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maximum-energy-boost-from-two-drinks | Java | O(n) 100% | With Explanation | java-on-100-with-explanation-by-zell_dev-sdd1 | Step 1: Initialize Dynamic Programming Arrays\nInitialize two arrays, dpA and dpB, to store the maximum energy boosts for energyDrinkA and energyDrinkB respecti | zell_dev | NORMAL | 2024-08-18T04:08:41.395011+00:00 | 2024-08-18T04:08:41.395029+00:00 | 81 | false | **Step 1: Initialize Dynamic Programming Arrays**\nInitialize two arrays, dpA and dpB, to store the maximum energy boosts for energyDrinkA and energyDrinkB respectively. Set up the base cases for the last two positions based on the given energy drinks.\n\n**Step 2: Compute Maximum Energy Boosts**\nIterate backward thro... | 2 | 0 | [] | 0 |
maximum-energy-boost-from-two-drinks | Easy DP Solution || Python, JavaScript, C++ ✅ | easy-dp-solution-python-javascript-c-by-5cweb | Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(1)\n\n# Code\nPython3 []\nclass Solution:\n def maxEnergyBoost(self, energyDrinkA: List[int], en | 101rror | NORMAL | 2024-08-18T04:01:52.727446+00:00 | 2024-08-18T04:23:36.458480+00:00 | 163 | false | # Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(1)\n\n# Code\n```Python3 []\nclass Solution:\n def maxEnergyBoost(self, energyDrinkA: List[int], energyDrinkB: List[int]) -> int:\n n = len(energyDrinkA)\n\n dpA, dpB = [0] * n, [0] * n\n\n dpA[0], dpB[0] = energyDrinkA[0], energyDri... | 2 | 0 | ['Dynamic Programming', 'C++', 'Python3', 'JavaScript'] | 0 |
maximum-energy-boost-from-two-drinks | Easy Python Solution | Recursion + DP | easy-python-solution-recursion-dp-by-pra-azus | Intuition\nThe problem involves making a series of choices between two options at each step: selecting from energyDrinkA or energyDrinkB. To maximize the total | pranav743 | NORMAL | 2024-08-27T16:22:08.284367+00:00 | 2024-08-27T16:22:08.284399+00:00 | 90 | false | # Intuition\nThe problem involves making a series of choices between two options at each step: selecting from `energyDrinkA` or `energyDrinkB`. To maximize the total energy boost, you need to carefully decide which choice to make at each step while considering the future consequences of each choice.\n\n# Complexity\n\n... | 1 | 0 | ['Dynamic Programming', 'Recursion', 'Python3'] | 0 |
maximum-energy-boost-from-two-drinks | Recursive dp solution || Beats 49% | recursive-dp-solution-beats-49-by-meet_p-f2wv | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nState: max sum i can ge | meet_p | NORMAL | 2024-08-26T12:10:11.836393+00:00 | 2024-08-26T12:10:11.836421+00:00 | 5 | 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**State: max sum i can get from either arrays starting from level to a.size()**\n\n**Transitions: only two possibilities either take from the current arrays or switch ... | 1 | 0 | ['C++'] | 0 |
maximum-energy-boost-from-two-drinks | Easy C++ Solution || DP Memoization | easy-c-solution-dp-memoization-by-akshat-0kk6 | Intuition -: Check out following link for complete explanation-:\nhttps://github.com/akshatsh0610/Data-Structures-and-Algorithms-Problems-Solution/tree/main/Lee | akshat0610 | NORMAL | 2024-08-19T15:57:45.706737+00:00 | 2024-08-19T15:57:45.706774+00:00 | 1 | false | Intuition -: Check out following link for complete explanation-:\nhttps://github.com/akshatsh0610/Data-Structures-and-Algorithms-Problems-Solution/tree/main/Leetcode%20Problems/Dynamic%20Programming/Medium/3259.%20Maximum%20Energy%20Boost%20From%20Two%20Drinks\n\n# Code\n```cpp []\nclass Solution {\npublic:\n vector... | 1 | 0 | ['Array', 'Dynamic Programming', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | The Shortest Possible One Line Solution | the-shortest-possible-one-line-solution-j4bju | \n\nconst maxEnergyBoost = ([headA, ...drinkA], [headB, ...drinkB]) =>\n Math.max(...drinkA.reduce(([a, b], vA, i) => [Math.max(a + vA, b), Math.max(b + drin | charnavoki | NORMAL | 2024-08-18T21:50:01.342423+00:00 | 2024-08-18T21:50:01.342445+00:00 | 18 | false | \n```\nconst maxEnergyBoost = ([headA, ...drinkA], [headB, ...drinkB]) =>\n Math.max(...drinkA.reduce(([a, b], vA, i) => [Math.max(a + vA, b), Math.max(b + drinkB[i], a)], [headA, headB]));\n\n```\n\n### it\'s a challenge for you to explain how it works\n### please upvote, you motivate me to solve problems in origin... | 1 | 0 | ['JavaScript'] | 0 |
maximum-energy-boost-from-two-drinks | Easy bottom up DP with memoization, 100% runtime | easy-bottom-up-dp-with-memoization-100-r-efto | Intuition\nEasy bottom up DP with memoization\n# Approach\nDefine a f(h,prev) where h is current hour and prev is the previous choice of energy drink (either a | worker-bee | NORMAL | 2024-08-18T21:42:39.195601+00:00 | 2024-08-18T21:42:39.195626+00:00 | 27 | false | # Intuition\nEasy bottom up DP with memoization\n# Approach\nDefine a `f(h,prev)` where `h` is current hour and `prev` is the previous choice of energy drink (either `a` or `b`). The function returns the max energy at hour `h` given `prev` choice at hour `h-1`.\n\nAt every hour `h` we have a choice to either choose `a`... | 1 | 0 | ['Dynamic Programming', 'Memoization', 'Python3'] | 0 |
maximum-energy-boost-from-two-drinks | Recursion -> Memoization -> Tablulation -> Space Optimization | recursion-memoization-tablulation-space-jajra | Intuition\n\nThe problem asks us to maximize the energy boost by selecting either nums1[i] or nums2[i] at each index of the array. To solve this problem, we can | _adeeb_ | NORMAL | 2024-08-18T15:31:11.787231+00:00 | 2024-08-18T15:31:11.787249+00:00 | 25 | false | # Intuition\n\nThe problem asks us to maximize the energy boost by selecting either nums1[i] or nums2[i] at each index of the array. To solve this problem, we can start with a simple recursive approach, where we explore all possible paths. However, this leads to redundant calculations, which can be optimized using dyna... | 1 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | JAVA | Recursion | Memo | Bottom-Up | java-recursion-memo-bottom-up-by-priyans-f16n | Code\n\nclass Solution {\n public long maxEnergyBoost(int[] energyDrinkA, int[] energyDrinkB) {\n return Math.max(helper(true, 0, energyDrinkA, energy | priyanshuawasthi14feb | NORMAL | 2024-08-18T15:13:18.936496+00:00 | 2024-08-18T15:13:18.936531+00:00 | 3 | false | # Code\n```\nclass Solution {\n public long maxEnergyBoost(int[] energyDrinkA, int[] energyDrinkB) {\n return Math.max(helper(true, 0, energyDrinkA, energyDrinkB), \n helper(false, 0, energyDrinkA, energyDrinkB));\n }\n\n public long helper (boolean takeA, int index, int []A, int ... | 1 | 0 | ['Java'] | 0 |
maximum-energy-boost-from-two-drinks | ✅ Java Solution | java-solution-by-harsh__005-s781 | CODE\nJava []\npublic long maxEnergyBoost(int[] energyDrinkA, int[] energyDrinkB) {\n\tint n = energyDrinkA.length;\n\n\t// 0 -> A, 1 -> B\n\tlong dp[][] = new | Harsh__005 | NORMAL | 2024-08-18T09:39:44.173979+00:00 | 2024-08-18T09:39:44.174008+00:00 | 50 | false | ## **CODE**\n```Java []\npublic long maxEnergyBoost(int[] energyDrinkA, int[] energyDrinkB) {\n\tint n = energyDrinkA.length;\n\n\t// 0 -> A, 1 -> B\n\tlong dp[][] = new long[n+1][2];\n\tdp[1][0] = energyDrinkA[0];\n\tdp[1][1] = energyDrinkB[0];\n\n\tfor(int i=1; i<n; i++) {\n\t\tdp[i+1][0] = Math.max(dp[i-1][1], dp[i]... | 1 | 0 | ['Java'] | 1 |
maximum-energy-boost-from-two-drinks | Easy to Understand | easy-to-understand-by-twasim-fsvw | 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 | twasim | NORMAL | 2024-08-18T08:52:38.644656+00:00 | 2024-08-18T08:52:38.644687+00:00 | 49 | 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. $$O(n)$$ -->\nO(n)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$... | 1 | 0 | ['Dynamic Programming', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | Easy to understand!!! Memoization ! Using 2D DP | easy-to-understand-memoization-using-2d-xrlu8 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | manvi_010 | NORMAL | 2024-08-18T07:39:13.113336+00:00 | 2024-08-18T07:39:13.113363+00:00 | 72 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Recursion', 'Memoization', 'Java'] | 0 |
maximum-energy-boost-from-two-drinks | Simple recursion + Memoization | simple-recursion-memoization-by-spacexgr-ogfl | Intuition\nEvery index has 2 possibilities either it comes from A or from B so it\'s DP problem as we don\'t what will yield us the best solution without knowin | spacexgragonrye3008507 | NORMAL | 2024-08-18T07:38:58.992237+00:00 | 2024-08-18T07:38:58.992263+00:00 | 37 | false | # Intuition\nEvery index has 2 possibilities either it comes from A or from B so it\'s DP problem as we don\'t what will yield us the best solution without knowing the answer of subproblem.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nIf we switch then we have to leave the current... | 1 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'Python3'] | 0 |
maximum-energy-boost-from-two-drinks | Maximum Energy Boost From Two Drinks using Dynamic programming | maximum-energy-boost-from-two-drinks-usi-qh5g | Intuition\nTo solve the problem of maximizing the total energy boost over n hours with two energy drinks, you can use dynamic programming. The key insight is to | giriprasath | NORMAL | 2024-08-18T07:17:54.534969+00:00 | 2024-08-18T07:17:54.535001+00:00 | 27 | false | # Intuition\nTo solve the problem of maximizing the total energy boost over n hours with two energy drinks, you can use dynamic programming. The key insight is to track the maximum energy boost you can achieve by either continuing with the same drink or switching drinks, considering the constraint that switching drinks... | 1 | 0 | ['Array', 'Dynamic Programming', 'Java'] | 0 |
maximum-energy-boost-from-two-drinks | Optimized Dynamic Programming with Constant Space (O(n) Time, O(1) Space Complexity) | optimized-dynamic-programming-with-const-llu9 | Intuition\nThe problem is about finding the maximum energy boost you can get by choosing energy drinks from two different arrays (energyDrinkA and energyDrinkB) | Tammali_Amulya | NORMAL | 2024-08-18T06:20:28.005037+00:00 | 2024-08-18T06:20:28.005068+00:00 | 60 | false | # Intuition\nThe problem is about finding the maximum energy boost you can get by choosing energy drinks from two different arrays (energyDrinkA and energyDrinkB). You can either select an energy drink from the current array or skip to the next one, potentially switching to the other array. The goal is to maximize the ... | 1 | 0 | ['Python3'] | 0 |
maximum-energy-boost-from-two-drinks | Recursion + Memoization | Beats 100% of users ❤️🔥 | recursion-memoization-beats-100-of-users-kfbm | \n\n# Complexity\n- Time complexity: O(n) \n Add your time complexity here, e.g.\n\n- Space complexity: O(n) \n Add your space complexity here, e.g. O(n) \n\n# | rashid_sid | NORMAL | 2024-08-18T04:27:56.411298+00:00 | 2024-08-18T04:27:56.411321+00:00 | 24 | false | \n\n# Complexity\n- Time complexity: $$O(n)$$ \n<!-- Add your time complexity here, e.g.-->\n\n- Space complexity: $$O(n)$$ \n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n using ll = long long;\n\nvector<vector<ll>> dp;\n ll solve(vector<int>& a, vector<int>& b,in... | 1 | 0 | ['Dynamic Programming', 'Recursion', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | ✨💯EASY SOLUTION.. direct and simple implementation, super easy to understand...✨✨💯💯 | easy-solution-direct-and-simple-implemen-o225 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | manyaagargg | NORMAL | 2024-08-18T04:24:54.204333+00:00 | 2024-08-18T04:24:54.204351+00:00 | 37 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Dynamic Programming', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | Dynamic Programming | dynamic-programming-by-kaluginpeter-ww6d | \n\n# Complexity\n- Time complexity: O(N)\n\n- Space complexity: O(N)\n\n# Code\n\nclass Solution:\n def maxEnergyBoost(self, energyDrinkA: List[int], energy | kaluginpeter | NORMAL | 2024-08-18T04:23:39.331311+00:00 | 2024-08-18T04:23:39.331329+00:00 | 89 | false | \n\n# Complexity\n- Time complexity: O(N)\n\n- Space complexity: O(N)\n\n# Code\n```\nclass Solution:\n def maxEnergyBoost(self, energyDrinkA: List[int], energyDrinkB: List[int]) -> int:\n n: int = len(energyDrinkA)\n\n dpA: list[int] = [0] * n\n dpB: list[int] = [0] * n\n\n dpA[0] = ener... | 1 | 1 | ['Dynamic Programming', 'Python', 'Python3'] | 0 |
maximum-energy-boost-from-two-drinks | Java || Memoization | java-memoization-by-viper__66-sruo | \nclass Solution {\n Long[][] memo;\n\n long rec(int level, int choosed, int[] a, int[] b) {\n if (level == a.length) {\n return 0;\n | Viper__66 | NORMAL | 2024-08-18T04:23:08.907870+00:00 | 2024-08-18T04:23:08.907897+00:00 | 36 | false | ```\nclass Solution {\n Long[][] memo;\n\n long rec(int level, int choosed, int[] a, int[] b) {\n if (level == a.length) {\n return 0;\n }\n\n\n if (memo[level][choosed + 1] != null) {\n return memo[level][choosed + 1];\n }\n\n long f = Long.MIN_VALUE;\n\t\... | 1 | 0 | ['Dynamic Programming', 'Memoization', 'Java'] | 0 |
maximum-energy-boost-from-two-drinks | Space Optimised DP Approach | space-optimised-dp-approach-by-mainframe-6x2t | Intuition\n Describe your first thoughts on how to solve this problem. \nUse dynamic programming approach to keep track of the maximum energy boost achievable a | MainFrameKuznetSov | NORMAL | 2024-08-18T04:13:20.390694+00:00 | 2024-08-18T04:13:20.390727+00:00 | 6 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUse dynamic programming approach to keep track of the maximum energy boost achievable at each step.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Iterate through arrays\n\n2. Calculate cA as the maximum of pA + ... | 1 | 0 | ['C++'] | 0 |
maximum-energy-boost-from-two-drinks | Recursion->Top Down DP || Beginner's guide to Boost Ratings || Easiest Solution || Beats 100% | recursion-top-down-dp-beginners-guide-to-pvga | \n\n# Code\n\n#include <bits/stdc++.h>\nusing namespace std;\n\n#define ll long long\n#define memo(a) memset(a, -1, sizeof(a))\n#define len(x) ((ll)x.size()) \n | suvro_datta | NORMAL | 2024-08-18T04:05:09.996552+00:00 | 2024-08-18T04:17:57.618309+00:00 | 104 | false | \n\n# Code\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\n#define ll long long\n#define memo(a) memset(a, -1, sizeof(a))\n#define len(x) ((ll)x.size()) \n#define pb push_back\n\nclass Solution {\npublic:\n ll dp[100005][4];\n ll rec(ll idx, vector<int>& A, vector<int>& B, char last) {\n if(idx >= ... | 1 | 0 | ['Dynamic Programming', 'Memoization', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | C++ 100% Dynamic Programming (Memo -> Tabulation) | c-100-dynamic-programming-memo-tabulatio-62ah | Dynamic programming\n## Memoization is in the commented code lambda rec()\n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(n)\n# Code\n\n#include | bramar2 | NORMAL | 2024-08-18T04:03:39.062666+00:00 | 2024-08-18T04:04:12.445770+00:00 | 8 | false | ## Dynamic programming\n## Memoization is in the commented code lambda rec()\n# Complexity\n- Time complexity: $$O(n)$$\n\n- Space complexity: $$O(n)$$\n# Code\n```\n#include <bits/stdc++.h>\n#pragma GCC optimize ("Ofast")\n#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx2,tune=native")\nusing namespace ... | 1 | 0 | ['Dynamic Programming', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | DP | Memoisation | C++ | dp-memoisation-c-by-sirius_108-jbfc | \n# Code\n\nclass Solution {\npublic:\n long long calculateMaxEnergy(int N, int X, vector<int>& A, vector<int>& B, vector<vector<long long>>& memo) {\n | sirius_108 | NORMAL | 2024-08-18T04:01:14.707636+00:00 | 2024-08-18T04:01:14.707712+00:00 | 32 | false | \n# Code\n```\nclass Solution {\npublic:\n long long calculateMaxEnergy(int N, int X, vector<int>& A, vector<int>& B, vector<vector<long long>>& memo) {\n if (N < 0) return 0; \n if (memo[N][X] != -1) return memo[N][X];\n\n long long maxE = 0;\n\n if (X == 0)\n maxE = max(calcu... | 1 | 0 | ['C++'] | 0 |
maximum-energy-boost-from-two-drinks | ✅Simple Solution ||Beat 100% ✅||DP | simple-solution-beat-100-dp-by-ashgiri49-p7tq | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ashgiri49455 | NORMAL | 2024-08-18T04:01:08.591756+00:00 | 2024-08-18T04:01:08.591792+00:00 | 71 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Dynamic Programming', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | DP (Memoization) | dp-memoization-by-saisuveer-ho2a | IntuitionGiven two arrays representing energy drinks from two sources, we can either pick a drink from A or B at each index.
Since each choice affects subsequen | SaiSuveer | NORMAL | 2025-04-03T06:50:31.316518+00:00 | 2025-04-03T06:50:31.316518+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Given two arrays representing energy drinks from two sources, we can either pick a drink from A or B at each index.
Since each choice affects subsequent choices, this suggests a recursive approach to try all possible ways.
However, a purely... | 0 | 0 | ['C++'] | 0 |
maximum-energy-boost-from-two-drinks | Beginner Friendly Pythonic Solution 🔥🔥🔥 | beginner-friendly-pythonic-solution-by-s-fe4q | IntuitionSimple, just get the maximum boost among the 2 choices.Complexity
Time complexity:
O(N)
Space complexity:
O(N)Code | Sherpy | NORMAL | 2025-03-18T23:57:05.033558+00:00 | 2025-03-18T23:57:05.033558+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Simple, just get the maximum boost among the 2 choices.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
O(N)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(N)
# Code
``... | 0 | 0 | ['Array', 'Dynamic Programming', 'Greedy', 'Python3'] | 0 |
maximum-energy-boost-from-two-drinks | DP brute force | dp-brute-force-by-awuxiaoqi-gdet | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | labao | NORMAL | 2025-03-04T04:06:51.646391+00:00 | 2025-03-04T04:06:51.646391+00:00 | 1 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Java'] | 0 |
maximum-energy-boost-from-two-drinks | Recursive Solution for dummy Like me🃏 | recursive-solution-for-dummy-like-me-by-8ifo3 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | shashanksaroj | NORMAL | 2025-02-08T09:54:50.400499+00:00 | 2025-02-08T09:54:50.400499+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Java'] | 0 |
maximum-energy-boost-from-two-drinks | DP || C++ || maxEnergyBoost. | dp-c-maxenergyboost-by-rishiinsane-bnhr | IntuitionApproachComplexity
Time complexity:
O(n)
Space complexity:
O(n)
Code | RishiINSANE | NORMAL | 2025-02-04T19:27:50.618596+00:00 | 2025-02-04T19:27:50.618596+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
O(n)
- Space complexity:
O(n)
# Code
```cpp []
class Solution {
public:
long long maxEnergyBoost(vector<int>& A, vector<int>& B) {
... | 0 | 0 | ['C++'] | 0 |
maximum-energy-boost-from-two-drinks | Python3 DP approach | explained | simple solution | python3-dp-approach-explained-simple-sol-pi26 | IntuitionDynamic programming.ApproachCreate 2 dp arrays to store the values for each current state, one in the A drinks array and the other in the B drinks arra | FlorinnC1 | NORMAL | 2025-01-21T20:00:20.869824+00:00 | 2025-01-21T20:00:20.869824+00:00 | 3 | false | # Intuition
Dynamic programming.
# Approach
Create 2 dp arrays to store the values for each current state, one in the A drinks array and the other in the B drinks array. Eg: a state ( dpA[i] ) in A drinks array would mean we got in here by either going from a position back from dpA such as dpA[i-1] or skipped a drink ... | 0 | 0 | ['Array', 'Dynamic Programming', 'Python3'] | 0 |
maximum-energy-boost-from-two-drinks | Clean | Logical | clean-logical-by-richardleee-pr9c | \nclass Solution {\n public long maxEnergyBoost(int[] energyDrinkA, int[] energyDrinkB) {\n //dp[i][0]: ending with A[i] = Math.max(dp[i - 1][0], dp[i | RichardLeee | NORMAL | 2024-12-18T11:13:36.640581+00:00 | 2024-12-18T11:13:36.640607+00:00 | 0 | false | ```\nclass Solution {\n public long maxEnergyBoost(int[] energyDrinkA, int[] energyDrinkB) {\n //dp[i][0]: ending with A[i] = Math.max(dp[i - 1][0], dp[i - 2][1]) + A[i]\n //dp[i][1]: ending with B[i] = Math.max(dp[i - 1][1], dp[i - 2][0]) + B[i]\n \n int n = energyDrinkA.length;\n ... | 0 | 0 | ['Dynamic Programming', 'Java'] | 0 |
maximum-energy-boost-from-two-drinks | Top Class | Ground Breaking | O(N) Solution | top-class-ground-breaking-on-solution-by-vo3g | Complexity
Time complexity:
O(N)
Space complexity:
O(N)
Code | shreet_123 | NORMAL | 2024-12-29T10:02:13.884573+00:00 | 2024-12-29T10:02:13.884573+00:00 | 5 | false |
# Complexity
- Time complexity:
O(N)
- Space complexity:
O(N)
# Code
```cpp []
class Solution {
public:
long long maxEnergyBoost(vector<int>& energyDrinkA, vector<int>& energyDrinkB) {
int n = energyDrinkA.size();
vector<long long> dp1(n, 0);
vector<long long> dp2(n, 0);
dp1[0] ... | 0 | 0 | ['Array', 'Dynamic Programming', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | Top Class | Ground Breaking | O(N) Solution | top-class-ground-breaking-on-solution-by-bown | Complexity
Time complexity:
O(N)
Space complexity:
O(N)
Code | shreet_123 | NORMAL | 2024-12-29T10:02:10.580357+00:00 | 2024-12-29T10:02:10.580357+00:00 | 1 | false |
# Complexity
- Time complexity:
O(N)
- Space complexity:
O(N)
# Code
```cpp []
class Solution {
public:
long long maxEnergyBoost(vector<int>& energyDrinkA, vector<int>& energyDrinkB) {
int n = energyDrinkA.size();
vector<long long> dp1(n, 0);
vector<long long> dp2(n, 0);
dp1[0] ... | 0 | 0 | ['Array', 'Dynamic Programming', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | Simple backtracking solution || Memoization || c++ | simple-backtracking-solution-memoization-nn3y | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | vikash_kumar_dsa2 | NORMAL | 2024-12-24T18:39:30.963796+00:00 | 2024-12-24T18:39:30.963796+00:00 | 3 | 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
`... | 0 | 0 | ['Dynamic Programming', 'Memoization', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | Java Optimized Solution | java-optimized-solution-by-heyysankalp-u0t8 | IntuitionThe problem involves maximizing the energy boost you can collect by picking values from two arrays A and B, subject to the constraint that consecutive | heyysankalp | NORMAL | 2024-12-24T18:22:07.831860+00:00 | 2024-12-24T18:22:07.831860+00:00 | 2 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem involves maximizing the energy boost you can collect by picking values from two arrays A and B, subject to the constraint that consecutive values cannot be picked from the same array. This can be solved using dynamic programming... | 0 | 0 | ['Array', 'Dynamic Programming', 'Java'] | 0 |
maximum-energy-boost-from-two-drinks | DP TABULATION | dp-tabulation-by-prachikumari-1blc | IntuitionYou can notice from ques you choice for being with currChoice either A/B depends on subsequents steps a/b : in order to maximise energyboost overAlldp[ | PrachiKumari | NORMAL | 2024-12-24T15:59:51.632489+00:00 | 2024-12-24T15:59:51.632489+00:00 | 1 | false | # Intuition
You can notice from ques you choice for being with currChoice either A/B depends on subsequents steps a/b : in order to maximise energyboost overAll
dp[i][2] i-> day & 2 choice {0->A & 1->B} hence
base case last day : it no choice (just take the choice you have ie to take the energyboost rather than think... | 0 | 0 | ['Java'] | 0 |
maximum-energy-boost-from-two-drinks | c++ solution | using memoization | c-solution-using-memoization-by-amit_207-5npj | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Amit_207 | NORMAL | 2024-12-16T12:50:59.086150+00:00 | 2024-12-16T12:50:59.086150+00:00 | 3 | 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)$$ --... | 0 | 0 | ['Dynamic Programming', 'Memoization', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | JAVA DP Easy Solution | java-dp-easy-solution-by-shivangi1-9a5h | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | shivangi1 | NORMAL | 2024-12-16T05:52:09.226798+00:00 | 2024-12-16T05:52:09.226798+00:00 | 2 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
maximum-energy-boost-from-two-drinks | A super easy Java solution with DP | a-super-easy-java-solution-with-dp-by-mi-mzwp | dpA[i+2] contains max boost possible after consuming ith drink from energy drink A.\ndpB[i+2] contains max boost possible after consuming ith drink from energy | mindmay | NORMAL | 2024-11-28T19:30:24.933633+00:00 | 2024-11-28T19:30:24.933655+00:00 | 3 | false | dpA[i+2] contains max boost possible after consuming ith drink from energy drink A.\ndpB[i+2] contains max boost possible after consuming ith drink from energy drink B.\n\ndpA[i+2] = Math.max(dpA[i+1]+ energyDrinkA[i], dpB[i] + energyDrinkA[i]); \n\t\t\tlogic for above state calculation is, either continue from A or sw... | 0 | 0 | ['Dynamic Programming', 'Java'] | 0 |
maximum-energy-boost-from-two-drinks | Python - State Machine | python-state-machine-by-jerryji040506-rwat | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | jerryji040506 | NORMAL | 2024-11-25T00:28:01.310902+00:00 | 2024-11-25T00:28:01.310934+00:00 | 0 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Dynamic Programming', 'Python3'] | 0 |
maximum-energy-boost-from-two-drinks | The simplest Python solution that beats 76% of submissions, DP, O(N) | the-simplest-python-solution-that-beats-9e9l8 | Approach\nSimple dynamic programmic solution without recursions, functions calls, etc.\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\ | alexeis | NORMAL | 2024-11-19T02:01:05.981852+00:00 | 2024-11-19T02:01:05.981878+00:00 | 2 | false | # Approach\nSimple dynamic programmic solution without recursions, functions calls, etc.\n\n# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(n)$$\n\n# Code\n```python3 []\nclass Solution:\n def maxEnergyBoost(self, energyDrinkA: List[int], energyDrinkB: List[int]) -> int:\n\n mbA=[energy... | 0 | 0 | ['Python3'] | 0 |
maximum-energy-boost-from-two-drinks | Simple solution using include exclude principle | simple-solution-using-include-exclude-pr-rs9m | Code\njava []\nclass Solution {\n public long maxEnergyBoost(int[] a, int[] b) {\n int n = a.length;\n long[][] dp = new long[n+1][3];\n | _jyoti_geek | NORMAL | 2024-11-15T18:01:24.827973+00:00 | 2024-11-15T18:01:24.828017+00:00 | 2 | false | # Code\n```java []\nclass Solution {\n public long maxEnergyBoost(int[] a, int[] b) {\n int n = a.length;\n long[][] dp = new long[n+1][3];\n for(int i=0;i<dp.length;i++){\n for(int j=0;j<=2;j++){\n dp[i][j] = -1;\n }\n }\n return helper(a, b, 0... | 0 | 0 | ['Array', 'Dynamic Programming', 'Java'] | 0 |
maximum-energy-boost-from-two-drinks | Straight Forward DP with Transition States | straight-forward-dp-with-transition-stat-r7dz | Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space complexity here, e.g. O(n) \n\n# Co | iitjsagar | NORMAL | 2024-11-09T23:01:59.527485+00:00 | 2024-11-09T23:01:59.527516+00:00 | 3 | false | # Complexity\n- Time complexity: $$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```python []\nclass Solution(object):\n def maxEnergyBoost(self, energyDrinkA, energyDrinkB):\n """\n :type... | 0 | 0 | ['Python'] | 0 |
maximum-energy-boost-from-two-drinks | dp | dp-by-user5285zn-hxli | rust []\nimpl Solution {\n pub fn max_energy_boost(a: Vec<i32>, b: Vec<i32>) -> i64 {\n let n = a.len();\n let mut dp : Vec<Vec<i64>> = vec![ve | user5285Zn | NORMAL | 2024-11-09T11:21:54.003810+00:00 | 2024-11-09T11:21:54.003835+00:00 | 3 | false | ```rust []\nimpl Solution {\n pub fn max_energy_boost(a: Vec<i32>, b: Vec<i32>) -> i64 {\n let n = a.len();\n let mut dp : Vec<Vec<i64>> = vec![vec![0; n]; 2];\n for i in 0..n {\n let i1 = i.wrapping_sub(1);\n let i2 = i.wrapping_sub(2);\n dp[0][i] = a[i] as i64 ... | 0 | 0 | ['Rust'] | 0 |
maximum-energy-boost-from-two-drinks | Single Pass ^ Dynamic Programm Python beats 90% solutions | single-pass-dynamic-programm-python-beat-eem3 | Intuition\n Describe your first thoughts on how to solve this problem. \nUsing DP for each tray to store maximum score achievable as far\n# Approach\n Describe | narendrakummara_9 | NORMAL | 2024-10-19T09:58:14.446206+00:00 | 2024-10-19T09:58:14.446228+00:00 | 0 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUsing DP for each tray to store maximum score achievable as far\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe can take current drink with previous drink of same tray or current drink with prev previous drink fro... | 0 | 0 | ['Python3'] | 0 |
maximum-energy-boost-from-two-drinks | Recursion - Memoization :) | recursion-memoization-by-roy_b-nox9 | 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 | roy_b | NORMAL | 2024-10-13T20:10:03.562241+00:00 | 2024-10-13T20:10:03.562275+00:00 | 2 | 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)$$ -->\n O(N)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O... | 0 | 0 | ['Java'] | 0 |
maximum-energy-boost-from-two-drinks | Beats 1000 % Easy C++ Solution | beats-1000-easy-c-solution-by-skale5747-jnxe | 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 | skale5747 | NORMAL | 2024-10-12T09:29:06.336037+00:00 | 2024-10-12T09:29:06.336060+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 |
maximum-energy-boost-from-two-drinks | scala oneliner | scala-oneliner-by-vititov-djgl | scala []\nobject Solution {\n def maxEnergyBoost(energyDrinkA: Array[Int], energyDrinkB: Array[Int]): Long =\n (energyDrinkA.iterator zip energyDrinkB.itera | vititov | NORMAL | 2024-10-10T20:04:12.358745+00:00 | 2024-10-10T20:04:12.358774+00:00 | 0 | false | ```scala []\nobject Solution {\n def maxEnergyBoost(energyDrinkA: Array[Int], energyDrinkB: Array[Int]): Long =\n (energyDrinkA.iterator zip energyDrinkB.iterator)\n .foldLeft((0L, 0L)){case ((c,d),(a,b)) =>\n ((c+a) max d, (d+b) max c)\n }.productIterator.map(_.asInstanceOf[Long]).max\n}\n``` | 0 | 0 | ['Linked List', 'Recursion', 'Scala'] | 0 |
maximum-energy-boost-from-two-drinks | 🔥TOP-DOWN DP || 🔥C++ || 🔥O(n) || 🔥SHORT CODE || 🔥EASIEST AND SHORT EXPLANATION | top-down-dp-c-on-short-code-easiest-and-yhi4i | dp[idx][toggle] defines currently you are on which index and in which array.\n\nsuppose, you are on array A, then you can either take drink from here and go nex | Omitul | NORMAL | 2024-10-04T05:53:32.104512+00:00 | 2024-10-04T05:54:05.712233+00:00 | 1 | false | **dp[idx][toggle] defines currently you are on which index and in which array.**\n\nsuppose, you are on array A, then you can either take drink from here and go next in this same array OR you can drink from here and take rest for one hour and go to the Array B.\nsame Goes for array B. Walk on both ways and get the maxi... | 0 | 0 | ['Dynamic Programming', 'C'] | 0 |
maximum-energy-boost-from-two-drinks | C++ top down recursion, memoisation, Elegant , super easy, well explained | c-top-down-recursion-memoisation-elegant-wn58 | Intuition\nAt any idx, two possibilities, whether to be on same current enery drink array(idx+1) or switch to another energy drink array, so while switch, we ha | nikhiljaiswal | NORMAL | 2024-10-02T06:20:55.121191+00:00 | 2024-10-02T06:20:55.121212+00:00 | 2 | false | # Intuition\nAt any idx, two possibilities, whether to be on same current enery drink array(idx+1) or switch to another energy drink array, so while switch, we have to skip the next hr(idx+2).\n\n\n# Approach\nAt any particular index idx and turn(A or B), we can find max boost can be collected from that idx to end and ... | 0 | 0 | ['C++'] | 0 |
maximum-energy-boost-from-two-drinks | JS DP - Very Simple Solution | js-dp-very-simple-solution-by-nebojsazak-nz7l | Intuition\n Describe your first thoughts on how to solve this problem. \nEach max hour sum for either drink is either previous hour of the same drink + current | nebojsazak | NORMAL | 2024-09-28T14:08:13.069902+00:00 | 2024-09-28T14:08:13.069919+00:00 | 7 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nEach max hour sum for either drink is either previous hour of the same drink + current side or other side hour - 2 + current side.\n\nSolution is max the value of the both drinks for the last hour.\n\n# Approach\nDP - number values holdin... | 0 | 0 | ['JavaScript'] | 0 |
maximum-energy-boost-from-two-drinks | Simple take/not_take solution | Easy | DP | C++ | simple-takenot_take-solution-easy-dp-c-b-ushj | \n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(2n)\n# Code\ncpp []\nclass Solution {\npublic:\n long long maxEnergyBoost(vector<int>& A, ve | tusharjain88954 | NORMAL | 2024-09-24T07:08:57.072223+00:00 | 2024-09-24T07:08:57.072253+00:00 | 2 | false | \n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(2n)\n# Code\n```cpp []\nclass Solution {\npublic:\n long long maxEnergyBoost(vector<int>& A, vector<int>& B) {\n // either start from A or B drink\n\n vector<vector<long long> > dp(A.size(), vector<long long> (3, -1));\n\n return r... | 0 | 0 | ['C++'] | 0 |
maximum-energy-boost-from-two-drinks | Simplest C++ Solution. | simplest-c-solution-by-pushpendra_singh-sgjf | 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 | 2024-09-24T04:46:52.010009+00:00 | 2024-09-24T04:46:52.010035+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 |
maximum-energy-boost-from-two-drinks | C++||EASIEST SOLUTION||RECURSIVE+MEMO|| | ceasiest-solutionrecursivememo-by-diksha-dd6m | \n\n# Code\ncpp []\nclass Solution {\npublic:\n int n;\n vector<vector<long long>> memo; // Memoization table\n \n long long solve(vector<int>& ene | diksha_uniyal | NORMAL | 2024-09-21T18:14:50.083560+00:00 | 2024-09-21T18:14:50.083586+00:00 | 2 | false | \n\n# Code\n```cpp []\nclass Solution {\npublic:\n int n;\n vector<vector<long long>> memo; // Memoization table\n \n long long solve(vector<int>& energyDrinkA, vector<int>& energyDrinkB, int idx, int isDrinkA) {\n if (idx >= n) return 0; // Base case: no more drinks\n \n if (memo[idx... | 0 | 0 | ['C++'] | 0 |
maximum-energy-boost-from-two-drinks | Python3 Linear Time, Linear Space Solution Leveraging BUDP : Bottom-Up Dynamic Programming | python3-linear-time-linear-space-solutio-5dhu | Intuition and Approach\nAlways positive integral energy\n\n# Complexity\nLet N := #-hours we consume energy drinks for \n- Time complexity:\nO(N)\n\n- Space co | 2018hsridhar | NORMAL | 2024-09-15T20:16:52.246098+00:00 | 2024-09-15T20:16:52.246117+00:00 | 0 | false | # Intuition and Approach\nAlways positive integral energy\n\n# Complexity\nLet $$N := $$ #-hours we consume energy drinks for \n- Time complexity:\n$$O(N)$$\n\n- Space complexity:\n$$O(N)$$ ( E ) $$O(1)$$ ( I ) \n\n# Code\n```python3 []\n\'\'\'\n3259. Maximum Energy Boost From Two Drinks\nURL := https://leetcode.com/pr... | 0 | 0 | ['Array', 'Dynamic Programming', 'C++', 'Python3'] | 0 |
maximum-energy-boost-from-two-drinks | C++ || Memoization || Tabulation | c-memoization-tabulation-by-its_rahul-opl2 | \n# Code\ncpp []\n#define ll long long\nclass Solution {\nprivate:\n ll solve(int ind, int f, vector<int>& a, vector<int>& b, vector<vector<ll>>& t) {\n | its_rahul | NORMAL | 2024-09-14T07:52:41.632285+00:00 | 2024-09-14T07:52:41.632318+00:00 | 2 | false | \n# Code\n```cpp []\n#define ll long long\nclass Solution {\nprivate:\n ll solve(int ind, int f, vector<int>& a, vector<int>& b, vector<vector<ll>>& t) {\n if(ind == a.size()) return 0;\n if(t[ind][f] != -1) return t[ind][f];\n ll x = 0, y = 0, z = 0;\n if(f) x = b[ind] + solve(ind+1, f, ... | 0 | 0 | ['C++'] | 0 |
maximum-energy-boost-from-two-drinks | JavaScript DP solution with straightforward thinking | javascript-dp-solution-with-straightforw-hqpl | Code\njavascript []\n/**\n * @param {number[]} energyDrinkA\n * @param {number[]} energyDrinkB\n * @return {number}\n */\nvar maxEnergyBoost = function(energyDr | jyun-han | NORMAL | 2024-09-11T00:21:57.169035+00:00 | 2024-09-11T00:21:57.169059+00:00 | 7 | false | # Code\n```javascript []\n/**\n * @param {number[]} energyDrinkA\n * @param {number[]} energyDrinkB\n * @return {number}\n */\nvar maxEnergyBoost = function(energyDrinkA, energyDrinkB) {\n const n = energyDrinkA.length;\n\n const dpA = new Array(n).fill(0);\n const dpB = new Array(n).fill(0);\n\n dpA[0] = e... | 0 | 0 | ['Dynamic Programming', 'JavaScript'] | 0 |
maximum-energy-boost-from-two-drinks | DP | dp-by-sk2102-m64q | 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 | SK2102 | NORMAL | 2024-09-10T19:17:18.852974+00:00 | 2024-09-10T19:17:18.853004+00:00 | 2 | 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:\nO(n)\n\n- Space complexity:\nO(n)\n\n# Code\n```cpp []\nclass Solution {\n long long energyBoostCal(vector<int>& energyDrinkA, v... | 0 | 0 | ['C++'] | 0 |
maximum-energy-boost-from-two-drinks | Easiest and most efficient approach explained | easiest-and-most-efficient-approach-expl-gdzx | Intuition\nTo consume A, you were either already drinking A in the previous hour, or you were consumming B and you skipped the previous. Same thing for B. \n\n# | jeehaytch | NORMAL | 2024-09-10T04:07:40.855797+00:00 | 2024-09-10T04:07:40.855818+00:00 | 3 | false | # Intuition\nTo consume A, you were either already drinking A in the previous hour, or you were consumming B and you skipped the previous. Same thing for B. \n\n# Approach\nIn each step, we calculate the 3 possibilities, i.e Drinking A, drinking B, or skipping. \nTo drink A, we either skipped before (take the value of ... | 0 | 0 | ['Python3'] | 0 |
maximum-energy-boost-from-two-drinks | Python || Dynamic Programming || Simple solution | python-dynamic-programming-simple-soluti-xfpy | 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 | vilaparthibhaskar | NORMAL | 2024-09-05T18:59:08.746825+00:00 | 2024-09-05T18:59:08.746855+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. $$O(n)$$ -->\no(n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n... | 0 | 0 | ['Array', 'Dynamic Programming', 'Python3'] | 0 |
maximum-energy-boost-from-two-drinks | Easy sol^n | easy-soln-by-singhgolu933600-z0nx | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | singhgolu933600 | NORMAL | 2024-09-03T06:13:38.522915+00:00 | 2024-09-03T06:13:38.522945+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Array', 'Dynamic Programming', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | See the python solution | see-the-python-solution-by-testcasefail-tauy | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | TestCaseFail | NORMAL | 2024-09-02T19:27:30.653718+00:00 | 2024-09-02T19:27:30.653753+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Python3'] | 0 |
maximum-energy-boost-from-two-drinks | Top Down approach C++ | top-down-approach-c-by-ankursingh10-ch66 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ankursingh10 | NORMAL | 2024-09-02T16:55:08.608857+00:00 | 2024-09-02T16:56:05.623273+00:00 | 6 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(2n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n^2)$$ for memoization\n<!-- Add your sp... | 0 | 0 | ['C++'] | 0 |
maximum-energy-boost-from-two-drinks | ESASY SOLUTION USING Dp | esasy-solution-using-dp-by-pranjali21384-3jr1 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | pranjali213847 | NORMAL | 2024-09-01T09:28:52.338651+00:00 | 2024-09-01T09:28:52.338675+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
maximum-energy-boost-from-two-drinks | "DP" solution, "for loop" might be more proper | Python | dp-solution-for-loop-might-be-more-prope-p8rh | \n"""\nDP \n\nstep-0. define drink_A and drink_B both of len(A) such that:\ndrink_A[i] := what is the maximal energy boost if we drink A[i]\ndrink_B[i] := what | wxy0925 | NORMAL | 2024-08-31T23:41:58.947150+00:00 | 2024-08-31T23:41:58.947174+00:00 | 0 | false | ```\n"""\nDP \n\nstep-0. define drink_A and drink_B both of len(A) such that:\ndrink_A[i] := what is the maximal energy boost if we drink A[i]\ndrink_B[i] := what is the maximal energy boost if we drink B[i]\n\nstep-1. init drink_A[0] = A[0], drink_B[0] = B[0]\n\nstep-2 dp update. Take drink_A[i] as example. If we can ... | 0 | 0 | ['Dynamic Programming', 'Python'] | 0 |
maximum-energy-boost-from-two-drinks | my best soln | my-best-soln-by-sachinab-ysdn | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | sachinab | NORMAL | 2024-08-31T17:41:45.631590+00:00 | 2024-08-31T17:41:45.631614+00:00 | 0 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
maximum-energy-boost-from-two-drinks | C++|| DP || Clean code | c-dp-clean-code-by-satyamshivam366-ppza | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | satyamshivam366 | NORMAL | 2024-08-31T14:38:10.383376+00:00 | 2024-08-31T14:38:10.383397+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
maximum-energy-boost-from-two-drinks | Easy to Understand || cpp || java | easy-to-understand-cpp-java-by-bellmanfo-aj97 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n create dp array of two row and n+1 columns.Code is self explanatory , ju | BellmanFord_25 | NORMAL | 2024-08-31T10:45:40.448878+00:00 | 2024-08-31T10:45:40.448910+00:00 | 2 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- create dp array of two row and n+1 columns.Code is self explanatory , just use first row for energyDrinkA and row 2 for energyDrinkB. If you are taking energyDrinkA then look for it\'s two previous places from where you... | 0 | 0 | ['C++'] | 0 |
maximum-energy-boost-from-two-drinks | Iterative DP | iterative-dp-by-eugenemsv-un6e | Intuition\nThe Dp(K) = max from:\n 1. a[k]+ DpA[k-1] \n 2. a[k] + DpB[k-2]\n 3. b[k] + DpB[k-1]\n 4. b[k] + DpA[k-2]\n\nby DpA we mean if the last state is A, | eugenemsv | NORMAL | 2024-08-30T21:19:26.165343+00:00 | 2024-08-30T21:19:26.165367+00:00 | 3 | false | # Intuition\nThe Dp(K) = max from:\n 1. a[k]+ DpA[k-1] \n 2. a[k] + DpB[k-2]\n 3. b[k] + DpB[k-1]\n 4. b[k] + DpA[k-2]\n\nby DpA we mean if the last state is A, by DpB -> b\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(... | 0 | 0 | ['Dynamic Programming', 'Java'] | 0 |
maximum-energy-boost-from-two-drinks | DP, 💽: O(1) only | dp-o1-only-by-ndr0216-7vz9 | 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 | ndr0216 | NORMAL | 2024-08-30T01:55:55.426595+00:00 | 2024-08-30T01:55:55.426632+00:00 | 4 | 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: $$O(1)$$\n<!-- Add your space complexity here... | 0 | 0 | ['Array', 'Dynamic Programming', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | dp solution | dp-solution-by-somaycoder-7oge | 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 | somaycoder | NORMAL | 2024-08-29T20:56:37.889212+00:00 | 2024-08-29T20:56:37.889229+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 | ['Array', 'Dynamic Programming', 'C++'] | 0 |
maximum-energy-boost-from-two-drinks | Easy one if you think this way . | easy-one-if-you-think-this-way-by-prasad-k8od | Intuition\n Describe your first thoughts on how to solve this problem. \nNo specific pattern means trying all possibilties , finding overlappinng subproblems, t | prasad_shewale | NORMAL | 2024-08-29T14:56:33.362377+00:00 | 2024-08-29T14:56:33.362412+00:00 | 0 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nNo specific pattern means trying all possibilties , finding overlappinng subproblems, thus thinking of DP.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThink of a recursion first then convert it to tabulation.\n# ... | 0 | 0 | ['C++'] | 0 |
minimum-cost-to-equalize-array | [Java/C++/Python] 4 cases, O(n) solution | javacpython-4-cases-on-solution-by-lee21-qnzq | Intuition\n- If cost1 is cheaper, all use cost1\n- Increment all to max(A), use cost2 as many as possible\n- Increment to bigger than max(A), use less cost1\n- | lee215 | NORMAL | 2024-05-05T05:01:39.613181+00:00 | 2024-05-06T05:29:54.127148+00:00 | 5,532 | false | # **Intuition**\n- If `cost1` is cheaper, all use `cost1`\n- Increment all to `max(A)`, use `cost2` as many as possible\n- Increment to bigger than `max(A)`, use less `cost1`\n- cost1 is too expensive, increment untill we won\'t use `cost1`\n<br>\n\n\n# **Explanation**\nCalculate the `min(A)` and `max(A)`\n`total = max... | 59 | 2 | ['C', 'Python', 'Java'] | 23 |
minimum-cost-to-equalize-array | Binary Search on Piecewise Linear Function | binary-search-on-piecewise-linear-functi-w577 | Intuition\nOne approach is to first find the value that we should make all the array elements equal to, then find the cost of doing that.\n\nHowever, directly f | jeffreyhu8 | NORMAL | 2024-05-05T04:01:05.535495+00:00 | 2024-05-05T04:05:16.650090+00:00 | 3,424 | false | # Intuition\nOne approach is to first find the value that we should make all the array elements equal to, then find the cost of doing that.\n\nHowever, directly finding a formula for the optimal value is not trivial. For example, the optimal value may not always be `max(nums)`, as shown by the case `nums = [1, 100_000,... | 45 | 5 | ['Binary Search', 'Python3'] | 7 |
minimum-cost-to-equalize-array | Deep Dive into Lee's O(n) Solution | deep-dive-into-lees-on-solution-by-rexch-q2ey | Click to view lee215\'s O(n) solution\n\nDisclaimer: This post is just my explanation of @lee215 solution and I do NOT own the solution.\n\nFor ease of reading/ | rexcheng | NORMAL | 2024-05-05T22:32:31.149899+00:00 | 2024-05-09T15:04:20.035635+00:00 | 897 | false | [Click to view lee215\'s O(n) solution](https://leetcode.com/problems/minimum-cost-to-equalize-array/solutions/5114202/java-c-python-4-cases-o-n-solution)\n\n***Disclaimer: This post is just my explanation of @lee215 solution and I do NOT own the solution.***\n\nFor ease of reading/understanding, let\'s call\n- `op1`: ... | 16 | 1 | ['Python', 'Python3', 'JavaScript'] | 5 |
minimum-cost-to-equalize-array | C++ Easy Solution || Line by Line Explanation (Brute Force Thinking) | c-easy-solution-line-by-line-explanation-knbp | \nGiven : Make all the array elements as equal\n\nThoughts :\nOur first thought after seeing the given examples is \nmake all the array elements to maximum elem | manoj-22 | NORMAL | 2024-05-05T07:58:05.346842+00:00 | 2024-05-05T08:38:32.625432+00:00 | 909 | false | ```\nGiven : Make all the array elements as equal\n\nThoughts :\nOur first thought after seeing the given examples is \nmake all the array elements to maximum element in array.\n\nWhat if, making all the arr elements to maximum is costlier than \nmaking all the arr elements to x (which is greater than max element in ar... | 15 | 0 | ['C++'] | 3 |
minimum-cost-to-equalize-array | Python O(n) fastest solution explained visually | python-on-fastest-solution-explained-vis-llq7 | Approach\n Describe your approach to solving the problem. \nThe easiest way to approach this problem is to think of each entry in nums as a stack of blocks (min | KellerWheat | NORMAL | 2024-05-05T05:25:28.143838+00:00 | 2024-05-05T17:29:46.664693+00:00 | 369 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nThe easiest way to approach this problem is to think of each entry in nums as a stack of blocks (minecraft style). The goal is to make all of the block stacks the same height. For example, the array [1, 3, 3, 4, 5] would look like this:\n Besides min/max/sum | python-3-o1-besides-minmaxsum-by-zsq007-frql | Intuition\n1. If cost1 is no more than a half of cost2, we can always choose Op1.\n2. If the maxGap is no more than a half of the ttlGap, we can always find pai | zsq007 | NORMAL | 2024-05-05T04:04:22.845430+00:00 | 2024-05-08T18:37:03.613598+00:00 | 2,346 | false | # Intuition\n1. If `cost1` is no more than a half of `cost2`, we can always choose Op1.\n2. If the `maxGap` is no more than a half of the `ttlGap`, we can always find pairs for Op2 (Here\'s why: [LC1953](https://leetcode.com/problems/maximum-number-of-weeks-for-which-you-can-work/)), except for the corner case that the... | 14 | 1 | ['Greedy', 'Python3'] | 5 |
minimum-cost-to-equalize-array | ✅ Detailed Explanation : Checking all combinations | detailed-explanation-checking-all-combin-9qi8 | Intuition\n\nThe range of the equalised value will be max(a)....2 * max(a)\n\nWhy ?\nWe need to allow all possible equal max values to be the final equal value | rbssmtkr | NORMAL | 2024-05-05T05:58:31.122688+00:00 | 2024-05-06T05:16:57.947228+00:00 | 1,126 | false | ## Intuition\n\nThe `range` of the equalised value will be `max(a)....2 * max(a)`\n\nWhy ?\nWe need to allow all possible equal max values to be the final equal value here as we need to consider all possible count of `cost2` operations as much as possible.\nIts possible that cost2 << cost1 and for using cost2 as much a... | 9 | 0 | ['Math', 'Combinatorics', 'C++'] | 4 |
minimum-cost-to-equalize-array | Constructive solution in O(n log n) | constructive-solution-in-on-log-n-by-ply-udg2 | Approach\nWe\'ll operate on a sorted diff array, where diff[i] = maxElement - arr[i].\nNote that if cost1 * 2 <= cost2, it is enough to just do single operation | plyusnovdmitrii | NORMAL | 2024-05-05T04:27:51.405330+00:00 | 2024-05-05T04:33:57.376177+00:00 | 1,419 | false | # Approach\nWe\'ll operate on a sorted `diff` array, where `diff[i] = maxElement - arr[i]`.\nNote that if `cost1 * 2 <= cost2`, it is enough to just do single operations, so the answer would be `sum(diff[i]) * cost1`.\nIn case `cost2 < cost1 * 2`, we\'ll try to use `cost2` operations as much as possible.\n\nFirst, note... | 7 | 1 | ['C++'] | 2 |
minimum-cost-to-equalize-array | Java Clean Solution | java-clean-solution-by-shree_govind_jee-6q90 | Code\nApproach-1\n\nclass Solution {\n private static final int MOD = (int)1e9 + 7;\n public int minCostToEqualizeArray(int[] nums, int cost1, int cost2) | Shree_Govind_Jee | NORMAL | 2024-05-05T04:12:23.065621+00:00 | 2024-05-05T04:13:55.544248+00:00 | 941 | false | # Code\n**Approach-1**\n```\nclass Solution {\n private static final int MOD = (int)1e9 + 7;\n public int minCostToEqualizeArray(int[] nums, int cost1, int cost2) {\n Arrays.sort(nums);\n int n = nums.length;\n if (n == 1) return 0;\n if (n == 2) return (int)((long)(nums[1] - nums[0]) ... | 6 | 0 | ['Array', 'Math', 'Java'] | 1 |
minimum-cost-to-equalize-array | Video Explanation (Dissecting problem into smaller sub-problems & building upwards) | video-explanation-dissecting-problem-int-nutj | Explanation\n\nClick here for the video\n\n# Code\n\nconst int M = 1e6;\nconst int MOD = 1e9+7;\n\ntypedef long long int ll;\n\nclass Solution {\npublic:\n i | codingmohan | NORMAL | 2024-05-05T12:41:23.137012+00:00 | 2024-05-05T12:41:23.137043+00:00 | 197 | false | # Explanation\n\n[Click here for the video](https://youtu.be/MqPaU6JAv9E)\n\n# Code\n```\nconst int M = 1e6;\nconst int MOD = 1e9+7;\n\ntypedef long long int ll;\n\nclass Solution {\npublic:\n int minCostToEqualizeArray(vector<int>& nums, int cost1, int cost2) {\n vector<int> req(M+1, 0);\n \n i... | 5 | 1 | ['C++'] | 2 |
minimum-cost-to-equalize-array | $O(9n)$ Solution, No Binary Search. Mathematical proofs for all needed properties. | o9n-solution-no-binary-search-mathematic-t7yl | This answer will focus on rigorously proving the required Lemmas to show correctness of the algorithm. \n\n# Intuition & Approach\n\nFor $M\geq \max(nums)$, def | eyfmharb | NORMAL | 2024-05-05T08:03:38.617531+00:00 | 2024-05-05T08:04:58.456671+00:00 | 339 | false | This answer will focus on **rigorously** proving the required Lemmas to show correctness of the algorithm. \n\n# Intuition & Approach\n\nFor $M\\geq \\max(nums)$, define $f(M)$ as the minimum cost to transform the entire array to have value $M$. Let us handle some easy cases first. \n\n**Lemma 1**: If $c_2 > 2c_1$, the... | 5 | 0 | ['C++'] | 1 |
minimum-cost-to-equalize-array | [Java] Commented & Explained Linear Solution O(N), with O(1) space | java-commented-explained-linear-solution-7e7i | Intuition\nWe only care about the Deltas between nums, so we want to find the sum of all Deltas, and max Delta (from minValue to maxValue) in nums. We can brin | bamboo168 | NORMAL | 2024-05-05T21:47:36.763479+00:00 | 2024-05-06T21:25:32.951496+00:00 | 625 | false | # Intuition\nWe only care about the Deltas between nums, so we want to find the **sum** of all Deltas, and **max** Delta (from `minValue` to `maxValue`) in `nums`. We can bring all nums up to max, or exceed max. Let\'s first try to solve for reaching max:\n\n`solve` helper function (to reach `maxValue`)\nIf we want t... | 4 | 0 | ['Array', 'Greedy', 'Geometry', 'Java'] | 1 |
minimum-cost-to-equalize-array | HARD||explain line by line|| c++ | hardexplain-line-by-line-c-by-sujalgupta-u2u5 | \n# Code\n\n// Define a class named Solution\n\nclass Solution {\npublic:\n // Define a member function named minCostToEqualizeArray that takes a vector of i | sujalgupta09 | NORMAL | 2024-05-05T05:55:13.095656+00:00 | 2024-05-05T05:55:13.095681+00:00 | 755 | false | \n# Code\n```\n// Define a class named Solution\n\nclass Solution {\npublic:\n // Define a member function named minCostToEqualizeArray that takes a vector of integers (nums), and two integers (cost1 and cost2) as parameters and returns an integer.\n int minCostToEqualizeArray(vector<int>& nums, int cost1, int co... | 4 | 2 | ['C++'] | 1 |
minimum-cost-to-equalize-array | Bruteforce + Greedy | bruteforce-greedy-by-penguinzzz-0x79 | Intuition\ngreedy\n\n# Approach\nPAIRING\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\no(1)\n\n# Code\n\n#define MOD 1000000007\nclass Solut | penguinzzz | NORMAL | 2024-05-05T05:12:47.472422+00:00 | 2024-05-05T05:12:47.472441+00:00 | 485 | false | # Intuition\ngreedy\n\n# Approach\nPAIRING\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\no(1)\n\n# Code\n```\n#define MOD 1000000007\nclass Solution {\npublic:\n long long func(vector<int>& a, int& cost1, int& cost2, long long t,long long mn,long long summ){\n int n=a.size();\n long... | 4 | 0 | ['C++'] | 1 |
minimum-cost-to-equalize-array | [Python3] Easy understand, clean code, no confusing branches (greedy, O(n) time, O(1) space) | python3-easy-understand-clean-code-no-co-ldpz | Intuition\n Describe your first thoughts on how to solve this problem. \nJust test all the possible optimal cases instead of using lots of branches\n\n# Approac | zzjjbb | NORMAL | 2024-05-05T04:50:45.603666+00:00 | 2024-05-26T04:47:36.808732+00:00 | 288 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nJust test all the possible optimal cases instead of using lots of branches\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nFirst, if `n <= 2` or `cost1 * 2 <= cost2`, we only use the first operation to get the opti... | 4 | 0 | ['Math', 'Greedy', 'Python3'] | 2 |
minimum-cost-to-equalize-array | [Python3] Ternary Search | python3-ternary-search-by-awice-2cpf | Let\'s call the moves single and double. First, c2 = min(c2, 2*c1) if two single moves are cheaper than a double.\n\nSuppose the final position has every eleme | awice | NORMAL | 2024-05-05T23:28:59.799962+00:00 | 2024-05-05T23:28:59.799981+00:00 | 61 | false | Let\'s call the moves single and double. First, `c2 = min(c2, 2*c1)` if two single moves are cheaper than a double.\n\nSuppose the final position has every element equal to `height`, what will be the `cost(height)`?\n\n- Say we have a total of `volume` increments to do. If no element requires a lot of work, we could ... | 3 | 0 | ['Python3'] | 0 |
minimum-cost-to-equalize-array | Detailed guide to Minimum Cost to Equalize Array | detailed-guide-to-minimum-cost-to-equali-r692 | Approach\nat each step, we can either increment single value at the cost of cost1 or two values at the cost of cost2.\nNow, if 2*cost1<=cost2 then we don\'t nee | miamimagna | NORMAL | 2024-05-05T06:39:37.876774+00:00 | 2024-05-05T07:01:15.338642+00:00 | 457 | false | # Approach\nat each step, we can either increment single value at the cost of `cost1` or two values at the cost of `cost2`.\nNow, if `2*cost1<=cost2` then we don\'t need to think too much, increment values singly is better\nThe problem arises when cost2 is better...\nNow, when cost2 is better, we would need to maximize... | 3 | 0 | ['Greedy', 'Sorting', 'C++'] | 2 |
minimum-cost-to-equalize-array | O(MAX ELEMENT) Greedy (No need of cases on cost1 and cost2) | omax-element-greedy-no-need-of-cases-on-ia9rc | Intuition\nTry thinking about the minimum operations required to make all the elements in the array equal to a particular value.\n\n# Approach\n\n\n# Complexity | the_halfblood_prince | NORMAL | 2024-05-05T04:35:56.814890+00:00 | 2024-05-05T06:19:55.272939+00:00 | 353 | false | # Intuition\nTry thinking about the minimum operations required to make all the elements in the array equal to a particular value.\n\n# Approach\n\n\n# Complexity\n- Time complexity: O(max Element)\n\n- Space complexity: O(1)\n# Code\n```\nusing ll = long long;\nconst ll N= 1e9 + 7;\nclass Solution {\npublic:\n int ... | 3 | 0 | ['Greedy', 'C++'] | 1 |
minimum-cost-to-equalize-array | Not mine solution But probably the best solution on rankings | not-mine-solution-but-probably-the-best-xhnnc | 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 | Harshitcode | NORMAL | 2024-05-05T04:25:16.473863+00:00 | 2024-05-05T04:25:16.473896+00:00 | 783 | 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. $$O(n)$$ --... | 3 | 3 | ['C++'] | 3 |
minimum-cost-to-equalize-array | O(n) Solution | Easy to understand with comments | on-solution-easy-to-understand-with-comm-95za | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nhttps://www.youtube.com/watch?v=1DRxNPliy0Q\nThis video helped me to unde | mj1609 | NORMAL | 2024-05-14T11:53:19.103502+00:00 | 2024-05-14T11:53:19.103534+00:00 | 79 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nhttps://www.youtube.com/watch?v=1DRxNPliy0Q\nThis video helped me to understand the logic. Go through the video once. \n\n\n# Code\n```\nconst int MOD = 1e9+7;\nclass Solution {\npublic:\n int minCostToEqualizeArray(vecto... | 2 | 0 | ['C++'] | 0 |
minimum-cost-to-equalize-array | Tricky Concept | Brute Force | c++ | tricky-concept-brute-force-c-by-allround-4c9e | Intuition\nif 2*cost1 <= cost2 than we simply use cost1 for all operation.\nelse we will maximise the use of cost2. \n\n# Approach\nwe check for all maxi elemen | allrounderankit | NORMAL | 2024-05-05T05:05:12.001512+00:00 | 2024-05-05T05:05:12.001528+00:00 | 665 | false | # Intuition\nif 2*cost1 <= cost2 than we simply use cost1 for all operation.\nelse we will maximise the use of cost2. \n\n# Approach\nwe check for all maxi element from which we can take minimum cost as answer\n\nStuck in Ternary search in contest :( but later realise it is not perfect concave structure.\n\n# Complexit... | 2 | 2 | ['C++'] | 2 |
minimum-cost-to-equalize-array | Brute Force + Greedy | Linear Time O(N) & Constant Space O(1) | brute-force-greedy-linear-time-on-consta-z97u | Intuition:\n1. If 2 * cost1 <= cost2 we will always leverage cost1\n2. We can make all elements equal to the max element at the very least. Hence, we will try | shahsb | NORMAL | 2024-05-15T05:18:38.328150+00:00 | 2024-05-15T05:28:36.181513+00:00 | 22 | false | # Intuition:\n1. If `2 * cost1 <= cost2` we will always leverage cost1\n2. We can make all elements equal to the max element at the very least. Hence, we will try out all possibilities from `max_element` to `2 * max_element`\n# CODE:\n```\n#define ll long long\nconst int MOD = 1e9 + 7;\n\nclass Solution {\nprivate:\n... | 1 | 0 | ['Greedy'] | 1 |
minimum-cost-to-equalize-array | Tricky but Understandable Solution | tricky-but-understandable-solution-by-ja-8ih1 | 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 | jainamit130 | NORMAL | 2024-05-09T11:52:34.330327+00:00 | 2024-05-09T11:52:34.330349+00:00 | 81 | 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(1)\n<!-- Add your space complexity here, e.g. $... | 1 | 0 | ['Array', 'Enumeration', 'C++'] | 0 |
minimum-cost-to-equalize-array | [HELP] Final value will be within [max, 2 * max], Why ? | help-final-value-will-be-within-max-2-ma-5pyd | This post is mainly to understand why would the final value be between [max, 2 * max] ?\n\nI believe this is the crux of the problem. Can someone help me unders | ankurd87 | NORMAL | 2024-05-06T04:40:32.160922+00:00 | 2024-05-06T04:40:32.160954+00:00 | 358 | false | This post is mainly to understand why would the final value be between [max, 2 * max] ?\n\nI believe this is the crux of the problem. Can someone help me understand this ? May be a mathematical proof ?\n\nI will try to understand it myself and then will update the post if I get an answer myself sooner.\n\nNote* the cod... | 1 | 0 | ['C++'] | 0 |
minimum-cost-to-equalize-array | BEST INTUITIVE SOLUTION ✅✅✅ 🥇 | best-intuitive-solution-by-rishu_raj5938-ss4f | 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 | Rishu_Raj5938 | NORMAL | 2024-05-05T12:58:14.401191+00:00 | 2024-05-05T12:58:14.401225+00:00 | 60 | 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 |
minimum-cost-to-equalize-array | [Python] Brutforce Works O(MAX(A)+N) | Explained | python-brutforce-works-omaxan-explained-e0k1q | Approach\n Describe your approach to solving the problem. \nThis is how i solved it (main observations)\n\n1. trivial case using only cost1 when 2*c1 < c2\n2. w | tr1ten | NORMAL | 2024-05-05T08:15:41.179383+00:00 | 2024-05-08T06:02:50.862941+00:00 | 287 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nThis is how i solved it (main observations)\n\n1. trivial case using only cost1 when 2*c1 < c2\n2. we need to pair up max elements so to use cost2 more number of times.\n\n\nLets solve another problem: \nf(mx): cost to make all element equal to mx usi... | 1 | 0 | ['Python3'] | 2 |
minimum-cost-to-equalize-array | 💥💥 Beats 100% on runtime and memory [EXPLAINED] | beats-100-on-runtime-and-memory-explaine-ifee | IntuitionMake all elements in the array equal with the minimum cost, so I need to find an optimal target value and decide the best way to reach it using single | r9n | NORMAL | 2025-01-30T15:47:43.695355+00:00 | 2025-01-30T15:47:43.695355+00:00 | 7 | false | # Intuition
Make all elements in the array equal with the minimum cost, so I need to find an optimal target value and decide the best way to reach it using single or double increment operations.
# Approach
Iterate through possible target values from the maximum element to twice its value while calculating the total in... | 0 | 0 | ['Array', 'Math', 'Binary Search', 'Greedy', 'Combinatorics', 'Enumeration', 'Kotlin'] | 0 |
minimum-cost-to-equalize-array | Optimal Pairing of Increments with Cost Analysis | 100% ✅ | optimal-pairing-of-increments-with-cost-4cajc | IntuitionMinimize cost by prioritizing the cheaper operation (c2 for pairing two elements) while handling cases where pairing is not possible.Approach
Calculat | DeepakPatil26 | NORMAL | 2025-01-04T16:30:04.444175+00:00 | 2025-01-04T16:30:04.444175+00:00 | 11 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Minimize cost by prioritizing the cheaper operation (`c2` for pairing two elements) while handling cases where pairing is not possible.
# Approach
<!-- Describe your approach to solving the problem. -->
1. Calculate the total deficit (sum ... | 0 | 0 | ['Array', 'Greedy', 'Enumeration', 'JavaScript'] | 0 |
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