
if (nums.size() == 1) return nums.front();
vector<long long> dp(nums.size());
dp[0] = nums[0];
dp[1] = max(nums[0] + nums[1], nums[0] - nums[1]);
for (int i = 2; i < nums.size(); i++) {
... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | scala oneliner | scala-oneliner-by-vititov-8m2g | null | vititov | NORMAL | 2025-02-16T23:45:10.551544+00:00 | 2025-02-16T23:45:10.551544+00:00 | 3 | false | ```scala []
object Solution {
def maximumTotalCost(nums: Array[Int]): Long =
nums.reverseIterator.foldLeft((0L,0L)){case ((p,n),num) =>
(num+(p max n), -num+p)
}._1
}
``` | 0 | 0 | ['Greedy', 'Scala'] | 0 |
maximize-total-cost-of-alternating-subarrays | 100% run time and 100 memory C# | 100-run-time-and-100-memory-c-by-c0ngtha-0jan | IntuitionUsing dynamic programming with pre and next for sum to i-2 number and i-1 numberApproachComplexity
Time complexity:
Space complexity:
Code | c0ngthanh | NORMAL | 2025-01-08T11:26:37.601565+00:00 | 2025-01-08T11:26:37.601565+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Using dynamic programming with pre and next for sum to i-2 number and i-1 number
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Sp... | 0 | 0 | ['Dynamic Programming', 'C#'] | 0 |
maximize-total-cost-of-alternating-subarrays | C++ | c-by-tinachien-r65j | \nusing LL = long long;\nclass Solution {\n LL dp[100005][2];\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n dp[0][0] = nums[0];\n | TinaChien | NORMAL | 2024-12-07T06:04:52.630024+00:00 | 2024-12-07T06:04:52.630066+00:00 | 3 | false | ```\nusing LL = long long;\nclass Solution {\n LL dp[100005][2];\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n dp[0][0] = nums[0];\n dp[0][1] = LLONG_MIN/2;\n int n = nums.size();\n for(int i = 1; i < n; i++){\n dp[i][0] = max(dp[i-1][0], dp[i-1][1]) + nums[i]... | 0 | 0 | [] | 0 |
maximize-total-cost-of-alternating-subarrays | DP? | dp-by-iitjsagar-u88u | Approach\n Describe your approach to solving the problem. \nEvery number can be either have a negative multiplier or not: \n1. If nums[i] has negative multiplie | iitjsagar | NORMAL | 2024-11-27T05:35:49.823097+00:00 | 2024-11-27T05:35:49.823132+00:00 | 1 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nEvery number can be either have a negative multiplier or not: \n1. If nums[i] has negative multiplier, then current cost = (max cost till nums[i-2]) + nums[i-1] - nums[i]\n2. If nums[i] has positive multiplier, then current cost = (max cost till nums[... | 0 | 0 | ['Dynamic Programming', 'Python'] | 0 |
maximize-total-cost-of-alternating-subarrays | 10 Lines|O(1) Space|Faster than 100% | 10-lineso1-spacefaster-than-100-by-saiki-3aut | 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 | saikiranpatil | NORMAL | 2024-11-07T18:36:48.340193+00:00 | 2024-11-07T18:59:04.606926+00:00 | 10 | 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```cpp []\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n long long pp, pm, cp, cm;... | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Straightforward DP solution with concise explanation. Time O(n) and Space O(1). | straightforward-dp-solution-with-concise-w0pw | Intuition\nIt is a dynamic programming problem. Each element in the input will be either taken as is or taken with a flipped sign in the final cost. \n\n# Appro | 00_11_00 | NORMAL | 2024-10-24T00:29:39.648426+00:00 | 2024-10-24T00:29:39.648468+00:00 | 2 | false | # Intuition\nIt is a dynamic programming problem. Each element in the input will be either taken as is or taken with a flipped sign in the final cost. \n\n# Approach\nWe can assume that DP_p[i] is the maximum cost of the arry [0, i] with the number at index i taken as is and DP_p[i] is the maximum cost of the arry [0, ... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Detailed explanation using 1D dp | detailed-explanation-using-1d-dp-by-raja-ojai | Intuition\nThe idea is to use dynamic programming.\n\n# Approach\nLet dp[i] stores the answer to the array from [i...n - 1] where n is the size of the array. Ou | rajat8020 | NORMAL | 2024-10-17T20:19:41.811032+00:00 | 2024-10-17T20:20:30.304239+00:00 | 3 | false | # Intuition\nThe idea is to use dynamic programming.\n\n# Approach\nLet dp[i] stores the answer to the array from [i...n - 1] where n is the size of the array. Our answer will be $$dp[0]$$. \n\n# Observation\nIt is never necessary to partition the array into size more than 2. In other words the size of the partition wi... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Beat 100% in both || O(1) space solution || easy to understand | beat-100-in-both-o1-space-solution-easy-bmbfz | Intuition\n- a means flip nums[i-1]\n- b means not flip nums[i-1]\n- i can flip if only i-1 not flip\n- i can flip or not flip, dependent from the state of i-1, | dnanper | NORMAL | 2024-10-12T09:19:59.538369+00:00 | 2024-10-12T09:22:45.216156+00:00 | 2 | false | # Intuition\n- a means flip nums[i-1]\n- b means not flip nums[i-1]\n- i can flip if only i-1 not flip\n- i can flip or not flip, dependent from the state of i-1, so we need to find max value of that two states.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | TC : O(n), SC: O(1) DP | tc-on-sc-o1-dp-by-anikets2002-s18a | \n# Code\ntypescript []\nfunction maximumTotalCost(nums: number[]): number {\n // we only take positive this mean we are startign a new sub array from here\n | anikets2002 | NORMAL | 2024-10-09T07:23:55.960832+00:00 | 2024-10-09T07:23:55.960869+00:00 | 0 | false | \n# Code\n```typescript []\nfunction maximumTotalCost(nums: number[]): number {\n // we only take positive this mean we are startign a new sub array from here\n let n = nums.length;\n if(n === 1)return nums[0];\n let maxYet = nums[0];\n let posPrev = 0, posCurr = 0, negPrev = 0, negCurr = 0;\n for(let... | 0 | 0 | ['TypeScript'] | 0 |
maximize-total-cost-of-alternating-subarrays | Simplest C++ Solution | simplest-c-solution-by-pushpendra_singh-1ugi | 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 | pushpendra_singh_ | NORMAL | 2024-09-24T18:36:31.323134+00:00 | 2024-09-24T18:36:31.323158+00:00 | 1 | 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. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | coded with very hit and trial process || cpp | coded-with-very-hit-and-trial-process-cp-bl8s | 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 | chanderveersinghchauhan08 | NORMAL | 2024-09-12T17:55:46.200602+00:00 | 2024-09-12T17:55:46.200656+00:00 | 0 | 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. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Positive OR negative sign | positive-or-negative-sign-by-tejasuwarad-974o | 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 | Tejasuwarade | NORMAL | 2024-09-12T16:30:03.063819+00:00 | 2024-09-12T16:30:03.063857+00:00 | 0 | 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. $$O(n)$$ -->\nTime Complexity: O(n)\n\n\n- Space complexity:\n<!-- Add your space complexi... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Most Optimized O(n) time and O(1) Space | most-optimized-on-time-and-o1-space-by-m-0wam | Intuition\nThe last subarray can end with subtraction or addition, If the last subarray was addition your option for the next element is to either continue that | mkshv | NORMAL | 2024-09-03T16:22:01.114911+00:00 | 2024-09-03T16:22:01.114946+00:00 | 2 | false | # Intuition\nThe last subarray can end with subtraction or addition, If the last subarray was addition your option for the next element is to either continue that subarray with a subtraction, OR start a new subarray with addition. \n\nIf the last subarray was a subtraction, you only have one option, that you can only a... | 0 | 0 | ['Python3'] | 0 |
maximize-total-cost-of-alternating-subarrays | 4 lines Java Easy | 4-lines-java-easy-by-parishithada-u62v | Complexity\n- Time complexity:\n Add your time complexity here, e.g. O(n) \nO(n2) \n\n- Space complexity:\n Add your space complexity here, e.g. O(n) \nO(n2) + | parishithada | NORMAL | 2024-08-09T17:51:39.601164+00:00 | 2024-08-09T17:51:39.601187+00:00 | 4 | false | # 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$$O(n*2) + (Stack Space- O(n))$$\n\n# Code\n```\nclass Solution {\n public long maximumTotalCost(int[] nums) {\n t=new Long[nums... | 0 | 0 | ['Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | dp ^^ | dp-by-andrii_khlevniuk-nbty | time: O(N); space: O(1)\n\n\n\n\nlong long maximumTotalCost(vector<int>& n)\n{ \n\tlong long pp{}, p{n[0]};\n\tfor(int i{1}; i<size(n); ++i)\n\t{\n\t\taut | andrii_khlevniuk | NORMAL | 2024-08-06T09:38:11.256811+00:00 | 2024-08-06T12:26:55.738130+00:00 | 3 | false | **time: `O(N)`; space: `O(1)`**\n\n\n\n```\nlong long maximumTotalCost(vector<int>& n)\n{ \n\tlong long pp{}, p{n[0]};\n\tfor(int i{1}; i<size(n); ++i)\n\t{\n\t\tauto t = max(p+n[i], pp+n[i-1]-n[i]);\n\t\... | 0 | 0 | ['C', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Max Cost of Alt Subarrays | max-cost-of-alt-subarrays-by-jain47040-7mqr | \n# Code\n\nclass Solution {\npublic:\nlong long maximumTotalCost(vector<int>& nums){\n\n int n = nums.size();\n vector<vector<long long>> Dp(n,vector<lon | jain47040 | NORMAL | 2024-08-04T13:18:39.374494+00:00 | 2024-08-04T13:18:39.374528+00:00 | 0 | false | \n# Code\n```\nclass Solution {\npublic:\nlong long maximumTotalCost(vector<int>& nums){\n\n int n = nums.size();\n vector<vector<long long>> Dp(n,vector<long long>(2));\n\n Dp[0][0] = nums[0];\n Dp[0][1] = nums[0];\n\n if(n == 1){\n return Dp[0][0];\n }\n\n Dp[1][0] = nums[1] + max(Dp[0][0... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Very concise and simple O(n) time/O(1) Space Solution | very-concise-and-simple-on-timeo1-space-4o55o | Intuition\nNo matter what happened before any num x, we can always add x to our running total. Conversely, we can only subtract x, if we added the previous num. | lifran | NORMAL | 2024-08-02T13:04:07.154250+00:00 | 2024-08-02T13:04:07.154288+00:00 | 2 | false | # Intuition\nNo matter what happened before any num `x`, we can always *add* `x` to our running total. Conversely, we can only *subtract* `x`, if we added the previous num.\nSo if we can keep track of the maximum possible cost for the cases when we *add* and when we *subtract* the *last* num, we can then calculate the ... | 0 | 0 | ['Dynamic Programming', 'Kotlin'] | 0 |
maximize-total-cost-of-alternating-subarrays | DP+Space optimized .....||Beats 99.90% ✅✅ | dpspace-optimized-beats-9990-by-soy_yo_m-5brh | \n\n# Intuition\n Describe your first thoughts on how to solve this problem. \nGoal was to maximise the total cost where cost is computed using alternative sum. | Soy_yo_mani | NORMAL | 2024-07-31T18:42:19.161441+00:00 | 2024-07-31T18:42:19.161479+00:00 | 1 | false | \n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nGoal was to maximise the total cost where cost is computed using alternative sum...us... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Simple Recursive DP | simple-recursive-dp-by-smartchuza-34ox | Complexity\n- Time complexity: O(2N)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(2N)\n Add your space complexity here, e.g. O(n) \n\n# | smartchuza | NORMAL | 2024-07-27T17:15:35.319828+00:00 | 2024-07-27T17:15:35.319883+00:00 | 0 | false | # Complexity\n- Time complexity: O(2*N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(2*N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n long long dp[100001][2];\n long long go(int i, int pos, vector<int>& nums)\n {\n ... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Java Dp and without Dp solution with explanation | java-dp-and-without-dp-solution-with-exp-9pj1 | Intuition\nAt each index we can \n-> start a new subarray with the current index or\n-> it would be part of the previous subarray. \n\nWe find maximum sum at ea | prathihaspodduturi | NORMAL | 2024-07-26T23:19:52.075935+00:00 | 2024-07-26T23:19:52.076000+00:00 | 3 | false | # Intuition\nAt each index we can \n-> start a new subarray with the current index or\n-> it would be part of the previous subarray. \n\nWe find maximum sum at each index using the above two points.\n\n# Approach\nlet dp[n] be the maximum sum at each index of the array.\n\nusing the two points in the intutuion :\n\nsta... | 0 | 0 | ['Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | Kotlin | Dynamic Programming | Iterative | O(n) time | O(1) space | kotlin-dynamic-programming-iterative-on-m35ss | 1. DP array\n\nfun maximumTotalCost(nums: IntArray): Long {\n val dp = LongArray(nums.size + 1)\n\n for (i in nums.lastIndex downTo 0) {\n dp[i] = | IrregularOne | NORMAL | 2024-07-22T18:52:51.874099+00:00 | 2024-07-22T19:05:05.999073+00:00 | 3 | false | # 1. DP array\n```\nfun maximumTotalCost(nums: IntArray): Long {\n val dp = LongArray(nums.size + 1)\n\n for (i in nums.lastIndex downTo 0) {\n dp[i] = nums[i].toLong() + dp[i + 1]\n\n if (i < nums.lastIndex) {\n dp[i] = maxOf(dp[i], (nums[i] - nums[i + 1]).toLong() + dp[i + 2])\n ... | 0 | 0 | ['Kotlin'] | 0 |
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