question_slug stringlengths 3 77 | title stringlengths 1 183 | slug stringlengths 12 45 | summary stringlengths 1 160 ⌀ | author stringlengths 2 30 | certification stringclasses 2
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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reverse-degree-of-a-string | Scala solution with implicit, zip and map | scala-solution-with-implicit-zip-and-map-y8c6 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | iyIeO99AmH | NORMAL | 2025-04-03T21:10:20.377309+00:00 | 2025-04-03T21:10:20.377309+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 | ['Scala'] | 0 |
reverse-degree-of-a-string | Intuitive solution with array to map character to value in reversed alphabet | intuitive-solution-with-array-to-map-cha-sekc | IntuitionWe'll use a simple array to map each lowercase character to its value in the reversed alphabet.ApproachWe'll set up a simple array to map each lowercas | mnjk | NORMAL | 2025-04-03T20:05:21.275617+00:00 | 2025-04-03T20:05:21.275617+00:00 | 3 | false | # Intuition
We'll use a simple array to map each lowercase character to its value in the reversed alphabet.
# Approach
We'll set up a simple array to map each lowercase character to its value in the reversed alphabet:
```ts
const m = [26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5,... | 0 | 0 | ['Array', 'String', 'TypeScript'] | 0 |
reverse-degree-of-a-string | python | python-by-binkswang-u2xp | Code | BinksWang | NORMAL | 2025-04-03T19:12:43.709857+00:00 | 2025-04-03T19:12:43.709857+00:00 | 2 | false | # Code
```python3 []
class Solution:
def reverseDegree(self, s: str) -> int:
arr = []
val = 26
rs = 0
while val > 0:
arr.append(val)
val -= 1
for i in range (0, len(s)):
rs += arr[ord(s[i]) - ord('a')] * (i + 1)
return rs
`... | 0 | 0 | ['Python3'] | 0 |
reverse-degree-of-a-string | c# | c-by-binkswang-89ua | Code | BinksWang | NORMAL | 2025-04-03T19:08:36.004979+00:00 | 2025-04-03T19:08:36.004979+00:00 | 5 | false | # Code
```csharp []
public class Solution {
public int ReverseDegree(string s) {
var arr = new int[26];
var val = 26;
var rs = 0;
for(var i = 0; i < arr.Length; i++){
arr[i] = val;
val--;
}
for(var i = 0; i < s.Length; i++){
rs += a... | 0 | 0 | ['C#'] | 0 |
reverse-degree-of-a-string | reverse-degree-of-a-string || very easy solution || JS | reverse-degree-of-a-string-very-easy-sol-rxn4 | Complexity
Time complexity:O(N)
Space complexity:O(1)
Code | shevykumar26 | NORMAL | 2025-04-03T17:56:38.958169+00:00 | 2025-04-03T17:56:38.958169+00:00 | 8 | false | # Complexity
- Time complexity:O(N)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```javascript []
/**
* @param {string} s
* @return {number}
*/
var reverseDegree = function(s) {
let sum=0
for(let i=0;i<s.length... | 0 | 0 | ['JavaScript'] | 0 |
reverse-degree-of-a-string | Java | java-by-soumya_699-x1hm | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Soumya_699 | NORMAL | 2025-04-03T04:29:35.207615+00:00 | 2025-04-03T04:29:35.207615+00:00 | 13 | 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 |
reverse-degree-of-a-string | Java | java-by-soumya_699-lr6d | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Soumya_699 | NORMAL | 2025-04-03T04:29:31.499443+00:00 | 2025-04-03T04:29:31.499443+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 |
reverse-degree-of-a-string | Reverse Degree of a String - 💯Beats 100.00% || C++✅✔️ | reverse-degree-of-a-string-beats-10000-c-izsb | Complexity
Time complexity:
Space complexity:
Code | itz_raju | NORMAL | 2025-04-02T20:06:15.581196+00:00 | 2025-04-02T20:06:15.581196+00:00 | 1 | false | 
# 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(1)
# Code
```cpp []
class So... | 0 | 0 | ['String', 'C++'] | 0 |
reverse-degree-of-a-string | Unique Approach | hashing | unique-approach-hashing-by-mr_ramzan-n77a | Code | mr_ramzan | NORMAL | 2025-04-02T17:17:14.427525+00:00 | 2025-04-02T17:17:14.427525+00:00 | 1 | false |
# Code
```cpp []
class Solution {
public:
int reverseDegree(string s) {
unordered_map<char,int>map;
for(char ch='a';ch<='z';ch++) map[ch]= 'z'-ch+1;
int sum=0;
for(int i=0;i<s.length();i++){
sum+=(map[s[i]]*(i+1));
}
return sum;
}
};
``` | 0 | 0 | ['Hash Table', 'String', 'C++'] | 0 |
reverse-degree-of-a-string | C++ Solution for Reverse Degree of a String | c-solution-for-reverse-degree-of-a-strin-mxqa | IntuitionThe problem likely involves transforming a string by calculating a "reverse degree." The goal is to map each character to its reverse position in the E | _tarunkr_chauhan | NORMAL | 2025-04-02T10:36:01.629945+00:00 | 2025-04-02T10:36:01.629945+00:00 | 2 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem likely involves transforming a string by calculating a "reverse degree." The goal is to map each character to its reverse position in the English alphabet ('a' becomes 26, 'b' becomes 25, etc.) and weight each character's revers... | 0 | 0 | ['C++'] | 0 |
reverse-degree-of-a-string | esay java code of beginers!!!!!!!!!!!!!!!! | esay-java-code-of-beginers-by-churchill0-js0k | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | CHURCHILL04 | NORMAL | 2025-04-02T10:00:48.302676+00:00 | 2025-04-02T10:00:48.302676+00:00 | 4 | 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 |
reverse-degree-of-a-string | Python fast one-line solution | python-fast-one-line-solution-by-redberr-l9oz | Code | redberry33 | NORMAL | 2025-04-02T07:08:52.987723+00:00 | 2025-04-02T07:08:52.987723+00:00 | 6 | false |
# Code
```python3 []
class Solution:
def reverseDegree(self, s: str) -> int:
return reduce(lambda acc, enum: acc + (enum[0] + 1) * (123 - ord(enum[1])), enumerate(s), 0)
``` | 0 | 0 | ['Python3'] | 0 |
reverse-degree-of-a-string | 3498 - Reverse Degree Of A String | 3498-reverse-degree-of-a-string-by-prati-chxn | Intuitionresult += (26 - (s.charAt(i) - 'a')) * (i + 1);Approachresult += (s.charAt(i) - 123) * -(i + 1);Complexity
Time complexity:
Space complexity:
Code | pratik_solanki | NORMAL | 2025-04-02T06:11:13.172980+00:00 | 2025-04-02T06:11:13.172980+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
result += (26 - (s.charAt(i) - 'a')) * (i + 1);
# Approach
<!-- Describe your approach to solving the problem. -->
result += (s.charAt(i) - 123) * -(i + 1);
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$... | 0 | 0 | ['Java'] | 0 |
reverse-degree-of-a-string | Java Easy Solution | java-easy-solution-by-iamsd-szpc | Code | iamsd_ | NORMAL | 2025-04-02T05:31:25.396453+00:00 | 2025-04-02T05:31:25.396453+00:00 | 3 | false | # Code
```java []
class Solution {
public int reverseDegree(String s) {
int i = 26;
Map<String, Integer> map = new HashMap<>();
for (char c = 'a'; c <= 'z'; c++) {
map.put(String.valueOf(c), i);
i--;
}
int sum = 0;
String[] srr = s.split(... | 0 | 0 | ['Java'] | 0 |
reverse-degree-of-a-string | 🔄📖 Reverse Alphabet Math Madness! 🔢🤯Simple Logic Behind the Code 🤓➡️💡PYTHON | reverse-alphabet-math-madness-simple-log-hp51 | IntuitionThe problem requires calculating a weighted sum where each letter's reverse alphabet value is multiplied by its 1-based index in the string.
Since lett | Shahin1212 | NORMAL | 2025-04-02T04:41:08.554637+00:00 | 2025-04-02T04:41:08.554637+00:00 | 3 | false | # Intuition
The problem requires calculating a weighted sum where each letter's **reverse alphabet value** is multiplied by its **1-based index** in the string.
Since letter values are fixed, we can **precompute** them in a dictionary and efficiently compute the sum in a **single pass**.
# Approach
1. **Create a dic... | 0 | 0 | ['Python3'] | 0 |
reverse-degree-of-a-string | Java 100% Faster Solution | java-100-faster-solution-by-tbekpro-p8vb | Complexity
Time complexity: O(N)
Space complexity: O(1)
Code | tbekpro | NORMAL | 2025-04-01T19:11:57.318045+00:00 | 2025-04-01T19:11:57.318045+00:00 | 4 | false | # Complexity
- Time complexity: O(N)
- Space complexity: O(1)
# Code
```java []
class Solution {
public int reverseDegree(String s) {
int result = 0;
for (int i = 0; i < s.length(); i++) {
int score = (26 - (s.charAt(i) - 'a')) * (i + 1);
result += score;
}
r... | 0 | 0 | ['Java'] | 0 |
reverse-degree-of-a-string | reverseIndexString | reverseindexstring-by-aditya_211-gpo4 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Aditya_211 | NORMAL | 2025-04-01T17:35:43.710724+00:00 | 2025-04-01T17:35:43.710724+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-product-of-three-numbers | Java O(1) space O(n) time solution beat 100% | java-o1-space-on-time-solution-beat-100-wung3 | Simply find out the three largest numbers and the two smallest numbers using one pass.\n\n public int maximumProduct(int[] nums) {\n int max1 = Intege | dreamchase | NORMAL | 2017-06-25T17:10:41.283000+00:00 | 2018-10-16T15:01:25.567757+00:00 | 53,928 | false | Simply find out the three largest numbers and the two smallest numbers using one pass.\n```\n public int maximumProduct(int[] nums) {\n int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE, min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;\n for (int n : nums) {\n ... | 315 | 2 | [] | 47 |
maximum-product-of-three-numbers | [Python] O(N) and 1 line | python-on-and-1-line-by-lee215-g80f | My first solution using sort\n\ndef maximumProduct(self, nums):\n nums.sort()\n return max(nums[-1] * nums[-2] * nums[-3], nums[0] * nums[1] * num | lee215 | NORMAL | 2017-06-28T19:15:59.247000+00:00 | 2018-09-02T20:50:07.670914+00:00 | 29,058 | false | My first solution using ```sort```\n````\ndef maximumProduct(self, nums):\n nums.sort()\n return max(nums[-1] * nums[-2] * nums[-3], nums[0] * nums[1] * nums[-1])\n`````\nI found a exactly same solution in discuss. Anyway, O(NlogN) is not adorable and O(N) is possible.\n````\ndef maximumProduct(self, nums... | 148 | 8 | [] | 21 |
maximum-product-of-three-numbers | General solution for any K | general-solution-for-any-k-by-kale_kale_-gyt1 | This is the first solution that can be used to solve the problem, Maximum Prodoct of K Numbers.\nHope it can help others when encountering a follow up question. | kale_kale_kale | NORMAL | 2018-05-05T02:49:19.018294+00:00 | 2018-09-25T06:07:12.515578+00:00 | 6,183 | false | This is the first solution that can be used to solve the problem, Maximum Prodoct of K Numbers.\nHope it can help others when encountering a follow up question.\n```\n public int maximumProduct(int[] nums) {\n Arrays.sort(nums);\n // if signs of the minimum and maximum number are the same, simply retur... | 74 | 0 | [] | 11 |
maximum-product-of-three-numbers | [C++] Two Solutions | c-two-solutions-by-pankajgupta20-6wyf | Sorting Solution\n\tclass Solution {\n\tpublic:\n\t\tint maximumProduct(vector& nums) {\n\t\t\tint n = nums.size();\n\t\t\tsort(nums.begin(), nums.end());\n\t\t | pankajgupta20 | NORMAL | 2021-05-13T12:42:16.670192+00:00 | 2021-05-13T12:43:19.470746+00:00 | 11,171 | false | ##### Sorting Solution\n\tclass Solution {\n\tpublic:\n\t\tint maximumProduct(vector<int>& nums) {\n\t\t\tint n = nums.size();\n\t\t\tsort(nums.begin(), nums.end());\n\t\t\treturn max(nums[n - 1] * nums[n - 2] * nums[n - 3], nums[0] * nums[1] * nums[n - 1]);\n\t\t}\n\t};\n##### Without Sorting Solution\n\tclass Solutio... | 60 | 3 | ['C', 'Sorting', 'C++'] | 4 |
maximum-product-of-three-numbers | Java Easy AC... | java-easy-ac-by-leiyu-j91a | \n public int maximumProduct(int[] nums) {\n \n Arrays.sort(nums);\n //One of the Three Numbers is the maximum value in the array.\n\n | leiyu | NORMAL | 2017-06-25T03:00:34.488000+00:00 | 2018-10-16T04:15:09.312973+00:00 | 17,561 | false | ```\n public int maximumProduct(int[] nums) {\n \n Arrays.sort(nums);\n //One of the Three Numbers is the maximum value in the array.\n\n int a = nums[nums.length - 1] * nums[nums.length - 2] * nums[nums.length - 3];\n int b = nums[0] * nums[1] * nums[nums.length - 1];\n ... | 59 | 3 | [] | 11 |
maximum-product-of-three-numbers | easiest two lines of code#python3 | easiest-two-lines-of-codepython3-by-moha-m2dp | 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 | Mohammad_tanveer | NORMAL | 2023-03-18T05:04:19.034291+00:00 | 2023-03-18T05:04:19.034336+00:00 | 5,698 | 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. $$O(n)$$ --... | 50 | 0 | ['Python3'] | 4 |
maximum-product-of-three-numbers | Share my python solution: one pass, O(n) time O(1) space | share-my-python-solution-one-pass-on-tim-sl33 | \n max1,max2,max3,min1,min2 = float('-Inf'),float('-Inf'),float('-Inf'),float('Inf'),float('Inf')\n for num in nums:\n if num >= max1:\ | kiiiid | NORMAL | 2017-07-16T05:06:19.580000+00:00 | 2018-10-25T06:15:52.574047+00:00 | 5,208 | false | ```\n max1,max2,max3,min1,min2 = float('-Inf'),float('-Inf'),float('-Inf'),float('Inf'),float('Inf')\n for num in nums:\n if num >= max1:\n max3,max2,max1 = max2,max1,num\n elif num >= max2:\n max3,max2 = max2,num\n elif num > max3:\n ... | 44 | 1 | [] | 5 |
maximum-product-of-three-numbers | C++ DP solution | c-dp-solution-by-wofainta-i4q2 | O(n) time, O(1) space\n\nclass Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n int f[2][4], i, j;\n f[0][0] = f[1][0] = 1;\n | wofainta | NORMAL | 2018-09-24T21:59:08.737154+00:00 | 2018-10-14T18:49:25.282190+00:00 | 4,733 | false | O(n) time, O(1) space\n```\nclass Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n int f[2][4], i, j;\n f[0][0] = f[1][0] = 1;\n for (j = 3; j > 0; --j) {\n f[0][j] = INT_MIN;\n f[1][j] = INT_MAX;\n } \n \n for (i = 0; i < nums.siz... | 39 | 1 | [] | 6 |
maximum-product-of-three-numbers | Time: O(n), Space: O(1) | time-on-space-o1-by-riyakushwaha-iso1 | Base: Maximum product of three numbers is possible only when all three numbers are positive or two numbers are negative and one is positive.\n\nApproach1: \n1. | riyakushwaha | NORMAL | 2021-12-03T10:53:25.994318+00:00 | 2021-12-03T10:53:25.994347+00:00 | 3,360 | false | Base: Maximum product of three numbers is possible only when all three numbers are positive or two numbers are negative and one is positive.\n\nApproach1: \n1. Sort the array, so three max numbers are in the end of the array, two negative numbers whose product will be max will be on the start of the array.\n2. return m... | 33 | 1 | ['Java'] | 2 |
maximum-product-of-three-numbers | C++ Solution with explanation | c-solution-with-explanation-by-princeerj-881v | Either product of 3 biggest positive values will be maxProduct or if there are negative values then pick the 2 biggest negative values and multiply with biggest | princeerjain | NORMAL | 2017-06-25T03:03:49.183000+00:00 | 2018-08-17T04:37:24.140252+00:00 | 9,395 | false | Either product of 3 biggest positive values will be maxProduct or if there are negative values then pick the 2 biggest negative values and multiply with biggest positive value\n\nSort the Array and compare above mentioned products\n\n\n int maximumProduct(vector<int>& nums) {\n sort(nums.begin(), nums.end());... | 31 | 2 | [] | 7 |
maximum-product-of-three-numbers | Begineer Friendly || O(nlogn) -> O(n) | begineer-friendly-onlogn-on-by-himanshuc-06t1 | IDEA : Product is going to be maximum if we take product of top 3 maximum number. and also negative*negative became Positive so we will also consider a combinat | himanshuchhikara | NORMAL | 2021-03-19T15:06:26.591642+00:00 | 2021-03-19T15:29:07.187379+00:00 | 1,215 | false | IDEA : Product is going to be maximum if we take product of top 3 maximum number. and also **negative*negative became Positive** so we will also consider a combination of 2 most negative with maximum .\nanswer = Math.max(max1*max2*max3,max1*min1*min2)\n**min1:** minimum element of array\n**min2:** 2nd minimum element o... | 30 | 0 | [] | 0 |
maximum-product-of-three-numbers | ✅✅Two Java Solution || Easy Explanation 🔥 🔥 | two-java-solution-easy-explanation-by-ri-ydyb | Approach 1\nThe code starts by sorting the input array, "nums," in ascending order using the Arrays.sort() method. Sorting the array allows us to consider diffe | rirefat | NORMAL | 2023-05-30T16:52:09.456002+00:00 | 2023-05-30T16:52:09.456047+00:00 | 6,349 | false | # Approach 1\nThe code starts by sorting the input array, "nums," in ascending order using the **Arrays.sort()** method. Sorting the array allows us to consider different cases and calculate the maximum product.\n\nThe next step involves calculating two potential maximum products:\n\n1. "case1" represents the product o... | 29 | 0 | ['Array', 'Math', 'Sorting', 'Java'] | 2 |
maximum-product-of-three-numbers | Beats 100% of users with C++ || Step By Step Explain || Easy to Understand || Best Approach || | beats-100-of-users-with-c-step-by-step-e-duu1 | Abhiraj Pratap Singh\n\n---\n\n# if you like the solution please UPVOTE it...\n\n---\n\n# Intuition\n Describe your first thoughts on how to solve this problem. | abhirajpratapsingh | NORMAL | 2024-02-16T18:41:02.047144+00:00 | 2024-02-16T18:41:02.047173+00:00 | 3,627 | false | # Abhiraj Pratap Singh\n\n---\n\n# if you like the solution please UPVOTE it...\n\n---\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- The problem asks for the maximum product that can be obtained by multiplying three integers from the given array. To achieve this, we can observe ... | 28 | 0 | ['Array', 'Math', 'Sorting', 'C++'] | 3 |
maximum-product-of-three-numbers | Python O(n) time O(1) space | python-on-time-o1-space-by-leeteatsleep-0nx7 | \nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n smallestTwo = [float(\'inf\')]*2\n largestThree = [float(\'-inf\')]*3\n | leeteatsleep | NORMAL | 2020-07-13T20:21:05.997910+00:00 | 2020-07-13T20:21:05.997946+00:00 | 4,851 | false | ```\nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n smallestTwo = [float(\'inf\')]*2\n largestThree = [float(\'-inf\')]*3\n for num in nums:\n if num <= smallestTwo[0]:\n smallestTwo[0] = num\n smallestTwo.sort(reverse=True)\n ... | 25 | 1 | ['Python', 'Python3'] | 2 |
maximum-product-of-three-numbers | Java solution with sorting | java-solution-with-sorting-by-2020-pkkg | In a sorted array, the maximum product will be either the last three largest elements or the first two (negative elements) multiplied by the last (largest) elem | 2020 | NORMAL | 2020-05-27T00:52:42.079298+00:00 | 2020-05-27T00:52:42.079345+00:00 | 2,020 | false | In a sorted array, the maximum product will be either the last three largest elements or the first two (negative elements) multiplied by the last (largest) element.\n```\nclass Solution {\n public int maximumProduct(int[] nums) {\n Arrays.sort(nums);\n int len = nums.length;\n \n return M... | 23 | 0 | ['Java'] | 3 |
maximum-product-of-three-numbers | 2-liner Python solution with explanations | 2-liner-python-solution-with-explanation-nxgw | There are two possible ways to get the largest number:\n1. biggest number * 2nd biggest * 3rd biggest\n2. biggest number * smallest number * 2nd smallest number | yangshun | NORMAL | 2017-06-25T04:41:36.936000+00:00 | 2018-10-14T18:51:34.465828+00:00 | 2,667 | false | There are two possible ways to get the largest number:\n1. biggest number * 2nd biggest * 3rd biggest\n2. biggest number * smallest number * 2nd smallest number (if the two smallest numbers are negative)\n\nThis formula will also work in the case there are all negative numbers, in which the smallest negative number wil... | 22 | 0 | [] | 1 |
maximum-product-of-three-numbers | JavaScript O(n) time (1) space | javascript-on-time-1-space-by-shengdade-zrdr | javascript\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar maximumProduct = function(nums) {\n let max1 = -Infinity;\n let max2 = -Infinity;\n | shengdade | NORMAL | 2020-02-06T21:59:07.898885+00:00 | 2020-02-06T21:59:07.898923+00:00 | 2,547 | false | ```javascript\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar maximumProduct = function(nums) {\n let max1 = -Infinity;\n let max2 = -Infinity;\n let max3 = -Infinity;\n let min2 = Infinity;\n let min1 = Infinity;\n for (let i = 0; i < nums.length; i++) {\n if (nums[i] > max1) {\n [max1, ma... | 20 | 0 | ['JavaScript'] | 2 |
maximum-product-of-three-numbers | ✅ EASIEST || 💯 BEAST || 🧑💻 3 LINES CODE || ⭐ JAVA | easiest-beast-3-lines-code-java-by-mukun-93wb | \n\n# Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal is to find the maximum product of three numbers in the array. There are | mukund_rakholiya | NORMAL | 2024-09-27T11:28:49.418833+00:00 | 2024-09-27T11:28:49.418870+00:00 | 4,341 | false | ```\n```\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n**The goal is to find the maximum product of three numbers in the array. There are two potential scenarios:**\n1. The three largest numbers yield the maximum product.\n2. In cases where there are negative numbers, the product of... | 19 | 0 | ['Array', 'Math', 'Sorting', 'Java'] | 2 |
maximum-product-of-three-numbers | c++ solution | with proper explaination | O(n) solution | c-solution-with-proper-explaination-on-s-uo5e | The basic approach that we all will first click is that just find the largest three numbers from the given vector nums and calculate the product-> and this wor | kritika_12 | NORMAL | 2021-07-03T15:31:53.497150+00:00 | 2021-07-03T15:59:01.277368+00:00 | 1,536 | false | The **basic approach** that we all will first click is that just find the largest three numbers from the given vector nums and calculate the product-> and this works to some extent only if the given vector nums is having positive elements\nbut what if there are negative elements too this basic approach will fail hence... | 16 | 0 | ['C', 'C++'] | 1 |
maximum-product-of-three-numbers | 👏Beats 98.57% of users with Java || ✅Simple & Easy Well Explained Solution without Sorting 🔥💥 | beats-9857-of-users-with-java-simple-eas-qls0 | Intuition\nFind First 3 Maximum number and 2 minimum numbers and return maximum of 3 max numbers or 2 min number multiply max number.\n\n# I Think This Can Help | Rutvik_Jasani | NORMAL | 2024-04-04T07:48:55.134122+00:00 | 2024-04-04T07:48:55.134147+00:00 | 2,287 | false | # Intuition\nFind First 3 Maximum number and 2 minimum numbers and return maximum of 3 max numbers or 2 min number multiply max number.\n\n# I Think This Can Help You(For Proof Click on the Image)\n[ {\n Arrays.sort(nums);\n int n = nums.length;\n int ans = Integer.MIN_VALUE;\ | rj0711 | NORMAL | 2022-09-08T18:00:21.166314+00:00 | 2022-09-12T10:02:33.125057+00:00 | 3,769 | false | ```\nclass Solution {\n public int maximumProduct(int[] nums) {\n Arrays.sort(nums);\n int n = nums.length;\n int ans = Integer.MIN_VALUE;\n \n ans = Math.max(nums[0]*nums[1]*nums[n-1], nums[n-1]*nums[n-2]*nums[n-3]);\n \n return ans;\n }\n}\n``` | 14 | 0 | ['Math', 'Java'] | 2 |
maximum-product-of-three-numbers | Python O( n log n ) sol. based on sorting. run-time 80%+ | python-o-n-log-n-sol-based-on-sorting-ru-lrqr | Python O( n log n ) sol. based on sorting\n\n\nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n\n # sort the nums with ascending | brianchiang_tw | NORMAL | 2020-01-06T06:11:22.197875+00:00 | 2020-01-06T06:11:22.197921+00:00 | 1,536 | false | Python O( n log n ) sol. based on sorting\n\n```\nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n\n # sort the nums with ascending order \n nums.sort()\n \n # for possible solution_#1: First three largest positive numbers\n candidate_1 = nums[-1] * nums... | 13 | 0 | ['Sorting', 'Python'] | 1 |
maximum-product-of-three-numbers | Easy solution for js | easy-solution-for-js-by-ahmedrithas48-1gvj | Approach\n Describe your approach to solving the problem. \n- First we have to sort it out by descending.\n- We have two case to get max number with three numbe | ahmedrithas48 | NORMAL | 2024-04-05T12:38:39.582654+00:00 | 2024-04-05T12:38:39.582679+00:00 | 700 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n- First we have to sort it out by descending.\n- We have two case to get max number with three numbers.\n- we have two cases -\n### If array contains only postive/negative numbers \n- take first 3 numbers .\n- for example if we sort this [4,2,15,6] de... | 12 | 0 | ['JavaScript'] | 0 |
maximum-product-of-three-numbers | Simple Java || C++ Code ☠️ | simple-java-c-code-by-abhinandannaik1717-9wy6 | java []\nclass Solution {\n public int maximumProduct(int[] nums) {\n int n = nums.length;\n Arrays.sort(nums);\n int m1 = nums[n-1]*num | abhinandannaik1717 | NORMAL | 2024-06-25T16:57:33.077575+00:00 | 2024-06-25T16:57:33.077601+00:00 | 3,059 | false | ```java []\nclass Solution {\n public int maximumProduct(int[] nums) {\n int n = nums.length;\n Arrays.sort(nums);\n int m1 = nums[n-1]*nums[n-2]*nums[n-3];\n int m2 = nums[0]*nums[1]*nums[n-1];\n if(m1>m2){\n return m1;\n }\n else{\n return m2;\... | 10 | 0 | ['Java'] | 4 |
maximum-product-of-three-numbers | 2 Lines of Code-->Powerful Heap Concept | 2-lines-of-code-powerful-heap-concept-by-uk9p | 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 | GANJINAVEEN | NORMAL | 2023-03-18T05:36:17.220860+00:00 | 2023-03-18T05:36:17.220895+00:00 | 2,056 | 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. $$O(n)$$ --... | 10 | 0 | ['Python3'] | 0 |
maximum-product-of-three-numbers | [Java] 3-pointer Method {99.73%} faster than orther solutions | java-3-pointer-method-9973-faster-than-o-2r7o | If you do like solution please upvote it , so it could be helpful to orthers\nSorting Method :\n\nclass Solution {\n public int maximumProduct(int[] nums) {\ | salmanshaikssk007 | NORMAL | 2021-12-09T06:44:30.211296+00:00 | 2021-12-09T07:04:42.539924+00:00 | 945 | false | **If you do like solution please upvote it , so it could be helpful to orthers**\nSorting Method :\n```\nclass Solution {\n public int maximumProduct(int[] nums) {\n \n Arrays.sort(nums);\n int n = nums.length;\n return Math.max(nums[n-1]*nums[n-2]*nums[n-3] , nums[0]*nums[1]*nums[n-1]);\n... | 10 | 0 | ['Sorting', 'Java'] | 0 |
maximum-product-of-three-numbers | 🏆💢💯 Faster✅💯 Lesser🧠 🎯 C++✅Python3🐍✅Java✅C✅Python🐍✅💥🔥💫Explained☠💥🔥 Beats 💯 | faster-lesser-cpython3javacpythonexplain-69nu | Intuition\n\n\n\n\n\nC++ []\nclass Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n sort(nums.begin(), nums.end());\n int n = nu | Edwards310 | NORMAL | 2024-05-29T03:01:36.719656+00:00 | 2024-05-29T03:01:36.719673+00:00 | 3,337 | false | # Intuition\n\n\n {\n nums = nums.sort((a,b) => b-a)\n return Math.max(nums[ | jagdish7908 | NORMAL | 2021-10-24T08:25:10.839421+00:00 | 2021-10-24T08:26:54.493698+00:00 | 1,491 | false | ```\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar maximumProduct = function(nums) {\n nums = nums.sort((a,b) => b-a)\n return Math.max(nums[0]*nums[1]*nums[2], nums[nums.length-1]*nums[nums.length-2]*nums[0])\n};\n``` | 9 | 0 | ['Sorting', 'JavaScript'] | 2 |
maximum-product-of-three-numbers | C++ {Speed,Mem} = {O(n), O(1)} simple | c-speedmem-on-o1-simple-by-crab_10legs-c2c1 | \nclass Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n \n \n int min_c = INT_MAX;\n int min_n = INT_MAX;\n | crab_10legs | NORMAL | 2020-07-19T05:31:17.445005+00:00 | 2020-07-19T05:35:34.442758+00:00 | 1,529 | false | ```\nclass Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n \n \n int min_c = INT_MAX;\n int min_n = INT_MAX;\n int max_c = INT_MIN;\n int max_n = INT_MIN;\n int max_n_n = INT_MIN;\n \n for(int i=0; i < nums.size(); i++){\n // 2... | 9 | 1 | ['C', 'C++'] | 3 |
maximum-product-of-three-numbers | Best Solution in C++ for Trees 🚀 || 100% working | best-solution-in-c-for-trees-100-working-8ykh | IntuitionTo find the maximum product, we should consider the largest numbers. But what if the array has negative numbers? 🤔 Negative numbers can flip the result | BladeRunner150 | NORMAL | 2024-12-31T06:05:32.222778+00:00 | 2024-12-31T06:05:32.222778+00:00 | 1,554 | false | # Intuition
To find the maximum product, we should consider the largest numbers. But what if the array has negative numbers? 🤔 Negative numbers can flip the result! So, we also check the smallest two negatives multiplied with the largest positive number. 🚀
# Approach
1. Sort the array. 🧹
2. Compare:
- Th... | 8 | 0 | ['Array', 'Math', 'Sorting', 'C++'] | 0 |
maximum-product-of-three-numbers | [Python3] O(N) and O(NlogN) solutions | python3-on-and-onlogn-solutions-by-ye15-yoif | The maximum can be one of below \n1) multiplying the three largest number;\n2) multiplying the largest number with two negative numbers largest in magnitude. \n | ye15 | NORMAL | 2019-08-12T03:02:50.896669+00:00 | 2020-06-29T19:49:20.721427+00:00 | 1,134 | false | The maximum can be one of below \n1) multiplying the three largest number;\n2) multiplying the largest number with two negative numbers largest in magnitude. \n\nSo to loop through the list and keep track of the three largest and two smallest numbers is enough to compute the maximum. \n\n```\nclass Solution:\n def m... | 8 | 0 | ['Python3'] | 1 |
maximum-product-of-three-numbers | Python, Straightforward with Explanation | python-straightforward-with-explanation-g0i4c | Sort the array. Any "middle" numbers not in the first 3 or last 3 cannot be used in the final answer. If we are using a middle number, it must have both a lef | awice | NORMAL | 2017-06-25T03:06:12.043000+00:00 | 2017-06-25T03:06:12.043000+00:00 | 4,850 | false | Sort the array. Any "middle" numbers not in the first 3 or last 3 cannot be used in the final answer. If we are using a middle number, it must have both a left-neighbor and a right-neighbor, and switching to one of these neighbors will increase the product. \n\n```\ndef maximumProduct(self, A):\n A.sort()\n if ... | 8 | 0 | [] | 2 |
maximum-product-of-three-numbers | 🎯SUPER EASY FOR BEGINNERS 👏 | super-easy-for-beginners-by-vigneshvaran-32qf | Code | vigneshvaran0101 | NORMAL | 2025-02-03T17:30:37.394114+00:00 | 2025-02-03T17:30:37.394114+00:00 | 1,115 | false |
# Code
```python3 []
class Solution:
def maximumProduct(self, nums: List[int]) -> int:
a = sorted(nums)
return max(a[-1]*a[-2]*a[-3], a[0]*a[1]*a[-1])
# nums[-1]*nums[-2]*nums[-3] gives the product of 3 maximum number (positive)
# nums[0]*nums[1]*nums[-1] gives the produ... | 7 | 0 | ['Python3'] | 0 |
maximum-product-of-three-numbers | Python O(NlogN) Solution Beat 100% | python-onlogn-solution-beat-100-by-tliu7-64os | Sort the array first, then check the product of (3 largest positive numbers) and (2 largest negative numbers and largest positive numbers).\n\n\nclass Solution: | tliu77 | NORMAL | 2019-11-29T02:55:45.130846+00:00 | 2019-11-29T02:56:02.189072+00:00 | 1,307 | false | Sort the array first, then check the product of (3 largest positive numbers) and (2 largest negative numbers and largest positive numbers).\n\n```\nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n nums.sort()\n return max(nums[0]*nums[1]*nums[-1], nums[-1]*nums[-2]*nums[-3])\n``` | 7 | 0 | ['Python', 'Python3'] | 0 |
maximum-product-of-three-numbers | Java sort, 3 lines | java-sort-3-lines-by-cubicon-otm8 | \npublic int maximumProduct(int[] a) {\n Arrays.sort(a);\n int len = a.length;\n return Math.max(a[0] * a[1] * a[len-1], a[len-1] * a[len-2 | Cubicon | NORMAL | 2017-06-25T03:07:39.140000+00:00 | 2017-06-25T03:07:39.140000+00:00 | 1,499 | false | ```\npublic int maximumProduct(int[] a) {\n Arrays.sort(a);\n int len = a.length;\n return Math.max(a[0] * a[1] * a[len-1], a[len-1] * a[len-2] * a[len-3]);\n\n }\n``` | 7 | 2 | [] | 3 |
maximum-product-of-three-numbers | Easy To Understand | 3 Lines 🚀🚀 | easy-to-understand-3-lines-by-sanket_des-07uh | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Use this test case for better understanding --> [-100,-98,-1,2,3,4]\n\n# | The_Eternal_Soul | NORMAL | 2023-10-17T13:03:42.620490+00:00 | 2023-10-17T13:03:42.620509+00:00 | 2,952 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n Use this test case for better understanding --> [-100,-98,-1,2,3,4]\n\n# Complexity\n- Time complexity: O(NlogN)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space comple... | 6 | 0 | ['Array', 'Math', 'C', 'Sorting', 'C++', 'Java', 'Python3'] | 0 |
maximum-product-of-three-numbers | Maximum Product of Three Numbers Solution in C++ | maximum-product-of-three-numbers-solutio-vrju | 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 | The_Kunal_Singh | NORMAL | 2023-05-02T16:52:49.674136+00:00 | 2023-05-02T16:52:49.674179+00:00 | 1,450 | 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(nlogn)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O... | 6 | 0 | ['C++'] | 0 |
maximum-product-of-three-numbers | Python O(n) solution || NO fancy one liner | python-on-solution-no-fancy-one-liner-by-gasn | \n\n# Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space complexity here, e.g. O(n) \n | Augus7 | NORMAL | 2023-01-31T16:06:52.400746+00:00 | 2023-01-31T16:09:57.733218+00:00 | 2,325 | false | \n\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def maximumProduct(self, vec: List[int]) -> int:\n if len(vec) == 3:\n retur... | 6 | 0 | ['Array', 'Math', 'Python3'] | 3 |
maximum-product-of-three-numbers | Most Efficient and Optimized || Fully Explained || O(n) || Java😊 | most-efficient-and-optimized-fully-expla-hsz6 | \n# Approach :\n\n1. Scan the array and compute Maximum, second maximum and third maximum element present in the array.\n2. Scan the array and compute Minimum a | N7_BLACKHAT | NORMAL | 2022-12-05T12:46:43.214826+00:00 | 2022-12-05T12:46:43.214867+00:00 | 1,850 | false | \n# Approach :\n\n1. Scan the array and compute Maximum, second maximum and third maximum element present in the array.\n2. Scan the array and compute Minimum and second minimum element present in the array.\n3. Return the maximum of product of Maximum, second maximum and third maximum and product of Minimum, second mi... | 6 | 0 | ['Array', 'Math', 'Java'] | 2 |
maximum-product-of-three-numbers | c++ | Two approaches | Fast solution | Asked in interview | c-two-approaches-fast-solution-asked-in-8ie8y | \n\n\n////Sorting method\n\nclass Solution {\npublic:\n int maximumProduct(vector<int>& a) {\n sort(a.begin(),a.end());\n int n=a.size();\n | srv-er | NORMAL | 2021-09-28T01:27:07.099334+00:00 | 2021-09-28T01:27:07.099460+00:00 | 916 | false | \n\n```\n////Sorting method\n\nclass Solution {\npublic:\n int maximumProduct(vector<int>& a) {\n sort(a.begin(),a.end());\n int n=a.size();\n return max(a[n-1]*a[n-2]*a[n-3],max(a[0]*a[1]*a[2],a[0]*a[1]*a[n-1]));\n \n }\n};\n\n////linear solution \n\nclass Solution {\npublic:\n int... | 6 | 0 | ['C', 'Sorting'] | 0 |
maximum-product-of-three-numbers | Java || Easy To Understand || nlog(n) | java-easy-to-understand-nlogn-by-vishal_-2u90 | Code\n\nclass Solution {\n public int maximumProduct(int[] nums) {\n int n=nums.length;\n Arrays.sort(nums);\n return Math.max(nums[0]*n | vishal_niet | NORMAL | 2023-03-21T05:27:20.946538+00:00 | 2023-03-21T05:27:20.946585+00:00 | 1,741 | false | # Code\n```\nclass Solution {\n public int maximumProduct(int[] nums) {\n int n=nums.length;\n Arrays.sort(nums);\n return Math.max(nums[0]*nums[1]*nums[n-1],nums[n-1]*nums[n-2]*nums[n-3]);\n // return max_product;\n \n }\n}\n``` | 5 | 0 | ['Sorting', 'Java'] | 0 |
maximum-product-of-three-numbers | 2 Approach 🔥 || Easy C++ ✅ Solution || Brute Force [Sorting] and Optimized Approach ⚡ | 2-approach-easy-c-solution-brute-force-s-q0lm | Hint\n\nHint 1: Scan the array and compute Maximum, second maximum, third maximum element, and minimum and second minimum element present in the array.\n\nHint | Vishal_Singh7 | NORMAL | 2023-03-10T09:21:38.183202+00:00 | 2023-03-10T09:21:38.183254+00:00 | 721 | false | # Hint\n\n**Hint 1:** Scan the array and compute Maximum, second maximum, third maximum element, and minimum and second minimum element present in the array.\n\n**Hint 2:** The answer will always be maximum of product of Maximum, second maximum and third maximum and product of Minimum, second minimum and Maximum elemen... | 5 | 0 | ['Sorting', 'C++'] | 1 |
maximum-product-of-three-numbers | Easiest C++ Solution | easiest-c-solution-by-harsh_1818-kvfk | Intuition\nwe will sort the vector.\n\n# Approach\nAfter sorting answer will always be product of first 2numbers and last number or last 3numbers.\n# Complexity | harsh_1818 | NORMAL | 2023-02-02T09:26:42.901282+00:00 | 2023-02-02T09:26:42.901335+00:00 | 2,132 | false | # Intuition\nwe will sort the vector.\n\n# Approach\nAfter sorting answer will always be product of first 2numbers and last number or last 3numbers.\n# Complexity\n- Time complexity:\nnlogn\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n\n sor... | 5 | 0 | ['C++'] | 1 |
maximum-product-of-three-numbers | 4 Lines of Python code | 4-lines-of-python-code-by-prernaarora221-132a | ```\nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n nums.sort()\n l1 = nums[-1]nums[-2]nums[-3]\n l2 = nums[0]num | prernaarora221 | NORMAL | 2022-06-06T18:23:04.847104+00:00 | 2022-06-06T18:23:04.847176+00:00 | 2,165 | false | ```\nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n nums.sort()\n l1 = nums[-1]*nums[-2]*nums[-3]\n l2 = nums[0]*nums[1]*nums[-1]\n return max(l1,l2) | 5 | 0 | ['Python', 'Python3'] | 1 |
maximum-product-of-three-numbers | C++ Simple and clean solution for Beginners || Brute force | c-simple-and-clean-solution-for-beginner-87kd | Please Upvote if it helped you !!!\nHappy Coding :) \n \n \n\nint maximumProduct(vector<int>& nums) \n {\n sort(nums.begin(),nums.end());\n int | rahulrajguru | NORMAL | 2022-01-13T10:48:25.632376+00:00 | 2022-07-13T12:35:50.318519+00:00 | 459 | false | Please Upvote if it helped you !!!\nHappy Coding :) \n \n \n```\nint maximumProduct(vector<int>& nums) \n {\n sort(nums.begin(),nums.end());\n int n=nums.size();\n int max= nums[n-1]*nums[n-2]*nums[n-3];\n if(max > nums[0]*nums[1]*nums[n-1])\n return max;\n return nums[... | 5 | 0 | ['C'] | 1 |
maximum-product-of-three-numbers | C++ code with Explanation | c-code-with-explanation-by-vishal_shady2-2f87 | Explanation:\nFirst we sort the array. Since the largest 3 numbers will make the largest product, we can look for the last 3 digits and multiply them. if the nu | Vishal_Shady23 | NORMAL | 2021-07-07T07:01:53.015764+00:00 | 2021-07-07T07:01:53.015792+00:00 | 612 | false | **Explanation:**\n*First we sort the array. Since the largest 3 numbers will make the largest product, we can look for the last 3 digits and multiply them. if the number contains more than 2 negative numbers with maximum magnitide. Then their product will yield maximum value along with the largest positive number. Exam... | 5 | 0 | ['C'] | 0 |
maximum-product-of-three-numbers | Java | 2ms | 99.33% faster | java-2ms-9933-faster-by-aish9-lerd | \n\nclass Solution {\n public int maximumProduct(int[] nums) {\n if(nums.length==3)\n return nums[0]*nums[1]*nums[2];\n int max1st=I | aish9 | NORMAL | 2021-06-09T15:14:51.174975+00:00 | 2021-06-09T15:14:51.175025+00:00 | 660 | false | \n```\nclass Solution {\n public int maximumProduct(int[] nums) {\n if(nums.length==3)\n return nums[0]*nums[1]*nums[2];\n int max1st=Integer.MIN_VALUE,max2nd=Integer.MIN_VALUE,max3rd... | 5 | 0 | ['Java'] | 1 |
maximum-product-of-three-numbers | python 3 with only 2 lines（easy to understand） | python-3-with-only-2-lineseasy-to-unders-ozq0 | \nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n a = sorted(nums)\n return(max(a[-1] *a[-2]*a[-3],a[0]*a[1]*a[-1]))\n | zhuanhuan | NORMAL | 2019-08-14T02:40:19.062308+00:00 | 2019-08-14T02:40:19.065222+00:00 | 1,520 | false | ```\nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n a = sorted(nums)\n return(max(a[-1] *a[-2]*a[-3],a[0]*a[1]*a[-1]))\n``` | 5 | 0 | ['Python', 'Python3'] | 1 |
maximum-product-of-three-numbers | JavaScript solution beat 100% | javascript-solution-beat-100-by-zhenximi-una9 | \n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar maximumProduct = function(nums) {\n let max1 = max2 = max3 = Number.NEGATIVE_INFINITY;\n l | zhenximi | NORMAL | 2018-06-13T02:29:38.697194+00:00 | 2018-06-13T02:29:38.697194+00:00 | 747 | false | ```\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar maximumProduct = function(nums) {\n let max1 = max2 = max3 = Number.NEGATIVE_INFINITY;\n let min1 = min2 = Number.POSITIVE_INFINITY;\n for (var i = 0; i < nums.length; i++) {\n var n = nums[i];\n if(n >= max1) {\n max3 ... | 5 | 0 | [] | 1 |
maximum-product-of-three-numbers | Max Product Trio: Navigating Negatives for Maximum Gain Using C++ | max-product-trio-navigating-negatives-fo-lmkb | Intuition\nThe problem requires finding the maximum product of three numbers from the list. The key observation is that the maximum product could either come fr | Krishnaa2004 | NORMAL | 2024-09-07T13:33:39.189013+00:00 | 2024-09-07T13:33:39.189043+00:00 | 742 | false | # Intuition\nThe problem requires finding the maximum product of three numbers from the list. The key observation is that the maximum product could either come from the three largest numbers or from two smallest (negative) numbers and the largest number.\n\n# Approach\n1. Sort the array.\n2. Calculate the product of th... | 4 | 0 | ['C++'] | 0 |
maximum-product-of-three-numbers | Superb Logic in java and Python(Heap concept) | superb-logic-in-java-and-pythonheap-conc-t9ng | \n\n# Superb Solution in JAVA\n\nclass Solution {\n public int maximumProduct(int[] nums) {\n Arrays.sort(nums);\n int n=nums.length;\n return | GPraveen | NORMAL | 2023-05-08T03:52:36.531127+00:00 | 2023-05-08T03:52:36.531169+00:00 | 1,161 | false | \n\n# Superb Solution in JAVA\n```\nclass Solution {\n public int maximumProduct(int[] nums) {\n Arrays.sort(nums);\n int n=nums.length;\n return Math.max(nums[0]*nums[1]*nums[n-1],nums[n-1]*nums[n-2]*nums[n-3]); \n }\n}\n```\n# Excellent Solution in Python Using Heap Concept\n```\nclass Solution:... | 4 | 0 | ['Heap (Priority Queue)', 'Java', 'Python3'] | 0 |
maximum-product-of-three-numbers | Simple C++ Solution | simple-c-solution-by-divyanshu_singh_cs-zbf3 | Intuition\n Describe your first thoughts on how to solve this problem. \nFirst we sort the vector then we return the product of first 2 element and last element | Divyanshu_singh_cs | NORMAL | 2023-03-10T04:19:15.737575+00:00 | 2023-03-10T04:19:15.737607+00:00 | 3,187 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirst we sort the vector then we return the product of first 2 element and last element or product of last three element\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add ... | 4 | 0 | ['C++'] | 1 |
maximum-product-of-three-numbers | Java Easy using sorting | java-easy-using-sorting-by-shailendranir-uppq | 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 | shailendraniranjan771 | NORMAL | 2023-02-20T15:17:10.420247+00:00 | 2023-02-20T15:17:10.420295+00:00 | 398 | 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)$$ --... | 4 | 0 | ['Java'] | 1 |
maximum-product-of-three-numbers | Maximum Product of 3 Numbers - O(n) Time Complexity | maximum-product-of-3-numbers-on-time-com-mjff | Intuition\nTo reduce the time complexity\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n- There are two approaches Brute Force App | Vanshika78 | NORMAL | 2023-02-02T20:44:22.105674+00:00 | 2023-02-02T20:44:22.105715+00:00 | 1,536 | false | # Intuition\nTo reduce the time complexity\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n- There are two approaches Brute Force Approach and the other one is the refined one that is using sorting.\n- I have used sorting approach to reduce the time complexity. \n<!-- Describe your a... | 4 | 0 | ['Array', 'Dynamic Programming', 'Sorting', 'C++'] | 0 |
maximum-product-of-three-numbers | Java Simple Solution, using Sorting | java-simple-solution-using-sorting-by-sh-01qh | Intuition\n Describe your first thoughts on how to solve this problem. \nIntuition was to sort the array and get the product\n\n# Approach\n Describe your appro | ShivaniSinghh | NORMAL | 2022-12-01T23:04:28.302165+00:00 | 2022-12-01T23:04:28.302201+00:00 | 977 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIntuition was to sort the array and get the product\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nI used sorting because we know that after sorting largest 3 elements will have maximum product or 2 leftmost eleme... | 4 | 0 | ['Java'] | 0 |
maximum-product-of-three-numbers | easiest c++ soution | easiest-c-soution-by-anish25-d08e | class Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n sort(nums.rbegin(),nums.rend());\n int l=nums.size();\n if(nums[0]*n | Anish25 | NORMAL | 2022-11-21T07:52:10.187334+00:00 | 2022-11-21T07:52:10.187363+00:00 | 720 | false | ```class Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n sort(nums.rbegin(),nums.rend());\n int l=nums.size();\n if(nums[0]*nums[1]*nums[2]>nums[0]*nums[l-1]*nums[l-2]){\n return nums[0]*nums[1]*nums[2];\n }\n else{\n return nums[0]*nums[l-1]*nums[l-... | 4 | 0 | [] | 1 |
maximum-product-of-three-numbers | python||233ms||simplest approach | python233mssimplest-approach-by-2stellon-4ocz | \t\tclass Solution:\n\t\t\tdef maximumProduct(self, a: List[int]) -> int:\n\t\t\t\ta.sort(reverse=True)\n\t\t\t\tv1=a[0]a[1]a[2]\n\t\t\t\tv2=a[0]a[-1]a[-2]\n\t\ | 2stellon7 | NORMAL | 2022-10-01T05:14:22.348948+00:00 | 2022-10-01T05:14:22.348982+00:00 | 1,316 | false | \t\tclass Solution:\n\t\t\tdef maximumProduct(self, a: List[int]) -> int:\n\t\t\t\ta.sort(reverse=True)\n\t\t\t\tv1=a[0]*a[1]*a[2]\n\t\t\t\tv2=a[0]*a[-1]*a[-2]\n\t\t\t\treturn max(v1,v2)\n | 4 | 0 | ['Python'] | 2 |
maximum-product-of-three-numbers | 100% easy || Maximum Product of Three Numbers || C++ Solution | 100-easy-maximum-product-of-three-number-13x3 | \nclass Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n sort(nums.begin(),nums.end());\n int a=nums[nums.size()-1]*nums[nums.siz | StArK19 | NORMAL | 2022-08-25T21:16:52.734466+00:00 | 2022-08-25T21:16:52.734514+00:00 | 958 | false | ```\nclass Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n sort(nums.begin(),nums.end());\n int a=nums[nums.size()-1]*nums[nums.size()-2]*nums[nums.size()-3];\n int b=nums[0]*nums[1]*nums[nums.size()-1];\n return max(a,b);\n }\n \n};\n``` | 4 | 0 | ['C', 'C++'] | 0 |
maximum-product-of-three-numbers | C++ simplest, O(N), easy understand, no sorting | c-simplest-on-easy-understand-no-sorting-8hz4 | \n//We should keep 3 biggest and 2 smallest elements of the array\n//max product is either product of 3 bigger or 2 smallest(negatives) and the biggest\nclass S | Moldoteck | NORMAL | 2022-05-12T08:46:14.916492+00:00 | 2022-05-12T08:46:14.916518+00:00 | 268 | false | ```\n//We should keep 3 biggest and 2 smallest elements of the array\n//max product is either product of 3 bigger or 2 smallest(negatives) and the biggest\nclass Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n int max1=INT_MIN, max2=INT_MIN, max3=INT_MIN;\n int min1=INT_MAX, min2=INT_MA... | 4 | 0 | [] | 0 |
maximum-product-of-three-numbers | C++ || 100% faster O(n) without sorting || 2 solutions ✔ | c-100-faster-on-without-sorting-2-soluti-8bbi | Logic :\nMax product of three no. come from\n\t\t1. product of 3 maximum positive no.\n\t\t2. product of 2 minimum negative no. and 1 maximum positive no.\n\nSo | AJAY_MAKVANA | NORMAL | 2021-07-30T09:30:30.233970+00:00 | 2021-08-12T15:29:42.824781+00:00 | 243 | false | ### **Logic :**\n**Max product of three no. come from**\n\t\t**1. product of `3 maximum positive no.`**\n\t\t**2. product of `2 minimum negative no.` and `1 maximum positive no.`**\n\n*So bellow **two** solution is based on this logic :)*\n\n### **1. Sort based solution**\n**TC = O(nlogn)\nSC = O(1)**\n```\nclass Solut... | 4 | 0 | [] | 1 |
maximum-product-of-three-numbers | Javascript Solution With Explanation | javascript-solution-with-explanation-by-pi24l | Explanation :-\nExample : - [-4 ,-3, -2, -1, 60]\nWe first sort the array in increasing order\nIf the above example had all positive integers, we would have sim | gaurang9 | NORMAL | 2020-09-25T17:54:37.471172+00:00 | 2020-09-25T17:54:37.471218+00:00 | 451 | false | Explanation :-\nExample : - [-4 ,-3, -2, -1, 60]\nWe first sort the array in increasing order\nIf the above example had all positive integers, we would have simply done the product of 3 largest numbers.\nBut since it also has negative numbers , and we know negative multiplied by negative is positive , we can consider t... | 4 | 0 | ['JavaScript'] | 2 |
maximum-product-of-three-numbers | Easy JS Solution | easy-js-solution-by-hbjorbj-d8hq | \n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar maximumProduct = function(nums) {\n let sorted = nums.sort((a,b) => a-b), len = nums.length;\ | hbjorbj | NORMAL | 2020-06-22T09:46:40.807051+00:00 | 2020-06-22T09:46:40.807084+00:00 | 555 | false | ```\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar maximumProduct = function(nums) {\n let sorted = nums.sort((a,b) => a-b), len = nums.length;\n let res1 = sorted[0]*sorted[1]*sorted[len-1],\n res2 = sorted[len-1]*sorted[len-2]*sorted[len-3]\n return Math.max(res1,res2);\n}; \n``` | 4 | 0 | ['JavaScript'] | 0 |
maximum-product-of-three-numbers | C# solutions w/ explaination | c-solutions-w-explaination-by-newbiecode-sno3 | There are 3 cases of the input array. And the conclusion is the maximun product is generated by either the 3 largest number or the largest number and 2 smallest | newbiecoder1 | NORMAL | 2020-03-27T20:10:42.685673+00:00 | 2020-04-23T04:17:12.596244+00:00 | 254 | false | There are 3 cases of the input array. And the conclusion is **the maximun product is generated by either the 3 largest number or the largest number and 2 smallest numbers.**\n1. All numbers are negative.\n\tmax product is generated by **the 3 largest negative numbers**.\n\te.g. [-4,-3,-2,-1]. max product = -3 * -2 * -... | 4 | 0 | [] | 0 |
maximum-product-of-three-numbers | Javascript - O(n) time & O(1) space | javascript-on-time-o1-space-by-achandel-cz9d | \nvar maximumProduct = function(nums) {\n // get top 3 max and top 2 min in one pass\n let max1 = -Infinity, \n max2 = -Infinity,\n max3 = - | achandel | NORMAL | 2019-11-28T14:00:30.094690+00:00 | 2019-11-28T14:00:30.094741+00:00 | 806 | false | ```\nvar maximumProduct = function(nums) {\n // get top 3 max and top 2 min in one pass\n let max1 = -Infinity, \n max2 = -Infinity,\n max3 = -Infinity,\n min1 = Infinity,\n min2 = Infinity;\n \n for(let num of nums){\n // fetch max\n if(num > max1){\n ... | 4 | 0 | ['JavaScript'] | 0 |
maximum-product-of-three-numbers | Share my Python solution | share-my-python-solution-by-wangruinju-68tv | Sort the array, the maximum product happens in two situations:\nthe last three or the first two with the last one.\nHere is the Python code!\n\nclass Solution(o | wangruinju | NORMAL | 2017-06-25T18:49:38.269000+00:00 | 2017-06-25T18:49:38.269000+00:00 | 1,259 | false | Sort the array, the maximum product happens in two situations:\nthe last three or the first two with the last one.\nHere is the Python code!\n```\nclass Solution(object):\n def maximumProduct(self, nums):\n """\n :type nums: List[int]\n :rtype: int\n """\n nums.sort()\n retu... | 4 | 0 | [] | 2 |
maximum-product-of-three-numbers | Very simple Javascript and typescript solution. Do check it out👍 | very-simple-javascript-and-typescript-so-qjqf | Intuition\nThe maximum product can either come from the three largest numbers or from two smallest negative numbers and the largest positive number.\n\n# Approa | meathirarosejohn | NORMAL | 2024-10-20T05:40:06.575578+00:00 | 2024-10-20T05:41:36.507311+00:00 | 281 | false | # Intuition\nThe maximum product can either come from the three largest numbers or from two smallest negative numbers and the largest positive number.\n\n# Approach\nTrack the three largest and two smallest numbers while iterating through the array. The result is the maximum product of either the three largest numbers ... | 3 | 0 | ['Array', 'Math', 'TypeScript', 'JavaScript'] | 1 |
maximum-product-of-three-numbers | 🔍Very Easy, Simple and Short Solution🚀 | very-easy-simple-and-short-solution-by-s-4l1k | Intuition\nThe key intuition is that the maximum product can arise from:\n\nThree largest positive numbers.\nTwo smallest negative numbers and the largest posit | shoaibkhalid65 | NORMAL | 2024-09-10T00:03:15.839947+00:00 | 2024-09-10T00:03:15.839975+00:00 | 731 | false | # Intuition\nThe key intuition is that the maximum product can arise from:\n\nThree largest positive numbers.\nTwo smallest negative numbers and the largest positive number.\n\n# Approach\nSort the array: Sort the array in ascending order.\n\nCalculate products:\n\nCalculate the product of the three largest elements.\n... | 3 | 0 | ['Java'] | 2 |
maximum-product-of-three-numbers | Python Easy Solution || Sorting | python-easy-solution-sorting-by-sumedh07-ojge | Code\n\nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n nums.sort()\n n=len(nums)\n return max(nums[n-1]*nums[n-2] | Sumedh0706 | NORMAL | 2023-07-17T05:45:05.030156+00:00 | 2023-07-17T05:45:05.030182+00:00 | 858 | false | # Code\n```\nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n nums.sort()\n n=len(nums)\n return max(nums[n-1]*nums[n-2]*nums[n-3],nums[0]*nums[1]*nums[n-1])\n \n``` | 3 | 0 | ['Sorting', 'Python3'] | 0 |
maximum-product-of-three-numbers | Super Logical Solution Using Java | super-logical-solution-using-java-by-gpr-333c | \n\n# Superb Solution using JAVA\n\nclass Solution {\n public int maximumProduct(int[] nums) {\n Arrays.sort(nums);\n int n=nums.length;\n ret | GPraveen | NORMAL | 2023-05-08T03:07:48.487576+00:00 | 2023-05-08T03:07:48.487621+00:00 | 1,834 | false | \n\n# Superb Solution using JAVA\n```\nclass Solution {\n public int maximumProduct(int[] nums) {\n Arrays.sort(nums);\n int n=nums.length;\n return Math.max(nums[0]*nums[1]*nums[n-1],nums[n-1]*nums[n-2]*nums[n-3]); \n }\n}\n``` | 3 | 0 | ['Java'] | 3 |
maximum-product-of-three-numbers | C++ Solution || Easy || Brute Force ✔✔ | c-solution-easy-brute-force-by-akanksha9-fi4f | Complexity\n- Time complexity: O(nlogn)\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 | akanksha984 | NORMAL | 2023-05-06T11:27:53.014835+00:00 | 2023-05-13T05:21:43.459184+00:00 | 2,274 | false | # Complexity\n- Time complexity: $$O(nlogn)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n int n= nums.size();\n sort(n... | 3 | 0 | ['Array', 'Math', 'Sorting', 'C++'] | 2 |
maximum-product-of-three-numbers | Solution | solution-by-deleted_user-jqed | C++ []\nclass Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n int max1 = INT32_MIN, max2 = INT32_MIN, max3 = INT32_MIN;\n int m | deleted_user | NORMAL | 2023-04-14T13:16:33.554001+00:00 | 2023-04-14T14:34:13.241539+00:00 | 4,005 | false | ```C++ []\nclass Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n int max1 = INT32_MIN, max2 = INT32_MIN, max3 = INT32_MIN;\n int min1 = INT32_MAX, min2 = INT32_MAX;\n for(int& num : nums) {\n if(num > max1) {\n max3 = max2, max2 = max1, max1 = num;\n ... | 3 | 0 | ['C++', 'Java', 'Python3'] | 0 |
maximum-product-of-three-numbers | Simple solution in python using sort | simple-solution-in-python-using-sort-by-bgfzt | \n\n# Code\n\nclass Solution:\n def maximumProduct(self, A: List[int]) -> int:\n A.sort()\n return max(A[0]*A[1]*A[-1], A[-1]*A[-2]*A[-3])\n | tryingall | NORMAL | 2023-03-30T07:23:34.055965+00:00 | 2023-03-30T07:23:34.055997+00:00 | 1,082 | false | \n\n# Code\n```\nclass Solution:\n def maximumProduct(self, A: List[int]) -> int:\n A.sort()\n return max(A[0]*A[1]*A[-1], A[-1]*A[-2]*A[-3])\n``` | 3 | 0 | ['Python3'] | 1 |
maximum-product-of-three-numbers | Brute Solution | brute-solution-by-anurag_rathore-rfpp | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Anurag_Rathore | NORMAL | 2023-03-01T11:52:47.228429+00:00 | 2023-03-01T11:52:47.228470+00:00 | 1,287 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | 0 | ['Java'] | 0 |
maximum-product-of-three-numbers | python 2 liner code faster | python-2-liner-code-faster-by-raghunath_-c3yc | \n def maximumProduct(self, nums: List[int]) -> int:\n nums.sort()\n return max((nums[-1]*nums[-2]*nums[-3]),(nums[0]*nums[1]*nums[-1]))\n | Raghunath_Reddy | NORMAL | 2022-09-27T11:39:26.624394+00:00 | 2022-09-27T11:39:26.624436+00:00 | 1,421 | false | ```\n def maximumProduct(self, nums: List[int]) -> int:\n nums.sort()\n return max((nums[-1]*nums[-2]*nums[-3]),(nums[0]*nums[1]*nums[-1]))\n``` | 3 | 0 | ['Python', 'Python3'] | 1 |
maximum-product-of-three-numbers | JS simple solution with sorting | js-simple-solution-with-sorting-by-kunkk-lexv | \nvar maximumProduct = function(nums) {\n nums = nums.sort((a,b) => a - b);\n \n return Math.max(nums[0] * nums[1] * nums[nums.length - 1], nums[nums.l | kunkka1996 | NORMAL | 2022-08-26T10:05:57.751759+00:00 | 2022-08-26T10:05:57.751794+00:00 | 624 | false | ```\nvar maximumProduct = function(nums) {\n nums = nums.sort((a,b) => a - b);\n \n return Math.max(nums[0] * nums[1] * nums[nums.length - 1], nums[nums.length - 1] * nums[nums.length - 2] * nums[nums.length - 3]);\n};\n``` | 3 | 0 | ['Sorting', 'JavaScript'] | 1 |
maximum-product-of-three-numbers | Python - Easy to understand on the basis of testcases!! | python-easy-to-understand-on-the-basis-o-w2u1 | \ndef maximumProduct(self, nums):\n n = len(nums)\n nums.sort()\n return max(nums[0] * nums[1] * nums[n - 1], nums[n-3] * nums[n - 2] * num | kartikkulshreshtha | NORMAL | 2022-08-23T06:33:49.712253+00:00 | 2022-08-23T06:33:49.712299+00:00 | 1,084 | false | ```\ndef maximumProduct(self, nums):\n n = len(nums)\n nums.sort()\n return max(nums[0] * nums[1] * nums[n - 1], nums[n-3] * nums[n - 2] * nums[n-1])\n``` | 3 | 0 | ['Python'] | 0 |
maximum-product-of-three-numbers | my python solution with clear explanation | my-python-solution-with-clear-explanatio-nznm | \nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n nums.sort()\n max1 = nums[-1]*nums[-2]*nums[-3]\n max2 = nums[0] | CodeHunter_Sankar | NORMAL | 2022-07-08T12:36:06.992255+00:00 | 2022-07-08T12:36:06.992302+00:00 | 278 | false | ```\nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n nums.sort()\n max1 = nums[-1]*nums[-2]*nums[-3]\n max2 = nums[0]*nums[1]*nums[-1]\n if max1 >= max2:\n return max1\n else:\n return max2\n""""\nexplanation: example \n list1=[1,2,3,4... | 3 | 0 | ['Python'] | 0 |
maximum-product-of-three-numbers | Maximum product of three Numbers || simple || efficient || sorting | maximum-product-of-three-numbers-simple-cfm1n | \nclass Solution {\npublic:\n \n static int cmp(int a,int b) // This cmp function to sort in descending order\n { // To wri | babasaheb256 | NORMAL | 2022-06-10T03:52:28.894480+00:00 | 2022-06-10T03:54:41.336600+00:00 | 891 | false | ```\nclass Solution {\npublic:\n \n static int cmp(int a,int b) // This cmp function to sort in descending order\n { // To write this type of function concept is simple\n \n\t\treturn a>b; // Try to think what you want like you basically \n ... | 3 | 0 | ['C', 'Sorting', 'C++'] | 2 |
maximum-product-of-three-numbers | Two lines of Code || Java || C++ | two-lines-of-code-java-c-by-chiragvohra-p8kn | class Solution {\n public int maximumProduct(int[] nums) {\n\t\n\t//As we know there is only two possibilties \n\t//1st possibilty: All numbers are positive | chiragvohra | NORMAL | 2022-05-13T21:39:35.001248+00:00 | 2022-05-13T21:39:35.001274+00:00 | 183 | false | class Solution {\n public int maximumProduct(int[] nums) {\n\t\n\t//As we know there is only two possibilties \n\t//1st possibilty: All numbers are positive in that case we have return product of last three number of sorted array.\n \t// 2nd possibilty : There may be negative numbers also in that case we have to ret... | 3 | 0 | ['Array', 'Sorting'] | 0 |
maximum-product-of-three-numbers | c++ || max product of 3 no.|| Basic || n* logn | c-max-product-of-3-no-basic-n-logn-by-sh-uw08 | Time complexity= n*log(n)\nn=size of given vector\n\nclass Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n int ans=INT_MIN,n=nums.size | sheetaljoshi | NORMAL | 2021-12-17T11:09:44.143630+00:00 | 2021-12-17T11:16:43.525334+00:00 | 352 | false | Time complexity= n*log(n)\nn=size of given vector\n```\nclass Solution {\npublic:\n int maximumProduct(vector<int>& nums) {\n int ans=INT_MIN,n=nums.size();\n sort(nums.begin(),nums.end());\n \n int x=nums[n-3];\n int y=nums[n-2];\n int z=nums[n-1];\n int a=nums[0];\... | 3 | 1 | ['C', 'C++'] | 1 |
maximum-product-of-three-numbers | python3 | python3-by-saurabht462-971z | ```\nimport math\nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n if len(nums)==3:return math.prod(nums)\n nums.sort(reve | saurabht462 | NORMAL | 2021-04-16T14:17:45.878546+00:00 | 2021-04-16T14:17:45.878583+00:00 | 159 | false | ```\nimport math\nclass Solution:\n def maximumProduct(self, nums: List[int]) -> int:\n if len(nums)==3:return math.prod(nums)\n nums.sort(reverse=True)\n print(nums)\n return max(math.prod(nums[:3]),math.prod([nums[-1],nums[-2],nums[0]])) | 3 | 0 | [] | 0 |
maximum-product-of-three-numbers | python || very simple solution | python-very-simple-solution-by-risky_cod-5qm0 | \n# Written by : Dhruv Vavliya\nnums = [1,2,3,4]\n\ndef max_product(nums):\n nums.sort(reverse=True)\n a = nums[0]*nums[1]*nums[2]\n b = nums[0]*nums[- | risky_coder | NORMAL | 2020-12-29T11:36:07.300888+00:00 | 2020-12-29T11:36:07.300930+00:00 | 727 | false | ```\n# Written by : Dhruv Vavliya\nnums = [1,2,3,4]\n\ndef max_product(nums):\n nums.sort(reverse=True)\n a = nums[0]*nums[1]*nums[2]\n b = nums[0]*nums[-1]*nums[-2]\n return max(a,b)\n\nprint(max_product(nums))\n``` | 3 | 0 | ['Math', 'Python', 'Python3'] | 1 |
maximum-product-of-three-numbers | 628-Maximum Product of Three Numbers-Py All-in-One By Talse | 628-maximum-product-of-three-numbers-py-a00z2 | Get it Done, Make it Better, Share the Best -- Talse\nI). Naive Sort\n| O(T): O(nlgn) | O(S): O(1) | Rt: 284ms | \npython\n def maximumProduct(self, nums: Li | talse | NORMAL | 2019-12-03T18:59:47.272360+00:00 | 2019-12-03T19:00:11.348926+00:00 | 177 | false | **Get it Done, Make it Better, Share the Best -- Talse**\n**I). Naive Sort**\n| O(T): O(nlgn) | O(S): O(1) | Rt: 284ms | \n```python\n def maximumProduct(self, nums: List[int]) -> int:\n nums.sort()\n return max(nums[-1] * nums[-2] * nums[-3], nums[0] * nums[1] * nums[-1])\n```\n\n\n**II). Limit State*... | 3 | 0 | [] | 0 |
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