#Maximum Coins Codechef Problem

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#Maximum Coins Codechef Problem Reels - @ridwanulhoquekawsar tarafından paylaşılan video - Monday , 16 February , 2026 ! ☑ 2nd problem of the day done ! ╰┈➤ (Codechef Starters 155, Div 4) • Problem No. C (GCD to 1 (Easy) ) ! • Problem Li
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Monday , 16 February , 2026 ! ☑ 2nd problem of the day done ! ╰┈➤ (Codechef Starters 155, Div 4) • Problem No. C (GCD to 1 (Easy) ) ! • Problem Link : ⤵︎ https://www.codechef.com/problems/GCD_1?tab=statement Observation & Intuition : 🕵🏻‍♂️ • Here, We can just put one 3 and in all the other places we'll put 2 ! • One thing we have noticed that here two scenario can happen. That is either Row will grow and Column will be fixed or Column will be grow and Row will be fixed ! Solution Approach : 🎯 • When we'll find a place where both index are same then we'll print here 2. • If (i == row - 1, && j >= row) Or (i >= col && j = col - 1), then we will print 2 here also. Otherwise we'll print 3. Time Complexity : O(n²) 📝 Implementation Uses : Adhoc </> 👨🏻‍💻 ! C++ Code Link : ╰┈જ⁀➴ https://github.com/Ridwanulhaquekawsar/Codechef-Problems/blob/main/C-GCD_to_1-Codechef-starters-155-Div-4.cpp
#Maximum Coins Codechef Problem Reels - @ridwanulhoquekawsar tarafından paylaşılan video - Sunday , 22 February , 2026 ! ☑ 2nd problem of the day done ! ╰┈➤ (Codechef Starters , Div 4) • Problem No. C (Make my array equal) ! • Problem Li
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Sunday , 22 February , 2026 ! ☑ 2nd problem of the day done ! ╰┈➤ (Codechef Starters , Div 4) • Problem No. C (Make my array equal) ! • Problem Link : ⤵︎ https://www.codechef.com/problems/MEBA?tab=statement Observation & Intuition : 🕵🏻‍♂️ • In this problem we are asked to determine whether it is possible to make the array equal or not ! • It is only possible to make array equal when there is only one non-zero element in this array, Otherwise not ! Solution Approach : 🎯 • At first we'll check how many non zero are given in this array ! • If non-zero element number is less than 2 then only it is possible to make array elements equal ! Time Complexity : O(n & nlogn) 📝 Implementation Uses : Set / Adhoc </> 👨🏻‍💻 ! C++ Code Link : ╰┈જ⁀➴ https://github.com/Ridwanulhaquekawsar/Codechef-Problems/blob/main/C-Make_My_Array_Equal-codechef-starters-145-Div-4.cpp #problemsolving #programming #coding #consistency #analyticalthinking
#Maximum Coins Codechef Problem Reels - @ridwanulhoquekawsar tarafından paylaşılan video - Wednesday , 18 February , 2026 ! ☑ 2nd problem of the day done ! ╰┈➤ (Codechef Starters 226, Div 4) • Problem No. D (Minimum Distinct) ! • Problem
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Wednesday , 18 February , 2026 ! ☑ 2nd problem of the day done ! ╰┈➤ (Codechef Starters 226, Div 4) • Problem No. D (Minimum Distinct) ! • Problem Link : ⤵︎ https://www.codechef.com/START226D/problems/MINDIS6?tab=statement Observation & Intuition : 🕵🏻‍♂️ • Here we're allowed to perform certain number operations on the given array and also we're asked to minimize the number of distinct elements in the array ! • To do that if we find the lower frequency elements and replace them first and then larger frequency if it is possible to do ! Solution Approach : 🎯 • At first we'll find all lower frequency elements first and then we'll reduce them one by one by using maximum operation we can make ! • And then we'll print the remaining distinct elements after performing the required operations ! Time Complexity : O(nlogn) 📝 Implementation Uses : map, set, Adhoc </> 👨🏻‍💻 ! C++ Code Link : ╰┈જ⁀➴ https://github.com/Ridwanulhaquekawsar/Codechef-Problems/blob/main/D-Minimum_Distinct-codechef-starters-226-Div-4.cpp #programming #problemsolving #coding
#Maximum Coins Codechef Problem Reels - @ridwanulhoquekawsar tarafından paylaşılan video - Tuesday , 3 March , 2026 ! ☑ 1st problem of the day done ! ╰┈➤ (Codechef Starters 209, Div 4) • Problem No. D (Binary Love) ! • Problem Link : ⤵︎
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Tuesday , 3 March , 2026 ! ☑ 1st problem of the day done ! ╰┈➤ (Codechef Starters 209, Div 4) • Problem No. D (Binary Love) ! • Problem Link : ⤵︎ https://www.codechef.com/problems/P5209?tab=statement Observation & Intuition : 🕵🏻‍♂️ 🧠 • In this Problem, We're Given a string and asked whether it is possible to make equal the count of substring "01" and "10" by deleting some character from either begining of the string or end of the string within few operation ! • If we can make equal then Alice will win otherwise Bob will win. • One thing is noticable here, that is count of "10" and "01" of if count of transition is greater than 1 or if after sorting the string ascending or descending order both of the string original and modified become equal then Alice will win. Because If count of transition is less than 2 then it will behave like this way : • 000000111 or 11110000 or 00000 or 11111 ! Solution Approach : 🎯 • We approached it in 4 different ways. One is just counting the substring of "01" and "10" and then check that whether both of the string count is greater or equal than 1 or not. • Second is by counting transition number. Here we just counting that in how many position it changed or converted from '1' to '0' or '0' to '1'. If transition count is greater than 1 then Alice will win otherwise Bob will win ! • Third is just sorting the original array once in ascending and then descending way. After the sort if original string become equal with one of this ascending or descending or then Bob will win otherwise Alice will win. You might be wondering why it works here ! It works because as here it changed its character at most 1 times then this string already in a sorted order that why our condition hit the Correct condition ! Time Complexity : O(n) 📝 Implementation Uses : Adhoc </> 👨🏻‍💻 ! C++ Code Link : ╰┈જ⁀➴ https://github.com/Ridwanulhaquekawsar/Codechef-Problems/blob/main/D-Binary_Love-codechef-starters-209-Div-4.cpp #problemsolving #programming #coding #consistency #analyticalthinking
#Maximum Coins Codechef Problem Reels - @ridwanulhoquekawsar tarafından paylaşılan video - Monday , 16 February , 2026 ! ☑ 1st problem of the day done ! ╰┈➤ (Codechef Starters 155, Div 4) • Problem No. B (Rectangled) ! • Problem Link : ⤵
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Monday , 16 February , 2026 ! ☑ 1st problem of the day done ! ╰┈➤ (Codechef Starters 155, Div 4) • Problem No. B (Rectangled) ! • Problem Link : ⤵︎ https://www.codechef.com/problems/RCTGLD?tab=statement Observation & Intuition : 🕵🏻‍♂️ • We know that if we want draw a rectangle then we have consider 4 integral sides and also if we want to maximize the area then we need to maximize the length and with both ! • Here, our main task is to calculate largest length and width using this available unit ! Solution Approach : 🎯 • At first we'll divide it into 4 parts so that we can distribute it to 4 sides ! • And then if remainder has more than 1 we can simply make pair up and we'll put it in length sides or width side as we like ! • Here, if we divide the available unit in two parts then remainder of 4 division will be goes equally to the both parts. • As both parts has length and width and also a extra 1, then we can take this one with width side by subtracting the length ! • length = n/4. Width = n/2 - n/4; Time Complexity : O(1) 📝 Implementation Uses : Adhoc </> 👨🏻‍💻 ! C++ Code Link : ╰┈જ⁀➴ https://github.com/Ridwanulhaquekawsar/Codechef-Problems/blob/main/B-Rectangled-Codechef-starters-155-Div-4.cpp
#Maximum Coins Codechef Problem Reels - @ridwanulhoquekawsar tarafından paylaşılan video - Wednesday , 4 March , 2026 ! ☑ 1st problem of the day done ! ╰┈➤ (Codechef Starters 209, Div 4) • Problem No. C (Small GCD Sort) ! • Problem Link
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Wednesday , 4 March , 2026 ! ☑ 1st problem of the day done ! ╰┈➤ (Codechef Starters 209, Div 4) • Problem No. C (Small GCD Sort) ! • Problem Link : ⤵︎ https://www.codechef.com/problems/P3209?tab=statement Observation & Intuition : 🕵🏻‍♂️ 🧠 • In this problem, We're given an N and we're asked find gcd for each 1 to N with N like : GCD(i, N) where {i = 1, 2, 3.....N}. And it instructs us to keep the i of Greater GCD in the begining position of the array and whose GCD of i same then we have to keep smaller index i to the left position in the array. Solution Approach : 🎯 • At first we'll calculate GCD for each i from 1 to N, and then we'll insert them into a vector. • And then we'll sort them according to the largest gcd of each i and then if any two of i has same gcd then we'll place the smaller i to the left position ! Time Complexity : O(n) 📝 Implementation Uses : Adhoc, Sorting </> 👨🏻‍💻 ! C++ Code Link : ╰┈જ⁀➴ https://github.com/Ridwanulhaquekawsar/Codechef-Problems/blob/main/C-Small_GCD_Sort-Codechef-starters-209-Div-4.cpp #Coding #programming #problemsolving #analyticalthinking #consistency
#Maximum Coins Codechef Problem Reels - @ridwanulhoquekawsar tarafından paylaşılan video - Monday , 23 February , 2026 ! ☑ 1st problem of the day done ! ╰┈➤ (Codechef Starters 143, Div 4) • Problem No. B (Maximum Coins) ! • Problem Link
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Monday , 23 February , 2026 ! ☑ 1st problem of the day done ! ╰┈➤ (Codechef Starters 143, Div 4) • Problem No. B (Maximum Coins) ! • Problem Link : ⤵︎ https://www.codechef.com/problems/MAXCOIN?tab=statement Observation & Intuition : 🕵🏻‍♂️ • There are a total n of matches ! From 1 to (n - x)'th matches we'll lose and we have to pay 2^i amount of score to opponent who wins the game from (n-x) to n ! Solution Approach : 🎯 • At first we'll calculate the score from 1 to (n-x)'th index ! • And then we'll subtract them from opponent who wins x matches ! Time Complexity : O(n) 📝 Implementation Uses : Adhoc </> 👨🏻‍💻 ! C++ Code Link : ╰┈જ⁀➴ https://github.com/Ridwanulhaquekawsar/Codechef-Problems/blob/main/B-Maximum_Coins-codechef-starters-143-Div-4.cpp #problemsolving #programming #coding #consistency #analyticalthinking
#Maximum Coins Codechef Problem Reels - @ridwanulhoquekawsar tarafından paylaşılan video - Friday , 27 February , 2026 ! ☑ 2nd problem of the day done ! ╰┈➤ (Codechef Starters 137, Div 4) • Problem No. B (Spell Splice) ! • Problem Link :
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Friday , 27 February , 2026 ! ☑ 2nd problem of the day done ! ╰┈➤ (Codechef Starters 137, Div 4) • Problem No. B (Spell Splice) ! • Problem Link : ⤵︎ https://www.codechef.com/problems/CMIX?tab=statement Observation & Intuition : 🕵🏻‍♂️ • In this Problem we're given N pairs of Numbers. Here, We can take two different pairs, And our task is basically to choose such two pairs where their diagonal or in crossline numbers multiples sum becomes as big as possible, That means we have to maximize the diagonal multiples sum ! • To do that we can take help of nested loop trick ! Solution Approach : 🎯 • At first we'll run a nested loop. • And then we'll choose two different pairs and then we'll add their dional number multiples. • And then we'll store this value after comparing to the max variable answer ! Time Complexity : O(n²) 📝 Implementation Uses : Nested Loop </> 👨🏻‍💻 ! C++ Code Link : ╰┈જ⁀➴ https://github.com/Ridwanulhaquekawsar/Codechef-Problems/blob/main/B-Spell_Splice-codechef-starters-137-Div-4.cpp #programming #problemSolving #coding #consistency #codechef
#Maximum Coins Codechef Problem Reels - @ridwanulhoquekawsar tarafından paylaşılan video - Sunday , 15 February , 2026 ! ☑ 1st problem of the day done ! ╰┈➤ (Codechef Starters 157, Div 4) • Problem No. C (Not Too Far Replacement) ! • Pro
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Sunday , 15 February , 2026 ! ☑ 1st problem of the day done ! ╰┈➤ (Codechef Starters 157, Div 4) • Problem No. C (Not Too Far Replacement) ! • Problem Link : ⤵︎ https://www.codechef.com/problems/TWOCLOSE?tab=statement Observation & Intuition : 🕵🏻‍♂️ • Here, our task is to place the largest possible element to the last index bucket as large as we can. • After sorting 1 to N we'll continuously check whether we can swap the the current value with the last elements of the array arr[N + 1], If current element is big and also it less than twice or equal then we can swap it. • if (current <= 2 * arr[N + 1]) Swap (curr, arr[N + 1]) ! Solution Approach : 🎯 • We'll take the array first, • And then we'll sort the array from 1 to N. • After that we'll run a loop from 1 to N and constantly by pushing largest element to last index we'll add the element to our answer variable within this process ! Time Complexity : O(nlong) 📝 Implementation Uses : Adhoc </> 👨🏻‍💻 ! C++ Code Link : ╰┈જ⁀➴ https://github.com/Ridwanulhaquekawsar/Codechef-Problems/blob/main/C-Not_Too_far_Replacement-Codechef-starters-157-Div-4.cpp #Coding #codinglife #coding
#Maximum Coins Codechef Problem Reels - @ridwanulhoquekawsar tarafından paylaşılan video - Tuesday , 24 February , 2026 ! ☑ 2nd problem of the day done ! ╰┈➤ (Codechef Starters 141, Div 4) • Problem No. C (Redundant Array) ! • Problem Li
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Tuesday , 24 February , 2026 ! ☑ 2nd problem of the day done ! ╰┈➤ (Codechef Starters 141, Div 4) • Problem No. C (Redundant Array) ! • Problem Link : ⤵︎ https://www.codechef.com/problems/REDUARRAY?tab=statement Observation & Intuition : 🕵🏻‍♂️ • In this problem we are given an array and we're asked to say what's minimum cost to make array elements equal . • Here, we're allowed to select a subarray portion of the array from L(left) bound to R(right) Bound and replace these portion elements by another random chosen integer Number ! • If we check for every occurence number by relacing all the elements of the array except this one ! We won't replace it where it already exists ! • Additionally we'll check whether by replacing all elements by 1, is it possible or not to get minimum operation cost that we need ! Solution Approach : 🎯 • At first, we'll find every elements frequency and then we'll try to find which occurence of elements will give the minimum operation cost if we replace remaining elements with it ! • Lastly we'll check occurence of 1 are present or not and then if we replace all the remaining elements by 1 then we'll check whether it's possible or not to get minimum operation Cost ! Time Complexity : O(nlogn) 📝 Implementation Uses : Ad-hoc, Map </> 👨🏻‍💻 ! C++ Code Link : ╰┈જ⁀➴ https://github.com/Ridwanulhaquekawsar/Codechef-Problems/blob/main/C-Redundant_Array-codechef-starters-141-Div-4.cpp #coding #consistency #problemsolving #programming #analyticalthinking
#Maximum Coins Codechef Problem Reels - @ridwanulhoquekawsar tarafından paylaşılan video - Saturday , 21 February , 2026 ! ☑ 1st problem of the day done ! ╰┈➤ (Codechef Starters 148, Div 4) • Problem No. C (FightBots) ! • Problem Link :
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Saturday , 21 February , 2026 ! ☑ 1st problem of the day done ! ╰┈➤ (Codechef Starters 148, Div 4) • Problem No. C (FightBots) ! • Problem Link : ⤵︎ https://www.codechef.com/problems/FIGBOT?tab=statement Observation & Intuition : 🕵🏻‍♂️ • In this problem, it says that is there any sequence where both cordinate will be equal. • Here one thing we can observe that if in a certain moment after few steps of alice robot when it will stand on last position of manhattan distance of Bob's robot then it will be possible become equal both of this cordinate ! Solution Approach : 🎯 • At first we'll calculate the manhattan distance of bob's robot between bob's and alice robots end position ! • And then we'll check that the distance is whether equal or not with the number of steps both of they have made in this process ! Time Complexity : O(n) 📝 Implementation Uses : Adhoc </> 👨🏻‍💻 ! C++ Code Link : ╰┈જ⁀➴ https://github.com/Ridwanulhaquekawsar/Codechef-Problems/blob/main/C-FightBots-codechef-starters-148-Div-4.cpp #ProblemSolving #coding #programming #analyticalthinking
#Maximum Coins Codechef Problem Reels - @ridwanulhoquekawsar tarafından paylaşılan video - Sunday , 15 February , 2026 ! ☑ 2nd problem of the day done ! ╰┈➤ (Codechef Starters 156, Div 4) • Problem No. C (Even Numbers Hate) ! • Problem L
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Sunday , 15 February , 2026 ! ☑ 2nd problem of the day done ! ╰┈➤ (Codechef Starters 156, Div 4) • Problem No. C (Even Numbers Hate) ! • Problem Link : ⤵︎ https://www.codechef.com/problems/EVENHATE?tab=statement Observation & Intuition : 🕵🏻‍♂️ • Since, we're allowed to rearrange the array so that it's prefix sum array contains maximum number of odd element. • So, here our main task is to maximize the odd prefix sum elements. • To do that, we can use one odd element and then rest of the elements have to choose even, After that if still have some odd elements remains then we can pair up these odd elements so that odd + odd = Even. Then final prefix sum will be odd + even = odd ! Solution Approach : 🎯 • At first we'll take an odd element and after that odd element we well place all the even numbers. • And then the remaining odd simply will place in the end of this arrangement, if odd elements we find after the process ! Time Complexity : O(n) 📝 Implementation Uses : Adhoc </> 👨🏻‍💻 ! C++ Code Link : ╰┈જ⁀➴ https://github.com/Ridwanulhaquekawsar/Codechef-Problems/blob/main/C-Even_Numbers_Hate-Codechef-starters-156-Div-4.cpp #coding #ProblemSolving

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