Leetcode 3101: Count Alternating Subarrays
You are given a binary array nums. A subarray is called alternating if no two adjacent elements in the subarray have the same value. Return the total number of alternating subarrays in the given binary array.
Problem
Approach
Steps
Complexity
Input: The input consists of a binary array nums of length n.
Example: nums = [0, 1, 1, 0]
Constraints:
• 1 <= nums.length <= 10^5
• nums[i] is either 0 or 1
Output: Return the total number of alternating subarrays in the binary array nums.
Example: Output: 6
Constraints:
Goal: Count all possible alternating subarrays in the given binary array.
Steps:
• 1. Initialize a dynamic programming array to store the lengths of alternating subarrays.
• 2. Iterate through the array, and for each element, calculate the length of the alternating subarray ending at that element.
• 3. Add the length of the alternating subarray to the result for each element.
Goal: The problem constraints limit the length of the input array.
Steps:
• 1 <= nums.length <= 10^5
• nums[i] is either 0 or 1
Assumptions:
• The input will always consist of a valid binary array of size between 1 and 100,000.
• Input: nums = [0, 1, 1, 0]
• Explanation: The alternating subarrays are: [0], [1], [0, 1], [1], [1, 0], and [0], which gives a total of 6.
• Input: nums = [1, 0, 1, 0, 1]
• Explanation: All subarrays in this binary array are alternating, resulting in 15 alternating subarrays.
Approach: To solve this problem, we can use dynamic programming to keep track of the lengths of alternating subarrays as we iterate through the input array.
Observations:
• The problem is related to counting alternating subarrays, which can be efficiently done by storing the length of each alternating subarray.
• We can use dynamic programming to store the lengths of alternating subarrays and calculate the result in linear time.
Steps:
• 1. Initialize an array dp of size n, where dp[i] stores the length of the alternating subarray ending at index i.
• 2. Iterate through the array, and for each element, if it differs from the previous element, increment dp[i] based on dp[i-1].
• 3. Sum the values of dp[i] to get the total number of alternating subarrays.
Empty Inputs:
• The input will always contain at least one element (nums.length >= 1).
Large Inputs:
• The solution should efficiently handle arrays with up to 100,000 elements.
Special Values:
• When the array contains only one element, the only possible alternating subarray is the element itself.
Constraints:
• The solution must run efficiently within the input constraints (1 <= nums.length <= 10^5).
long long countAlternatingSubarrays(vector<int>& nums) {
int n = nums.size();
vector<long long> dp(n, 0);
dp[0] = 1;
long long res = 1;
for(int i = 1; i < n; i++) {
dp[i] = nums[i] == nums[i - 1]? 1: dp[i - 1] + 1;
// cout << dp[i];
res += dp[i];
}
return res;
}
1 : Function Definition
long long countAlternatingSubarrays(vector<int>& nums) {
Defines the function `countAlternatingSubarrays` that takes a vector of integers `nums` and returns the total number of alternating subarrays.
2 : Variable Initialization
int n = nums.size();
Initializes `n` to the size of the input vector `nums`, representing the number of elements in the array.
3 : Variable Initialization
vector<long long> dp(n, 0);
Creates a dynamic programming array `dp` of size `n`, initializing all values to 0. This array will store the length of alternating subarrays ending at each position.
4 : Base Case Initialization
dp[0] = 1;
Sets the first element of the `dp` array to 1, representing the base case where the first element is an alternating subarray of length 1.
5 : Variable Initialization
long long res = 1;
Initializes `res` to 1, which will accumulate the total number of alternating subarrays.
6 : For Loop Setup
for(int i = 1; i < n; i++) {
Begins a `for` loop starting from index 1, iterating through the array to calculate the length of alternating subarrays ending at each position.
7 : DP Calculation
dp[i] = nums[i] == nums[i - 1]? 1: dp[i - 1] + 1;
For each element, checks if it is equal to the previous one. If so, sets `dp[i]` to 1 (indicating the start of a new subarray); otherwise, increments `dp[i]` by 1 to extend the alternating subarray.
8 : Accumulating Results
res += dp[i];
Adds `dp[i]` to `res`, accumulating the total number of alternating subarrays.
9 : Return Statement
return res;
Returns the total count of alternating subarrays stored in `res`.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n), where n is the length of the input array. This is because we iterate through the array once, performing constant-time operations.
Best Case: O(1)
Worst Case: O(n)
Description: The space complexity is O(n) due to the dynamic programming array used to store the lengths of alternating subarrays.
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