Leetcode 2588: Count the Number of Beautiful Subarrays
You are given a 0-indexed integer array
nums
. In one operation, you can choose two indices and subtract a power of two from both elements at those indices. A subarray is considered beautiful if you can make all of its elements equal to zero by applying the operation any number of times. Your task is to return the number of beautiful subarrays in nums
.Problem
Approach
Steps
Complexity
Input: The input consists of an integer array `nums`.
Example: For example, `nums = [6, 4, 5, 3, 4]`.
Constraints:
• 1 <= nums.length <= 10^5
• 0 <= nums[i] <= 10^6
Output: The output is an integer that represents the number of beautiful subarrays in the array `nums`.
Example: For `nums = [6, 4, 5, 3, 4]`, the output is `3`.
Constraints:
• The result will always be a valid integer.
Goal: The goal is to find the number of subarrays that can be made equal to 0 by applying the specified operations.
Steps:
• 1. Loop through all possible subarrays in `nums`.
• 2. For each subarray, check if it can be transformed into an array of zeros by applying the operations.
• 3. Count the number of such subarrays and return the count.
Goal: The array contains between 1 and 10^5 integers, and each integer is between 0 and 10^6.
Steps:
• 1 <= nums.length <= 10^5
• 0 <= nums[i] <= 10^6
Assumptions:
• The input array contains valid integers within the given range.
• Input: For `nums = [6, 4, 5, 3, 4]`
• Explanation: The beautiful subarrays are `[6, 4]`, `[6, 4, 5, 3, 4]`, and `[4, 5, 3]` as each can become all zeros after applying the operations.
Approach: To solve this problem, we need to find the subarrays where it is possible to make all elements equal to zero by applying the operation any number of times. A strategy would be to check each subarray and verify if it satisfies the condition for becoming a beautiful subarray.
Observations:
• For each subarray, check if it can be made to have all elements equal to zero using the given operations.
• This problem seems to involve checking the binary representation of the numbers and ensuring we can make them equal.
Steps:
• 1. Iterate through each subarray in the given array `nums`.
• 2. For each subarray, attempt to apply the subtraction operation to check if all elements can be reduced to zero.
• 3. Count how many subarrays are beautiful and return the count.
Empty Inputs:
• The problem guarantees that the array will not be empty.
Large Inputs:
• The solution should be optimized to handle large inputs up to 10^5 elements.
Special Values:
• If all numbers are already zero, all subarrays are beautiful.
Constraints:
• The length of the array is guaranteed to be within the specified limits.
long long beautifulSubarrays(vector<int>& nums) {
int n= nums.size();
long long cnt = 0;
map<int, int> mp;
mp[0] = 1;
int tmp = 0;
for(int i = 0; i < n; i++) {
tmp ^= nums[i];
if(mp.count(tmp)) cnt+= mp[tmp];
mp[tmp]++;
}
return cnt;
}
1 : Function Definition
long long beautifulSubarrays(vector<int>& nums) {
Defines the 'beautifulSubarrays' function, which takes a vector of integers as input and returns the count of beautiful subarrays.
2 : Variable Initialization
int n= nums.size();
Initializes 'n' to the size of the 'nums' array.
3 : Variable Initialization
long long cnt = 0;
Initializes the 'cnt' variable to store the count of beautiful subarrays.
4 : Data Structure Initialization
map<int, int> mp;
Declares a map 'mp' to store the frequency of each XOR value encountered during the iteration.
5 : Data Structure Initialization
mp[0] = 1;
Initializes the map with the value 0 mapped to 1, representing the XOR value of an empty subarray.
6 : Variable Initialization
int tmp = 0;
Initializes 'tmp' to 0, which will be used to accumulate the XOR of the elements in the array.
7 : Loop
for(int i = 0; i < n; i++) {
Begins a loop that iterates over the array 'nums'.
8 : XOR Operation
tmp ^= nums[i];
Applies the XOR operation on the current element of 'nums' with 'tmp', updating 'tmp' at each step.
9 : Condition Check
if(mp.count(tmp)) cnt+= mp[tmp];
Checks if the current XOR value 'tmp' has been encountered before. If so, adds the frequency of 'tmp' from the map to 'cnt'.
10 : Data Structure Update
mp[tmp]++;
Increments the frequency of the current XOR value 'tmp' in the map.
11 : Return
return cnt;
Returns the total count of beautiful subarrays found.
Best Case: O(n)
Average Case: O(n^2)
Worst Case: O(n^3)
Description: In the worst case, we may need to check each subarray, which can lead to a time complexity of O(n^3). Optimizing this solution is essential.
Best Case: O(n)
Worst Case: O(n^2)
Description: The space complexity depends on the storage of subarrays being checked.
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