Leetcode 1589: Maximum Sum Obtained of Any Permutation
You are given an array of integers nums and a list of requests, where each request specifies a range in the array. The ith request asks for the sum of all elements from nums[starti] to nums[endi], inclusive. Your task is to determine the maximum total sum of all requests for any permutation of nums.
Problem
Approach
Steps
Complexity
Input: You are given an array of integers nums, and a list of requests. Each request consists of two integers [starti, endi] representing a range in nums. The length of nums can be up to 10^5, and the number of requests can be as large as 10^5.
Example: Input: nums = [1, 3, 2, 4], requests = [[0, 1], [1, 3]]
Constraints:
• 1 <= nums.length <= 10^5
• 0 <= nums[i] <= 10^5
• 1 <= requests.length <= 10^5
• 0 <= starti <= endi < n
Output: Return the maximum possible sum of all request sums for any permutation of the nums array, modulo 10^9 + 7.
Example: Output: 21
Constraints:
• The output should be an integer representing the maximum possible total sum modulo 10^9 + 7.
Goal: The goal is to find the maximum total sum of the request sums for any permutation of nums by efficiently calculating the contributions of each element to the total sum based on the request ranges.
Steps:
• 1. Create an array of length n, where each element corresponds to the number of times it is included in a request range.
• 2. Sort the nums array in increasing order and the request contributions in decreasing order.
• 3. Multiply each element of nums with its corresponding contribution and sum the results.
Goal: The problem requires an efficient approach due to the large input size constraints.
Steps:
• 1 <= nums.length <= 10^5
• 0 <= nums[i] <= 10^5
• 1 <= requests.length <= 10^5
• 0 <= starti <= endi < n
Assumptions:
• The requests are well-formed, with starti <= endi.
• All elements in nums are distinct and non-negative.
• Input: Input: nums = [1, 3, 2, 4], requests = [[0, 1], [1, 3]]
• Explanation: The permutation [4, 3, 2, 1] gives the maximum total sum for the requests. The first request sums nums[0] + nums[1] = 4 + 3 = 7 and the second request sums nums[1] + nums[2] + nums[3] = 3 + 2 + 1 = 6. The total sum is 7 + 6 = 13.
Approach: We can solve this problem by treating the requests as intervals that increase the frequency of certain elements in nums. The elements of nums that appear in many requests should be assigned larger values to maximize the total sum. By sorting both the nums array and the request contributions, we can efficiently compute the result.
Observations:
• We need to calculate how often each element of nums is included in the request ranges.
• Sorting both nums and the request contributions in a way that matches higher frequencies with higher values will give the maximum sum.
• The problem boils down to sorting the frequency of requests and the values in nums to pair them optimally.
Steps:
• 1. Initialize an array 'cnt' to track how many times each element in nums is used across the requests.
• 2. Process the requests by incrementing cnt[starti] and decrementing cnt[endi + 1] to mark the start and end of each range.
• 3. Compute the prefix sum of cnt to get the actual frequency of each index in nums.
• 4. Sort nums and cnt in descending order and multiply corresponding elements.
• 5. Sum the products and return the result modulo 10^9 + 7.
Empty Inputs:
• Handle the case where nums has only one element or there are no requests.
Large Inputs:
• Ensure the solution can handle the largest possible input sizes efficiently.
Special Values:
• Handle cases where all values in nums are zero, or where the ranges in requests are minimal (e.g., starti == endi).
Constraints:
• Ensure the solution works within the provided time and space constraints.
int maxSumRangeQuery(vector<int>& nums, vector<vector<int>>& req) {
int n = nums.size();
long long res = 0;
vector<long long> cnt(n, 0);
for(auto& r: req) {
cnt[r[0]]++;
if(r[1] + 1 < n)
cnt[r[1] +1]--;
}
for(int i = 1; i < n; i++)
cnt[i] += cnt[i -1];
sort(nums.begin(), nums.end());
sort(cnt.begin(), cnt.end());
for(int i = 0; i < n; i++)
res += ((long long) nums[i] * cnt[i]);
return res % 1000000007;
}
1 : Function Declaration
int maxSumRangeQuery(vector<int>& nums, vector<vector<int>>& req) {
This line defines the function `maxSumRangeQuery` that takes two arguments: a vector of integers `nums` and a 2D vector `req` representing the ranges for computation.
2 : Variable Initialization
int n = nums.size();
This initializes the variable `n` to store the size of the `nums` array.
3 : Variable Initialization
long long res = 0;
Here, we initialize `res` to 0. This will hold the final result of the sum after performing the necessary calculations.
4 : Array Initialization
vector<long long> cnt(n, 0);
We create a vector `cnt` initialized to zeros. This will keep track of the frequency of range updates for each index in `nums`.
5 : Loop (Outer)
for(auto& r: req) {
This loop iterates over each range in the `req` vector. Each range is represented by the indices `r[0]` and `r[1]`.
6 : Range Update
cnt[r[0]]++;
This line increments the value at index `r[0]` in the `cnt` vector, indicating that the range starting at `r[0]` is updated.
7 : Conditional Update
if(r[1] + 1 < n)
This checks if the end of the range `r[1]` is within bounds. If true, we need to decrement the value just after the range.
8 : Range Update
cnt[r[1] +1]--;
This line decrements the value just after the range `r[1] + 1`, signaling the end of the range effect.
9 : Prefix Sum
for(int i = 1; i < n; i++)
This loop starts the prefix sum operation on the `cnt` array, which accumulates the range counts.
10 : Prefix Sum Update
cnt[i] += cnt[i -1];
This line adds the previous value of `cnt[i-1]` to `cnt[i]`, effectively accumulating the range counts.
11 : Sorting
sort(nums.begin(), nums.end());
This sorts the `nums` array in ascending order to align the highest values with the most frequent ranges.
12 : Sorting
sort(cnt.begin(), cnt.end());
This sorts the `cnt` array to match the corresponding frequencies with the sorted `nums` array.
13 : Final Computation Loop
for(int i = 0; i < n; i++)
This loop computes the final result by multiplying each value in `nums` by the corresponding frequency in `cnt`.
14 : Final Computation
res += ((long long) nums[i] * cnt[i]);
This line adds the product of the sorted values in `nums` and their corresponding frequencies in `cnt` to the result `res`.
15 : Modulo Operation
return res % 1000000007;
After computing the total sum, we return the result modulo 1000000007 to avoid overflow and meet the problem's requirements.
Best Case: O(n log n) - Sorting nums and cnt arrays takes O(n log n).
Average Case: O(n log n)
Worst Case: O(n log n)
Description: The most expensive operation is sorting the nums and cnt arrays.
Best Case: O(n) - The space complexity is primarily determined by the storage of the cnt array and nums.
Worst Case: O(n) - Storing the cnt array and intermediate results requires O(n) space.
Description: The space complexity is linear, proportional to the size of nums and the request list.
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