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
classes | content stringlengths 4 576k | upvotes int64 0 11.5k | downvotes int64 0 358 | tags stringlengths 2 193 | comments int64 0 2.56k |
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maximum-total-damage-with-spell-casting | Neat comprehensible efficient solution | neat-comprehensible-efficient-solution-b-d164 | IntuitionApproachComplexity
Time complexity: O(n log n)
Space complexity: O(n)
Code | shmirrakhimov | NORMAL | 2025-03-23T17:01:58.084387+00:00 | 2025-03-23T17:01:58.084387+00:00 | 2 | 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 log n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(n)
<!-- Add your space complexity here, e.g. $$O(n)... | 0 | 0 | ['JavaScript'] | 0 |
maximum-total-damage-with-spell-casting | O(Nlog(N)) sort and traverse | onlogn-sort-and-traverse-by-awuxiaoqi-4iz2 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | labao | NORMAL | 2025-03-06T16:40:52.587752+00:00 | 2025-03-06T16:40:52.587752+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Java'] | 0 |
maximum-total-damage-with-spell-casting | Maximum Total Damage with Spell Casting | maximum-total-damage-with-spell-casting-s82js | Code | _jyoti_geek | NORMAL | 2025-02-17T06:35:33.402610+00:00 | 2025-02-17T06:35:33.402610+00:00 | 8 | false | # Code
```java []
class Solution {
public long maximumTotalDamage(int[] arr) {
Arrays.sort(arr);
HashMap<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < arr.length; i++) {
map.put(arr[i], map.getOrDefault(arr[i], 0) + 1);
}
long[] dp = new long[arr.le... | 0 | 0 | ['Array', 'Hash Table', 'Two Pointers', 'Dynamic Programming', 'Sorting', 'Java'] | 0 |
maximum-total-damage-with-spell-casting | Easy to understand DP java | easy-to-understand-dp-java-by-zpw1337-spyo | IntuitionSort the input list into key value pairs, where key is the spell power, and value is its occurences.let dp[k] denote most power casting only from first | zpw1337 | NORMAL | 2025-02-13T00:45:39.706654+00:00 | 2025-02-13T00:45:39.706654+00:00 | 5 | false | # Intuition
Sort the input list into key value pairs, where key is the spell power, and value is its occurences.
let dp[k] denote most power casting only from first k spells with unique power.
Then we build dp from 0 to n-1.
At each step, we can choose current spell or not.
If we don't, we try to find the best answer... | 0 | 0 | ['Java'] | 0 |
maximum-total-damage-with-spell-casting | 79 ms Beats 93.56% DP | 79-ms-beats-9356-dp-by-dinhson-nguyen-9cv4 | IntuitionApproachComplexity
Time complexity:
O(n)
Space complexity:
Code | sondinh1701_ | NORMAL | 2025-02-10T08:44:54.481911+00:00 | 2025-02-10T08:44:54.481911+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
$$O(n)$$
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
... | 0 | 0 | ['C++'] | 0 |
maximum-total-damage-with-spell-casting | Sorting + 2 Pointers | sorting-2-pointers-by-yelleti-18-08av | IntuitionApproachComplexity
Time complexity:O(n∗logn)
Space complexity:O(n)
Code | yelleti-18 | NORMAL | 2025-01-29T13:24:28.270357+00:00 | 2025-01-29T13:24:28.270357+00:00 | 5 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:$$O(n*logn)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:$$O(n)$$
<!-- Add your space complexity here, e.g.... | 0 | 0 | ['C++'] | 0 |
maximum-total-damage-with-spell-casting | scala greedy oneliner | scala-greedy-oneliner-by-vititov-p0fc | null | vititov | NORMAL | 2025-01-28T00:59:40.532534+00:00 | 2025-01-28T00:59:40.532534+00:00 | 5 | false | ```scala []
object Solution {
import scala.util.chaining._
def maximumTotalDamage(power: Array[Int]): Long =
power.groupMapReduce(_.toLong)(_ => 1L)(_ + _).to(List).sorted.iterator
.foldLeft(Map(-3L -> 0L)){ case (aMap,(p,cnt)) =>
aMap.iterator.filter{case (k,_) => k+2<p}.maxBy(_._2)
.pipe... | 0 | 0 | ['Hash Table', 'Greedy', 'Sorting', 'Scala'] | 0 |
maximum-total-damage-with-spell-casting | Simple Solution O(NlogN) | simple-solution-onlogn-by-kush-sahu-ud3a | IntuitionApproachComplexity
Time complexity:O(NlogN)
Space complexity:O(N)
Code | kush-sahu | NORMAL | 2025-01-19T21:31:21.975046+00:00 | 2025-01-19T21:31:21.975046+00:00 | 5 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:O(NlogN)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:O(N)
<!-- Add your space complexity here, e.g. $$O(n)$$ -... | 0 | 0 | ['Java'] | 0 |
maximum-total-damage-with-spell-casting | Best Solution Ever || Think Bottom-Up || Easiest Explanation || Easiest Implementation || | best-solution-ever-think-bottom-up-easie-82c6 | IntuitionCode | krishna_6431 | NORMAL | 2025-01-18T18:55:06.663780+00:00 | 2025-01-18T18:55:06.663780+00:00 | 3 | false | # Intuition
https://youtu.be/z14TI29ATQ0
# Code
```cpp []
using ll = long long;
class Solution {
public:
long long maximumTotalDamage(vector<int>& nums) {
map<ll,ll>dp;
map<ll,ll>mp;
for(auto x : nums){
mp[x]++;
}
// for(auto y : mp){
// cout << y.fir... | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
maximum-total-damage-with-spell-casting | Dynamic Programming and Binary Search | dynamic-programming-and-binary-search-by-qz1q | IntuitionApproachBinary Search and Dynamic ProgrammingComplexity
Time complexity: O(nlogn)
Space complexity: O(n)
Code | Akash-96 | NORMAL | 2024-12-30T07:05:24.857715+00:00 | 2024-12-30T07:05:24.857715+00:00 | 5 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
Binary Search and Dynamic Programming
# Complexity
- Time complexity: O(nlogn)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(n)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```... | 0 | 0 | ['C++'] | 0 |
maximum-total-damage-with-spell-casting | Easy C++ Solution | Must SEE | Hash Table | easy-c-solution-must-see-hash-table-by-n-o3wb | Code | NehaGupta_09 | NORMAL | 2024-12-15T10:56:10.707876+00:00 | 2024-12-15T10:56:10.707876+00:00 | 3 | false | \n\n# Code\n```cpp []\nclass Solution {\npublic:\n unordered_map<int,vector<int>>mp;\n vector<long long>dp;\n long long maximumTotalDamage(vector<int>& power) {\n dp.resize(power.size(),-1);\n sort(power.begin(),power.end());\n int i=0;\n while(i<power.size()){\n int ele=... | 0 | 0 | ['Array', 'Hash Table', 'Dynamic Programming', 'Sorting', 'C++'] | 0 |
maximum-total-damage-with-spell-casting | Use priority_queue(max heap) after sorting | use-priority_queuemax-heap-after-sorting-kq5q | Intuition\n Describe your first thoughts on how to solve this problem. \nFirst sort the array then use a max heap [sum, -biggest number];\nIn each iteration fir | zsnwd | NORMAL | 2024-12-06T06:58:26.817461+00:00 | 2024-12-06T06:58:26.817503+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirst sort the array then use a max heap [sum, -biggest number];\nIn each iteration first pop the potentially invalid nodes;\nThen get the max sum for the current number(deduplicatoin included).\n\n# Complexity\n- Time complexity:\n<!-- A... | 0 | 0 | ['C++'] | 0 |
maximum-total-damage-with-spell-casting | Two Intuitive DP Approaches, beats 90% | two-intuitive-dp-approaches-beats-90-by-e221n | Approach 1: Overkill DP -> Memory Limit Exceeded\nPerform dynamic programming over all possible power caps p_cap in the range $[0, \max(\text{powers})]$; this i | PeterFavero | NORMAL | 2024-12-03T01:27:09.694157+00:00 | 2024-12-03T01:27:09.694179+00:00 | 11 | false | # Approach 1: Overkill DP -> Memory Limit Exceeded\nPerform dynamic programming over all possible power caps `p_cap` in the range $[0, \\max(\\text{powers})]$; this is innefficient and exceeds memory limitations because the only power caps that matter are the unique spell powers in the `powers`.\n\nTime Complexity: $O(... | 0 | 0 | ['Python3'] | 0 |
maximum-total-damage-with-spell-casting | 1D DP | 1d-dp-by-mohith_jain-76yc | 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 | mohith_jain | NORMAL | 2024-11-21T13:16:18.420711+00:00 | 2024-11-21T13:16:18.420745+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:\nO(nlogn)\n\n- Space complexity:\nO(n)\n\n# Code\n```java []\nclass Solution {\n public long maximumTotalDamage(int[] power) {\n ... | 0 | 0 | ['Java'] | 0 |
maximum-total-damage-with-spell-casting | Sort and DP | sort-and-dp-by-lambdacode-dev-dcyn | Intuition\n- Sort and then consider each spell with increasing power. \n- Of all choice history (with combinatorial count) made so far, the effective states are | lambdacode-dev | NORMAL | 2024-11-11T16:19:34.856407+00:00 | 2024-11-11T16:19:34.856468+00:00 | 2 | false | # Intuition\n- Sort and then consider each spell with increasing power. \n- Of all choice history (with combinatorial count) made so far, the effective states are polynomial, because only the largest power spell matter for future choices.\n- Only maixmum 3 such states are needed due to the specific constraints.\n\n\n# ... | 0 | 0 | ['C++'] | 0 |
maximum-total-damage-with-spell-casting | 100% easy code | 100-easy-code-by-alokpratapsing_cs22-5he6 | 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 | alokpratapsing_cs22 | NORMAL | 2024-11-08T18:02:30.507821+00:00 | 2024-11-08T18:02:30.507858+00:00 | 1 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
maximum-total-damage-with-spell-casting | C++ sort process then iterate with deque | c-sort-process-then-iterate-with-deque-b-061z | cpp []\nclass Solution {\npublic:\n long long maximumTotalDamage(vector<int>& power) {\n vector<pair<long long, long long>> p;\n sort(power.beg | user5976fh | NORMAL | 2024-11-05T07:29:52.557639+00:00 | 2024-11-05T07:29:52.557666+00:00 | 4 | false | ```cpp []\nclass Solution {\npublic:\n long long maximumTotalDamage(vector<int>& power) {\n vector<pair<long long, long long>> p;\n sort(power.begin(), power.end());\n long long n = power[0], f = 0;\n for (auto& num : power) {\n if (num == n) ++f;\n else {\n ... | 0 | 0 | ['C++'] | 0 |
maximum-total-damage-with-spell-casting | DP | Similar to House Robber | dp-similar-to-house-robber-by-yimang-z23p | Intuition\nSimilar to https://leetcode.com/problems/house-robber/\n\n# Complexity\n- Time complexity: O(n log n)\n\n- Space complexity: O(n)\n Add your space co | yimang | NORMAL | 2024-10-26T06:27:54.345279+00:00 | 2024-10-26T06:32:22.563542+00:00 | 8 | false | # Intuition\nSimilar to https://leetcode.com/problems/house-robber/\n\n# Complexity\n- Time complexity: O(n log n)\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n def maximumTotalDamage(self, power: List[int]) -> int:\n count = Co... | 0 | 0 | ['Python3'] | 0 |
maximum-total-damage-with-spell-casting | Python Solution, Beats 99.47%, Bottom-up DP | python-solution-beats-9947-bottom-up-dp-ozb39 | This solution uses a bottom up dynamic programming approach and beats 99.47% of other approaches runtime wise and 71.01% in memory use.\n\n## Idea\nTo solve thi | fayyazbasil | NORMAL | 2024-10-24T22:06:40.621834+00:00 | 2024-10-24T22:06:40.621856+00:00 | 4 | false | This solution uses a bottom up dynamic programming approach and beats 99.47% of other approaches runtime wise and 71.01% in memory use.\n\n## Idea\nTo solve this we will first get a count of all powers with the same damage (as multiple powers can have the same damage) and then sort all the unqiue damages of the powers.... | 0 | 0 | ['Dynamic Programming', 'Python'] | 0 |
maximum-total-damage-with-spell-casting | Beats 100% of solutions. Python DP | beats-100-of-solutions-python-dp-by-ohud-ywjc | Intuition\n Describe your first thoughts on how to solve this problem. \nDynamic Programming\n\n# Approach\n Describe your approach to solving the problem. \n1. | ohudson | NORMAL | 2024-10-19T17:31:21.408806+00:00 | 2024-10-19T17:31:21.408827+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nDynamic Programming\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Store frequency of powers in hashmap\n2. Remove duplicates from power array\n3. Sort power array\n4. 1-D DP where dp[i] represents maximum powe... | 0 | 0 | ['Python3'] | 0 |
maximum-total-damage-with-spell-casting | Beats 100% as of 10/19/24 | beats-100-as-of-101924-by-ohudson-no4v | 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 | ohudson | NORMAL | 2024-10-19T17:22:32.160610+00:00 | 2024-10-19T17:22:32.160633+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)$$ --... | 0 | 0 | ['Python3'] | 0 |
maximum-total-damage-with-spell-casting | short code with deep understanding | short-code-with-deep-understanding-by-ra-jg5q | Pre-requisite\nSolve this problem: https://leetcode.com/problems/house-robber/description/\n\n\n# Approach\nif the product of frequency of power and power were | rajat8020 | NORMAL | 2024-10-17T23:27:51.202147+00:00 | 2024-10-17T23:27:51.202174+00:00 | 1 | false | # Pre-requisite\nSolve this problem: https://leetcode.com/problems/house-robber/description/\n\n\n# Approach\nif the product of frequency of power and power were $$<=$$ $$10^6$$ then this problem would be converted to the problem mentioned in pre-requisute. But give the constraints we won\'t be able to take an array of... | 0 | 0 | ['C++'] | 0 |
maximum-total-damage-with-spell-casting | Memoization & Recursion | memoization-recursion-by-roy_b-1pc7 | Intuition\n Describe your first thoughts on how to solve this problem. \nThis could be done with a hashmap to make it look cleaner but were memoizaing it so it | roy_b | NORMAL | 2024-10-11T18:01:31.616748+00:00 | 2024-10-11T18:01:31.616790+00:00 | 5 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis could be done with a hashmap to make it look cleaner but were memoizaing it so it doesn\'t make any difference.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nI avoided running loops and including prev and just... | 0 | 0 | ['Java'] | 0 |
maximum-total-damage-with-spell-casting | Java Solution || TC SC O(N) || Sorting | java-solution-tc-sc-on-sorting-by-harsha-pdij | \n\n# Code\njava []\nclass Solution {\n public long maximumTotalDamage(int[] power) {\n\n Arrays.sort(power);\n long[] dp = new long[power.leng | harshal_yallewar | NORMAL | 2024-10-09T10:30:20.877842+00:00 | 2024-10-09T10:30:20.877869+00:00 | 4 | false | \n\n# Code\n```java []\nclass Solution {\n public long maximumTotalDamage(int[] power) {\n\n Arrays.sort(power);\n long[] dp = new long[power.length];\n HashMap<Integer, Integer> map = new HashMap<>();\n\n for(int num: power){\n map.put(num, map.getOrDefault(num,0)+1);\n ... | 0 | 0 | ['Array', 'Hash Table', 'Dynamic Programming', 'Sorting', 'Java'] | 0 |
find-the-count-of-monotonic-pairs-i | Easy Video Solution 🔥 || How to 🤔 in Interview || 3D DP || O(50*50*N) | easy-video-solution-how-to-in-interview-topbf | If you like the solution Please Upvote and subscribe to my youtube channel\n\n# Intuition\n Describe your first thoughts on how to solve this problem. \nDry Run | ayushnemmaniwar12 | NORMAL | 2024-08-11T05:23:52.559061+00:00 | 2024-08-11T09:52:56.104239+00:00 | 2,089 | false | ***If you like the solution Please Upvote and subscribe to my youtube channel***\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nDry Run few examples and observe how you can build the logic\n\n\n\n***Easy Video Explanation***\n \nhttps://youtu.be/W9Job3hNmvA\n\n# Code\n\n\n```C++... | 26 | 3 | ['Dynamic Programming', 'Combinatorics', 'C++'] | 3 |
find-the-count-of-monotonic-pairs-i | [Java/C++/Python] 1D DP, O(n) Space | javacpython-1d-dp-on-space-by-lee215-n5jj | [Java/C++/Python] DP, O(n) Solution\n----------------------------------------------\n\n# Explanation\ndp[i][j] means the count of first i + 1 pairs and arr1[i] | lee215 | NORMAL | 2024-08-11T04:01:40.155237+00:00 | 2024-08-18T13:24:06.145913+00:00 | 2,645 | false | [Java/C++/Python] DP, O(n) Solution\n----------------------------------------------\n\n# **Explanation**\n`dp[i][j]` means the count of first `i + 1` pairs and `arr1[i] = j`.\n\nSince arr1[0] can be any value that `0 <= arr1[0] <= A[0]`,\nwe initilize `d[0][j] = 1`\n\nFrom the problem we have\n`A[i - 1] - arr1[i - 1] >... | 22 | 3 | ['C', 'Python', 'Java'] | 7 |
find-the-count-of-monotonic-pairs-i | Short | Simple | 3D DP | C++| Video Solution | O(50*50*N) | short-simple-3d-dp-c-video-solution-o505-juv0 | Intuition\n Describe your first thoughts on how to solve this problem. \nwe will go recursively and check for every possible pair of numbers.\nwe have x and y r | gavnish_kumar | NORMAL | 2024-08-11T04:08:49.990876+00:00 | 2024-08-11T07:34:50.567591+00:00 | 833 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nwe will go recursively and check for every possible pair of numbers.\nwe have x and y representing arr1 current index value and similarly y of arr2.\nnow for 0th index we can have range ```0<=x<=nums[0]``` and ```0<=y<=nums[0]``` so just ... | 14 | 0 | ['C++'] | 4 |
find-the-count-of-monotonic-pairs-i | Intuition + Approach || Using Memoization + Pair Sum Algorithm | intuition-approach-using-memoization-pai-z2jq | Intuition and Approach\n\n\nFor example:\n\n nums = [2,3,2]\n Conditions for good monotonic pairs:\n 1. equal to size n\n 2. arr1 increasing\n 3. | langesicht | NORMAL | 2024-08-11T04:50:13.035429+00:00 | 2024-08-11T09:32:42.843270+00:00 | 1,902 | false | # Intuition and Approach\n\n\nFor example:\n\n nums = [2,3,2]\n Conditions for good monotonic pairs:\n 1. equal to size n\n 2. arr1 increasing\n 3. arr2 decreasing\n 4. arr1[i]+arr2[i]=nums[i]\n\n\nSuppose :\n\n arr1 = [ _ , _ , _ ],\n arr2 = [ _ , _ , _ ]\n\n We have to fill these positions ... | 13 | 1 | ['Python', 'C++', 'Java'] | 2 |
find-the-count-of-monotonic-pairs-i | Super simple solution using combinatorics | super-simple-solution-using-combinatoric-7bvc | Key Insights\n\nThe problem can be viewed as a way of distributing the values in nums between two arrays while maintaining their monotonic properties.\nThe mini | Mudit-B | NORMAL | 2024-08-11T04:15:49.344067+00:00 | 2024-08-11T04:15:49.344090+00:00 | 913 | false | # Key Insights\n\nThe problem can be viewed as a way of distributing the values in nums between two arrays while maintaining their monotonic properties.\nThe minimum possible value for arr2 at any position constrains the possible distributions.\nOnce we find the minimum possible value for arr2, the problem transforms i... | 10 | 0 | ['Python3'] | 5 |
find-the-count-of-monotonic-pairs-i | DP Based Solution | Memorization | Tabulation | dp-based-solution-memorization-tabulatio-ry2i | Intuition\nIf we observe carefully observe we don\'t really need to form arrays we just need to play with the conditions. Look at the arr1 it should be non-decr | tanishq0_0 | NORMAL | 2024-08-11T04:04:45.331427+00:00 | 2024-08-11T04:04:45.331448+00:00 | 1,020 | false | # Intuition\nIf we observe carefully observe we don\'t really need to form arrays we just need to play with the conditions. Look at the arr1 it should be non-decreasing means current element should be ATLEAST previous, (that\'s our first condition for finding the pairs for arr1). Now similarly we can check condition fo... | 10 | 1 | ['Dynamic Programming', 'C++'] | 3 |
find-the-count-of-monotonic-pairs-i | Python 3 || 9 lines, no arrays || T/S: 99% / 89% | python-3-9-lines-no-arrays-ts-99-89-by-s-w4z6 | \n\nclass Solution:\n def countOfPairs(self, nums: List[int]) -> int:\n\n n, mn = len(nums), nums.pop(0)\n num1, num2 = 0, mn\n \n fo | Spaulding_ | NORMAL | 2024-08-11T20:12:52.443516+00:00 | 2024-10-16T22:12:44.934175+00:00 | 152 | false | \n```\nclass Solution:\n def countOfPairs(self, nums: List[int]) -> int:\n\n n, mn = len(nums), nums.pop(0)\n num1, num2 = 0, mn\n \n for num in nums:\n if num < num1: return 0\n\n if num > num1 + num2: \n num1 = num - num2 \n\n num2 = num - nu... | 9 | 0 | ['Python3'] | 0 |
find-the-count-of-monotonic-pairs-i | Video Explanation - Recursion Journey from O(N*X*X*X) --> O(N*X*X) --> O(N*X) | video-explanation-recursion-journey-from-y7rw | Explanation\n\nClick here for the video\n\n# Code\n\nconst int N = 2001;\nconst int M = 1e9+7;\n\nclass Solution {\npublic:\n int countOfPairs(vector<int>& n | codingmohan | NORMAL | 2024-08-11T06:32:04.569625+00:00 | 2024-08-11T06:32:04.569652+00:00 | 280 | false | # Explanation\n\n[Click here for the video](https://youtu.be/37S9Debk8Bg)\n\n# Code\n```\nconst int N = 2001;\nconst int M = 1e9+7;\n\nclass Solution {\npublic:\n int countOfPairs(vector<int>& nums) {\n int n = nums.size();\n vector<int> dp(N, 1); // dp[i+1]\n \n for (int i = n-1; i >= ... | 6 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Simple JAVA Memo || Traditional DP (Striver) || ✅ | simple-java-memo-traditional-dp-striver-qpr2x | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Harshsharma08 | NORMAL | 2024-08-11T04:09:21.955872+00:00 | 2024-08-11T04:09:21.955912+00:00 | 419 | 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)$$ --... | 6 | 0 | ['Array', 'Dynamic Programming', 'Memoization', 'Java'] | 3 |
find-the-count-of-monotonic-pairs-i | Simple short Top Down solution. 50*N, 2D Beats 90% | simple-short-top-down-solution-50n-2d-be-ubfs | Code\n\nclass Solution {\npublic:\n\n int n;\n int mod=1e9+7;\n int dp[2000][51];\n int dfs(vector<int>& nums,int indx,int previous)\n {\n | Michael_Teng6 | NORMAL | 2024-08-18T17:39:09.121153+00:00 | 2024-09-14T16:22:06.105709+00:00 | 133 | false | # Code\n```\nclass Solution {\npublic:\n\n int n;\n int mod=1e9+7;\n int dp[2000][51];\n int dfs(vector<int>& nums,int indx,int previous)\n {\n if(indx==n) return 1;\n if(dp[indx][previous]!=-1) return dp[indx][previous];\n long long ans=0;\n for(int i=previous;i<=nums[indx];i... | 5 | 0 | ['Recursion', 'Memoization', 'C++'] | 1 |
find-the-count-of-monotonic-pairs-i | C++ solution using dp | c-solution-using-dp-by-sachin_kumar_shar-443h | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Sachin_Kumar_Sharma | NORMAL | 2024-08-11T05:20:43.993688+00:00 | 2024-08-11T05:20:43.993725+00:00 | 107 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(N*50)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(N*50)\n<!-- Add your space complexity here, ... | 5 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Mastering 3D Dynamic Programming: Efficient Pair Counting in Arrays | mastering-3d-dynamic-programming-efficie-4srf | Intuition\nThe problem at hand is about counting valid pairs from an array of integers under specific constraints. The first thought is that this problem can be | Saurabh_1602 | NORMAL | 2024-08-11T05:05:16.123526+00:00 | 2024-08-11T05:05:16.123555+00:00 | 524 | false | # Intuition\nThe problem at hand is about counting valid pairs from an array of integers under specific constraints. The first thought is that this problem can be approached using dynamic programming, as it involves breaking down the problem into smaller overlapping subproblems. By keeping track of the minimum and maxi... | 5 | 0 | ['Dynamic Programming', 'C++'] | 1 |
find-the-count-of-monotonic-pairs-i | C++ Top down dp with memoization | c-top-down-dp-with-memoization-by-ss_sks-x0p9 | \n# Code\n\nclass Solution {\npublic:\n int n;\n int memo[2000][51];\n long long mod = 1e9+7;\n int recursion(vector<int>& nums,int idx,int prev1){\ | ss_sks_4851 | NORMAL | 2024-08-11T04:05:11.318680+00:00 | 2024-08-11T04:27:59.737646+00:00 | 236 | false | \n# Code\n```\nclass Solution {\npublic:\n int n;\n int memo[2000][51];\n long long mod = 1e9+7;\n int recursion(vector<int>& nums,int idx,int prev1){\n if(idx>=n) return 1;\n if(memo[idx][prev1]!=-1) return memo[idx][prev1];\n long long ans = 0;\n int prev2=51;\n if(idx>... | 4 | 0 | ['Dynamic Programming', 'Memoization', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | 🔥 |🔥 | You will understand it, I guarantee.🔥 |🔥 | | you-will-understand-it-i-guarantee-by-pr-39op | Intuition\n Describe your first thoughts on how to solve this problem. \nTo tackle the problem of counting monotonic pairs (arr1, arr2), we need to construct tw | prajwal_nimbalkar | NORMAL | 2024-08-13T17:24:16.840918+00:00 | 2024-08-13T17:24:16.840944+00:00 | 144 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo tackle the problem of counting monotonic pairs (arr1, arr2), we need to construct two arrays:\n\n* arr1 which is monotonically non-decreasing.\n* arr2 which is monotonically non-increasing.\nThe sum of corresponding elements from arr1 ... | 3 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Beginner friendly solution || Memoization || C++ | beginner-friendly-solution-memoization-c-1ba4 | Approach\nValue of elements in nums array ranges from 0 to 50.\n\narr1[i] + arr2[i] = nums[i]\n\n\nwe know the value of nums[i].If we fix the value of arr1[i] t | naman65 | NORMAL | 2024-08-11T13:45:10.919839+00:00 | 2024-08-11T13:45:10.919870+00:00 | 123 | false | # Approach\nValue of elements in nums array ranges from 0 to 50.\n```\narr1[i] + arr2[i] = nums[i]\n\n```\nwe know the value of nums[i].If we fix the value of arr1[i] then we can easly find the value of arr2[i].\n```\narr2[i] = nums[i] - arr1[i];\n\n```\nSo we take all possible values of arr1[i] and the possible value ... | 3 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 1 |
find-the-count-of-monotonic-pairs-i | 3D-dp easy to understand solution !! | 3d-dp-easy-to-understand-solution-by-suj-k6dl | \n\n# Approach\nUse previous values to check if curr value of i and j can be inserted in the arrays or not. (there is no need to make 2 arrays we just need to | sujal-08 | NORMAL | 2024-08-15T13:47:52.055869+00:00 | 2024-08-15T13:47:52.055904+00:00 | 56 | false | \n\n# Approach\nUse previous values to check if curr value of i and j can be inserted in the arrays or not. (there is no need to make 2 arrays we just need to keep track of prev value of both arraays)\n\n\n# Code\n```\nclass Solution {\npublic:\n\n int MOD = 1e9 + 7;\n int solve(vector<int>& nums, int ind, int p... | 2 | 0 | ['Dynamic Programming', 'C++'] | 1 |
find-the-count-of-monotonic-pairs-i | Python3 - Combinatorics Explanation O(n) | python3-combinatorics-explanation-on-by-ye3ow | Intuition\nI originally solved this question using 2D DP. I saw the other combinatorics solutions and found them very interesting. It took me a bit to understan | b4aprogrammer123 | NORMAL | 2024-08-14T05:43:47.450155+00:00 | 2024-08-14T05:43:47.450177+00:00 | 60 | false | # Intuition\nI originally solved this question using 2D DP. I saw the other combinatorics solutions and found them very interesting. It took me a bit to understand the intuition behind the counting, so thought I would share my attempt at an explanation for anyone else trying to understand the solution.\n\n# Approach\n#... | 2 | 0 | ['Python3'] | 1 |
find-the-count-of-monotonic-pairs-i | Easy to Understand C++ Solution || Very Intuitive DP(Memomization) | easy-to-understand-c-solution-very-intui-ipmz | Intuition\n Describe your first thoughts on how to solve this problem. \nWhen approaching this problem, think about how to count valid sequences where each elem | jitenagarwal20 | NORMAL | 2024-08-12T07:19:37.222894+00:00 | 2024-08-12T07:19:37.222915+00:00 | 213 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWhen approaching this problem, think about how to count valid sequences where each element in the sequence must satisfy certain constraints relative to the previous elements. \nThe key idea is to use dynamic programming to explore all pos... | 2 | 0 | ['Dynamic Programming', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | 92%Beats in C++|| Easy understandable DP Tabulation solution for absolute Beginners 💯✅ | 92beats-in-c-easy-understandable-dp-tabu-oxmz | \n\n# Approach\n Describe your approach to solving the problem. \nFirst we need to define the state dp[i][j] :\nThe State dp[i][j] means if we want number of po | srimukheshsuru | NORMAL | 2024-08-11T18:28:03.692044+00:00 | 2024-08-11T19:17:54.861467+00:00 | 122 | false | \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nFirst we need to define the state **```dp[i][j]```** :\nThe State dp[i][j] means if we want number of possible pairs of arr1 and arr2 ... | 2 | 0 | ['Dynamic Programming', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Dynamic Programming 2D array (Python) | dynamic-programming-2d-array-python-by-r-epzk | 2d array Dynamic Programming\n\n## Problem Statement\n\nWe are given an array of positive integers nums of length n. The goal is to find the number of monotonic | ragharao2001 | NORMAL | 2024-08-11T04:04:56.520564+00:00 | 2024-08-11T04:18:00.278665+00:00 | 106 | false | # 2d array Dynamic Programming\n\n## Problem Statement\n\nWe are given an array of positive integers `nums` of length `n`. The goal is to find the number of monotonic pairs \\((arr1, arr2)\\) such that:\n\n- `arr1` is monotonically non-decreasing: `arr1[0] <= arr1[1] <= ... <= arr1[n-1]`.\n- `arr2` is monotonically... | 2 | 0 | ['Python'] | 1 |
find-the-count-of-monotonic-pairs-i | Java Clean DP + Memoization Solution | java-clean-dp-memoization-solution-by-sh-s4p8 | Complexity\n- Time complexity:\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:\n Add your space complexity here, e.g. O(n) \n\n# Code\n\nclas | Shree_Govind_Jee | NORMAL | 2024-08-11T04:04:40.574515+00:00 | 2024-08-11T04:04:40.574539+00:00 | 265 | false | # Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n \n int MOD = 1_000_000_007;\n \n private long solveMemo(int[] nums, int idx1, int idx2, int num, long[][][... | 2 | 1 | ['Array', 'Math', 'Dynamic Programming', 'Recursion', 'Memoization', 'Java'] | 1 |
find-the-count-of-monotonic-pairs-i | [Python3] dp | python3-dp-by-ye15-1w44 | \nclass Solution:\n def countOfPairs(self, nums: List[int]) -> int:\n n = len(nums)\n dp = [[0]*51 for _ in range(n+1)]\n dp[n] = [1]*51 | ye15 | NORMAL | 2024-08-11T04:01:37.917060+00:00 | 2024-08-14T18:38:27.423107+00:00 | 334 | false | ```\nclass Solution:\n def countOfPairs(self, nums: List[int]) -> int:\n n = len(nums)\n dp = [[0]*51 for _ in range(n+1)]\n dp[n] = [1]*51\n for i in range(n-1, -1, -1): \n diff = 0 \n if i: diff = max(0, nums[i] - nums[i-1])\n for j in range(50, -1, -1):... | 2 | 0 | ['Python3'] | 0 |
find-the-count-of-monotonic-pairs-i | C++ | 2D DP Solution | Clean solution | c-2d-dp-solution-clean-solution-by-kena7-py3h | Code | kenA7 | NORMAL | 2025-04-11T11:28:19.841068+00:00 | 2025-04-11T11:28:19.841068+00:00 | 1 | false |
# Code
```cpp []
class Solution {
public:
#define ll long long
int dp[2001][51];
ll mod=1000000007;
int find(int i,int prev1,vector<int> &nums)
{
if(i>=nums.size())
return 1;
if(dp[i][prev1]!=-1)
return dp[i][prev1];
ll res=0;
for(int ar1=... | 1 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | [Python3] 🤔 Thought process from "3d DP" to 1d DP with prefix sum, bottom-up | python3-thought-process-from-3d-dp-to-1d-7qir | Intuition\n Describe your first thoughts on how to solve this problem. \nSince there\'s a linear relation on the index, arr1 must be non-decreasing, arr2 must b | vvave | NORMAL | 2024-09-03T04:07:03.844878+00:00 | 2024-09-03T04:14:33.888103+00:00 | 13 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSince there\'s a linear relation on the index, arr1 must be non-decreasing, arr2 must be non-increasing. If we can\'t think of a combinatorial approach first go, DP seems appropriate.\n\n**On determining the dimension of DP array:**\n\nTh... | 1 | 0 | ['Dynamic Programming', 'Prefix Sum', 'Python3'] | 0 |
find-the-count-of-monotonic-pairs-i | Python DP with explanation | python-dp-with-explanation-by-pchen36-snhm | 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 | pchen36 | NORMAL | 2024-08-11T21:47:12.987471+00:00 | 2024-08-16T17:29:12.407725+00:00 | 96 | 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 *51 *51)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n*51)\n<!-- Add your space complexity h... | 1 | 0 | ['Python3'] | 0 |
find-the-count-of-monotonic-pairs-i | Simple | Intuitive | Top - Down DP | C++ | simple-intuitive-top-down-dp-c-by-rishab-48g0 | Code\n\n#define MOD 1000000007\nclass Solution {\npublic:\n int helper(vector<int>&arr,int idx,int num1,int num2,vector<vector<int>>&dp){\n if(idx==ar | rishabhteli14 | NORMAL | 2024-08-11T07:37:47.443457+00:00 | 2024-08-11T07:37:47.443485+00:00 | 68 | false | # Code\n```\n#define MOD 1000000007\nclass Solution {\npublic:\n int helper(vector<int>&arr,int idx,int num1,int num2,vector<vector<int>>&dp){\n if(idx==arr.size())return 1;\n int ans = 0;\n if(dp[idx][num1]!=-1)return dp[idx][num1] % MOD;\n for(int i=num1;i<=arr[idx];i++){\n i... | 1 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Memoized Solution | Python | memoized-solution-python-by-sheshankkuma-wc7u | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | sheshankkumarsingh28 | NORMAL | 2024-08-11T05:23:26.709500+00:00 | 2024-08-11T12:37:16.557940+00:00 | 38 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n*max(nums)^2)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nfrom functools import cache\nclass Solution:... | 1 | 0 | ['Recursion', 'Memoization', 'Python', 'Python3'] | 0 |
find-the-count-of-monotonic-pairs-i | C++ | O(n x max(nums)^3) -> O(n x max(nums)^2) | c-on-x-maxnums3-on-x-maxnums2-by-abg_001-uv9p | Intuition\nThe core idea is to use dynamic programming (DP) to break down the problem into smaller subproblems. The key challenge is to ensure that:\n\n- arr1 r | abg_001 | NORMAL | 2024-08-11T05:16:40.028704+00:00 | 2024-08-11T05:16:40.028727+00:00 | 45 | false | # Intuition\nThe core idea is to use dynamic programming (DP) to break down the problem into smaller subproblems. The key challenge is to ensure that:\n\n- arr1 remains non-decreasing as we move from left to right in the array.\n- arr2 remains non-increasing as we move from left to right in the array.\n\nThis requires ... | 1 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | C++ | Easy DP Memoization with unordered_map | c-easy-dp-memoization-with-unordered_map-olqb | \n# Complexity\n- Time complexity: O(nm^2) where n = length of nums, and m is how high the value in nums[i] is\n\n- Space complexity: O(nm^2)\n\nFaster solution | bramar2 | NORMAL | 2024-08-11T04:45:17.724870+00:00 | 2024-08-11T04:45:49.541366+00:00 | 25 | false | \n# Complexity\n- Time complexity: $$O(n*m^2)$$ where n = length of nums, and m is how high the value in nums[i] is\n\n- Space complexity: $$O(n*m^2)$$\n\nFaster solution at Q4 sol: https://leetcode.com/problems/find-the-count-of-monotonic-pairs-ii/solutions/5619514/c-dp-tabulation-table-complement-comments/\n# Code\n`... | 1 | 0 | ['Hash Table', 'Dynamic Programming', 'Bit Manipulation', 'Memoization', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | DP - simple MEMOIZATION [beats 100%] | dp-simple-memoization-beats-100-by-zaid5-b9u5 | Intuition\nThe goal of the problem is to count the number of valid pairs in a given list nums where certain conditions are met. The specific conditions involve | zaid5496 | NORMAL | 2024-08-11T04:25:05.033975+00:00 | 2024-08-11T04:26:19.082485+00:00 | 29 | false | # Intuition\nThe goal of the problem is to count the number of valid pairs in a given list nums where certain conditions are met. The specific conditions involve partitioning each element of the list into two parts and ensuring that the pairs meet specific criteria defined by the indices and the constraints\n\n# Approa... | 1 | 0 | ['Dynamic Programming', 'Python', 'Java'] | 0 |
find-the-count-of-monotonic-pairs-i | C++ || DP || Recursion+Memoization | c-dp-recursionmemoization-by-akash_ktr-lji7 | Intuition\n Describe your first thoughts on how to solve this problem. \nwe need to count all the pairs of arrays, that means we need to check for all the pairs | Akash_Ktr | NORMAL | 2024-08-11T04:19:15.742868+00:00 | 2024-08-11T04:19:15.742888+00:00 | 87 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nwe need to count all the pairs of arrays, that means we need to check for all the pairs of elements at every index. This can be done by solving the question dynamically\n\n# Approach\n<!-- Describe your approach to solving the problem. --... | 1 | 0 | ['Dynamic Programming', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | ✅ 🎯 📌 Simple Solution || Beats 100% in Time || Dynamic Programming ✅ 🎯 📌 | simple-solution-beats-100-in-time-dynami-4g12 | Initialization:\n - We start by initializing an array curr for the first element of nums.\n - curr[i] stores the number of ways to form valid pairs where ar | vvnpais | NORMAL | 2024-08-11T04:06:48.889815+00:00 | 2024-08-11T04:14:46.898305+00:00 | 94 | false | 1. **Initialization**:\n - We start by initializing an array `curr` for the first element of `nums`.\n - `curr[i]` stores the number of ways to form valid pairs where `arr1[0] = i` and `arr2[0] = nums[0] - i`. Each index in `curr` is initially set to `1` because any valid split of `nums[0]` has exactly one valid pa... | 1 | 0 | ['Dynamic Programming', 'Python3'] | 0 |
find-the-count-of-monotonic-pairs-i | C++ | 2D Dp solution | c-2d-dp-solution-by-pranjal_dubey-np72 | Intuition\nApply 2D Dp using index and the last element(for the decreasing array) as the states.\n\n# Complexity\n- Time complexity:\nO(n 50 * 50)\n\n- Space co | Pranjal_Dubey_ | NORMAL | 2024-08-11T04:03:00.000793+00:00 | 2024-08-11T04:03:00.000824+00:00 | 124 | false | # Intuition\nApply 2D Dp using index and the last element(for the decreasing array) as the states.\n\n# Complexity\n- Time complexity:\nO(n* 50 * 50)\n\n- Space complexity:\nO(n*50) + O(n)\n\n# Code\n```\nclass Solution {\npublic:\n int mod = 1e9 + 7;\n \n int solve(vector<int>& nums, int idx, int last_ele, ve... | 1 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | straightforward dp | straightforward-dp-by-zinchse-2czh | Observation. It\'s sufficient to control only arr_1, as the arr_2 is automatically determined by the sum condition (nums[i] = arr_1[i] + arr_2[i])\n\n# Approach | zinchse | NORMAL | 2024-08-11T04:02:38.550196+00:00 | 2024-08-13T16:06:39.719604+00:00 | 67 | false | **Observation.** It\'s sufficient to control only `arr_1`, as the `arr_2` is automatically determined by the sum condition (`nums[i] = arr_1[i] + arr_2[i]`)\n\n# Approach\n\nLet `m = max(nums)`, which bounds the values in the arrays due to the non-negativity constraint. We introduce a `dp` array, where `dp[i]` represen... | 1 | 0 | ['Python3'] | 0 |
find-the-count-of-monotonic-pairs-i | C++ | Bootom up 2D DP | optimized space - O(50), time - O(n * 50 * 50) | c-bootom-up-2d-dp-optimized-space-o50-ti-to0r | We count all the possible ending value combination at each index,\nand keep results saved in dp for easy calculation for next step.\n\n\nclass Solution {\npubli | zhhackk | NORMAL | 2024-08-11T04:00:58.960012+00:00 | 2024-08-11T05:35:18.727997+00:00 | 169 | false | We count all the possible ending value combination at each index,\nand keep results saved in dp for easy calculation for next step.\n\n```\nclass Solution {\npublic:\n int countOfPairs(vector<int>& nums) {\n long long kMod = 1e9 + 7;\n int n = nums.size();\n int max_num_size = 50;\n // dp... | 1 | 0 | ['Dynamic Programming', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | JS Beats 100% DP Top Bottom : Look Behind and continue | js-beats-100-dp-top-bottom-look-behind-a-qlf3 | IntuitionAt each state you can choose which combination is valid to continue the previous oneApproachComplexity
Time complexity:
5 * 10e6
Space complexity:
1 | tonitannoury01 | NORMAL | 2025-03-16T17:01:08.907588+00:00 | 2025-03-16T17:01:08.907588+00:00 | 1 | false | # Intuition
At each state you can choose which combination is valid to continue the previous one
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
5 * 10e6
- Space complexity:
10e5
# Code
```javascript []
/**
* @param {number[]} nums
* @return {number}
*/
var coun... | 0 | 0 | ['JavaScript'] | 0 |
find-the-count-of-monotonic-pairs-i | 💥 Beats 100% on runtime [EXPLAINED] | beats-100-on-runtime-explained-by-r9n-qe17 | IntuitionI first realize that the goal is to count all the possible pairs of numbers from the given list where each pair satisfies a specific condition. The mai | r9n | NORMAL | 2025-02-04T02:08:34.690175+00:00 | 2025-02-04T02:08:34.690175+00:00 | 5 | false | # Intuition
I first realize that the goal is to count all the possible pairs of numbers from the given list where each pair satisfies a specific condition. The main idea is to explore each number and try to find valid combinations based on the previous choices, and keep track of the results to avoid recalculating overl... | 0 | 0 | ['Array', 'Hash Table', 'Math', 'Dynamic Programming', 'Bit Manipulation', 'Memoization', 'Combinatorics', 'Kotlin'] | 0 |
find-the-count-of-monotonic-pairs-i | DP, O(n (max(nums))^2) | dp-on-maxnums2-by-filinovsky-x53e | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Filinovsky | NORMAL | 2024-12-25T23:35:40.443360+00:00 | 2024-12-25T23:35:40.443360+00:00 | 6 | 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 | ['Python3'] | 0 |
find-the-count-of-monotonic-pairs-i | 3d dp | 3d-dp-by-thamanbharti-cqk8 | Intuitionwe have choices for arr1[i] && arr2[i] also we need to memoize the prev element choosen for both arr1 and arr2 since arr1 should be increasing hence ou | thamanbharti | NORMAL | 2024-12-16T07:06:54.703259+00:00 | 2024-12-16T07:06:54.703259+00:00 | 8 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nwe have choices for arr1[i] && arr2[i] also we need to memoize the prev element choosen for both arr1 and arr2 since arr1 should be increasing hence out of two parts of nums[i] nums[i]==arr1[i]+arr2[i]\n maxel is choosen for arr1 and mine... | 0 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | 3d dp | 3d-dp-by-thamanbharti-d7vv | Intuitionwe have choices for arr1[i] && arr2[i] also we need to memoize the prev element choosen for both arr1 and arr2 since arr1 should be increasing hence ou | thamanbharti | NORMAL | 2024-12-16T07:06:50.437683+00:00 | 2024-12-16T07:06:50.437683+00:00 | 1 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nwe have choices for arr1[i] && arr2[i] also we need to memoize the prev element choosen for both arr1 and arr2 since arr1 should be increasing hence out of two parts of nums[i] nums[i]==arr1[i]+arr2[i]\n maxel is choosen for arr1 and mine... | 0 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | C++ Simple Bottom-up DP, O(1) space | c-simple-bottom-up-dp-o1-space-by-pradyu-4tq1 | Code\ncpp []\n/*\nNOTE: the k-loop can be optimised further using a prefix array. \n*/\nclass Solution {\npublic:\n int countOfPairs(vector<int>& nums) {\n | pradyumnaym | NORMAL | 2024-11-26T11:52:53.714989+00:00 | 2024-11-26T11:52:53.715015+00:00 | 4 | false | # Code\n```cpp []\n/*\nNOTE: the k-loop can be optimised further using a prefix array. \n*/\nclass Solution {\npublic:\n int countOfPairs(vector<int>& nums) {\n int n = nums.size(), MOD = 1e9 + 7;\n vector<int> prev(51, 0), next(51);\n\n for (int i = 0; i < 51; i++) {\n if (nums[0] -... | 0 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Typescript | 3D memoization | typescript-3d-memoization-by-deejaydee-rrus | \nfunction countOfPairs(nums: number[]): number {\n \n const mod = 1e9 + 7;\n const dp = Array.from({length: nums.length}, () => Array.from({length: 5 | winterarc22 | NORMAL | 2024-11-15T19:59:53.242615+00:00 | 2024-11-15T19:59:53.242653+00:00 | 2 | false | ```\nfunction countOfPairs(nums: number[]): number {\n \n const mod = 1e9 + 7;\n const dp = Array.from({length: nums.length}, () => Array.from({length: 54}, () => Array(54).fill(-1)));\n \n const solve = (index, prev1, prev2) : number => {\n \n if(index === nums.length) {\n ret... | 0 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization'] | 0 |
find-the-count-of-monotonic-pairs-i | [C++] Bottom up DP. | c-bottom-up-dp-by-lovebaonvwu-6oqj | \ncpp []\nclass Solution {\npublic:\n int countOfPairs(vector<int>& nums) {\n int dp[2001][51] = {0};\n int mod = 1e9 + 7;\n for (int j | lovebaonvwu | NORMAL | 2024-10-18T01:45:33.564441+00:00 | 2024-10-18T01:45:33.564464+00:00 | 2 | false | \n```cpp []\nclass Solution {\npublic:\n int countOfPairs(vector<int>& nums) {\n int dp[2001][51] = {0};\n int mod = 1e9 + 7;\n for (int j = 0; j <= nums[0]; ++j) {\n dp[0][j] = 1;\n }\n\n int n = nums.size();\n for (int i = 1; i < n; ++i) {\n for (int ... | 0 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Clean C++ Solution using DP | clean-c-solution-using-dp-by-ritikraj26-tjwx | Intuition\n Describe your first thoughts on how to solve this problem. \nDynamice Programming since we have to find the number of arrays.\n\n# Approach\n Descri | ritikraj26 | NORMAL | 2024-10-10T17:22:03.553818+00:00 | 2024-10-10T17:26:14.229939+00:00 | 2 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nDynamice Programming since we have to find the number of arrays.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe are always iterating from the previously picked element up to nums[i], so Case 1 is already satisi... | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Python , Dynamic programming | python-dynamic-programming-by-ashish3804-6b0u | Problem\n\nYou are given an array nums where each element nums[i] represents the total number of pairs to split. For each element, you can split it into two non | ashish380431 | NORMAL | 2024-09-21T19:48:41.504254+00:00 | 2024-09-21T19:48:41.504278+00:00 | 5 | false | ## Problem\n\nYou are given an array `nums` where each element `nums[i]` represents the total number of pairs to split. For each element, you can split it into two non-negative values `a` and `b` such that `a + b = nums[i]`. The goal is to find the number of valid ways to split all elements such that for every pair `(a... | 0 | 0 | ['Dynamic Programming', 'Python3'] | 0 |
find-the-count-of-monotonic-pairs-i | Recursive two-pointer / prefix sum 94.82% on Monotonic Pairs II, 99.70% on Monotonic Pairs I | recursive-two-pointer-prefix-sum-9482-on-ibtn | Intuition\n\nIf [$k_1$,$k_2$,...$k_{i-1}$, $k_i$] is a part of a monotonic pair for the first $i$ element prefix array, \n[$k_1$,$k_2$,...$k_{i-1}$] is one for | glugglug | NORMAL | 2024-09-12T17:28:20.840120+00:00 | 2024-09-12T17:28:20.840152+00:00 | 2 | false | # Intuition\n\nIf [$k_1$,$k_2$,...$k_{i-1}$, $k_i$] is a part of a monotonic pair for the first $i$ element prefix array, \n[$k_1$,$k_2$,...$k_{i-1}$] is one for the first $i-1$ element prefix array.\n\nThe count of monotonic pairs will also be the count of pairs ending in $x_1$ + the count ending in $x_2$ + the count ... | 0 | 0 | ['Two Pointers', 'Dynamic Programming', 'Recursion', 'Prefix Sum', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Commented solution. You will definitely understand guys | commented-solution-you-will-definitely-u-vhaf | 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 | parth_gujral_ | NORMAL | 2024-09-08T05:25:34.310945+00:00 | 2024-09-08T05:25:34.310975+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)$$ --... | 0 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Detailed explanation, recursion with memoization, Python3 | detailed-explanation-recursion-with-memo-jbp3 | Let\'s start with the most brute force solution possible and then optimize it.\n\nThe brute force solution is to generate all the valid arrays and then count th | barris | NORMAL | 2024-09-04T14:52:15.838943+00:00 | 2024-09-04T14:52:15.838979+00:00 | 13 | false | Let\'s start with the most brute force solution possible and then optimize it.\n\nThe brute force solution is to generate all the valid arrays and then count them all. I get that it may seem a little silly to do this since our complexity will be exponential, but going through this step will help us understand the probl... | 0 | 0 | ['Python3'] | 0 |
find-the-count-of-monotonic-pairs-i | CPP || DP (Rec+Memo) | cpp-dp-recmemo-by-satyamshivam366-3ptd | Intuition\n Describe your first thoughts on how to solve this problem. \ncheck all ways == dp\nalso we only have to determine the number of ways\n\n# Approach\n | satyamshivam366 | NORMAL | 2024-09-03T06:53:53.320118+00:00 | 2024-09-03T06:53:53.320150+00:00 | 5 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\ncheck all ways == dp\nalso we only have to determine the number of ways\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n-... | 0 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | SIMPLEST SOLUTION | simplest-solution-by-animesh_maurya-oztg | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Animesh_Maurya | NORMAL | 2024-08-28T04:18:43.907232+00:00 | 2024-08-28T04:18:43.907261+00:00 | 5 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Array', 'Math', 'Dynamic Programming', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Python || Dynamic Programming | python-dynamic-programming-by-vilaparthi-ahyt | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | vilaparthibhaskar | NORMAL | 2024-08-23T15:31:31.390518+00:00 | 2024-08-23T15:31:31.390556+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nm = max element\nn = len(nums)\nO(n * m * m)\n\n- Space complexity:\n<!-- Ad... | 0 | 0 | ['Array', 'Math', 'Dynamic Programming', 'Combinatorics', 'Python3'] | 0 |
find-the-count-of-monotonic-pairs-i | Beats 95% users 🔥🔥 | beats-95-users-by-arnavsoney-7puu | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | arnavsoney | NORMAL | 2024-08-22T12:06:27.885023+00:00 | 2024-08-22T12:06:27.885051+00:00 | 2 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(N^3)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O... | 0 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Memoization Recursion | memoization-recursion-by-ducanh123-02zm | Intuition\n Describe your first thoughts on how to solve this problem. \nmemorizaton recursion f(curr_idx,pre_el_ar1,pre_el_ar2)\nkeep track 1 current idx becau | ducanh123 | NORMAL | 2024-08-19T05:43:34.267072+00:00 | 2024-08-19T05:43:34.267105+00:00 | 10 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nmemorizaton recursion f(curr_idx,pre_el_ar1,pre_el_ar2)\nkeep track 1 current idx because each recursion we take element for ar1 and ar2, and pre_el_ar1, pre_el_ar2 for keep track array condition that question mention \n# Approach\n<!-- D... | 0 | 0 | ['Recursion', 'Memoization', 'Python3'] | 0 |
find-the-count-of-monotonic-pairs-i | Dynamic Programming Solution (2D DP C++) | dynamic-programming-solution-2d-dp-c-by-pczj9 | Intuition\nThe problem requires us to find the number of ways to split an array into two subarrays: one in increasing order and the other in decreasing order. S | Yash-mehra | NORMAL | 2024-08-17T12:37:04.903390+00:00 | 2024-08-17T12:37:04.903423+00:00 | 5 | false | # Intuition\nThe problem requires us to find the number of ways to split an array into two subarrays: one in increasing order and the other in decreasing order. Specifically:\n\n- We need to create two arrays, first_array and second_array, from the given array nums.\n- The element at index i in first_array should be be... | 0 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | beats 81% in time, and 98 %in space | beats-81-in-time-and-98-in-space-by-morn-e5b3 | 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 | Morningstar_30 | NORMAL | 2024-08-17T10:23:35.999810+00:00 | 2024-08-17T10:23:35.999840+00:00 | 1 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Recursion -> DP, Intuitive, Memoization, Python3, 3D DP | recursion-dp-intuitive-memoization-pytho-5bgv | Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition is similar to longest consequetive subarray problem where we pass the las | stiffmeister1122 | NORMAL | 2024-08-17T08:52:12.799847+00:00 | 2024-08-17T08:52:12.799871+00:00 | 15 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition is similar to longest consequetive subarray problem where we pass the last element to the function.\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nSend prev1 and prev2 for arr1 and arr2 respetively... | 0 | 0 | ['Array', 'Dynamic Programming', 'Python3'] | 0 |
find-the-count-of-monotonic-pairs-i | DP + Prefix in Time O(NM) || Space O(M), Beat 99% || 91%🔥 | dp-prefix-in-time-onm-space-om-beat-99-9-d0jw | Intuition\n Describe your first thoughts on how to solve this problem. \nassume M = Max_element\nUsing DP by n * M array\n# Approach\n Describe your approach to | hsu_1997 | NORMAL | 2024-08-17T07:26:23.327721+00:00 | 2024-08-17T07:30:42.095383+00:00 | 6 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nassume M = Max_element\nUsing DP by n * M array\n# Approach\n<!-- Describe your approach to solving the problem. -->\nEach amount in dp[i][j] is insert j at the last monotonically non-decreasing arr1.\nWe can count the number in monotonic... | 0 | 0 | ['Dynamic Programming', 'Prefix Sum', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | [C++] DP | c-dp-by-ericyxing-pp26 | Code\n\nclass Solution {\npublic:\n int countOfPairs(vector<int>& nums) {\n long long mod = 1e9 + 7, ans = 0;\n int n = nums.size();\n v | EricYXing | NORMAL | 2024-08-16T22:36:52.531002+00:00 | 2024-08-16T22:36:52.531033+00:00 | 0 | false | # Code\n```\nclass Solution {\npublic:\n int countOfPairs(vector<int>& nums) {\n long long mod = 1e9 + 7, ans = 0;\n int n = nums.size();\n vector<vector<long long>> dp(n, vector<long long>(51, 0));\n for (int j = 0; j <= nums[0]; j++)\n dp[0][j] = 1;\n for (int i = 1; i... | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | C++ optimal Solution✅✅|Space optimization | c-optimal-solutionspace-optimization-by-9tikk | 3D DP SOLUTION \n\nclass Solution {\npublic:\n const long long MOD = 1e9 + 7;\n vector<vector<vector<long long>>> memo;\n\n long long solve(int ind, lo | Jayesh_06 | NORMAL | 2024-08-16T10:06:45.174286+00:00 | 2024-08-16T10:06:45.174322+00:00 | 10 | false | # 3D DP SOLUTION \n```\nclass Solution {\npublic:\n const long long MOD = 1e9 + 7;\n vector<vector<vector<long long>>> memo;\n\n long long solve(int ind, long long mn, long long mx, vector<int>& nums) {\n if (ind >= nums.size()) {\n return 1; }\n \n \n if (memo[ind][... | 0 | 0 | ['Array', 'Math', 'Dynamic Programming', 'Combinatorics', 'Prefix Sum', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | DP Memorization Solution C++✅✅ | dp-memorization-solution-c-by-jayesh_06-8mna | \n# Code\n\nclass Solution {\npublic:\n int mod = 1e9 + 7;\n int dp[2001][52][52];\n int solve(int i, int n, vector<int>& v, int prev1, int prev2) {\n | Jayesh_06 | NORMAL | 2024-08-16T09:54:53.369283+00:00 | 2024-08-16T09:54:53.369320+00:00 | 1 | false | \n# Code\n```\nclass Solution {\npublic:\n int mod = 1e9 + 7;\n int dp[2001][52][52];\n int solve(int i, int n, vector<int>& v, int prev1, int prev2) {\n if (i == n)\n return 1;\n if (dp[i][prev1][prev2] != -1)\n return dp[i][prev1][prev2];\n int ans = 0;\n for... | 0 | 0 | ['Array', 'Math', 'Dynamic Programming', 'Combinatorics', 'Prefix Sum', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | [Java] Bottom up DP solution 50 * 50 * N solution | java-bottom-up-dp-solution-50-50-n-solut-0cut | \n\n# Approach\nSame as DFS Memo, for each dp[v1][v2] calculate sum of all previous valid pairs at i - 1 position \n\n# Complexity\n- Time complexity: O(51 * 51 | yrq | NORMAL | 2024-08-15T18:47:40.414118+00:00 | 2024-08-16T17:24:43.568494+00:00 | 10 | false | \n\n# Approach\nSame as DFS Memo, for each dp[v1][v2] calculate sum of all previous valid pairs at i - 1 position \n\n# Complexity\n- Time complexity: O(51 * 51 * N)\n\n- Space complexity: O(51 * 51)\n# Code\n```\nclass Solution {\n public int countOfPairs(int[] nums) {\n int n = nums.length;\n if(n ==... | 0 | 0 | ['Java'] | 0 |
find-the-count-of-monotonic-pairs-i | Find Count of Monotonic Pairs I [C++]: 2D-DP Approach, Beats 92% of all solutions | find-count-of-monotonic-pairs-i-c-2d-dp-wtk1o | Intuition\n Describe your first thoughts on how to solve this problem. \nYou need to divide the nums array into two monotonic arrays of left and right which sho | amrit2104 | NORMAL | 2024-08-15T14:42:48.128872+00:00 | 2024-08-15T14:42:48.128905+00:00 | 9 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nYou need to divide the nums array into two monotonic arrays of left and right which should be non-increasing and non-decreasing respectively. Think of how to do it!!\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n... | 0 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Dynamic Programming | dynamic-programming-by-codecarbon-e7o3 | Intuition\n Describe your first thoughts on how to solve this problem. The problem is essentially about counting pairs in a sequence while adhering to specific | codecarbon | NORMAL | 2024-08-15T11:36:41.320542+00:00 | 2024-08-15T11:36:41.320581+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->The problem is essentially about counting pairs in a sequence while adhering to specific constraints. This often hints at using dynamic programming (DP) to keep track of valid states as we iterate through the list. By using DP, we can break... | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Easy Dp solution . | easy-dp-solution-by-lcb_2022039-8baf | 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 | LCB_2022039 | NORMAL | 2024-08-15T11:33:37.896559+00:00 | 2024-08-15T11:33:37.896586+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:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(N*51*51) = O(N)\n\n- Space complexity:\n<!-- Add your space complexity her... | 0 | 0 | ['Array', 'Math', 'Dynamic Programming', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | using 3D DP | using-3d-dp-by-sauravjuyal001-qwkl | Intuition \n\n\n\n\n# Approach \n\npass two variable asc and desc to function with 0,0.\nThese two value will be used to check monotonic arrangement of array we | sauravjuyal001 | NORMAL | 2024-08-14T08:37:57.152011+00:00 | 2024-08-14T08:37:57.152030+00:00 | 9 | false | # Intuition \n\n\n\n\n# Approach \n\npass two variable asc and desc to function with 0,0.\nThese two value will be used to check monotonic arrangement of array we are creating.\nfor each value of num run a from 0 to num[i] loop and check if asc<=i\n and num[i]-i>=desc if true then you find a pair for a index then make ... | 0 | 0 | ['Array', 'Math', 'Dynamic Programming', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | 2d dp simple approach with recursion | 2d-dp-simple-approach-with-recursion-by-9jbs3 | \n Describe your first thoughts on how to solve this problem. \n\n# Approach\n So basically in most of the dp problem if we can \n find what are the values that | Arantx1 | NORMAL | 2024-08-13T18:07:30.972958+00:00 | 2024-08-13T18:07:30.972998+00:00 | 0 | false | \n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n So basically in most of the dp problem if we can \n find what are the values that are changing in every transition than we can easily find the solution.\nchange state are (index,number)\nwe can use the fact that if first number is x t... | 0 | 0 | ['C++'] | 0 |
find-the-count-of-monotonic-pairs-i | 2D DP || Recursion + Memoization || Super Easy || C++ | 2d-dp-recursion-memoization-super-easy-c-pukr | Code\n\nclass Solution \n{\npublic:\n int dp[51][2001];\n int mod=1e9+7;\n int f(int prev1, int i, vector<int> &nums)\n {\n if(i==nums.size() | lotus18 | NORMAL | 2024-08-13T06:22:49.613706+00:00 | 2024-08-13T06:32:06.062176+00:00 | 31 | false | # Code\n```\nclass Solution \n{\npublic:\n int dp[51][2001];\n int mod=1e9+7;\n int f(int prev1, int i, vector<int> &nums)\n {\n if(i==nums.size()) return 1;\n if(dp[prev1][i]!=-1) return dp[prev1][i];\n int prev2=(i?nums[i-1]-prev1:nums[0]);\n int ans=0;\n for(int x=prev1... | 0 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Easy Recursive 3d Dp Solution | easy-recursive-3d-dp-solution-by-kvivekc-2t3d | 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 | kvivekcodes | NORMAL | 2024-08-13T05:11:21.676770+00:00 | 2024-08-13T05:11:21.676797+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 |
find-the-count-of-monotonic-pairs-i | Use DP - Easy to Understand - ✅|| Java || 3 ms ||🔥|| Beats 100.00% ||🏆 | use-dp-easy-to-understand-java-3-ms-beat-udyv | Code\n\nclass Solution {\n private static final int MOD = 1_000_000_007;\n\n public int countOfPairs(int[] nums) {\n int[] maxShift = new int[nums. | djqtdj | NORMAL | 2024-08-13T00:56:38.355379+00:00 | 2024-08-13T00:56:38.355406+00:00 | 12 | false | # Code\n```\nclass Solution {\n private static final int MOD = 1_000_000_007;\n\n public int countOfPairs(int[] nums) {\n int[] maxShift = new int[nums.length];\n maxShift[0] = nums[0];\n int currShift = 0;\n\n for (int i = 1; i < nums.length; i++) {\n currShift = Math.max(c... | 0 | 0 | ['Java'] | 0 |
find-the-count-of-monotonic-pairs-i | C++ | DP | c-dp-by-ghost_tsushima-v124 | \n#define mod 1000000007\n\nclass Solution {\npublic:\n \n int getPrev2(int i, int prev1, vector<int> &nums) {\n if (i == 0) return 50;\n re | ghost_tsushima | NORMAL | 2024-08-12T18:07:44.513089+00:00 | 2024-08-12T18:07:44.513117+00:00 | 0 | false | ```\n#define mod 1000000007\n\nclass Solution {\npublic:\n \n int getPrev2(int i, int prev1, vector<int> &nums) {\n if (i == 0) return 50;\n return nums[i-1] - prev1;\n }\n \n int find(vector<vector<int> > &dp, int i, int prev, vector<int> &nums) {\n if(i == nums.size()) return 1;\n ... | 0 | 0 | [] | 0 |
find-the-count-of-monotonic-pairs-i | easy solution 👇👇 | easy-solution-by-aman2839-q053 | 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 | AMAN2839 | NORMAL | 2024-08-12T16:45:31.222599+00:00 | 2024-08-12T16:45:31.222634+00:00 | 7 | 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 |
find-the-count-of-monotonic-pairs-i | very easy java | very-easy-java-by-murariambofficial-f36j | 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 | murariambofficial | NORMAL | 2024-08-12T15:45:47.829631+00:00 | 2024-08-12T15:45:47.829668+00:00 | 12 | 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 |
find-the-count-of-monotonic-pairs-i | Beats 100% in Time Complexity | beats-100-in-time-complexity-by-kumarabh-j9gk | 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 | kumarabhinav002 | NORMAL | 2024-08-12T15:27:40.196304+00:00 | 2024-08-12T15:27:40.196337+00:00 | 1 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
find-the-count-of-monotonic-pairs-i | 76 ms Beats 100 % | 76-ms-beats-100-by-kavyanlavti-kn4l | 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 | KavyanLavti | NORMAL | 2024-08-12T13:15:45.598465+00:00 | 2024-08-12T13:15:45.598494+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:\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 |
find-the-count-of-monotonic-pairs-i | Recursion+ 2D DP | recursion-2d-dp-by-pushprog-yvgb | \nclass Solution {\npublic:\n using ll = long long;\n ll M=1e9+7;\n ll dp[2001][52];\n ll rec(int i,int aprev,int bprev,vector<int> &v, int mx){\n | Pushprog | NORMAL | 2024-08-12T12:29:14.953475+00:00 | 2024-08-12T12:29:14.953503+00:00 | 17 | false | ```\nclass Solution {\npublic:\n using ll = long long;\n ll M=1e9+7;\n ll dp[2001][52];\n ll rec(int i,int aprev,int bprev,vector<int> &v, int mx){\n int n=v.size();\n if(i==n){return 1;}\n \n if(dp[i][aprev]!=-1){return dp[i][aprev];}\n \n ll res=0;\n \n ... | 0 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C', 'C++'] | 0 |
find-the-count-of-monotonic-pairs-i | Using tabulation | using-tabulation-by-venkatarohit_p-l6j0 | 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 | venkatarohit_p | NORMAL | 2024-08-12T09:36:13.478312+00:00 | 2024-08-12T09:36:13.478332+00:00 | 10 | 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 |
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