Leetcode 2865: Beautiful Towers I
You are given an array heights representing the number of bricks in n consecutive towers. Your task is to remove some bricks to form a mountain-shaped tower arrangement. In this arrangement, the tower heights are non-decreasing, reaching a maximum peak value with one or multiple consecutive towers and then non-increasing. Return the maximum possible sum of heights of a mountain-shaped tower arrangement.
Problem
Approach
Steps
Complexity
Input: The input consists of an array of integers representing the heights of the towers. You are to remove some bricks to form a mountain-shaped arrangement.
Example: heights = [4, 6, 2, 3, 5]
Constraints:
• 1 <= n <= 1000
• 1 <= heights[i] <= 10^9
Output: Return the maximum possible sum of the heights of the towers after removing some bricks to form a mountain-shaped arrangement.
Example: For input heights = [4, 6, 2, 3, 5], the output is 15.
Constraints:
Goal: The goal is to rearrange the towers into a mountain shape by removing some bricks, maximizing the sum of heights in the process.
Steps:
• For each tower, calculate the possible sum by removing bricks on both sides to create a mountain-shaped arrangement.
• Iterate over all possible peaks and find the one that results in the maximum sum.
Goal: The solution must handle arrays with sizes up to 1000 and heights up to 10^9.
Steps:
• 1 <= n <= 1000
• 1 <= heights[i] <= 10^9
Assumptions:
• The array contains at least one tower.
• The number of bricks in each tower is always greater than 0.
• Input: For input heights = [4, 6, 2, 3, 5], the output is 15.
• Explanation: By removing bricks and forming a mountain shape, we maximize the sum of the tower heights to 15.
Approach: We will compute the maximum possible sum of heights by iterating through each possible peak and adjusting the tower heights accordingly.
Observations:
• The solution requires us to check every possible peak in the tower arrangement.
• For each peak, we will adjust the neighboring towers to form a valid mountain shape.
• The main challenge is to compute the sum efficiently for each possible peak and find the maximum sum.
Steps:
• For each possible peak, calculate the sum of heights in the non-decreasing and non-increasing sequences.
• Track the maximum sum encountered during the iteration.
Empty Inputs:
• The input array will never be empty as per the constraints.
Large Inputs:
• The solution must handle arrays of size 1000 efficiently.
Special Values:
• If all towers have the same height, the mountain shape will be symmetric with the peak at any index.
Constraints:
• Ensure that the sum is maximized after removing bricks to form the mountain shape.
#define ll long long
long long maximumSumOfHeights(vector<int>& a) {
int n=a.size();
ll int ans=0;
for(int i=0;i<n;i++){
ll int sum = a[i];
int prev=a[i];
for(int j=i-1;j>=0;j--){
prev = min(a[j],prev);
sum += prev;
}
prev=a[i];
for(int j=i+1;j<n;j++){
prev = min(a[j],prev);
sum += prev;
}
if(sum > ans) ans=sum;
}
return ans;
}
1 : Macro Definition
#define ll long long
Define a macro 'll' as an alias for 'long long' to simplify the code.
2 : Function Start
long long maximumSumOfHeights(vector<int>& a) {
Start of the function definition for maximumSumOfHeights, which takes a vector of integers as input.
3 : Variable Initialization
int n=a.size();
Initialize the variable 'n' to store the size of the input array 'a'.
4 : Variable Declaration
ll int ans=0;
Declare a variable 'ans' of type 'long long' to store the maximum sum of heights found.
5 : Outer Loop Start
for(int i=0;i<n;i++){
Start the outer loop that iterates over each element in the array 'a'.
6 : Inner Sum Initialization
ll int sum = a[i];
Initialize the variable 'sum' to the value of the current element of the array.
7 : Prev Variable Initialization
int prev=a[i];
Initialize the 'prev' variable to the current element 'a[i]', representing the previous minimum height.
8 : Leftward Loop Start
for(int j=i-1;j>=0;j--){
Start the inner loop that iterates leftwards from the current index 'i'.
9 : Leftward Min Calculation
prev = min(a[j],prev);
Update the 'prev' variable to the minimum value between 'prev' and 'a[j]'.
10 : Leftward Sum Update
sum += prev;
Add the current 'prev' value to 'sum' to accumulate the sum of heights from the left side.
11 : Rightward Loop Start
prev=a[i];
Reset the 'prev' variable to the current element 'a[i]' before starting the rightward loop.
12 : Rightward Loop Start
for(int j=i+1;j<n;j++){
Start the inner loop that iterates rightwards from the current index 'i'.
13 : Rightward Min Calculation
prev = min(a[j],prev);
Update the 'prev' variable to the minimum value between 'prev' and 'a[j]'.
14 : Rightward Sum Update
sum += prev;
Add the current 'prev' value to 'sum' to accumulate the sum of heights from the right side.
15 : Max Sum Check
if(sum > ans) ans=sum;
Check if the current 'sum' is greater than the current 'ans' value, and update 'ans' if true.
16 : Return Statement
return ans;
Return the value of 'ans', which holds the maximum sum of heights.
Best Case: O(n^2)
Average Case: O(n^2)
Worst Case: O(n^2)
Description: The algorithm involves iterating over all possible peaks and calculating the sum for each peak, resulting in a time complexity of O(n^2).
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant, O(1), as we only need a few extra variables to store the sum and indices.
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