Problem Link :- GFG POTD
Problem statement
You have a garden with n flowers lined up in a row. The height of ith flower is ai units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Easiest Explanation
You have a garden with n flowers lined up in a row, where each flower has a certain height. You want to water them for k days. In one day, you can water a continuous set of w flowers (you can only do this once in a day). Whenever you water a flower, its height increases by 1 unit.
Your goal is to maximize the height of the smallest flower throughout the entire k-day period. In other words, you want to make sure that the smallest flower is always as tall as possible, even after you water other flowers.
To solve this problem using binary search, we first need to define the range of values that we will search over. Since we want to maximize the height of the smallest flower, the range of values we will search over is between 0 and the maximum height of the flowers.
We will then use binary search to find the optimal height that we can increase all the flowers to while keeping the smallest flower at least that height. We will start by setting the lower bound of our binary search to 0 and the upper bound to the maximum height of the flowers.
In each iteration of the binary search, we will calculate the middle value between the lower and upper bounds. We will then simulate watering the flowers for k days with the assumption that all the flowers are increased to this middle value. If the smallest flower in this simulation is at least the middle value, we know that we can increase all the flowers to at least the middle value. We can then set the lower bound of the binary search to the middle value and continue the search in the upper half of the range.
On the other hand, if the smallest flower in the simulation is less than the middle value, we know that we cannot increase all the flowers to at least the middle value. We need to decrease our target height to a value lower than the current middle value, so we set the upper bound of the binary search to the middle value and continue the search in the lower half of the range.
We repeat this process until the lower and upper bounds converge to the same value, at which point we have found the optimal height that we can increase all the flowers to while keeping the smallest flower at least that height.
Once we have found this optimal height, we can simulate watering the flowers for k days with the assumption that all the flowers are increased to this height, and return the height of the smallest flower in this simulation as the answer to the problem.
I hope that explanation helps! If you find any difficulty in solving, feel free to comment with your doubts.
Your goal is to maximize the height of the smallest flower throughout the entire k-day period. In other words, you want to make sure that the smallest flower is always as tall as possible, even after you water other flowers.
To solve this problem using binary search, we first need to define the range of values that we will search over. Since we want to maximize the height of the smallest flower, the range of values we will search over is between 0 and the maximum height of the flowers.
We will then use binary search to find the optimal height that we can increase all the flowers to while keeping the smallest flower at least that height. We will start by setting the lower bound of our binary search to 0 and the upper bound to the maximum height of the flowers.
In each iteration of the binary search, we will calculate the middle value between the lower and upper bounds. We will then simulate watering the flowers for k days with the assumption that all the flowers are increased to this middle value. If the smallest flower in this simulation is at least the middle value, we know that we can increase all the flowers to at least the middle value. We can then set the lower bound of the binary search to the middle value and continue the search in the upper half of the range.
On the other hand, if the smallest flower in the simulation is less than the middle value, we know that we cannot increase all the flowers to at least the middle value. We need to decrease our target height to a value lower than the current middle value, so we set the upper bound of the binary search to the middle value and continue the search in the lower half of the range.
We repeat this process until the lower and upper bounds converge to the same value, at which point we have found the optimal height that we can increase all the flowers to while keeping the smallest flower at least that height.
Once we have found this optimal height, we can simulate watering the flowers for k days with the assumption that all the flowers are increased to this height, and return the height of the smallest flower in this simulation as the answer to the problem.
I hope that explanation helps! If you find any difficulty in solving, feel free to comment with your doubts.
class Solution { #define ll long long public: bool check(ll minHeight,int n,ll k,int w,vector <int> &a){ queue<pair<int,ll>> q; ll prev=0; for(int i=0;i<n;i++){ if(a[i]<minHeight){ while(q.size()>0 and i-q.front().first>=w){ prev-=q.front().second; q.pop(); } ll req=minHeight-a[i]-prev; if(req<=0) continue; q.push({i,req}); prev+=req; k-=req; } } return k>=0; } long long int maximizeMinHeight(int n,int k,int w,vector <int> &a) { ll lo=0,hi=1e14,ans; while(lo<=hi){ ll mid=(lo+hi)/2; if(check(mid,n,k,w,a)){ ans=mid; lo=mid+1; } else hi=mid-1; } return ans; } };
Here's how the function works:
The Solution class contains two functions: check and maximizeMinHeight.
The check function takes the minimum height minHeight, number of flowers n, number of days k, number of flowers that can be watered in a single day w, and an array of flower heights a as input. It checks if we can water the flowers to achieve a minimum height of minHeight. It does this by simulating watering the flowers for k days with the assumption that all the flowers are increased to minHeight. It uses a queue to keep track of the flowers that need to be watered and their required amount of water. If we run out of water before watering all the flowers, then we return false. Otherwise, we return true.
The maximizeMinHeight function takes the number of flowers n, number of days k, number of flowers that can be watered in a single day w, and an array of flower heights a as input. It uses binary search to find the maximum achievable minimum height. We start with the lower bound lo set to 0 and the upper bound hi set to a very large number, such as 1e14. In each iteration of the binary search, we calculate the middle value between lo and hi. We then call the check function with the middle value as the minimum height. If the check function returns true, then we update the answer ans to the middle value and continue the binary search in the upper half of the range. Otherwise, we continue the binary search in the lower half of the range. We repeat this process until the lower and upper bounds converge to the same value, at which point we have found the optimal minimum height. Finally, we return the optimal minimum height as the answer.
The check function takes the minimum height minHeight, number of flowers n, number of days k, number of flowers that can be watered in a single day w, and an array of flower heights a as input. It checks if we can water the flowers to achieve a minimum height of minHeight. It does this by simulating watering the flowers for k days with the assumption that all the flowers are increased to minHeight. It uses a queue to keep track of the flowers that need to be watered and their required amount of water. If we run out of water before watering all the flowers, then we return false. Otherwise, we return true.
The maximizeMinHeight function takes the number of flowers n, number of days k, number of flowers that can be watered in a single day w, and an array of flower heights a as input. It uses binary search to find the maximum achievable minimum height. We start with the lower bound lo set to 0 and the upper bound hi set to a very large number, such as 1e14. In each iteration of the binary search, we calculate the middle value between lo and hi. We then call the check function with the middle value as the minimum height. If the check function returns true, then we update the answer ans to the middle value and continue the binary search in the upper half of the range. Otherwise, we continue the binary search in the lower half of the range. We repeat this process until the lower and upper bounds converge to the same value, at which point we have found the optimal minimum height. Finally, we return the optimal minimum height as the answer.
Learning Outcomes
Solving the flower watering optimization problem provides crucial learning outcomes in resource management and algorithmic design. Firstly, it hones skills in efficiently allocating limited resources, reflecting real-life scenarios where optimizing resource usage is pivotal. The use of binary search not only introduces a powerful algorithmic concept but also instills a mindset of systematic exploration and refinement, applicable across diverse problem-solving contexts.
The queue-based mechanism underscores the importance of historical considerations in decision-making, emphasizing adaptability and iterative strategies. This problem cultivates a nuanced understanding of balancing priorities over time, valuable in scenarios beyond the realm of flower gardens.
Overall, tackling this challenge imparts practical skills in strategic resource allocation, algorithmic efficiency, and adaptable decision-making—skills that extend well beyond the coding environment, finding applications in everyday challenges where optimizing resources and making informed decisions are essential.
The queue-based mechanism underscores the importance of historical considerations in decision-making, emphasizing adaptability and iterative strategies. This problem cultivates a nuanced understanding of balancing priorities over time, valuable in scenarios beyond the realm of flower gardens.
Overall, tackling this challenge imparts practical skills in strategic resource allocation, algorithmic efficiency, and adaptable decision-making—skills that extend well beyond the coding environment, finding applications in everyday challenges where optimizing resources and making informed decisions are essential.
Conclusion
In conclusion, the flower watering optimization problem encapsulates fundamental principles of resource management and efficient decision-making. Through the lens of a garden with limited water over k days, the problem instills a mindset of strategic resource allocation, a skill crucial in real-world scenarios. The binary search algorithm employed in the solution not only demonstrates its versatility in coding challenges but also imparts a structured approach to refining solutions within a defined range.
Furthermore, the queue-based mechanism adds a layer of complexity by considering historical watering requirements. This highlights the importance of adaptability and iterative strategies in managing priorities over time, skills transferable to various practical situations.
Ultimately, while seemingly a botanical puzzle, this problem equips learners with valuable skills applicable beyond coding—skills in optimizing resources, making informed decisions, and developing efficient algorithms. It serves as a microcosm of challenges encountered in broader contexts, emphasizing the universal relevance of coding problems in nurturing problem-solving proficiency.
Furthermore, the queue-based mechanism adds a layer of complexity by considering historical watering requirements. This highlights the importance of adaptability and iterative strategies in managing priorities over time, skills transferable to various practical situations.
Ultimately, while seemingly a botanical puzzle, this problem equips learners with valuable skills applicable beyond coding—skills in optimizing resources, making informed decisions, and developing efficient algorithms. It serves as a microcosm of challenges encountered in broader contexts, emphasizing the universal relevance of coding problems in nurturing problem-solving proficiency.
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